Minimum percentage can always be determined by picking the lowest percentage (in this case, 🍩 at 65%), and then subtracting away all the "did not eat X" percentages. In this case, those are 20%, 20%, and 10%, for a total of 50%. 65% - 50% = 15%. The actual amount that ate all four items is probably higher, up to maximum of the lowest percentage, 65%.
@@knoxiegb1782 Formally speaking, what I did was just a specific application of the proof given near the end of the video. The union of WXYZ is no smaller than the whole space (100 percentage points) minus the complement of each individual set. Thing is, (whole space) - (complement of set W) = the set W, by definition. You can, as other replies have said, pick any starting point. I just find it most natural to start from the smallest one first.
For the practical application, you actually do NOT want to target the people who bought all 4. Bundles are typically cheaper than the sum of the individual items - that's the point of the bundle. If we only sell the bundle to people who bought all 4 anyway, we just lose money. The people we ACTUALLY want to target are the ones who bought less, but might be convinced to buy all 4. Most likely, the people who usually buy 3 out of 4 could be tempted by a bundle offer. The donut merchant would be the most likely to profit from this, as most people who only skipped out on 1 probably skipped the donut.
The information request is still useful though. We know, that with such bundle we loose with at least 15% of the customers, and up to 65%. So, looking for info about people who bought all 4 makes sense, but we should focus on the maximum number of people who did that. Then we can set up the pricing in such way, that even in worst case scenario extra spending of the part of 35% of customers will make up for the loss on 65%. This is of course a "safe bet" scenario, most companies probably could tolerate some risk and look for "most likely % of people who bought all 4" to use in estimations instead.
@@edsimnett The four numbers most certainly aren't independent variables. People have preferences and don't truly randomly buy food in the real world. Upper and lower boundaries however still apply.
You also have to measure the time people bought items. If people bought all four at once or 3 items at once (usually a family) that's different than an individual buying all four items but over a longer period of time.
@@willguggn2 true, and for a small sampling ..not very informative ..but on a larger sampling .. the errors smooth out... as in the exaple lets assume 20 ppl...the variance from day to day will be massive if we say saple 20000/day for a week ... i would say the error becomes so small .. we can with a quite high degree of certainty predict the sales... in the future... and what prferences ppl have .. we can then add other factors as weather, pricing etc... ...we can also just by using common sense do estimations (if its 10 deg and windy, the probability our outdoor ice cream stand sell alot of icecream is quite low)
The easiest way is calculating the inverse. The max amount of people that didn't eat all 4. You just add the percentages of people who didn't eat each food. 10 for ice-cream, 20 for pretzels and pizza, and 35 for donut. 10+20+20+35=85, which means at most 85% didn't eat all 4 foods, so at least 15% had to eat all 4.
You can make that mathematically sound by showing that the minimum percentage goes up if there are overlaps between the people who don't eat each food. It's still the best way though.
Those looking for a simpler method: I did this problem instantly in my head. 100-65 = 35, 100-80 = 20, 100-80 = 20 & 100-90 = 10. Add all these numbers up (35+20+20+10) = 85. 100 - 85 = 15%! Explanation: 35% didn’t eat a doughnut, 20% didn’t eat a pretzel etc. So 85% didn’t eat all 4 foods. So 15% did!
Here's what I did: 65% (Donut) + 80% (Pretzel) = 145%. 145% - 100% = *45% of people who ate a Donut/Pretzel ate both of them.* 45% (Donut+Pretzel) + 80% (Pizza) = 125%. 125% - 100% = *25% of people who ate a Donut, Pretzel, or Pizza ate all 3 of them.* 25% (Donut + Pretzel + Pizza) + 90% (Ice Cream) = 115%. 115% - 100% = *15% of people who ate any of the 4 foods ate all 4 of them.* Therefore, at least 15% of customers must have eaten all four foods.
My elementary school way is the simplest (and also proves it at the same time) * 65% eat A, 80% eat B, so "eat A and B" is at least 65% + 80% - 100% = 45% * now same logic, 45% eat AB, 80% eat C, therefore "eat ABC" is at least 25% * eat ABCD is at least 15% Each step itself also naively proved that it's a reachable lower bound (thus minimum) The calculation is even simpler as: 1. you remove the 100th so it's smaller number 2. you can early exit if the sum is less than 100 and claim that the min is already 0 without adding all other numbers.
I solved it the same way, but I would have put parentheses, otherwise the solution would be unclear without them. We assume that in our group of eaters it is precisely those people who did not eat other food; 65% - (100% - 80%) - (100% - 80%) - (100% - 90%) = 80% - (100% - 65%) - (100% - 80%) - (100% - 90%) = 90% - (100% - 65%) - (100% - 80%) - (100% - 80%) This is basically the same first method from the video... Although the second one is similar. I don’t see any difference between the solutions, we rely on the same thing
@@oqui7009 1. 65% - (100% - 80%) is no more clear than simply 65% + 80% - 100%, if you think about it. In fact I believe the entire elementary school "math" stuff is about thinking through these kind of logics. 2. you forgot about the possibility of "early exit". We do not need to write everything down at once. We can do the simple arithmetic in our mind and whenever we get negative number we know it's 0 and there is no need for further calculations.
In the first calculation about "A and B" how did you count for those who didn't eat either A or B ? Its the set intersection formula n(A or B) = n(A) + n(B) -n(A and B). But in this case how can one assume that n(A or B) is infact 100% of the total sample? Also there are a minimum of 10% of total sample who did not eat either A or B
I approached this by simplifying it to 2 foods. This can be visualized by two sliding bars (line segments) of various lengths which must be positioned in a specific space (100%). How would we minimize the overlap? Easy, we put one bar all the way to one side and the other bar all the way to the other. So for example with Ice Cream (90%) and Pizza (80%), the mininum overlap would be 70%. But now we know a minimum of 70% had both Ice Cream and Pizza, so I can repeat this with the Pretzel (80%). So mininum 50% had Ice Cream/Pizza/Pretzel. Apply this once more with Donut (65%) and the minimum overlap is 15%.
Yes, minimal possible overlap of the two lowest percentages, where you stack one from the "left" and one from the "right". That would also tell you right away, that if two food items were at maximum 50%, the minimum of people who ate everything is 0%.
Actually, if your two items with the lowest percentages would "sum up" and that combined result is less than 100%, your minimal possible overlap would always be 0%. In an extreme example, out of 1000 people you have 999 eating pizza, and 1 person eating brussel sprouts. Those 999 and 1 can be completely different groups of individuals - so you can have 0 overlap. Adding more food items to your combined considerarions here won't have any impact on the overall minimum across ALL food items then.
I tried to do something similar but my approach was very complicated, I tried to make a square with 10 smaller boxes along each arm, thus 100 of them inside. Then, I tried representing each beverage along each corner to minimize the overlap...but i forgot that the result will be skewed if I used multiple beverages at the same time, I think I had to start with two beverages and put them at the diagonally opposite end then calculate their overlap and continue like you, but yours is much better and simple. Also, what if the percentage of one of the beverages was way less like 2 or 3% then we wouldn't get any overlap, would that mean that but is possible that no one bought all four?
@@aditya37mviews3weeksago3 Yes, if you have one food item with like only 3%, any of the other food items would need to have over 97%, so that there could be a minimal, but none-zero overlap among all.
So that’s what my wrong calculation of 37.44% measured. You’re saying this percent measures the expected percent of people to eat all 4 foods. But why is the expected percent, the minimum percent, and the maximum percent all able to be calculated yet mean different things. I am not sure I actually understand what these three terms means in this contexts. All I know is I had thought I was finding the solution when I multiplied the 4 original percentages together to get 37.44% but found out I was mistaken. The 65% for the maximum makes intuitive sense to me as well I think but I don’t get why the expected is not the same as the minimum in this case?
@@kennyalbano1922 Expected is more akin to the average. The minimum is essentially assuming as much as possible that people didn't eat all four and getting the smallest number which satisfies that criteria. Two easy examples to illustrate. If you flip a fair coin 20 times, the expected number of heads is 10. The minimum is 0, and the maximum is 20; both are unlikely, but technically possible. If you roll 2 fair 6-sided dice, the expected total of roll of the dice is 7; as that's what you'll get on average. The minimum is 2, and the maximum is 12. Amusingly the expected roll of a single die is 3.5; even though that's not technically possible.
@@WombatMan64 are there any ways to measure spread with this type of data. Such as range and outliers, does it follow a bell curve and if so what are its standard deviation and variance? Can the data be described by an equation and displayed graphically? I mean where the 4 foods intersect as in the question by the data.
@@kennyalbano1922 not really with only the information supplied. If you had a table with every single customer and exactly what they purchased, and had that info every day then no doubt you could do some good analysis, but the information provided in this question is quite... minimal. Appropriate really.
@@edsimnett Actually it can be very useful. Businesses work on a daily basis. If you're gonna do business with a company, it's not only important to know how much someone sells on average and how much on the good days, but also how much they sell on the worst days (and also the factors contributing to the lowest and highest sales).
42 seconds in. "Amusement park: What is the minimum percentage of visitors that must have eaten all 4 foods?" 0% Odds are some did not buy any food. - Solved!
I did it by thinking of a group of 100 people, and taking it at groups of 2 at a time. If you take the first two groups, 80 people had the pretzel, which means minus the 35 that didnt have the donut you still need 45 that did. Then for pizza you take the 55 away that didnt have both donut and pretezel but you still need 25 that had both, so then you need 25 that had all 3. Then for icecream you take 75 away that didnt have all previous 3, but that still leaves you with 15 that did, so atleast 15 had to have all 4.
I did the second method but then I realized it's easier: Assuming that there are 100 people and no person ate the same item twice. If all of them ate all 4 items there would be 400 sales. But if we add up the numbers there has been 315 sales. So at most 85 people did not eat all 4 items. That means at minimum 15 people ate all 4 items.
> no person ate the same item twice. The thing is, in the problem description this is left implicit. Somewhere while explaining the answers, this assumption was suddenly put into place. Before that, nobody excluded a person eating, say, 4 slices of pizza.
@@metamud8686 Right. But maybe it's not important how much one person bought of each item. So for example if a person buys 1 donut or 10 doesn't matter. The question then would be, did they buy it or not.
35% didn't eat a donut, 20% didn't eat a pretzel, 20% didn't eat a pizza, 10% didn't eat an ice cream. Worst case scenario : they are not the same people, therefore 35 + 20 + 20 +10 = 85% of people didn't eat one item, so 15% ate all 4.
I loved seeing it worked in many different ways. This is the real skill, being able to approach things in multiple ways and gauge which one will be simplest.
The "No Doughnut" set contains 35% of the visitors. "No Pretzels" has 20% of them, "No Pizza" has 20% and "No Ice Cream" has 10%. Anyone not in any of those sets must have bought all four foods. To minimise these people you need to maximise the numbers in the above-mentioned four sets, and to do that you need to assume no sharing of people between them. In that case you simply sum the sets' percentages to get the total %age of visitors that did not buy all four foods, because in doing so no one gets counted twice: 35+20+20+10 = 85. Therefore 15% bought all four foods in this extreme case.
I got the first 65% 80% trick to arrive at 45% due to seeing this kind of puzzle but didn't think to extend it out further to all four, cool trick! It's also interesting to note that, so long as each set contains less than 100%, you can keep adding more and more orthogonal sets and eventually arrive at 0% minimum overlap Eg, for sets that contain 50% of a population, you need 2 such sets to have a minimum overlap of 0%, if they cover 75% then you need 4 sets, 90% coverage requires 10 such sets, etc etc but however many ways you have of grouping people, so long as you have enough of them you'll be able to show that the minimum possible overlap is 0%
Intuitive answer is 15%. There is an algorithm to execute which makes this exceptionally easy. Basic description: You take the two lowest percentages to start. You have to calculate their intersection so that is going to be 65-(100-80)=45 then you take that answer and intersect the 3rd. 45 - (100-80) = 25 Take that number and intersect the 4th. 25 - (100-90) = 15. Done; it should take you approximately 3 seconds to do in your head. Algorithm: var accumulator = 0 foreach(var p in list){ if accumulator ==0 accumulator = p else accumulator -= (100-p) } accumulator will contain the intersection as long as there is one. A defense is necessary to prevent a disjoint set from resetting the accumulator to a positive number, but I'm not presenting this as a general solution to the problem of all intersecting sets. You have to sort them lowest to highest. It would have been a better problem if he had presented them out of sort order
Seems like a great problem to illustrate key ideas like DeMorgan's law and the Union Bound, but I don't think the connection was made as clearly as it could have been. There's a lot of intuition that was brought out be each solution, I think the connection to formalized concepts could have been a lot stronger.
Maybe, unless part of the goal is to encourage those who were getting just three items to "just pay a little bit extra" to get all 4. Particularly if the difference between 3 items and the discounted 4 items is more than the unit cost.
another simple way to calculate it. First pick a starting food, and keep that percentage of people who had it. Then list out the percentage of people who havent eaten any of the other given foods. From there, just start subtracting. 65% of people ate a donut 20% of people did not eat a donut (65-20=45) 20% of people did not eat a pizza (45-20=25) 10% of people did not eat a ice cream (25-10=15) There for, the answer is 15%
neat problem with many ways to solve! i drew the sample space as a 4 by 5 grid, such that every cell in the grid represented 5%. I then filled the grids with markers for each food, each time adding the nth food starting by avoiding the cells that already had (n-1) foods in them. this leaves only three of the 20 cells with all four markers!
Solving Geometrically: make a square ABCD of sides 10 each. this represents 100% of visitors. Assign each vertex an food type. lets say A = Donut B = Pretzel C = Pizza D = Ice cream Draw a smaller square from the vertex A of side SQRT(65) Draw a smaller square from the vertex B of side SQRT(80) Draw a smaller square from the vertex C of side SQRT(80) Draw a smaller square from the vertex D of side SQRT(90) The area of rectangle represented by the intersection of all 4 smaller squares is the minimum percentage of people who ate all 4 foods.
First: 90% vs 80%, the 10% that didn’t eat from the 90% could of ate part of 80% Thus, there is a minimum of 70% overlap Second: 80% vs 80%, since one of the 80% has a 70% overlap from before, 30% of the people who didn’t overlap could of ate from the other 80% Thus, there is a minimum of 50% overlap Third: 65% vs 80%, since the 80% has a 50% overlap from before, 50% of people who didn’t overlap could of ate from the 65% Thus, there is a minimum of 15% overlap. Didn’t see many people point out the logic, hope this helps.
Actually you ask two different questions, how many ate all the foods and what is the min percentage that ate them all. I agree with you about the minimum, it is 15%. But I think a better answer to the question, "how many?" is 37.44% based on converting the percentages to probabilities and multiplying them to get the probability that a given customer ate all items.
@@pivotfever You are right that he does treat it as a question about minimum numbers, but he does start by saying that what they are really interested in is how many would buy all three, and the minimum number that ate all three in this sample is not the best statistic to use for that.
@@MichaelGowland well he says the marketing people wanna know how many people are interested, not interested in how many people. Thus the actual question, what’s the minimum percentage, not most likely
The opening statement is that "the food vendors tell you..." the percentage of foods eaten. But from their perspective, each of the people must buy at least *one* food item (otherwise the vendors would not be able to count the person). So the maximum percentage who eat all four foods should be
There is no 'lesser than' the maximum. The least eaten food item is eaten by exactly 65% of the people (who buy at least 1 food item). Therefor, the maximum amount of possible people that bought all 4 food items is 65%. It's not at most 65%. It is exactly 65%.
@@DanielFSmith I mean, it's just common sense. Here's a solution with 100 people. 65% of them eat at least 1 food type. So 65 people eat at least 1 food type. It also means 65 people can eat all 4 food types at most. The rest of the percentages don't matter. 65% is the lowest. Whether it's 80% or 90% of people that eat other food types, it's still 65 people at most who can eat all 4 since there's a good chunk of at least 35% that haven't eaten the least popular food and as such, cannot possibly have eaten all types.
@@thenonexistinghero Let's take an example with 60% and, say, 90% using 10 people. Pie. aaaaaa0000 Cake. bbbbbbbb0 Doesn't work... The last person didn't eat at least one. You can only get 50% max.
I either got lucky or found a much easier way to do this: -Adding up the percentages of those that didn't eat each food equals 85%. -Subtract that answer from 100%, which equals 15%. I can't explain why that worked, but it apparently did.
There's an easier way to solve this. Taking Doughnuts as the example, you add up the number of people that did not eat Pretzels (20%), Pizza (20%), or Ice Cream (10%); and then subtract that sum from the total number of people who ate Doughnuts. 20+20+10=50 and 65-50=15 so 15 is the answer. You can technically do this with any of the foods. Taking Pretzels for the next example, 35+20+10=65 and 80-65=15.
I tried at first to follow the method of option A with weird bar graphs but gave up and found myself doing naturally going with option B. I wondered if I had the right idea pairing donuts with pretzels and pizza and ice cream but it seemed to make sense to me. And I'm glad the video showed that I was right!
I just saw a video with a math puzzle on it. At the top of a paper write "I have $50" Now divide the paper into two columns, Spent Balance 20. 30 15. 15 09. 06 06. 00 For each of these you spend some amount of money, then you write your new balance in the balance column. Now total up each column. The spent adds up to 50, which we expected. Yet the balances all add up to 51...where did that extra one dollar come from?
So the minimum that bought all four is 15%, the max is 65%, and if all four of these variables are independent then the expected case (0.65 x 0.8 x 0.8 x 0.9) is ~37%.
In the future can you please use symbols when modeling data? The colors you used for the donut and the nothing tiles at 1:30 are nearly indecipherable to me, as i am colorblind. Thanks, much appreciated.
Thanks for giving us easy questions sometimes! I used the first method in my head and I feel like I am super smart right now. 🙂 I.e. 65 - 10 - 20 - 20 = 15.
My thought process at the start of the video: Take any of the percentages that ate _one_ food, and subtract the percentage that _didn't_ eat each of the other 3. Starting with donuts: 65 - 20 - 20 - 10 = 15. Starting with pretzels or pizza: 80 - 35 - 20 - 10 = 15. Starting with ice cream: 90 - 35 - 20 - 20 = 15. So the answer would be 15%.
My guess is 15%. Thinking about it geometrically, you can frame it as an unoptimization problem, where you want as many boxes of a 10x10 grid to be touched by a region of 3 and not 4 foods. Because we are minimizing, there isnt any necessary shape the regions must fill, so we can use that to our advantage. First, choose the 65% as our last food applied. Make a 'layer of the 90% food and 10% from the 65, leaving us with 55. Do this again with the 80%s, keeping in mind that the leftovers from the 80s and the 90 don't have to overlap, so the leftover from the first 80% is 55-20, or 35, then after the next 80%, we have 15. Numerically this looks like[ 65 - (100-90) - (100-80) - (100-80)] %, or 15%. Fun problem :)
If we assume that all the non eaters of any category were disjoint (if you don't eat pizza you eat all other 3), then the maximum when you add all up is 10% + 20% + 20% + 35% = 85% meaning at minimum 15% eat all 4, but that could be larger if the don't eat categories are overlapping. If we allow some people to eat none of them, then the max that could eat all 4 would be 65% equal to the smallest category, donuts.
Using the pigeonhole principle is trivial. Use a 100 cells pigeonhole, now distribute 80 pretzels, 80 slizes of pizza, 65 donuts and 90 icecreams and look for how many cells have necessary the 4 items
i took two items and took the inverse of the bigger and subtracted from the smaller. substituting the result and repeating yielded 15 percent. after i saw that any order yields the same solution, i am pretty confident
Another did it all in my head and got 15%, as follows - perhaps it's a variation on Venn diagram, but I find this much easier to visualize, We want minimum % that ate all four - keep that in mind, so, we'll work with the four food percentages, bottom up, 90%, 80%, 80%, 65%, respectively, as follows: think of a square now shade it starting at left, until it's shaded 90% of the way to the right. We want minimum that ate all, so ... now overlap that, minimizing overlap, starting from the right and shading to the left until shaded to 80% of the way back to the left. We minimize the overlap doing it this way, so to get the overlap take the first one and how short it comes of far right, so 100% - 90% = 10% now subtract that from the 2nd that shades right to left by 80%, 80% - 10% = 70%. That's minimum for first two foods. Keep that result, and discard our individual tracking of the first two foods, and similarly repeat, shade from left - 70% must've eaten first two, now shade 3rd food 80% from right to left. How much overlaps? Again, take the first (now combined from first two) of 70% from 100%, that leaves 30% (that may not have eaten the first two), subtract that from the right to left 80% shading of the 3rd food, and that leaves 50%, so 50% must've eaten the first 3 foods. And again, next iteration, shade 50% from left for all that must've eaten first three foods. Now shade 4th food from right to left to 65% - there's an overlap of 15% out of the 100%, so 15% must've eaten all four foods. Could also do similarly thinking along a line segment instead of square - unit length, or length 100, whatever's easiest to think of in one's head. But I found the mental imagery of square shading and thinking of that as % of whole easier to conceptualize, and would work for for any number of people (well, integer people if they're positive integer multiple of our 20 which is smallest that would fit positive integer number of people to all the given percentages). So ... 20 people, ... or 20 million, or 20 billion ... shade the square ... and in all cases, results in 15%.
I used a method similar to the second, take each percentage and minus it from 100 add all the remainders together then take the result away from 100. So it becomes 100-65=35, 100-80=20, 100-80=20, 100-90=10. 35+20+20+10=85. 100-85=15%
OK. Before watching, I thought about it and I was trying to come up with ways to figure out overlap. So if 90% of people ate one item, then only 10% ate something else. Those 10% could have been the people that ate food B, alone, and 20% of those that didn't eat B would have meant only 70% overlap. So 90%=10 80%=20 80%=20 65%=35 10+20+20+35=85. There's at LEAST 15% that ate all four foods
i thought this was actually pretty simple, proof your math videos DO help. i just added up the percentages of people who didnt eat(0.35+0.2+0.2+0.1) and did 1-0.85 to get 15%
@@thomasmaughan4798 we are technically looking for the minimum number of people who would be interested, and yes in practice there may be some overlap however thats not what we are looking for
I feel kind of silly now because I just multiplied the percentages of each of the chances assuming it'd show the probability of someone being selected amongst each group (meaning they ate all of the above food items) getting 37.44%. I am not exactly sure why our answers disagree. I think the difference comes down to me assuming that the amount of food purchased would have no correlation to how many more items would be purchased. But because someone would be less likely to buy food after eating it's not sufficient to merely find the chance, you have to find the minimum because you're looking for value already in the data.
the minimum percentage is 15%, because 35% didn't ate their donuts, 20% pretzels, 20% pizza, and 10% didn't ate their ice cream. While the maximum percentage is simply 65% ate all of their foods, and 10% didn't ate anything
The way I solved it was sort of a blend of the first and second method (basically the second approach but without thinking about it in terms of venn diagrams, and more like in the first approach instead). I started from the end, figuring out what percentage ate pizza and ice cream: ::::::::.. - pizza (80%) :::::::::. - ice cream (90%) Now the minimum possible overlap looks like this: ::::::::.. - pizza (80%) .::::::::: - ice cream (90%) .:::::::.. - overlap (70%) The I considered "pizza&ice cream" as one food and replaced them in the list with that, with 70% as amount, and repeated that until I was left with all 4 foods as one. So I realized that I only needed to figure out how to solve it for just 2 kinds of foods, and from that I could solve it for any amount. And for just 2 kinds of foods I could easily visualize it (as shown above)
If 65% ate donuts then in the "worst" case 65-100+80=45% ate donuts and pretzels, because we are trying to distribute the pretzels so, that we have minimal conjuction of those two sets (first filling up the non-donuts people and only after that filling up the donuts people set). Same is applied to pizza. 45% (those who ate donust and pretzel) => 45-100+80=25% of those who ate donuts, pretzels and pizza. In the end we apply the same logic again for ice cream. 25-100+90=15%. At least 15% of people ate all 4 foods.
Imagine it as 100 people. Start with the 90 that ate ice cream. Worst case scenario is that 20 of them didn't eat pizza, a different 20 didn't eat pretzels, and a different 35 didn't eat donuts. 90 - 20 - 20 - 35 = 15. That still leaves 15 people who must've eaten all 4, since they don't belong to any of the maximally distributed non-eater groups. You could start with any of the groups, do the same thing and arrive at the same answer. Start with the 65 that ate donuts. Worst case is 20 didn't eat pretzels, a different 20 didn't eat pizza, and a different 10 didn't eat ice cream. 65 - 20 - 20 - 10 = 15.
Exactly this problem was (in 1885!) subject of Knot X of 'Tangled Tales' by Lewis Carroll. In 'The Chelsea Pensioners', Carroll writes: _Problem_ - If 70 per cent have lost an eye, 75 per cent an ear, 80 per cent an arm, 85 per cent a leg: what percentage, _at least_ , must have lost all four? _Solution_ - (I adopt that of Polar Star, as being better then my own.) Adding the wounds together, we get 70+75+80+85=310, among 100 men; which gives 3 to each, and 4 to 10 men. Therefore the least percentage is 10."
I seem to have framed it a little differently. I imagined 100 diners and thought that of the 80 eating pizza, there were a maximum of 10 who did _not_ eat ice-cream. So, 70 of those eating pizza _also_ ate ice-cream. Then I thought that of the 70 eating ice-cream and pizza, there were a maximum of 20 who were _not_ eating pretzels. So, 50 of those eating ice-cream and pizza _also_ ate pretzels. Of that 50, there were a maximum of 35 who were not eating ice-cream, pizza _and_ pretzels which means that at least 15 of those eating doughnuts were _also_ eating ice-cream, pizza and pretzels. My method feels sort of similar to the methods shown in the video but it seems somehow different too.
A more visual, but also odd way imo is the following: Imagine a line of 100 people standing from left to right. Now take any of the number of percentages from above, say 80% (pretzel) and color the people from left with yellow. Now, take, say 90% (ice cream) and color the people from right with blue. The overlapping amount of people will be the ones who at minimum ate both pretzels and icecream, which is 70 people. Do that again with the other numbers and you will end up with 15 people = 15% This is by far more complicated than it should be but its the first idea that came to my mind because i wanted to visualize it. Edit: I wrote this comment before watching the video.
My method was similar to the first one. I thought "What would people have to do in order to get these numbers with the least people getting all 4?" And the answer is everyone who didn't buy 4 is buying 3 items exactly. So I'm assuming that everyone who didn't buy ice cream bought a pizza, a pretzel, and a donut. And everyone who didn't buy a pizza bought icecream, a pretzel and a donut and so on. So I took the inverse of every percentage and added them up since assuming everyone is behaving this way, they are mutually exclusive combinations. They added up to 85%. So 85% of people could buy only 3 items and still get these percentages, so the remaining 15% have to buy all 4.
The answer is simple: Let the numbers be named a, b, c, d. Max = Min (a, b, c, d) = Min (65, 80, 80, 90) = 65 Min = a +b + c + d - 300 = 65 + 80 + 80 + 90 - 300 = 15
Found it a lot simpler to inverse the calculations, and said No!Donut (35%) + No!Preztel (20%) + No!Pizza (20%) + No!Ice (10%), which adds up to 85% that can have not eaten all the food. and as such, since it's only possible for 85% of the guest to not have eaten all 4 things, that would mean that at least 15% doesn't fall in this group and as such have actually eaten all 4 things
couldnt you make a case where everyone that had a donut never ate anything else or the same case with any other type of food or is that getting outside of probabilities?
On the other hand what's the likeliest number for those who ate all four? Start by assuming that the probabilities are independent: that someone who eats one of the items isn't more or less likely to eat any one of the three others than someone who doesn't eat the first one. Then the probability of eating all four is 65%*80%*80%*90%=37.44%. Mind you, independent probability is a big if!
Look at that: 35 = no (a); 20 = no (b); 20 = no (c) and 10 = no (d). So, 35 + 20 + 20 + 10 = 85 persons who do not eat at least one of the four types. Thus, (a), and (b), and (c), and (d) do not belong to this set of 85 persons. Therefore, 15 people eat all four types. The union of the four sets may be less than 85. So, the complement may be greater than 15. Therefore 15 people are the minimum.
Here's how I did it and I think it's simpler than the solution given. Suppose we have 100 visitors. 65 had donuts and 80 had pretzels. That's a total of 145, so we know at least 45 people had both. 80 people had pizza and at least 45 had both pretzels and donuts for a total of 125. So at least 25 people had all three. 90 people had ice cream and at least 25 had the other three for a total of 115, so at least 15 people had all four. Yes, this is pretty close to the Venn diagram solution that Presh gave, but I didn't need a diagram, and quickly arrived at the same answer.
Here's how I solved it: *I took 100 people as a sample size to make the calculation easier. I took the 90% first, then that means 10% of the people didn't eat ice cream. *I took the pizza slice(80%) next, then that means 20% of them didn't eat pizza slice, and I carried over the 10% from earlier to here, so that means 30% neither ate the pizza slice nor ice cream(or both). *Same process with the donut(20%). That means 20% of the people didn't eat donuts and I carried the 30% from earlier again here. That means 50% of people didn't eat at least one of the following: Donuts, Pizza Slice and Ice Cream. *Finally, I took the pretzel(65%). That means 35% of the people didn't eat it then I carried over the 50% of the people from the previous proposition. That means 85% of people didn't eat at least one of these 4 items. That leaves me 15% of people who ate all of these 4 items. I don't know how this works, but hey, I got the correct answer and that's what matters. 😅😅😅 On a side note, I realized that 100 people isn't necessary here. 😂😂😂
I guess technically the answer would be somewhere between 15 and 65 percent because you can’t really know for certain with such limited info- but for the sake of the logic problem, yes at least 15 percent ate all 4 foods. Example of how to solve the problem “it’s possible that the 20% that did NOT eat pizza DID eat donuts, so we can remove their 20% from those who ate ALL 4 foods”. 65-20 = 45% then repeat this step in the problem for the for the other two foods and you get 65-20-20-10 = 15 (start with the donut value and subtract the DIFFERENCE between the other percentages of foods eaten and 100%. This tells you the lower limit of AT LEAST what percentage of people MUST have consumed all 4 foods - because that lowest percentile MUST apply to all 4 groups. This will tell you the POSSIBLE percentage of people that MIGHT have eaten the other 3 foods, but not the 4th. It is 50% that might not have eaten 3 or less foods, but not all 4. And so the most accurate answer would be a range of somewhere between 15% and the full 65% because albeit unlikely, you can not rule out that everyone who took a donut also took the other 3 foods as well. Highly unlikely maybe but still likely.
I solved it by finding the lowest common denominator if the percentiles, which is 20, which will be our population of park goers. 13 donuts were eaten (A), 16 pretzels were eaten (B), 16 pizza slices were eaten (C), and 18 ice creams were eaten (D). Then I made a table of 20 people, and distributed A, B, C, and D as evenly as possible -- only 3 people ate all four, or 15%. Not the simplest solution, but it got there.
@@flacsomtodosclas2165 Yes. Good job. You figured that out. Congrats. These videos frequently encourage people to solve the problem in the comments before watching the solutions, which is what I, and many other people commenting, did.
what i did was similar to method 1. I imagined 4 rings with gaps in them. a ring 65% complete (with a 35% gap in it), two 80%complete rings, and so on. Turning the rings so that all the gaps lined up, 65% of it was 4 rings thick (that'd be the maximum). Turning the rings so that none of the gaps line up, 15% of it was 4 rings thicc
of course, the real answer is to figure out that it's a relatively low amount of people and then propose a bundle/voucher with all four items for the price of a bit more than three items and running a good advertising campaign for that. the assumption is thta most bundle users would end up eating three or two items, with only few actually eating all four and getting their money's worth. imo the trick would be to advertise the convenience of it, only paying once to get all four items, which would (in the expectation of the customer) improve service time.
The real world conclusion of this story is that the amusement park sells the bundle and finds out that much lower than 15% buy it (probably even less than 5%). The reason being human behavior; no one wants to eat all four at the same time. Even if the bundle were like coupons to be used throughout the day, most people don't plan on eating all four of those items and/or are hesitant to part with their money up front to commit to eating all four. They'd rather just spend their money at POS when they have a craving for something.
The bundles might be interesting for groups that share the food. I don't think it's a very popular idea anyway, though. A third of the people in this scenario apparently just don't like donuts, so they will be skeptical with a bundle that forces them to buy one.
Before I get into your solutions, I'll take a stab. I reckon what matters is that none of the values are 50% or less. If they were, then the absolute minimum percent shared by all four would be zero. Because of this, the lowest values would determine the minimum percentage shared by all four, because the higher values could not overlap less. So 80% leaves 20% of the patrons uncounted. The 65%, if it accounts for all 20% of those not previously uncounted, must share 45% with the 80%. Thus, 45% must have had both donuts and pretzels, say. Of those 45% counted twice, a minimum of 25% must be counted a third time by the pizza's 80%. Similarly, if all 75% of those patrons only counted twice at this point had ice cream, then the remaining 15% of those who had icecream had all 4 items. So 15% of the patrons had all 4 items. If anyone had nothing, then even more would have had all 4.
I just subtracted the 20 20 10 from 65 to get min. You figure the worst case scenario is the people that didn't eat one thing would eat the other. It did confuse for a while, but ended up simple
Worse case scenario.... Put up a graph. The highest 90% goes one way (ltr). and the lowest 65%(rtl) The two 80% ones (pizza and pretzels) one goes from ltr and the other goes rtl, respectively. The two 80% ones turn into only a 40% overlap. This overlap is covered by the 90% by default, thus no change. However, the 65% can only cover 15% less space from the other side, thus the lowest possible amount that all four amounts can overlap is 25% That said, second best case is if both 80% are on the same side (ltr), while the 65% and 90% cover the other side (rtl) then it's only the 65% vs the 80% that we'd have to worry about, so that'd be 45% in the overlap. And of course the absolute best case is if all 65% of the donut eaters had all four, so that's be a max of 65%
I got the right answer but possibly by the incorrect method: 90% had ice cream, 10% didn’t 80% had pizza, assume the 20% that didn’t also didn’t have ice cream = 70% 80% had pretzels, 20% didn’t, assume they aren’t pizza-ice cream eaters = 50% 65% had donuts, 35% didn’t, assume they didn’t have any of the others = 50-35 = 15% I just evaluated it linearly, but I can’t figure out if this is different from evaluating it as a binary tree as shown in the video?
My intuitive guess was to imagine the 65% and the 80% filling a bar from opposite ends. Because 65% definitely ate doughnuts, and 80% definitely ate pretzels, then the overlap is the minimum people that ate both. Doing this for pizza and ice cream will net you the same number or higher so it had to be 15. Also side note, I just realised typing this out, this is my exact mental method for solving those nonogram puzzles
I guess I solved it the first way, though I didn't think of it like moving colors around. I just thought: 10% of guests didn't buy ice cream. Let's say a completely different 20% of guests didn't buy pizza, and a completely different 20% of guests didn't buy a pretzel, and a a completely different 35% of guests didn't buy a donut. That still leaves 15% who must've bought all four. 100 - 10 - 20 - 20 - 35 = 15.
That's the equivalent of saying 90% of people who ate ice cream ate pretzels, and 80% of those people ate pizza, etc. Which could be true, if we were looking at probability and the random chances a single person ate all 4 assuming all events are independent the PROBABILITY would be 37.44%. However we aren't looking at random probability a single person ate all 4, we are looking for the minimum number possible which is a completely different question. Just because 90% of the TOTAL # of people ate ice cream doesn't necessarily mean 90% of the people who ate pretzels also ate ice cream.
That sort of process works if the four percentages are independent variables! But in this case, the question requires us to assume that the variables are dependent on each other to calculate a worst case scenario.
Unfortunately no as you’re assuming that the portion of people who bought donuts are buying at the exact same ratios as the overall population which my not be true leading to an overestimate
my spin on this problem before watching: make a 10 by 10 grid 90 are for ice cream the 10 unused are used for pizza,then 70 overlap 30 spaces have only one food,fill them up with pretzels,50 rest overlap with the rest 50 spaces with only two foods are used for dough nuts,15 left for overlap with the rest -->15% guaranteed overlap yay,I was right!
Solved in 3 mins.. very satisfied, thank you! I solved by considering how many % of all people ever tried one of the food, then 2, then 3 and finally all 4.
I guess 15% my thought process is All - (people that didn't eat donut) - (people that didn't eat pretzel) - (people that didn't eat pizza) - (people that didn't eat ice cream) 100 - (100-65) - (100-80) - (100-80) - (100-90) 100 - 35 - 20 - 20 - 10 15 (method 1) huh didn't think to visualize it like that, basically my thought but easier
It's not as simple as raw math. Just introduce the 4-pack. If it sells out, raise the quantity. If you have excess at the end of the day, lower the quantity
Simple solution: negate everything (35 did not eat donut, 20 pretzel, 20 pizza, 10 ice cream), add them up and you get that 85% skipped a food, so 100-85=15% who didn't skip any, thus ate all.
You are forgetting about the fact that some visitors could have eaten NONE of the 4 foods ... that could be as high as 10% (ice cream max 90%). Minimum who ate all 4 foods then is actually 0% if you do a 4-circle Venn diagram with all possible intersections. Put a 0 in the very middle quadruple-intersection. Put 15 in the donut/pretzel/pizza triple-intersection. Put 25 in the donut/pizza/icecream triple-intersection and another 25 in the donut/pretzel/icecream triple-intersection. And finally put 40 in the pizza/pretzel/icecream triple-intersection. Put zero everywhere else in the diagram, and 10 outside all 4 circles. I solved it with a system of equations, with 10 outside the circles, so needing the locations within the circles to total 90. Then forced the 0 in the middle. Then some algebra from there. The symmetry of the pizza and pretzels helped some. There may be other solutions.
Take the lowest percentage .. 65% .. and now work out the 'did not eat' percentage for the other three .. 20% twice and 10% once .. add them together .. 20 + 20 + 10 = 50 .. and now subtract 50 from 65 .. giving you 15 .. and telling you that the minimum number of people who ate all four is 15% .. going up to a maximum of 65%.
I did the Venn diagram trick, but started with Ice cream and worked down. If 90/100 people ate Ice cream then 10/100 did not eat Ice Cream. If 80/100 ate Pizza Slices then at most 10 of those people also did not eat Ice Cream, therefore at least 70/100 people ate both Pizza and Ice Cream. If at least 70/100 people ate both Pizza and Ice Cream then at most 30/100 people did not eat both. If 80/100 people ate Pretzels then at most 30 of those people did not also eat both Pizza and Ice Cream, therefore at least 50/100 people at all three Pretzels, Pizza, and Ice Cream. If 50/100 people at all three then 50/100 did not eat all three. If 65/100 people ate donuts then at most 50 of those people did not also eat all three other options, therefore at least 15/100 people ate all four food items offered.
The scenario is surprisingly interresting though. The minimum (15%) and the maximum (65%) are very different, and both very unlikely. More probable (but still likely quite a bit from the real value) is the product of the percentages (0.65 x 0.8 x 0.8 x 0.9 ≈ 0.37%). That assumes that the numbers are independent, and they probably aren't. I wonder how close a statistician would get to the real answer having no statistics beyond the percentages given, but with all other food related statistics (general statisticcs and spesific statistics ralated to places like amusement parks) available. Some will allways go all out on food stuffs. Some avoid expensive tourist food at all costs. I guess most people will maybe try to keep to one or two items (It's still expensive, and we tent to get full after all), and that would drag the "ate all" group in direction of the minimum. I wonder if there is some statistical constant/formula that will give a good answer to cases like this.
Minimum percentage can always be determined by picking the lowest percentage (in this case, 🍩 at 65%), and then subtracting away all the "did not eat X" percentages. In this case, those are 20%, 20%, and 10%, for a total of 50%. 65% - 50% = 15%. The actual amount that ate all four items is probably higher, up to maximum of the lowest percentage, 65%.
Yeah I remember being in school and struggling with a variation of this problem. Is there any flaw with that reasoning?
That's how I solved it.
You don't have to start with the lowest percentage, any of them work as a starting point.
@@knoxiegb1782 Formally speaking, what I did was just a specific application of the proof given near the end of the video. The union of WXYZ is no smaller than the whole space (100 percentage points) minus the complement of each individual set. Thing is, (whole space) - (complement of set W) = the set W, by definition.
You can, as other replies have said, pick any starting point. I just find it most natural to start from the smallest one first.
max(0,100-sum(100-x for x in percentages))
simpler.
For the practical application, you actually do NOT want to target the people who bought all 4. Bundles are typically cheaper than the sum of the individual items - that's the point of the bundle. If we only sell the bundle to people who bought all 4 anyway, we just lose money.
The people we ACTUALLY want to target are the ones who bought less, but might be convinced to buy all 4. Most likely, the people who usually buy 3 out of 4 could be tempted by a bundle offer. The donut merchant would be the most likely to profit from this, as most people who only skipped out on 1 probably skipped the donut.
The information request is still useful though. We know, that with such bundle we loose with at least 15% of the customers, and up to 65%. So, looking for info about people who bought all 4 makes sense, but we should focus on the maximum number of people who did that. Then we can set up the pricing in such way, that even in worst case scenario extra spending of the part of 35% of customers will make up for the loss on 65%. This is of course a "safe bet" scenario, most companies probably could tolerate some risk and look for "most likely % of people who bought all 4" to use in estimations instead.
@@edsimnett The four numbers most certainly aren't independent variables. People have preferences and don't truly randomly buy food in the real world. Upper and lower boundaries however still apply.
You also have to measure the time people bought items. If people bought all four at once or 3 items at once (usually a family) that's different than an individual buying all four items but over a longer period of time.
Keep in mind that the *_"geniuses in marketing"_* was meant sarcastically.
@@willguggn2 true, and for a small sampling ..not very informative
..but on a larger sampling .. the errors smooth out...
as in the exaple lets assume 20 ppl...the variance from day to day will be massive
if we say saple 20000/day for a week ... i would say the error becomes so small .. we can with a quite high degree of certainty predict the sales... in the future... and what prferences ppl have .. we can then add other factors as weather, pricing etc...
...we can also just by using common sense do estimations (if its 10 deg and windy, the probability our outdoor ice cream stand sell alot of icecream is quite low)
The easiest way is calculating the inverse. The max amount of people that didn't eat all 4. You just add the percentages of people who didn't eat each food. 10 for ice-cream, 20 for pretzels and pizza, and 35 for donut. 10+20+20+35=85, which means at most 85% didn't eat all 4 foods, so at least 15% had to eat all 4.
Nice, but don't forget to also limit the answer to between 0-100% .
If it exceeds 100% subtracted, you can state that there is no portion of the populace that guaranteed ate all types of food.
Exactly! @@RoderickEtheria
@@ShaiFishmanwow i would have missed good content if i haven't come through comment section 😄
this is how I solved it
I solved it by inverting the percents (100-p)%, then adding it together, to figure out what percentage at most didn’t eat at least one food
Best solution
You can make that mathematically sound by showing that the minimum percentage goes up if there are overlaps between the people who don't eat each food. It's still the best way though.
same!
Or add them together and do (mod 100).
I did the same thing and arrived at 15%, but since I did it "intuitively", I didn't have confidence in the result.
Those looking for a simpler method: I did this problem instantly in my head. 100-65 = 35, 100-80 = 20, 100-80 = 20 & 100-90 = 10. Add all these numbers up (35+20+20+10) = 85. 100 - 85 = 15%! Explanation: 35% didn’t eat a doughnut, 20% didn’t eat a pretzel etc. So 85% didn’t eat all 4 foods. So 15% did!
Here's what I did:
65% (Donut) + 80% (Pretzel) = 145%. 145% - 100% = *45% of people who ate a Donut/Pretzel ate both of them.*
45% (Donut+Pretzel) + 80% (Pizza) = 125%. 125% - 100% = *25% of people who ate a Donut, Pretzel, or Pizza ate all 3 of them.*
25% (Donut + Pretzel + Pizza) + 90% (Ice Cream) = 115%. 115% - 100% = *15% of people who ate any of the 4 foods ate all 4 of them.*
Therefore, at least 15% of customers must have eaten all four foods.
My elementary school way is the simplest (and also proves it at the same time)
* 65% eat A, 80% eat B, so "eat A and B" is at least 65% + 80% - 100% = 45%
* now same logic, 45% eat AB, 80% eat C, therefore "eat ABC" is at least 25%
* eat ABCD is at least 15%
Each step itself also naively proved that it's a reachable lower bound (thus minimum)
The calculation is even simpler as:
1. you remove the 100th so it's smaller number
2. you can early exit if the sum is less than 100 and claim that the min is already 0 without adding all other numbers.
This is the one correct solution.
I solved it the same way, but I would have put parentheses, otherwise the solution would be unclear without them. We assume that in our group of eaters it is precisely those people who did not eat other food; 65% - (100% - 80%) - (100% - 80%) - (100% - 90%) = 80% - (100% - 65%) - (100% - 80%) - (100% - 90%) = 90% - (100% - 65%) - (100% - 80%) - (100% - 80%)
This is basically the same first method from the video... Although the second one is similar. I don’t see any difference between the solutions, we rely on the same thing
@@oqui7009
1. 65% - (100% - 80%) is no more clear than simply 65% + 80% - 100%, if you think about it. In fact I believe the entire elementary school "math" stuff is about thinking through these kind of logics.
2. you forgot about the possibility of "early exit". We do not need to write everything down at once. We can do the simple arithmetic in our mind and whenever we get negative number we know it's 0 and there is no need for further calculations.
That's how I did it. Took me about 20-30 seconds.
In the first calculation about "A and B" how did you count for those who didn't eat either A or B ?
Its the set intersection formula
n(A or B) = n(A) + n(B) -n(A and B). But in this case how can one assume that n(A or B) is infact 100% of the total sample?
Also there are a minimum of 10% of total sample who did not eat either A or B
I approached this by simplifying it to 2 foods. This can be visualized by two sliding bars (line segments) of various lengths which must be positioned in a specific space (100%). How would we minimize the overlap? Easy, we put one bar all the way to one side and the other bar all the way to the other. So for example with Ice Cream (90%) and Pizza (80%), the mininum overlap would be 70%. But now we know a minimum of 70% had both Ice Cream and Pizza, so I can repeat this with the Pretzel (80%). So mininum 50% had Ice Cream/Pizza/Pretzel. Apply this once more with Donut (65%) and the minimum overlap is 15%.
Yes, minimal possible overlap of the two lowest percentages, where you stack one from the "left" and one from the "right".
That would also tell you right away, that if two food items were at maximum 50%, the minimum of people who ate everything is 0%.
Actually, if your two items with the lowest percentages would "sum up" and that combined result is less than 100%, your minimal possible overlap would always be 0%.
In an extreme example, out of 1000 people you have 999 eating pizza, and 1 person eating brussel sprouts. Those 999 and 1 can be completely different groups of individuals - so you can have 0 overlap.
Adding more food items to your combined considerarions here won't have any impact on the overall minimum across ALL food items then.
That's exactly how I did it. Too many smart people in Presh's audience, LOL
I tried to do something similar but my approach was very complicated, I tried to make a square with 10 smaller boxes along each arm, thus 100 of them inside. Then, I tried representing each beverage along each corner to minimize the overlap...but i forgot that the result will be skewed if I used multiple beverages at the same time, I think I had to start with two beverages and put them at the diagonally opposite end then calculate their overlap and continue like you, but yours is much better and simple. Also, what if the percentage of one of the beverages was way less like 2 or 3% then we wouldn't get any overlap, would that mean that but is possible that no one bought all four?
@@aditya37mviews3weeksago3
Yes, if you have one food item with like only 3%, any of the other food items would need to have over 97%, so that there could be a minimal, but none-zero overlap among all.
Minimum 15%. Maximum 65%. Expected percent of people to eat all 4 foods 37.44%.
So that’s what my wrong calculation of 37.44% measured. You’re saying this percent measures the expected percent of people to eat all 4 foods. But why is the expected percent, the minimum percent, and the maximum percent all able to be calculated yet mean different things. I am not sure I actually understand what these three terms means in this contexts. All I know is I had thought I was finding the solution when I multiplied the 4 original percentages together to get 37.44% but found out I was mistaken. The 65% for the maximum makes intuitive sense to me as well I think but I don’t get why the expected is not the same as the minimum in this case?
@@kennyalbano1922 Expected is more akin to the average. The minimum is essentially assuming as much as possible that people didn't eat all four and getting the smallest number which satisfies that criteria.
Two easy examples to illustrate. If you flip a fair coin 20 times, the expected number of heads is 10. The minimum is 0, and the maximum is 20; both are unlikely, but technically possible.
If you roll 2 fair 6-sided dice, the expected total of roll of the dice is 7; as that's what you'll get on average. The minimum is 2, and the maximum is 12. Amusingly the expected roll of a single die is 3.5; even though that's not technically possible.
@@WombatMan64 are there any ways to measure spread with this type of data. Such as range and outliers, does it follow a bell curve and if so what are its standard deviation and variance? Can the data be described by an equation and displayed graphically? I mean where the 4 foods intersect as in the question by the data.
@@kennyalbano1922 not really with only the information supplied.
If you had a table with every single customer and exactly what they purchased, and had that info every day then no doubt you could do some good analysis, but the information provided in this question is quite... minimal. Appropriate really.
@@edsimnett Actually it can be very useful. Businesses work on a daily basis. If you're gonna do business with a company, it's not only important to know how much someone sells on average and how much on the good days, but also how much they sell on the worst days (and also the factors contributing to the lowest and highest sales).
42 seconds in. "Amusement park: What is the minimum percentage of visitors that must have eaten all 4 foods?" 0% Odds are some did not buy any food. - Solved!
I did it by thinking of a group of 100 people, and taking it at groups of 2 at a time. If you take the first two groups, 80 people had the pretzel, which means minus the 35 that didnt have the donut you still need 45 that did. Then for pizza you take the 55 away that didnt have both donut and pretezel but you still need 25 that had both, so then you need 25 that had all 3. Then for icecream you take 75 away that didnt have all previous 3, but that still leaves you with 15 that did, so atleast 15 had to have all 4.
I can never solve the puzzles in the channel because when I open the video the solution is right in front of me in the top comment.
I did the second method but then I realized it's easier:
Assuming that there are 100 people and no person ate the same item twice.
If all of them ate all 4 items there would be 400 sales.
But if we add up the numbers there has been 315 sales.
So at most 85 people did not eat all 4 items.
That means at minimum 15 people ate all 4 items.
> no person ate the same item twice.
The thing is, in the problem description this is left implicit. Somewhere while explaining the answers, this assumption was suddenly put into place.
Before that, nobody excluded a person eating, say, 4 slices of pizza.
@@metamud8686 Right. But maybe it's not important how much one person bought of each item. So for example if a person buys 1 donut or 10 doesn't matter. The question then would be, did they buy it or not.
35% didn't eat a donut, 20% didn't eat a pretzel, 20% didn't eat a pizza, 10% didn't eat an ice cream. Worst case scenario : they are not the same people, therefore 35 + 20 + 20 +10 = 85% of people didn't eat one item, so 15% ate all 4.
I loved seeing it worked in many different ways. This is the real skill, being able to approach things in multiple ways and gauge which one will be simplest.
Isnt this taking into account that there could be people who ate nothing? There is no data saying that everyone in the park ate something, right?
I was mentally thinking of Method 2, but was physically trying to write it out using Method 1, and using 100 people.
The "No Doughnut" set contains 35% of the visitors. "No Pretzels" has 20% of them, "No Pizza" has 20% and "No Ice Cream" has 10%. Anyone not in any of those sets must have bought all four foods. To minimise these people you need to maximise the numbers in the above-mentioned four sets, and to do that you need to assume no sharing of people between them. In that case you simply sum the sets' percentages to get the total %age of visitors that did not buy all four foods, because in doing so no one gets counted twice: 35+20+20+10 = 85. Therefore 15% bought all four foods in this extreme case.
I got the first 65% 80% trick to arrive at 45% due to seeing this kind of puzzle but didn't think to extend it out further to all four, cool trick!
It's also interesting to note that, so long as each set contains less than 100%, you can keep adding more and more orthogonal sets and eventually arrive at 0% minimum overlap
Eg, for sets that contain 50% of a population, you need 2 such sets to have a minimum overlap of 0%, if they cover 75% then you need 4 sets, 90% coverage requires 10 such sets, etc etc but however many ways you have of grouping people, so long as you have enough of them you'll be able to show that the minimum possible overlap is 0%
I just calculated extra after mini 65% overlapping in other three food found easy 15℅ to be minimum.
Intuitive answer is 15%. There is an algorithm to execute which makes this exceptionally easy.
Basic description:
You take the two lowest percentages to start. You have to calculate their intersection so that is going to be
65-(100-80)=45
then you take that answer and intersect the 3rd.
45 - (100-80) = 25
Take that number and intersect the 4th.
25 - (100-90) = 15.
Done; it should take you approximately 3 seconds to do in your head.
Algorithm:
var accumulator = 0
foreach(var p in list){
if accumulator ==0 accumulator = p
else accumulator -= (100-p)
}
accumulator will contain the intersection as long as there is one. A defense is necessary to prevent a disjoint set from resetting the accumulator to a positive number, but I'm not presenting this as a general solution to the problem of all intersecting sets.
You have to sort them lowest to highest. It would have been a better problem if he had presented them out of sort order
Seems like a great problem to illustrate key ideas like DeMorgan's law and the Union Bound, but I don't think the connection was made as clearly as it could have been. There's a lot of intuition that was brought out be each solution, I think the connection to formalized concepts could have been a lot stronger.
From a marketing perspective: if on average everyone is already buying at least three food items, there’s really no need to offer a combo discount.
Maybe, unless part of the goal is to encourage those who were getting just three items to "just pay a little bit extra" to get all 4. Particularly if the difference between 3 items and the discounted 4 items is more than the unit cost.
another simple way to calculate it.
First pick a starting food, and keep that percentage of people who had it.
Then list out the percentage of people who havent eaten any of the other given foods.
From there, just start subtracting.
65% of people ate a donut
20% of people did not eat a donut (65-20=45)
20% of people did not eat a pizza (45-20=25)
10% of people did not eat a ice cream (25-10=15)
There for, the answer is 15%
After my headache subsides, I'll watch this video again and see if I can follow the logic this time.
neat problem with many ways to solve! i drew the sample space as a 4 by 5 grid, such that every cell in the grid represented 5%. I then filled the grids with markers for each food, each time adding the nth food starting by avoiding the cells that already had (n-1) foods in them. this leaves only three of the 20 cells with all four markers!
I did exactly this!
Solving Geometrically:
make a square ABCD of sides 10 each. this represents 100% of visitors. Assign each vertex an food type. lets say
A = Donut
B = Pretzel
C = Pizza
D = Ice cream
Draw a smaller square from the vertex A of side SQRT(65)
Draw a smaller square from the vertex B of side SQRT(80)
Draw a smaller square from the vertex C of side SQRT(80)
Draw a smaller square from the vertex D of side SQRT(90)
The area of rectangle represented by the intersection of all 4 smaller squares is the minimum percentage of people who ate all 4 foods.
First: 90% vs 80%, the 10% that didn’t eat from the 90% could of ate part of 80%
Thus, there is a minimum of 70% overlap
Second: 80% vs 80%, since one of the 80% has a 70% overlap from before, 30% of the people who didn’t overlap could of ate from the other 80%
Thus, there is a minimum of 50% overlap
Third: 65% vs 80%, since the 80% has a 50% overlap from before, 50% of people who didn’t overlap could of ate from the 65%
Thus, there is a minimum of 15% overlap.
Didn’t see many people point out the logic, hope this helps.
Actually you ask two different questions, how many ate all the foods and what is the min percentage that ate them all. I agree with you about the minimum, it is 15%. But I think a better answer to the question, "how many?" is 37.44% based on converting the percentages to probabilities and multiplying them to get the probability that a given customer ate all items.
I don’t think he ever asked how many, just the minimum
@@pivotfever You are right that he does treat it as a question about minimum numbers, but he does start by saying that what they are really interested in is how many would buy all three, and the minimum number that ate all three in this sample is not the best statistic to use for that.
@@MichaelGowland well he says the marketing people wanna know how many people are interested, not interested in how many people. Thus the actual question, what’s the minimum percentage, not most likely
Well, this was an easy one.
Just imagine each one of the non-eaters don’t overlap, so its just 100%-35%-20%-20%-10%=15%
The opening statement is that "the food vendors tell you..." the percentage of foods eaten. But from their perspective, each of the people must buy at least *one* food item (otherwise the vendors would not be able to count the person). So the maximum percentage who eat all four foods should be
There is no 'lesser than' the maximum. The least eaten food item is eaten by exactly 65% of the people (who buy at least 1 food item). Therefor, the maximum amount of possible people that bought all 4 food items is 65%. It's not at most 65%. It is exactly 65%.
@@thenonexistinghero Show me a solution using the: 20 people example...
@@DanielFSmith I mean, it's just common sense. Here's a solution with 100 people.
65% of them eat at least 1 food type.
So 65 people eat at least 1 food type.
It also means 65 people can eat all 4 food types at most.
The rest of the percentages don't matter. 65% is the lowest. Whether it's 80% or 90% of people that eat other food types, it's still 65 people at most who can eat all 4 since there's a good chunk of at least 35% that haven't eaten the least popular food and as such, cannot possibly have eaten all types.
@@thenonexistinghero Let's take an example with 60% and, say, 90% using 10 people.
Pie. aaaaaa0000
Cake. bbbbbbbb0
Doesn't work... The last person didn't eat at least one. You can only get 50% max.
@@DanielFSmith They aren't eating from a single slice of pie. The amount of pies is limitless.
I either got lucky or found a much easier way to do this:
-Adding up the percentages of those that didn't eat each food equals 85%.
-Subtract that answer from 100%, which equals 15%.
I can't explain why that worked, but it apparently did.
There's an easier way to solve this. Taking Doughnuts as the example, you add up the number of people that did not eat Pretzels (20%), Pizza (20%), or Ice Cream (10%); and then subtract that sum from the total number of people who ate Doughnuts. 20+20+10=50 and 65-50=15 so 15 is the answer. You can technically do this with any of the foods. Taking Pretzels for the next example, 35+20+10=65 and 80-65=15.
I tried at first to follow the method of option A with weird bar graphs but gave up and found myself doing naturally going with option B.
I wondered if I had the right idea pairing donuts with pretzels and pizza and ice cream but it seemed to make sense to me. And I'm glad the video showed that I was right!
I just saw a video with a math puzzle on it.
At the top of a paper write "I have $50"
Now divide the paper into two columns, Spent Balance
20. 30
15. 15
09. 06
06. 00
For each of these you spend some amount of money, then you write your new balance in the balance column.
Now total up each column.
The spent adds up to 50, which we expected. Yet the balances all add up to 51...where did that extra one dollar come from?
So the minimum that bought all four is 15%, the max is 65%, and if all four of these variables are independent then the expected case (0.65 x 0.8 x 0.8 x 0.9) is ~37%.
In the future can you please use symbols when modeling data? The colors you used for the donut and the nothing tiles at 1:30 are nearly indecipherable to me, as i am colorblind. Thanks, much appreciated.
Another shortcut...
*Sum, then subtract 300%.*
that was easy. i found the answer in my head after 30sec thought and i'm average at these puzzles.
Thanks for giving us easy questions sometimes! I used the first method in my head and I feel like I am super smart right now. 🙂 I.e. 65 - 10 - 20 - 20 = 15.
Same! I started with figuring out 65 - 10 made sense. And then said, "wait, it's as easy as just subtracting the two 20s now."
My thought process at the start of the video: Take any of the percentages that ate _one_ food, and subtract the percentage that _didn't_ eat each of the other 3.
Starting with donuts: 65 - 20 - 20 - 10 = 15. Starting with pretzels or pizza: 80 - 35 - 20 - 10 = 15. Starting with ice cream: 90 - 35 - 20 - 20 = 15.
So the answer would be 15%.
My guess is 15%. Thinking about it geometrically, you can frame it as an unoptimization problem, where you want as many boxes of a 10x10 grid to be touched by a region of 3 and not 4 foods. Because we are minimizing, there isnt any necessary shape the regions must fill, so we can use that to our advantage. First, choose the 65% as our last food applied. Make a 'layer of the 90% food and 10% from the 65, leaving us with 55. Do this again with the 80%s, keeping in mind that the leftovers from the 80s and the 90 don't have to overlap, so the leftover from the first 80% is 55-20, or 35, then after the next 80%, we have 15. Numerically this looks like[ 65 - (100-90) - (100-80) - (100-80)] %, or 15%. Fun problem :)
If we assume that all the non eaters of any category were disjoint (if you don't eat pizza you eat all other 3), then the maximum when you add all up is 10% + 20% + 20% + 35% = 85% meaning at minimum 15% eat all 4, but that could be larger if the don't eat categories are overlapping. If we allow some people to eat none of them, then the max that could eat all 4 would be 65% equal to the smallest category, donuts.
Using the pigeonhole principle is trivial. Use a 100 cells pigeonhole, now distribute 80 pretzels, 80 slizes of pizza, 65 donuts and 90 icecreams and look for how many cells have necessary the 4 items
i took two items and took the inverse of the bigger and subtracted from the smaller. substituting the result and repeating yielded 15 percent. after i saw that any order yields the same solution, i am pretty confident
This is basically gerrymandering for food groups.
Another did it all in my head and got 15%, as follows - perhaps it's a variation on Venn diagram, but I find this much easier to visualize,
We want minimum % that ate all four - keep that in mind, so, we'll work with the four food percentages, bottom up, 90%, 80%, 80%, 65%, respectively, as follows:
think of a square
now shade it starting at left, until it's shaded 90% of the way to the right. We want minimum that ate all, so ...
now overlap that, minimizing overlap, starting from the right and shading to the left until shaded to 80% of the way back to the left.
We minimize the overlap doing it this way, so to get the overlap take the first one and how short it comes of far right, so 100% - 90% = 10%
now subtract that from the 2nd that shades right to left by 80%, 80% - 10% = 70%. That's minimum for first two foods.
Keep that result, and discard our individual tracking of the first two foods, and similarly repeat, shade from left - 70% must've eaten first two,
now shade 3rd food 80% from right to left. How much overlaps? Again, take the first (now combined from first two) of 70% from 100%, that leaves 30% (that may not have eaten the first two),
subtract that from the right to left 80% shading of the 3rd food, and that leaves 50%, so 50% must've eaten the first 3 foods.
And again, next iteration, shade 50% from left for all that must've eaten first three foods.
Now shade 4th food from right to left to 65% - there's an overlap of 15% out of the 100%,
so 15% must've eaten all four foods. Could also do similarly thinking along a line segment instead of square - unit length, or length 100, whatever's easiest to think of in one's head. But I found the mental imagery of square shading and thinking of that as % of whole easier to conceptualize, and would work for for any number of people (well, integer people if they're positive integer multiple of our 20 which is smallest that would fit positive integer number of people to all the given percentages). So ... 20 people, ... or 20 million, or 20 billion ... shade the square ... and in all cases, results in 15%.
I used a method similar to the second, take each percentage and minus it from 100 add all the remainders together then take the result away from 100. So it becomes 100-65=35, 100-80=20, 100-80=20, 100-90=10. 35+20+20+10=85. 100-85=15%
OK. Before watching, I thought about it and I was trying to come up with ways to figure out overlap.
So if 90% of people ate one item, then only 10% ate something else. Those 10% could have been the people that ate food B, alone, and 20% of those that didn't eat B would have meant only 70% overlap.
So 90%=10
80%=20
80%=20
65%=35
10+20+20+35=85.
There's at LEAST 15% that ate all four foods
i thought this was actually pretty simple, proof your math videos DO help. i just added up the percentages of people who didnt eat(0.35+0.2+0.2+0.1) and did 1-0.85 to get 15%
@@thomasmaughan4798 we are technically looking for the minimum number of people who would be interested, and yes in practice there may be some overlap however thats not what we are looking for
I feel kind of silly now because I just multiplied the percentages of each of the chances assuming it'd show the probability of someone being selected amongst each group (meaning they ate all of the above food items) getting 37.44%.
I am not exactly sure why our answers disagree.
I think the difference comes down to me assuming that the amount of food purchased would have no correlation to how many more items would be purchased.
But because someone would be less likely to buy food after eating it's not sufficient to merely find the chance, you have to find the minimum because you're looking for value already in the data.
the minimum percentage is 15%, because 35% didn't ate their donuts, 20% pretzels, 20% pizza, and 10% didn't ate their ice cream. While the maximum percentage is simply 65% ate all of their foods, and 10% didn't ate anything
The way I solved it was sort of a blend of the first and second method (basically the second approach but without thinking about it in terms of venn diagrams, and more like in the first approach instead). I started from the end, figuring out what percentage ate pizza and ice cream:
::::::::.. - pizza (80%)
:::::::::. - ice cream (90%)
Now the minimum possible overlap looks like this:
::::::::.. - pizza (80%)
.::::::::: - ice cream (90%)
.:::::::.. - overlap (70%)
The I considered "pizza&ice cream" as one food and replaced them in the list with that, with 70% as amount, and repeated that until I was left with all 4 foods as one.
So I realized that I only needed to figure out how to solve it for just 2 kinds of foods, and from that I could solve it for any amount. And for just 2 kinds of foods I could easily visualize it (as shown above)
If 65% ate donuts then in the "worst" case 65-100+80=45% ate donuts and pretzels, because we are trying to distribute the pretzels so, that we have minimal conjuction of those two sets (first filling up the non-donuts people and only after that filling up the donuts people set). Same is applied to pizza. 45% (those who ate donust and pretzel) => 45-100+80=25% of those who ate donuts, pretzels and pizza. In the end we apply the same logic again for ice cream. 25-100+90=15%. At least 15% of people ate all 4 foods.
Imagine it as 100 people. Start with the 90 that ate ice cream. Worst case scenario is that 20 of them didn't eat pizza, a different 20 didn't eat pretzels, and a different 35 didn't eat donuts. 90 - 20 - 20 - 35 = 15. That still leaves 15 people who must've eaten all 4, since they don't belong to any of the maximally distributed non-eater groups. You could start with any of the groups, do the same thing and arrive at the same answer. Start with the 65 that ate donuts. Worst case is 20 didn't eat pretzels, a different 20 didn't eat pizza, and a different 10 didn't eat ice cream. 65 - 20 - 20 - 10 = 15.
Exactly this problem was (in 1885!) subject of Knot X of 'Tangled Tales' by Lewis Carroll. In 'The Chelsea Pensioners', Carroll writes:
_Problem_ - If 70 per cent have lost an eye, 75 per cent an ear, 80 per cent an arm, 85 per cent a leg: what percentage, _at least_ , must have lost all four?
_Solution_ - (I adopt that of Polar Star, as being better then my own.) Adding the wounds together, we get 70+75+80+85=310, among 100 men; which gives 3 to each, and 4 to 10 men. Therefore the least percentage is 10."
The real question is, why is everyone eating so much food? The median guest ate at 3 distinct food stands during their trip, that's nuts
Given that they could just be getting a drink or a snack, it’s not crazy that they could be visiting 3 food stands in a day at an amusement park.
I seem to have framed it a little differently. I imagined 100 diners and thought that of the 80 eating pizza, there were a maximum of 10 who did _not_ eat ice-cream. So, 70 of those eating pizza _also_ ate ice-cream. Then I thought that of the 70 eating ice-cream and pizza, there were a maximum of 20 who were _not_ eating pretzels. So, 50 of those eating ice-cream and pizza _also_ ate pretzels. Of that 50, there were a maximum of 35 who were not eating ice-cream, pizza _and_ pretzels which means that at least 15 of those eating doughnuts were _also_ eating ice-cream, pizza and pretzels.
My method feels sort of similar to the methods shown in the video but it seems somehow different too.
A more visual, but also odd way imo is the following:
Imagine a line of 100 people standing from left to right.
Now take any of the number of percentages from above, say 80% (pretzel) and color the people from left with yellow. Now, take, say 90% (ice cream) and color the people from right with blue. The overlapping amount of people will be the ones who at minimum ate both pretzels and icecream, which is 70 people.
Do that again with the other numbers and you will end up with 15 people = 15%
This is by far more complicated than it should be but its the first idea that came to my mind because i wanted to visualize it.
Edit: I wrote this comment before watching the video.
My method was similar to the first one. I thought "What would people have to do in order to get these numbers with the least people getting all 4?" And the answer is everyone who didn't buy 4 is buying 3 items exactly. So I'm assuming that everyone who didn't buy ice cream bought a pizza, a pretzel, and a donut. And everyone who didn't buy a pizza bought icecream, a pretzel and a donut and so on. So I took the inverse of every percentage and added them up since assuming everyone is behaving this way, they are mutually exclusive combinations. They added up to 85%. So 85% of people could buy only 3 items and still get these percentages, so the remaining 15% have to buy all 4.
Nice to have an easy problem every so often.
The answer is simple:
Let the numbers be named a, b, c, d.
Max = Min (a, b, c, d) = Min (65, 80, 80, 90) = 65
Min = a +b + c + d - 300 = 65 + 80 + 80 + 90 - 300 = 15
Found it a lot simpler to inverse the calculations, and said No!Donut (35%) + No!Preztel (20%) + No!Pizza (20%) + No!Ice (10%), which adds up to 85% that can have not eaten all the food. and as such, since it's only possible for 85% of the guest to not have eaten all 4 things, that would mean that at least 15% doesn't fall in this group and as such have actually eaten all 4 things
couldnt you make a case where everyone that had a donut never ate anything else or the same case with any other type of food or is that getting outside of probabilities?
On the other hand what's the likeliest number for those who ate all four? Start by assuming that the probabilities are independent: that someone who eats one of the items isn't more or less likely to eat any one of the three others than someone who doesn't eat the first one. Then the probability of eating all four is 65%*80%*80%*90%=37.44%.
Mind you, independent probability is a big if!
I usually don't do this, but solved it in less than 10 seconds. More of a logic puzzle than math.
Another way to solve.
The percentage who didn't take the food is 35+20+20+10=85
So all food are taken by
100-85=15%
Look at that: 35 = no (a); 20 = no (b); 20 = no (c) and 10 = no (d). So, 35 + 20 + 20 + 10 = 85 persons who do not eat at least one of the four types. Thus, (a), and (b), and (c), and (d) do not belong to this set of 85 persons. Therefore, 15 people eat all four types. The union of the four sets may be less than 85. So, the complement may be greater than 15. Therefore 15 people are the minimum.
Here's how I did it and I think it's simpler than the solution given. Suppose we have 100 visitors. 65 had donuts and 80 had pretzels. That's a total of 145, so we know at least 45 people had both. 80 people had pizza and at least 45 had both pretzels and donuts for a total of 125. So at least 25 people had all three. 90 people had ice cream and at least 25 had the other three for a total of 115, so at least 15 people had all four. Yes, this is pretty close to the Venn diagram solution that Presh gave, but I didn't need a diagram, and quickly arrived at the same answer.
Here's how I solved it:
*I took 100 people as a sample size to make the calculation easier. I took the 90% first, then that means 10% of the people didn't eat ice cream.
*I took the pizza slice(80%) next, then that means 20% of them didn't eat pizza slice, and I carried over the 10% from earlier to here, so that means 30% neither ate the pizza slice nor ice cream(or both).
*Same process with the donut(20%). That means 20% of the people didn't eat donuts and I carried the 30% from earlier again here. That means 50% of people didn't eat at least one of the following: Donuts, Pizza Slice and Ice Cream.
*Finally, I took the pretzel(65%). That means 35% of the people didn't eat it then I carried over the 50% of the people from the previous proposition. That means 85% of people didn't eat at least one of these 4 items. That leaves me 15% of people who ate all of these 4 items.
I don't know how this works, but hey, I got the correct answer and that's what matters. 😅😅😅
On a side note, I realized that 100 people isn't necessary here. 😂😂😂
Sorry. I inverted the donut and the pretzel, but you get the point. 😅😅😅
Wow. That was more involved than I expected from my rule-of-thumb estimation.
I love the videos presenting plenty of methods!
I guess technically the answer would be somewhere between 15 and 65 percent because you can’t really know for certain with such limited info- but for the sake of the logic problem, yes at least 15 percent ate all 4 foods. Example of how to solve the problem “it’s possible that the 20% that did NOT eat pizza DID eat donuts, so we can remove their 20% from those who ate ALL 4 foods”. 65-20 = 45% then repeat this step in the problem for the for the other two foods and you get 65-20-20-10 = 15 (start with the donut value and subtract the DIFFERENCE between the other percentages of foods eaten and 100%. This tells you the lower limit of AT LEAST what percentage of people MUST have consumed all 4 foods - because that lowest percentile MUST apply to all 4 groups. This will tell you the POSSIBLE percentage of people that MIGHT have eaten the other 3 foods, but not the 4th. It is 50% that might not have eaten 3 or less foods, but not all 4. And so the most accurate answer would be a range of somewhere between 15% and the full 65% because albeit unlikely, you can not rule out that everyone who took a donut also took the other 3 foods as well. Highly unlikely maybe but still likely.
The answer is 15%, because it is the MINIMUM amount of people that have eaten all four foods. Not "How many did".
I solved it by finding the lowest common denominator if the percentiles, which is 20, which will be our population of park goers. 13 donuts were eaten (A), 16 pretzels were eaten (B), 16 pizza slices were eaten (C), and 18 ice creams were eaten (D). Then I made a table of 20 people, and distributed A, B, C, and D as evenly as possible -- only 3 people ate all four, or 15%. Not the simplest solution, but it got there.
This is the literraly the first method shown in the video
@@flacsomtodosclas2165 Yes. Good job. You figured that out. Congrats. These videos frequently encourage people to solve the problem in the comments before watching the solutions, which is what I, and many other people commenting, did.
what i did was similar to method 1. I imagined 4 rings with gaps in them. a ring 65% complete (with a 35% gap in it), two 80%complete rings, and so on. Turning the rings so that all the gaps lined up, 65% of it was 4 rings thick (that'd be the maximum). Turning the rings so that none of the gaps line up, 15% of it was 4 rings thicc
I think it would be 65% at the most and 15% at the least
of course, the real answer is to figure out that it's a relatively low amount of people and then propose a bundle/voucher with all four items for the price of a bit more than three items and running a good advertising campaign for that. the assumption is thta most bundle users would end up eating three or two items, with only few actually eating all four and getting their money's worth. imo the trick would be to advertise the convenience of it, only paying once to get all four items, which would (in the expectation of the customer) improve service time.
For not "ez" starting percentages, the circles diagram is the best solution
The real world conclusion of this story is that the amusement park sells the bundle and finds out that much lower than 15% buy it (probably even less than 5%). The reason being human behavior; no one wants to eat all four at the same time. Even if the bundle were like coupons to be used throughout the day, most people don't plan on eating all four of those items and/or are hesitant to part with their money up front to commit to eating all four. They'd rather just spend their money at POS when they have a craving for something.
Yeah. This is two mains and two desserts. A person might like a deal for one and one. But not all four.
The bundles might be interesting for groups that share the food.
I don't think it's a very popular idea anyway, though. A third of the people in this scenario apparently just don't like donuts, so they will be skeptical with a bundle that forces them to buy one.
Before I get into your solutions, I'll take a stab.
I reckon what matters is that none of the values are 50% or less. If they were, then the absolute minimum percent shared by all four would be zero.
Because of this, the lowest values would determine the minimum percentage shared by all four, because the higher values could not overlap less. So 80% leaves 20% of the patrons uncounted. The 65%, if it accounts for all 20% of those not previously uncounted, must share 45% with the 80%. Thus, 45% must have had both donuts and pretzels, say.
Of those 45% counted twice, a minimum of 25% must be counted a third time by the pizza's 80%.
Similarly, if all 75% of those patrons only counted twice at this point had ice cream, then the remaining 15% of those who had icecream had all 4 items.
So 15% of the patrons had all 4 items. If anyone had nothing, then even more would have had all 4.
I just subtracted the 20 20 10 from 65 to get min. You figure the worst case scenario is the people that didn't eat one thing would eat the other. It did confuse for a while, but ended up simple
Worse case scenario.... Put up a graph. The highest 90% goes one way (ltr). and the lowest 65%(rtl) The two 80% ones (pizza and pretzels) one goes from ltr and the other goes rtl, respectively.
The two 80% ones turn into only a 40% overlap. This overlap is covered by the 90% by default, thus no change. However, the 65% can only cover 15% less space from the other side, thus the lowest possible amount that all four amounts can overlap is 25%
That said, second best case is if both 80% are on the same side (ltr), while the 65% and 90% cover the other side (rtl) then it's only the 65% vs the 80% that we'd have to worry about, so that'd be 45% in the overlap.
And of course the absolute best case is if all 65% of the donut eaters had all four, so that's be a max of 65%
Darn, I got it wrong. Had the right idea, but wrong execution. Thanks!
I got the right answer but possibly by the incorrect method:
90% had ice cream, 10% didn’t
80% had pizza, assume the 20% that didn’t also didn’t have ice cream = 70%
80% had pretzels, 20% didn’t, assume they aren’t pizza-ice cream eaters = 50%
65% had donuts, 35% didn’t, assume they didn’t have any of the others = 50-35 = 15%
I just evaluated it linearly, but I can’t figure out if this is different from evaluating it as a binary tree as shown in the video?
My intuitive guess was to imagine the 65% and the 80% filling a bar from opposite ends. Because 65% definitely ate doughnuts, and 80% definitely ate pretzels, then the overlap is the minimum people that ate both. Doing this for pizza and ice cream will net you the same number or higher so it had to be 15. Also side note, I just realised typing this out, this is my exact mental method for solving those nonogram puzzles
Yeah did the same as you
General answer to this problem (given n percentages) is the sum of the percentages - (n-1), so in this case 0.65+0.8+0.8+0.9 - 3 = 0.15.
I guess I solved it the first way, though I didn't think of it like moving colors around. I just thought:
10% of guests didn't buy ice cream.
Let's say a completely different 20% of guests didn't buy pizza, and a completely different 20% of guests didn't buy a pretzel, and a a completely different 35% of guests didn't buy a donut. That still leaves 15% who must've bought all four. 100 - 10 - 20 - 20 - 35 = 15.
Why doesn't applying each percentage consecutively work?
90% of 80% of 80% of 65 gives me 37.44% overlap?
That's the equivalent of saying 90% of people who ate ice cream ate pretzels, and 80% of those people ate pizza, etc. Which could be true, if we were looking at probability and the random chances a single person ate all 4 assuming all events are independent the PROBABILITY would be 37.44%.
However we aren't looking at random probability a single person ate all 4, we are looking for the minimum number possible which is a completely different question. Just because 90% of the TOTAL # of people ate ice cream doesn't necessarily mean 90% of the people who ate pretzels also ate ice cream.
That sort of process works if the four percentages are independent variables! But in this case, the question requires us to assume that the variables are dependent on each other to calculate a worst case scenario.
Unfortunately no as you’re assuming that the portion of people who bought donuts are buying at the exact same ratios as the overall population which my not be true leading to an overestimate
my spin on this problem before watching:
make a 10 by 10 grid
90 are for ice cream
the 10 unused are used for pizza,then 70 overlap
30 spaces have only one food,fill them up with pretzels,50 rest overlap with the rest
50 spaces with only two foods are used for dough nuts,15 left for overlap with the rest
-->15% guaranteed overlap
yay,I was right!
Solved in 3 mins.. very satisfied, thank you!
I solved by considering how many % of all people ever tried one of the food, then 2, then 3 and finally all 4.
Let's say there are 100 people.
We need to filter out the guys item wise. Hence
1) Icecream - 90, non ice cream - 10
2)Pizza - 80, non pizza - 20
Thus ice cream & pizza =
90 - (100-80) = 70
3)Pretzel - 80
Thus ice & piz & przl = 80- (100 - 70)
= 50
4) donut - 65
This ice & piz & przl & donut =
65 - (100-50) = 15
I guess 15%
my thought process is
All - (people that didn't eat donut) - (people that didn't eat pretzel) - (people that didn't eat pizza) - (people that didn't eat ice cream)
100 - (100-65) - (100-80) - (100-80) - (100-90)
100 - 35 - 20 - 20 - 10
15
(method 1) huh didn't think to visualize it like that, basically my thought but easier
It's not as simple as raw math. Just introduce the 4-pack. If it sells out, raise the quantity. If you have excess at the end of the day, lower the quantity
Simple solution: negate everything (35 did not eat donut, 20 pretzel, 20 pizza, 10 ice cream), add them up and you get that 85% skipped a food, so 100-85=15% who didn't skip any, thus ate all.
1000th video!! Congrats man!
I can’t believe actually solved one for once. Just subtracted the percentage of each that didn’t eat that thing from 100 and got 15.
somewhere between 15% and 65%, we can also calculate average number which is 40%
You are forgetting about the fact that some visitors could have eaten NONE of the 4 foods ... that could be as high as 10% (ice cream max 90%). Minimum who ate all 4 foods then is actually 0% if you do a 4-circle Venn diagram with all possible intersections. Put a 0 in the very middle quadruple-intersection. Put 15 in the donut/pretzel/pizza triple-intersection. Put 25 in the donut/pizza/icecream triple-intersection and another 25 in the donut/pretzel/icecream triple-intersection. And finally put 40 in the pizza/pretzel/icecream triple-intersection. Put zero everywhere else in the diagram, and 10 outside all 4 circles. I solved it with a system of equations, with 10 outside the circles, so needing the locations within the circles to total 90. Then forced the 0 in the middle. Then some algebra from there. The symmetry of the pizza and pretzels helped some. There may be other solutions.
Take the lowest percentage .. 65% .. and now work out the 'did not eat' percentage for the other three .. 20% twice and 10% once .. add them together .. 20 + 20 + 10 = 50 .. and now subtract 50 from 65 .. giving you 15 .. and telling you that the minimum number of people who ate all four is 15% .. going up to a maximum of 65%.
I did the Venn diagram trick, but started with Ice cream and worked down. If 90/100 people ate Ice cream then 10/100 did not eat Ice Cream. If 80/100 ate Pizza Slices then at most 10 of those people also did not eat Ice Cream, therefore at least 70/100 people ate both Pizza and Ice Cream. If at least 70/100 people ate both Pizza and Ice Cream then at most 30/100 people did not eat both. If 80/100 people ate Pretzels then at most 30 of those people did not also eat both Pizza and Ice Cream, therefore at least 50/100 people at all three Pretzels, Pizza, and Ice Cream. If 50/100 people at all three then 50/100 did not eat all three. If 65/100 people ate donuts then at most 50 of those people did not also eat all three other options, therefore at least 15/100 people ate all four food items offered.
The scenario is surprisingly interresting though. The minimum (15%) and the maximum (65%) are very different, and both very unlikely. More probable (but still likely quite a bit from the real value) is the product of the percentages (0.65 x 0.8 x 0.8 x 0.9 ≈ 0.37%). That assumes that the numbers are independent, and they probably aren't. I wonder how close a statistician would get to the real answer having no statistics beyond the percentages given, but with all other food related statistics (general statisticcs and spesific statistics ralated to places like amusement parks) available.
Some will allways go all out on food stuffs. Some avoid expensive tourist food at all costs. I guess most people will maybe try to keep to one or two items (It's still expensive, and we tent to get full after all), and that would drag the "ate all" group in direction of the minimum. I wonder if there is some statistical constant/formula that will give a good answer to cases like this.