2/3 chance you will choose a goat, so when he reveals another goat that means you get 2/3 chance to get the car. If someone just explained this to me first time I would have got it.
Throughout the game, these are the possible combinations:- A. CAR GOAT GOAT B. GOAT CAR GOAT C. GOAT GOAT CAR Suppose you chose door 1. In Case A, switching will not win a car. In Case B, switching will win a car. In Case C, switching will win a car. As seen above, Switching gives 2/3 (66%) chance to win the car, and that my friend is the solution to the Monty Hall problem ;D
this is how i like to think of it. a) when you initially choose a card, your odds of having the car is 1/3. b) the dealer keeps two cards, the odds of the dealer having the car is 2/3. c) the dealer reveals one of his cards he knows to be a goat. then he asks you if you want to switch. d) you should switch because the odds that you have the car is still 1/3. the odds that the dealer has the car is still 2/3 because he will only ever reveal a goat. hope that helps!
Yeah same here ,that makes it way easier to understand , but at the same time feels wrong , cause we are adding a made up condition that the dealer wants the car. Idk it seems like this problem is just based on perspective or utterly specific conditions
Yes but that's not how the logic works, no one mentioned we are dealing with a dodgy dealer, the logic is and doesn't make sense. it's 2 in 3 chance each time or 50/50 regardless if you make a switch or not.
This doesn't work. If the dealer has 2/3 and then revealed the card which is 2/3 (dealers chance)x 1/2(chance of the card is correct) = 1/3 = your chance. Which is a 50 50
Nice video. It really demonstrates that your first pick only offers you a 1/3 chance. Therefore it is always better to choose from the other two (when Monty has made it an easy choice).
To the people that don't get it. Always Switching is the same as picking the other 2 doors. If he let you have the other 2 doors after 1st pick would you take them?
This is another easier explanation which continues from what I learnt from this video: There's an 2/3 (66%) chance that you've selected one of the goats from the beginning. Then Monty reveals one of the goats and eliminates it. Because of the 66% probability from the beginning you have more likely to have selected the other goat, whereas the car is only 33% chance, thus it makes sense to switch. (There's a high probability that you have selected the goat and the other goat has been removed, it makes sense to swtich)
I like to think of it as a choice between 2 doors, but Monty adds an extra goat to the beginning to increase the odds of you making that choice when you currently have a goat. After that has been achieved, he takes the extra goat away because it has served its purpose. So really, the game is a choice between 2 doors, but is rigged to have your starting door conceal a goat 2/3 of the time. Another way is to imagine a different game with the same outcome: Monty gives you a 3-sided die with numbers 1, 2, and 3 on it. He has a piece of paper in his pocket with one of those numbers on it. You roll the die and then Monty asks if you think the side you rolled matches the number in his pocket. If you answer correctly, you win.
Rules: There is only one red card, and you win if you end up with the red card. The easiest way to think about this type of problem is you must first pick a single card at random. Now before revealing anything about any of the three cards, you are given the option to (1) Keep the single card you selected, or (2) Trade your single card selected for BOTH of the two cards you didn't select. The best choice is obvious, and most everyone will immediately jump at the chance to exchange one card for the other two because they can "see" they have just doubled their chances of winning by having two chances of winning as opposed to only one. Anytime we "switch", we are in effect trading one card for two, and revealing one of the two cards doesn't change the fact that we are getting "two for the price of one" when we switch.
For all you people who say 50/50 you are wrong. If you pick a goat you will have a 66 percent chance of getting a goat. If you pick a goat Monty will be forced to show you the other goat and if you switch you will get a car. This means by picking a goat and switching to get a car you will always win the car if you pick a goat and switch, and the chances of you picking a goat is 66 percent. Like so everyone can see please.
Listen, Monty gives nothing away. That's why his name is Monty. 50-50 at the start; 50-50 at the end. www.onofframp.blogspot.com/2014/09/monty-hall-door-goat-and-car.html
John D Give it a rest, you're wrong. Mathematically, logically and empirically it's NOT 50/50 between staying and switching. You're just making yourself look foolish,
+Peter Thai oh shit, that makes sense now... so you are more likely to choose a goat in the first round, then if you switch in the second round, you are just as likely to get a car (if your first choice was in fact, a goat).
The single detail that is necessary for this to work is that the host (or in your case, the dealer) knows what lies behind each door/card, and knowingly chooses the wrong one to reveal. And that increases the possibility of the switch being correct because they purposefully avoided revealing the other one.
the reason it's not 5050 is because the host ALWAYS shows you a goat and never a car, but no one explains that. If the host could flip a door and show a car , then it'd be a fair 33 percent
Yes, only when you choose prize in your initial door, host will have opportunity to choose between two doors with the goat, while in other situation, when the goat is behind your door, host can only offer you to switch to the door with a prize. And that (host left with no option but to offer you door with the prize) happens twice as much (66%) than situation where host can choose between two doors with a goat behind (33%).
I feel bad for him... I hope he's not making a full out of himself. I would have been probably the same way, if I hadn't learned about it on my own... lol
Yes, fancy explanations are probably not needed. And yet we are still from time to time blessed with pearls from those of mighty intellect who *insist* the second choice is 50-50.
Kirk Bocek haha I've already done the cards 20 times 11 times switching worked 9 times it didnt . doesn't seem like switching is 2/3 chance to me. You are going by group mentaliy, everybody says switching increases chancesso you lose common sense and just agree haha there are only 2.
Well you don't know the probability here. Go read the comments on the RUclips posting. Seriously, don't think you can just *figure* this out. Do the cards. You'll be surprised. I promise.
I shall make it simple. At your first selection, there is 1 chance out of 3 to select the winning door, when you'd better keep your choice but there is 2 chances out of 3 to select a loosing door, when you'd better switch your choice. As you don't know the result of your first choice, odds of winning is larger (=2/3) by shifting.
When you pick a goat and then switch you'll get the car and when you pick the car and then switch you'll get the goat and since there are two goats and a one car in the beginning you will more likely pick a goat than a car, simple as that
And just for full disclosure, I myself struggled mightily with this when it was first presented to me, and eventually made a little computer algorithm to simulate the game, and to "prove" that switching didn't make a difference. Had it run for a few hundred trials, and to my astonishment, the switch proved to, indeed, double your chances of winning. Only after that did I put enough effort into understanding why this is, and finally "got" it. Now that I understand it, it seems obvious.
Does your 'algorithm' always bet on door "one"🚪(was it a 'constant' choice in the experiment) or every time a random (different) door was chosen to bet on, before you check what's behind the 2 remaining🚪?
The host opening one of the remaining doors reveals information about the likelihood of the prize being behind the remaining door, while giving you zero new information about the likelihood that the prize is behind your door (because the host could never open the door you picked). The problem is called a paradox not because both probabilities are correct, but because people overwhelmingly believe the probability is 1/2 when that is wrong. It's a paradox of behavior, not math.
It's quite simple to explain. There are 3 doors, 2 incorrect (goats), 1 correct (car!). You choose one of them. One of the incorrect doors is revealed to be incorrect. You can then switch if you want to. Switching will only work if at first you chose one of the wrong doors. Seeing there are 2/3 wrong doors at the beginning, switching will work 2/3 of the time.
Tracey-Ann Thompson-Carroll MGF 1106 The card you choose the odds of winning the car is 1/3, but the dealer has two cards, which gives him a 2/3 chance of still having the car. He showcases one of the cards to be a goat and tricks the contestant so he or she thinks about switching.
Part1 It's easier to understand if looked from the car perspective (the prize u want): 1) picking a car = 33.3% (one door) 2) picking a goat = 66.6% (two doors) If you stick to your 1st choice, your best is 33.3%, but by switching you're trading your 33.3% for the remainder--the 66.6% (in effect, you are choosing TWO doors vs. one door). So, staying = choosing one door = 33.3%. switching = choosing two doors (opened door and switched door) = 66.6%. Basically, opening two doors is better.
As I've said here, again and again, don't try to do this puzzle in your head. *Actually* do the cards and see what results you get. *Then* come back and tell us what you get.
Kirk Bocek done with paper, actually :-|. my point isn't that the monty hall problem is wrong or that kinda stuff. what i want to say is the trick is actually in the 1st assumption (en.wikipedia.org/wiki/Monty_Hall_problem#Standard_assumptions); and only through that assumption, the whole thing gets a paradoxical look. and when u get what this assumption does to this theory, sadly enough, all charms will just be wiped away.
Md. Abdullah Al Muyid Yes, but if you leave out those assumptions, it is no longer the "Monty Hall problem", and so it is irrelevant what the chances without those assumptions would be. You should not really think of those points as "assumptions", they are the rules which HAVE to be followed for any discussion to make sense. Same as that the host has to actually KNOW what is behind each door, so that can can ALWAYS show a goat and does not accidentally open the door with the car behind it.
Has anyone figured out what the odds are as to which door people are more likely to choose? Is it 33% for each door, or do people more often pick the door on the far left since in the west we read from left to right, for example?
"Would this work if there is no game show host? for example, if there are three doors and one of them is an exit and the other two have booby traps. If you pick one of the doors, but your friend comes out of one of the other 2 and confirms it's a booby trap, would switching your initial choice mean you have an advantage?" I wanted to reply to this comment, but google decided not to let me, for some reason. The Monty Hall Problem requires that the "host" *intentionally* open a door that reveals a goat. In this case, your friend is opening a door *at random*. In theory, they might've gotten lucky and picked the exit (as opposed to the host, who will *never* choose the car). When you take the host out of the equation, it becomes a variant known as the Monty *Fall* Problem. Since the door was opened completely at random, switching or staying makes no difference (50/50), unlike in the Monty Hall Problem where switching wins 66% of the time and staying wins 33% of the time. This variant is what likely makes a lot of people think it's a 50/50 shot - the fact that the host will not reveal the car is what changes the odds. Worth mentioning are two other variants to this problem. First is the unfriendly sounding "Monty Hell" variant. Let's say the bad guy who threw you into this booby-trapped dungeon is the host. He knows where the exit is, but he obviously won't show it to you. However, the bad guy "forgot" to place it in the rules that he must always give to the option to switch, so he will only give you the option to switch if you picked the winning door! Because of this, if you're playing Monty Hell, switching will win *0%* of the time. The flip side of this I'll refer to as "Monty Heaven". Let's say the bad guy is lazy and sends his assistant to be the host. But the assistant is sick of the bad guy and actually wants you to escape. She figures out the rules and uses the same exploit, but this time in your favor - you will only be offered the chance to switch if you picked a losing door. In this case, switching will win *100%* of the time.
Nipun, your question makes a valid point. Many, many persons on these Monty Hall videos make the claim that if Monty forgets where the goats are, takes a guess, and luckily picks a goat door, that that makes the option to switch a 50/50 proposition. Those who make that claim are absolutely wrong. It is still to the contestants clear advantage to switch. The problem above about booby-trapped doors is identical to the Monty Hall problem. The person trying to find the exit should switch when his friend shows him a bad door.
At the beginning, you have a 67% (2/3) chance of picking the wrong door. Once you pick a door and a wrong door is revealed to you, the chance that you first picked the wrong door doesn't change; all that changed is the information you have on the 3 choices. So it’s still a 67% chance you first picked the WRONG door, but now you know which of the two remaining doors is wrong so there is a 67% chance that SWITCHING will get you the RIGHT door.
Every time this problem is presented, you see people arguing to the last, absolutely CONVINCED that the chance is 50/50, and viewing this "fact" as so obvious that they really don't put enough effort to understanding the explanations of why it is not so. Some of the people who begin this way, do eventually get it through the various explanations, but some do require an actual test like in this video. That doesn't make them stupid, just a bit too sure of the accuracy of their own intuition.
...the probability, just like an uneven coin would scew the probability of heads or tails, despite there being just two options. If it helps, a mathematically equivalent way of looking at the choise of whether to switch or not, inthe Monty Hall problem, is if the host didn't open a door at all, but gave you an option to switch from your one door, to the two remaining doors. You still have two options - either stick with your door, or switch to the other two doors, but now it should be...
No, I did not mean "switch from your one door to choose one of the two remaining doors" - If I meant that, I would have written that. The bit in brackets what is what you read into my post, it was not intended, or implied by the text. What I wrote was "switch from your one door to the two remaining doors". In other words, instead of one door, now you have the option of simultaneously choosing BOTH the other doors.
Let's say the car is behind door #2. Now, pick a door: 1) If you pick door #1, Monty shows you a goat behind door #3. If you swap, you win. If you stay, you lose 2) If you pick door #2, Monty shows you a goat behind door #1 or door #3. If you swap, you lose. If you stay, you win. 3) If you pick door #3, Monty shows you a goat behind door #1. If you swap, you win. If you stay, you win. You will win about 66.6% of the time by swapping, and win about 33.3% of the time by staying.
Now if your actually saying "to switch from your one door, to the two remaining doors" then this would be a different scenario, and hence would even be applicable to the original premise of the monty hall problem of only being able to choose 1 door at a time. However, you should probably still swap in that scenario: 1) the probability that your first choice was correct is 1/3, which means you should most likely swap 2) swapping is a better probability because there's now more doors in 1 option
The percent can't be added. If a factor change, all the ecuation change too. If the guest show the car you still have the 33%? And 30/70(any side) is the normal result of the 50/50 odds.
Part2, it's not 50/50, and 66.6% represent the car b/c a goat was shown. E.g: Event A) Picked car and stayed = 33.3% of picking car Event B) Picked car and switched = 33.3% of picking car Event C) Picked goat and stayed = still 33.3% of picking car Event D) Picked goat and switched = still 33.3% of picking car plus the 33.3% of the shown goat resulting in 66.6% of picking car if you switched (if you stayed, you're back to 33.3% and NOT 50% because door1=33.3%, dr2=33.3%, dr3=33.3%).
Try it yourself: www.grand-illusions.com/simulator/montysim.htm Note: It says it won't work, but click and it does. This site lets you "play" the Monty Hall Problem and see the 1/3 and 2/3 split for yourself. You can make it run 100 or 1,000 times!
IE Only. Didn't work in Chrome for me. Didn't try Firefox. Very nice BTW. An excellent way for people to easily do this programmatically. Is your simulator an argument for 50-50? I received precisely the results expected. This is a simple problem with very well defined parameters. There are indeed *two* choices here. This first is which of three doors to choose. You can choose any of the three. That choice can result in a goat or a car. The second choice is to switch or not. Don't try to somehow remove the first choice. It's really there. If you don't switch, your odds are 1 out of 3. So if you *do* switch, how can that magically change you from a 1 out of 3 probability universe into a 1 out of 2 probability universe?
Kirk Bocek It says it doesn't work, but just click and then it does. The odds remain 1/3 for the first card, and increase to 2/3 for the unpicked card, the Monty Hall Problem is true! I spent quite a while arguing against it, but I was wrong. The "key" is that it is NOT random, the removed card is known, thus the second round isn't "random chance" at all!
5Cats After, what, 3 years, you are the first, the very first to allow his mind to be changed and to admit it. You are a god among men, an angel among demons. You are the very force of order in the universe. But I gush after having been hammered here for so long and watched others receive the same treatment. You are correct. The door being opened is not a random event. Monty knows which doors have goats and shows you one of them. There is always one available even if you have selected one with your initial choice. But again, it is not "removed." It is simply opened. It is information given to you. There are still three doors. That's why the odds remain 1/3 to 2/3.
Yes! The "it's not random" part threw me for quite a while. If it were random, the odds would be 50-50, BUT you'd get knocked out 1/3 of the time! Since you cannot get knocked out, the odds do indeed carry over from round 1 to round 2, giving you 1/3 and 2/3. The "Gambler's Fallacy" only applies to random games of chance, not the MHP, this also confused me. I am indeed a cat among the mice, Lolz!
5Cats At the outset you don't get knocked out. Guaranteed. So there is no probability of getting the first round wrong. Or right. It's neutral. It's a non-event with a guarantee that Monty will do what he always does-turn over a goat first. It exerts no control on the second choice, and does not filter down conditionally. Judgment is not passed from an event that's 100% not to knock you out. These are not linked events. After the first flip, there are only two doors to consider. You're guaranteed to get there because of the way Monty plays. No need for a pad and paper. Or cards.
...self evident that one wouldnt' even EXPECT the probability to be 50-50, despite there being just two options: stick with your door, or take the other TWO doors. That the host opens the goat-door of those two doors before you make the choise, does not change the fact that you might just as well have been offered to switch to the two doors, which eliminates the unintuitive nature of the problem. In this way, relevant information can make one outcome, in a two-possibility situation, more...
Then address my marble example, which I created to show that previous events can alter the probabilities of a binary choise between two hidden options. Do you think there's something paradoxical in that, rather than straight forward, completely non-problematic math?
It's heartwarming that nearly a year and a half after posting this video I *still* see comments trying to argue that the final choice is 50-50. If you think that's the answer, do the cards and keep the tally. You'll be surprised. I promise.
Would this work if there is no game show host? for example, if there are three doors and one of them is an exit and the other two have booby traps. If you pick one of the doors, but your friend comes out of one of the other 2 and confirms it's a booby trap, would switching your initial choice mean you have an advantage?
Amir Aribokill No. In that case, it's 50-50. The host of the quiz knows where the car is, so possibility of the host's opening the door where there's a car is zero. On the other hand, the friend knows nothing about what's behind the doors, which means possibility to open the door without booby traps is 1/3.
@@Zaekr2111 once a door is revealed to have the booby trap, then you should switch because your door always only has 33 percent chance of a good outcome. Switching doubles your chances to get a door without traps.
probable than the other. The 50-50 situation ONLY applies if no such probability-scewing information is present. There is absolutely nothing in mathematics that requires a two option situation like this to have equal probability for both options.
....Finally, if none of that is enough to convince you, do the experiment described in this video, and record the resuts for yourself. Or, do as I did way back when I struggled with this problem - make a short computer program to simulate the game, and run it through a few hundred iterations. You'll see the probabilities approach the 66% chance for switching and 33% chance for not switching.
It seems like it would be 50-50 because you are focused on the probability that a given piece of information is true (e.g., that the car is behind Door #1) and not the probability that an event will occur (e.g., that you will win the car). So once the wrong car is revealed, there IS a 50% chance that the car is behind either of the two remaining options, but there is still only a 33% chance that you actually PICKED the door that the car is behind. So it is in your favor to switch.
Because some people refuse to believe that it is beneficial to switch and require this experiment. Some people are so stubborn that absolutely nothing you say would convince them, even after experimenting on simulations.
if I try out for an NFL football team, there are two outcomes: I make the team or I don't. By your logic, I have a 1/2 of making the team, which is obviously not true. Having two options means the probabilities of the two options must sum to 1; it does not mean they must be equal.
"the host doesn't give you another opportunity to choose a door again" You originally said: "if the host didn't open a door at all, but gave you an option to switch from your one door, to the two remaining doors." Assuming you mean "switch from your one door [to choose one of the] two remaining doors" then you are left with the original choice you started with, which is choosing 1 out of the three options. Of course, that assuming that's what you meant to say, rather than what you wrote...
Don't think that this is a simple thing. For many years a lot of mathematicians thought that in the second instance there was 50% of chances. And many of the common people that act like they "get it" and "everyone who doesn't get it is a stupid" are, in my opinion, not really getting it, but only accepting the "truth" that is being said to them. It took me like months to accept that there was really 2/3 chances of winning with the switch.
1=goat 2=goat 3=car. I pick number 1, 2 is revealed. i switch to 3 to get car. I pick number 2, 1 is revealed. i switch to 3 to get car. I pick 3, the number is revealed, i switch to another number I get goat. therefore switching gives you 2/3 chance of winning the car.
Regarding the coin toss comparison - fine, if you don't like that, then how about this: There are two jars filled with with black and white marbles. One jar has a 50-50 ratio of black and white, and the other has a 30-70 black-white ratio. Now a game show host picks one marble from each jar, blindly - neither you nor the host sees what colors were picked. He then, without looking, places the marbles in two boxes, and the boxes are closed. You know only which box has a marble from which of...
...still would have ended up in a situation where you have one door that might hold the price. The timing of the reveal of one of the false doors does not matter - mathematically, the situation of either revealing one of the false doors and then letting you switch, is equivalent to the situation where you are allowed to switch to both of the remaining doors. That's why there's the 2/3 chance to win with switching - there's no real "paradox" there, just the counterintuitive feel to the solution.
My analysis already acknowledges that it's most likely better to swap, because the probability that your first choice was correct is only 1/3. So I don't see why you're telling me something that I've already acknowledged...Though, my first comment doesn't acknowledge the switch, as it was merely focused on explaining that the player is always left with 2 remaining options. If you would have bothered reading my comments you would have seen I've already acknowledged that it's better to swap.
@bucko386 "If he offered you the swap without opening any doors you would say YES" your comment is incorrect, without opening the door there is no variable change, therefore no point taking the swap, change your mind as many times as you want , with 3 doors its 33.3/33.3/33.3 chance on all doors. when monty shows you 1 incorrect door that brings in variable change, now the chance of the door you originally picked stays at 33% but the other remaining door is now 66%
Acknowledging that it is better to switch only makes you half right. The problem is that you have also asserted that a probability of 1/2 is also correct, when it is not. The "evidence" or "logic" you seem to be resting on is that if you have two possible outcomes, then the chance that one outcome is correct is 1/2. This is only true if the events are equiprobable. Let me give you an example,
...you swap to the other TWO doors instead of revealing where the goat is behind one of those two doors. The ONLY difference there is that in this latter scenario, the answer would be intuitive and obvious, that you should switch - two doors is better than one! But that's exactly what you get if you switch AFTER the one false door out of the two remaining is revealed. In the altered scenario that door would just have been opened, to reveal a goat, after you made the choise to switch - you....
I did address your marble example, inadvertently. It's essentially your 'swap opportunity of choosing 2 doors simultaneously' example. My analysis already concludes it's better to swap, because the odds that your 1st choice was correct is only 1/3. However, you're still left with 2 options - 1 goat, 1 car - & you're only aloud to pick 1 of the 2 remaining doors, thus giving a 1/2 probability, by default. Nevertheless, my analysis still concordantly acknowledges the likelihood for swapping.
It has been three years and I do not know if you keep thinking the same thing, but just in case. Think of the odds as the approximate fraction of times an event will be fulfilled after numerous repetitions. For example, a coin has probability 1/2 to fall heads or tails, so after 1000 attempts it should have left about 500 times heads, and about 500 times tails. The margin of error must be small. When we say that the probability of winning in the Monty Hall problem is 1/3, we refer when your choice NEVER changes, so after many attempts you would have won about a third of the time maintaining that same strategy. You will always keep the door of the initial election, you will not make again a random selection among the remaining two doors, and I think that is the greatest confusion. The probability of being successful at first was 1/3. If anyway you will not change the choice, in the end you have to win the prize many times as if the presenter had never opened the other door. That action does not affect the result. It's like if you have a box with 99 white balls and only 1 black; grab one without seeing it and keep covered in hand. The probability of having caught the black is 1/100. If someone removes all other balls in the box leaving only one, but you will not have the opportunity to change which is already in your hand, then the probability of having caught the black will not increase to 1/2. In the Monty Hall problem is the same: if you will never change, the proportion of times you win must be the same as if they never give you the opportunity to change. If this is not clear, another way to look at it is thinking that the probability of an event is equal to the number of favorable cases divided by the number of possible cases, only if the cases are equaly likely. For example, if a dice was not balanced, the probability to be out wouldn't be 1/6 for each number, despite they are still six possible outcomes. In this case, the fact that gives greater probability weight to a gate than the other is that yours was completely randomly selected from three possible doors, while the other was selected with the obligation that the car must remain covered. Think about this: when opening the other, does it give you additional information about your door? The answer is no, you could have chosen one with a goat ore one with a car, and anyway it was going to be an additional goat, so seeing that in another door there is a goat does not make yours more likely to have the car. Instead, you effectively have new information about the other door closed. Suppose you had chosen the door 1 and the presenter opened the 3. You know he is forced to leave the car covered. If he did not open 2, it is probably because the car was behind door 2 and not behind 3, and so he was forced to open 3. Or maybe the election between 2 and 3 was irrelevant because both had goats, but this is more difficult, because that implies you chose the correct door at first, which has probability 1/3. Confusion usually comes because some people see the two stages as independent games, when in fact they are not. If at first you had chosen the car (1 case), changing will make you lose. If at first you had chosen a goat (2 cases), changing will make you win. As seen, if changing makes you win or not depends on your first election. If in the second stage the choice was randomly, then the probability of winning would be 1/2. To understand why this doesn't contradict previous conclusion is that it is good to think the odds as a fraction of times you succeed after numerous attempts. Making randomly, about half of the time you pick the same door you chose at first, and about half of the time you pick the door the presenter left closed. This is because randomly there is nothing that forces to pick one door more times than the other. In the half you pick the same door you chose at first, 1/3 you should win, but that represents 1/2 * 1/3 = 1/6 of the total, and you should lose 2/3 of the time (2/6 of the total). In the half you chose the presenter's door, you should win 2/3 (2/6 of total) and lose 1/3 (1/6 of the total). C: Changing N: No changing N C Win---> 1/6 + 2/6 = 3/6 = 1/2 Lose--> 2/6 + 1/6 = 3/6 = 1/2
Wondering your thoughts on this? I think why people get confused is that after they pick the 1 of 3 cards, and then one goat is revealed, what they don't get is that they have already made the 1 in 3 choice, and that revealing the other goat does not change things as is. Just seeing one card that is not the desired one does not change the odds of what has already happened. But now, they are given the choice of choosing between the two cards not yet revealed thinking they have a 50/50 chance?
The only thing that matters is your initial choice. Two out of 3 times, you will pick a goat. Obviously, staying with your first choice will result in you getting a goat 66.666...% of the time, on average. You will only select the car 1 out of 3 times. Staying with your initial pick will result in you getting the car only 33.333...% of the time. The strategy is obvious - switch doors and you should win 2 out of 3 times. I really don't see why people have such a hard time understanding that.
@paddygilbert I made this video in response to those who *swear* up and down, on all that is holy, that the final probability of switching is 50/50 and not the correct 33/66. There is just nothing you can say to these people to convince them otherwise. I got so frustrated with one of these people that I made this video. If you ever get into this argument, send that person here and tell them to do this experiment.
No, you did not do any valid ananylsis, it never was valid and it isn't now. Two options means there is a 50/50 chance- unless you can explain how one option is more or less likely to occur than the other. And we have done so here. Two choices means 50-50 doesn't always apply, and this is one of those cases. You used a false statement to do your "analysis" so it was never valid at all. Now admit you were wrong.
*SIGH* 2 out of 3 times you will choose a goat initially, right? That means that if you stay with your initial choice, 2 out of 3 times you will get a goat, right? On the other hand, if you trade doors after Monty shows you a goat, you will win the car 2 out of 3 times, right? It doesn't matter if you play once, a hundred times, or a gazillion times - the odds stay the same and you will win about 66.6% of the time if you swap doors.
But I'm not sure if you get this or not by looking at one of your other comments. I won't be bothered with reading the book-worthy list of replies, so if you you get this, just ignore my comments and leave it for someone else to read in the hopes that it helps them understand; I rather not to go back and forth on this with you (or anyone else for that matter) unless you are confused/want clarification.
Oh! I missed that part! That's kind of funny, eh? Just came to look at that recent comment. It still "bends my brain" a little but it's still 1/3 and 2/3 after the 'extra' card is removed. It was 1/3 for your card in BOTH rounds? Perhaps it's easier to understand it like that? If it's 1/3 with 2 other cards, then each card is 1/3. If it's still 1/3 with ONE other card? That card other must be at 2/3 odds. That makes sense? I hope it can help someone grasp this slippery problem.
5Cats I think you have it. The mistake that I and most others make is to "remove" the first door when it is opened and somehow think that now only two doors are involved when we make our choice. Three doors are always involved.
What if there were TWO contestants and each of these two contestants were ardent mathematicians, both with an innate understanding of the principle behind the Hall paradox. After each choose a door, and the third unchosen door was opened to reveal a zonk, would each of the two players, if given the chance, leap at the opportunity to accept the other fellow's door? (And in this scenario the 2 scholars are rivals who would not agree to sell the car and split the proceeds) Hmm? Johnny Radionic ™
You missread, or misunderstood my comment regarding the altered scenario for the monty hall problem. The point was that the host doesn't give you another opportunity to choose a door again - he gives you an opportunity to choose between the door you originally chose, or the TWO doors that you didn't choose. This is a binary choise: stick with your door, or choose BOTH the other doors. No third option there. And that is exactly mathematically equivalent to how the game actually goes.
@paddygilbert, I didn't make this for you, you get it. You need to hook up with one of these knuckleheads who swear the answer is 50/50. *Then* you'll see why I made it.
The fact that there WAS ONCE 2/3 doors that are goats becomes irrelevant as soon as 1 of the those 2 doors are revealed, and hence eliminated from the option, because then, there are only 2 options left - 1 goat, 1 car - to choose from during the swap opportunity. It doesn't matter if there were 3 goats and 1 car, if 2 goats are revealed before the opportunity to swap, because either way, you're still left with only 2 options; 1 car, 1 goat; it's always 50/50
....And are you saying that since this is a binary choise between two options, and one choise gives you better odds than the other, this should somehow be considered paradoxical? Of course there are no paradoxes involved. No more than there are in the monty hall problem - the probability calculations are straight forward, and there's no requirement AT ALL that having a blinded choise between two options must imply 50-50 chances.
...And even then he'll probably just tell you that it doesn't solve the paradox that supposely exists. I don't know if anything we say will ever convince him that there is no paradox. Two options doesn't make it 50/50. Well, sometimes it does, and if you can logically demonstrate why one is more likely to occur than the other (as done here), then it's NOT 50/50. And that's exactly what happened here.
"This is only true if the events are equiprobable." False. Equiprobability would only determine the relevancy of the probabilities; not whether or not they're true. As I've already denoted previously, you could argue that the 1/3 probability is more relevant than the 1/2 probability, as the 1/3 probability information can produce better tangible results than the 1/2 probability information, but that doesn't make the other probability untrue; it's entirely erroneous to assume it does.
"No, if you are only left choose 1 out of 2 options, then you will have a 1/2 probability of choosing the correct option... So both probabiliteis are correct." WRONG.
..the original jars, and you know the ratios of black-and-white marbles in those jars. Let's call the 50-50 jar "Jar A", and the box with the marble out of that jar "Box A", and the other jar and box "Jar B and "Box B" respectively. Now you are given a choise between the boxes - if the box you pick has a white marble in it, you win a million dollars, if it has a black marble in it, you win nothing. Are you now saying that there somehow should be a 50/50 chance of winning with either box?...
"No actually it's wrong" No, it's right. In fact I have logically demonstrated my analysis through deductive reasoning; so yeah, it's reasonable. The onus is now for you to disprove me; this has yet to occur. "a 50/50 chance you're going to die tommorow[sic]" False comparison, unless you're talking about suicide/euthanasia, because a suicide/euthanasia are choices to make, like in the monty hall choices; yes or no; 1/2... Either something happens or it doesn't; that's the basis of reality.
what do you think "If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal?"
Sorry, got the wrong impression from the argument you're having with WhyTauIsRight AndPiisWrong - if you're clear on the issue of the probabilities, I don't know what the argument is about.
"The percent can't be added" Yes they can. "If the guest show the car you still have the 33%?" No. "And 30/70(any side) is the normal result of the 50/50 odds." Wrong.
Yes it is 50% after you are left with two choices and I'll prove it quite simply. If half the time you switch and half the time you stay you will win 50% of the time. LOL I am so damn smart. Who is smarter than me? No one. Of course if you switch 100% of the time you will win a lot more cars. So people are basically arguing two different points. Some people say that after one choice is eliminated you have a 50/50 chance of winning. They are right. Other people say that if you switch every time you will win 2/3 of the time. They are right too.
CacacacaYEAH - I recommend searching RUclips for "The Monty Hall Problem Explained" - watch the video by axmurderer27. That's perhaps the clearest explanation of the problem, and why switching does double your chances from aprox. 33% to aprox. 66%. If that's not enough, look up the Wikipedia article on "Monty Hall Problem"; it explains it in various ways, including a mathematical proof. The article also gives an interesting history of the problem, and how it's fooled even mathematicians.
No, if you are only left choose 1 out of 2 options, then you will have a 1/2 probability of choosing the correct option. But simultaneously, the probability that your first choice was correct is only 1/3, which means you should most likely swap. So both probabilities are correct. Just for fun, here's another probability that's also true. The probability that all your choices are correct is 2/5. 5 is the total number of options you've had throughout, and 2 is the number of possible right choices.
Relevancy? A probability is only relevant if it is true. 1/3 is the more relevant probability here because it is the mathematically derivable and empirically provable true probability of the event. 1/2 is just something you made up with zero mathematical or empirical evidence. It is not an assumption to say that 1/2 is untrue; it is a provable fact.
Ahh, so that's what the argument is about. Of course there isn't any real paradox - there's only the conflict between "common sense" expectation, and the math. The probabilities aren't in any sense "paradoxical".
Yes, you would have an advantage. But your friend is still acting as the game show host. He is still looking behind the doors the way the game show host would.
"if you don't like that" It simply wasn't comparable to the original premise. It's apparent to me now that you meant what you wrote before, so I will only reiterate that I understand what you're saying, but it's still a false comparison. There's simply nothing in the original monty hall premise which causes the doors to have an irregularity. You don't get "2 doors for one choice", there's no '1 transparent door' (or anything else that would cause an irregularity in the doors) ...etc
Having two choises does NOT mean that the probability is automatically 50-50. This is your basic assumption, and it is simply false. A fair coin flip gives a 50-50 probability of heads or tails, but an irregular coin may produce, say, a 55-45 outcome probability. If there were just two doors originally, a goat behind one, and a car behind one, placed randomly, then the chance of picking a car WOULD be 50-50. But you cannot ignore the relevant details of the problem, which scew....
2/3 chance you will choose a goat, so when he reveals another goat that means you get 2/3 chance to get the car.
If someone just explained this to me first time I would have got it.
Throughout the game, these are the possible combinations:-
A. CAR GOAT GOAT
B. GOAT CAR GOAT
C. GOAT GOAT CAR
Suppose you chose door 1.
In Case A, switching will not win a car.
In Case B, switching will win a car.
In Case C, switching will win a car.
As seen above, Switching gives 2/3 (66%) chance to win the car, and that my friend is the solution to the Monty Hall problem ;D
Pseudo mathematics
smart
this is the only acceptable explanation in the entire youtube
@@basiert7701no he is right
It's a 2 door proposition 1 car and 1 goat
1:1 odds
this is how i like to think of it.
a) when you initially choose a card, your odds of having the car is 1/3.
b) the dealer keeps two cards, the odds of the dealer having the car is 2/3.
c) the dealer reveals one of his cards he knows to be a goat. then he asks you if you want to switch.
d) you should switch because the odds that you have the car is still 1/3. the odds that the dealer has the car is still 2/3 because he will only ever reveal a goat.
hope that helps!
Yeah same here ,that makes it way easier to understand , but at the same time feels wrong , cause we are adding a made up condition that the dealer wants the car. Idk it seems like this problem is just based on perspective or utterly specific conditions
That was a MUCH better…probably the best and most simplified explanation I’ve read.
Yes but that's not how the logic works, no one mentioned we are dealing with a dodgy dealer, the logic is and doesn't make sense. it's 2 in 3 chance each time or 50/50 regardless if you make a switch or not.
This doesn't work.
If the dealer has 2/3 and then revealed the card which is 2/3 (dealers chance)x 1/2(chance of the card is correct) = 1/3 = your chance. Which is a 50 50
Nice video. It really demonstrates that your first pick only offers you a 1/3 chance. Therefore it is always better to choose from the other two (when Monty has made it an easy choice).
To the people that don't get it. Always Switching is the same as picking the other 2 doors. If he let you have the other 2 doors after 1st pick would you take them?
Brooooo i ve been looking everything and it just didnt make sense only helpful comment is yours thanks
This is another easier explanation which continues from what I learnt from this video: There's an 2/3 (66%) chance that you've selected one of the goats from the beginning. Then Monty reveals one of the goats and eliminates it. Because of the 66% probability from the beginning you have more likely to have selected the other goat, whereas the car is only 33% chance, thus it makes sense to switch. (There's a high probability that you have selected the goat and the other goat has been removed, it makes sense to swtich)
I like to think of it as a choice between 2 doors, but Monty adds an extra goat to the beginning to increase the odds of you making that choice when you currently have a goat. After that has been achieved, he takes the extra goat away because it has served its purpose.
So really, the game is a choice between 2 doors, but is rigged to have your starting door conceal a goat 2/3 of the time.
Another way is to imagine a different game with the same outcome: Monty gives you a 3-sided die with numbers 1, 2, and 3 on it. He has a piece of paper in his pocket with one of those numbers on it. You roll the die and then Monty asks if you think the side you rolled matches the number in his pocket. If you answer correctly, you win.
That really helps me to understand this.
Only way I was able to understand it, thank you
Rules: There is only one red card, and you win if you end up with the red card. The easiest way to think about this type of problem is you must first pick a single card at random. Now before revealing anything about any of the three cards, you are given the option to (1) Keep the single card you selected, or (2) Trade your single card selected for BOTH of the two cards you didn't select.
The best choice is obvious, and most everyone will immediately jump at the chance to exchange one card for the other two because they can "see" they have just doubled their chances of winning by having two chances of winning as opposed to only one. Anytime we "switch", we are in effect trading one card for two, and revealing one of the two cards doesn't change the fact that we are getting "two for the price of one" when we switch.
EXACTLY! I have described it that way many times.
For all you people who say 50/50 you are wrong. If you pick a goat you will have a 66 percent chance of getting a goat. If you pick a goat Monty will be forced to show you the other goat and if you switch you will get a car. This means by picking a goat and switching to get a car you will always win the car if you pick a goat and switch, and the chances of you picking a goat is 66 percent. Like so everyone can see please.
Listen, Monty gives nothing away. That's why his name is Monty. 50-50 at the start; 50-50 at the end.
www.onofframp.blogspot.com/2014/09/monty-hall-door-goat-and-car.html
John D
Give it a rest, you're wrong. Mathematically, logically and empirically it's NOT 50/50 between staying and switching. You're just making yourself look foolish,
+Peter Thai oh shit, that makes sense now...
so you are more likely to choose a goat in the first round, then if you switch in the second round, you are just as likely to get a car (if your first choice was in fact, a goat).
The single detail that is necessary for this to work is that the host (or in your case, the dealer) knows what lies behind each door/card, and knowingly chooses the wrong one to reveal. And that increases the possibility of the switch being correct because they purposefully avoided revealing the other one.
the reason it's not 5050 is because the host ALWAYS shows you a goat and never a car, but no one explains that.
If the host could flip a door and show a car , then it'd be a fair 33 percent
Yes, only when you choose prize in your initial door, host will have opportunity to choose between two doors with the goat, while in other situation, when the goat is behind your door, host can only offer you to switch to the door with a prize.
And that (host left with no option but to offer you door with the prize) happens twice as much (66%) than situation where host can choose between two doors with a goat behind (33%).
Good point.
Having proof of this is great, now the truth of this is solidified for me, thanks for making this video
Your first pick is always most likely wrong, so always switch.
Got a guy at work who insists wrongly, it's 50/50. He just doesn't get it.
I feel bad for him... I hope he's not making a full out of himself. I would have been probably the same way, if I hadn't learned about it on my own... lol
Yes, fancy explanations are probably not needed. And yet we are still from time to time blessed with pearls from those of mighty intellect who *insist* the second choice is 50-50.
Your stupid. it is only 2 choices left 50% chance 3rd grade math
When Bob has done the cards, Bob is allowed to make comments. This is the purpose of the cards.
Kirk Bocek
haha I've already done the cards 20 times 11 times switching worked 9 times it didnt . doesn't seem like switching is 2/3 chance to me. You are going by group mentaliy, everybody says switching increases chancesso you lose common sense and just agree haha there are only 2.
Kirk Bocek
LOL I am an Actuary went to Northwestern if you don't know what that is look it up. I think I wud know probability haha.
Well you don't know the probability here. Go read the comments on the RUclips posting. Seriously, don't think you can just *figure* this out. Do the cards. You'll be surprised. I promise.
I shall make it simple.
At your first selection, there is 1 chance out of 3 to select the winning door, when you'd better keep your choice
but there is 2 chances out of 3 to select a loosing door, when you'd better switch your choice.
As you don't know the result of your first choice, odds of winning is larger (=2/3) by shifting.
Yes.
When you pick a goat and then switch you'll get the car and when you pick the car and then switch you'll get the goat and since there are two goats and a one car in the beginning you will more likely pick a goat than a car, simple as that
And just for full disclosure, I myself struggled mightily with this when it was first presented to me, and eventually made a little computer algorithm to simulate the game, and to "prove" that switching didn't make a difference. Had it run for a few hundred trials, and to my astonishment, the switch proved to, indeed, double your chances of winning. Only after that did I put enough effort into understanding why this is, and finally "got" it. Now that I understand it, it seems obvious.
Does your 'algorithm' always bet on door "one"🚪(was it a 'constant' choice in the experiment) or every time a random (different) door was chosen to bet on, before you check what's behind the 2 remaining🚪?
The host opening one of the remaining doors reveals information about the likelihood of the prize being behind the remaining door, while giving you zero new information about the likelihood that the prize is behind your door (because the host could never open the door you picked). The problem is called a paradox not because both probabilities are correct, but because people overwhelmingly believe the probability is 1/2 when that is wrong. It's a paradox of behavior, not math.
It's quite simple to explain.
There are 3 doors, 2 incorrect (goats), 1 correct (car!). You choose one of them. One of the incorrect doors is revealed to be incorrect. You can then switch if you want to. Switching will only work if at first you chose one of the wrong doors. Seeing there are 2/3 wrong doors at the beginning, switching will work 2/3 of the time.
Tracey-Ann Thompson-Carroll
MGF 1106
The card you choose the odds of winning the car is 1/3, but the dealer has two cards, which gives him a 2/3 chance of still having the car.
He showcases one of the cards to be a goat and tricks the contestant so he or she thinks about switching.
Part1
It's easier to understand if looked from the car perspective (the prize u want):
1) picking a car = 33.3% (one door)
2) picking a goat = 66.6% (two doors)
If you stick to your 1st choice, your best is 33.3%,
but by switching you're trading your 33.3% for the remainder--the 66.6% (in effect, you are choosing TWO doors vs. one door).
So, staying = choosing one door = 33.3%.
switching = choosing two doors (opened door and switched door) = 66.6%.
Basically, opening two doors is better.
it'll ALWAYS BE 50-50 if u consider the chance that "the door u initially choose, the host may open that."
As I've said here, again and again, don't try to do this puzzle in your head. *Actually* do the cards and see what results you get. *Then* come back and tell us what you get.
Kirk Bocek done with paper, actually :-|.
my point isn't that the monty hall problem is wrong or that kinda stuff. what i want to say is the trick is actually in the 1st assumption (en.wikipedia.org/wiki/Monty_Hall_problem#Standard_assumptions); and only through that assumption, the whole thing gets a paradoxical look. and when u get what this assumption does to this theory, sadly enough, all charms will just be wiped away.
Md. Abdullah Al Muyid Yes, but if you leave out those assumptions, it is no longer the "Monty Hall problem", and so it is irrelevant what the chances without those assumptions would be. You should not really think of those points as "assumptions", they are the rules which HAVE to be followed for any discussion to make sense. Same as that the host has to actually KNOW what is behind each door, so that can can ALWAYS show a goat and does not accidentally open the door with the car behind it.
Has anyone figured out what the odds are as to which door people are more likely to choose? Is it 33% for each door, or do people more often pick the door on the far left since in the west we read from left to right, for example?
"Would this work if there is no game show host? for example, if there are three doors and one of them is an exit and the other two have booby traps. If you pick one of the doors, but your friend comes out of one of the other 2 and confirms it's a booby trap, would switching your initial choice mean you have an advantage?"
I wanted to reply to this comment, but google decided not to let me, for some reason.
The Monty Hall Problem requires that the "host" *intentionally* open a door that reveals a goat. In this case, your friend is opening a door *at random*. In theory, they might've gotten lucky and picked the exit (as opposed to the host, who will *never* choose the car).
When you take the host out of the equation, it becomes a variant known as the Monty *Fall* Problem. Since the door was opened completely at random, switching or staying makes no difference (50/50), unlike in the Monty Hall Problem where switching wins 66% of the time and staying wins 33% of the time. This variant is what likely makes a lot of people think it's a 50/50 shot - the fact that the host will not reveal the car is what changes the odds.
Worth mentioning are two other variants to this problem. First is the unfriendly sounding "Monty Hell" variant. Let's say the bad guy who threw you into this booby-trapped dungeon is the host. He knows where the exit is, but he obviously won't show it to you. However, the bad guy "forgot" to place it in the rules that he must always give to the option to switch, so he will only give you the option to switch if you picked the winning door! Because of this, if you're playing Monty Hell, switching will win *0%* of the time.
The flip side of this I'll refer to as "Monty Heaven". Let's say the bad guy is lazy and sends his assistant to be the host. But the assistant is sick of the bad guy and actually wants you to escape. She figures out the rules and uses the same exploit, but this time in your favor - you will only be offered the chance to switch if you picked a losing door. In this case, switching will win *100%* of the time.
I dont understand how intention can affect the probability. Could you explain it a bit more?
Nipun, your question makes a valid point. Many, many persons on these Monty Hall videos make the claim that if Monty forgets where the goats are, takes a guess, and luckily picks a goat door, that that makes the option to switch a 50/50 proposition. Those who make that claim are absolutely wrong. It is still to the contestants clear advantage to switch.
The problem above about booby-trapped doors is identical to the Monty Hall problem. The person trying to find the exit should switch when his friend shows him a bad door.
This has blown my mind
At the beginning, you have a 67% (2/3) chance of picking the wrong door. Once you pick a door and a wrong door is revealed to you, the chance that you first picked the wrong door doesn't change; all that changed is the information you have on the 3 choices. So it’s still a 67% chance you first picked the WRONG door, but now you know which of the two remaining doors is wrong so there is a 67% chance that SWITCHING will get you the RIGHT door.
Every time this problem is presented, you see people arguing to the last, absolutely CONVINCED that the chance is 50/50, and viewing this "fact" as so obvious that they really don't put enough effort to understanding the explanations of why it is not so. Some of the people who begin this way, do eventually get it through the various explanations, but some do require an actual test like in this video. That doesn't make them stupid, just a bit too sure of the accuracy of their own intuition.
Thanks this helps,
Best way this has been shown to me, i get it now
...the probability, just like an uneven coin would scew the probability of heads or tails, despite there being just two options.
If it helps, a mathematically equivalent way of looking at the choise of whether to switch or not, inthe Monty Hall problem, is if the host didn't open a door at all, but gave you an option to switch from your one door, to the two remaining doors. You still have two options - either stick with your door, or switch to the other two doors, but now it should be...
No, I did not mean "switch from your one door to choose one of the two remaining doors" - If I meant that, I would have written that. The bit in brackets what is what you read into my post, it was not intended, or implied by the text. What I wrote was "switch from your one door to the two remaining doors". In other words, instead of one door, now you have the option of simultaneously choosing BOTH the other doors.
Let's say the car is behind door #2. Now, pick a door:
1) If you pick door #1, Monty shows you a goat behind door #3. If you swap, you win. If you stay, you lose
2) If you pick door #2, Monty shows you a goat behind door #1 or door #3. If you swap, you lose. If you stay, you win.
3) If you pick door #3, Monty shows you a goat behind door #1. If you swap, you win. If you stay, you win.
You will win about 66.6% of the time by swapping, and win about 33.3% of the time by staying.
Very helpful for my engineer father thank you!
Now if your actually saying "to switch from your one door, to the two remaining doors" then this would be a different scenario, and hence would even be applicable to the original premise of the monty hall problem of only being able to choose 1 door at a time. However, you should probably still swap in that scenario:
1) the probability that your first choice was correct is 1/3, which means you should most likely swap
2) swapping is a better probability because there's now more doors in 1 option
The percent can't be added.
If a factor change, all the ecuation change too.
If the guest show the car you still have the 33%?
And 30/70(any side) is the normal result of the 50/50 odds.
Part2, it's not 50/50, and 66.6% represent the car b/c a goat was shown. E.g:
Event A) Picked car and stayed = 33.3% of picking car
Event B) Picked car and switched = 33.3% of picking car
Event C) Picked goat and stayed = still 33.3% of picking car
Event D) Picked goat and switched = still 33.3% of picking car plus the 33.3% of the shown goat resulting in 66.6% of picking car if you switched (if you stayed, you're back to 33.3% and NOT 50% because door1=33.3%, dr2=33.3%, dr3=33.3%).
Try it yourself:
www.grand-illusions.com/simulator/montysim.htm
Note: It says it won't work, but click and it does.
This site lets you "play" the Monty Hall Problem and see the 1/3 and 2/3 split for yourself. You can make it run 100 or 1,000 times!
IE Only. Didn't work in Chrome for me. Didn't try Firefox. Very nice BTW. An excellent way for people to easily do this programmatically.
Is your simulator an argument for 50-50? I received precisely the results expected.
This is a simple problem with very well defined parameters. There are indeed *two* choices here. This first is which of three doors to choose. You can choose any of the three. That choice can result in a goat or a car. The second choice is to switch or not. Don't try to somehow remove the first choice. It's really there.
If you don't switch, your odds are 1 out of 3. So if you *do* switch, how can that magically change you from a 1 out of 3 probability universe into a 1 out of 2 probability universe?
Kirk Bocek It says it doesn't work, but just click and then it does.
The odds remain 1/3 for the first card, and increase to 2/3 for the unpicked card, the Monty Hall Problem is true! I spent quite a while arguing against it, but I was wrong.
The "key" is that it is NOT random, the removed card is known, thus the second round isn't "random chance" at all!
5Cats After, what, 3 years, you are the first, the very first to allow his mind to be changed and to admit it. You are a god among men, an angel among demons. You are the very force of order in the universe.
But I gush after having been hammered here for so long and watched others receive the same treatment.
You are correct. The door being opened is not a random event. Monty knows which doors have goats and shows you one of them. There is always one available even if you have selected one with your initial choice.
But again, it is not "removed." It is simply opened. It is information given to you. There are still three doors. That's why the odds remain 1/3 to 2/3.
Yes! The "it's not random" part threw me for quite a while. If it were random, the odds would be 50-50, BUT you'd get knocked out 1/3 of the time! Since you cannot get knocked out, the odds do indeed carry over from round 1 to round 2, giving you 1/3 and 2/3.
The "Gambler's Fallacy" only applies to random games of chance, not the MHP, this also confused me.
I am indeed a cat among the mice, Lolz!
5Cats At the outset you don't get knocked out. Guaranteed. So there is no probability of getting the first round wrong. Or right. It's neutral. It's a non-event with a guarantee that Monty will do what he always does-turn over a goat first. It exerts no control on the second choice, and does not filter down conditionally. Judgment is not passed from an event that's 100% not to knock you out. These are not linked events. After the first flip, there are only two doors to consider. You're guaranteed to get there because of the way Monty plays. No need for a pad and paper. Or cards.
...self evident that one wouldnt' even EXPECT the probability to be 50-50, despite there being just two options: stick with your door, or take the other TWO doors.
That the host opens the goat-door of those two doors before you make the choise, does not change the fact that you might just as well have been offered to switch to the two doors, which eliminates the unintuitive nature of the problem.
In this way, relevant information can make one outcome, in a two-possibility situation, more...
I would stay no matter what. I'd rather lose picking wrong than being tricked to change if i guessed right. Lol
Then pick two doors but say the door that you don't want. Then after he opens a goat door, switch.
Then address my marble example, which I created to show that previous events can alter the probabilities of a binary choise between two hidden options. Do you think there's something paradoxical in that, rather than straight forward, completely non-problematic math?
It's heartwarming that nearly a year and a half after posting this video I *still* see comments trying to argue that the final choice is 50-50. If you think that's the answer, do the cards and keep the tally. You'll be surprised. I promise.
Because they only do it 100 times and a freak occurence happens and they declare they don't trust it and quit.
. A diagram on wiki is very helpful to understand this.
Would this work if there is no game show host? for example, if there are three doors and one of them is an exit and the other two have booby traps. If you pick one of the doors, but your friend comes out of one of the other 2 and confirms it's a booby trap, would switching your initial choice mean you have an advantage?
Amir Aribokill No. In that case, it's 50-50. The host of the quiz knows where the car is, so possibility of the host's opening the door where there's a car is zero. On the other hand, the friend knows nothing about what's behind the doors, which means possibility to open the door without booby traps is 1/3.
But if a door is revealed to be a booby trap by a friend, doesnt it become the same as Monty showing you the goat?
@@Zaekr2111 once a door is revealed to have the booby trap, then you should switch because your door always only has 33 percent chance of a good outcome. Switching doubles your chances to get a door without traps.
probable than the other. The 50-50 situation ONLY applies if no such probability-scewing information is present. There is absolutely nothing in mathematics that requires a two option situation like this to have equal probability for both options.
Why do you always pick the same card number 1 all of the time? Isn't that non-random?
....Finally, if none of that is enough to convince you, do the experiment described in this video, and record the resuts for yourself. Or, do as I did way back when I struggled with this problem - make a short computer program to simulate the game, and run it through a few hundred iterations. You'll see the probabilities approach the 66% chance for switching and 33% chance for not switching.
It seems like it would be 50-50 because you are focused on the probability that a given piece of information is true (e.g., that the car is behind Door #1) and not the probability that an event will occur (e.g., that you will win the car). So once the wrong car is revealed, there IS a 50% chance that the car is behind either of the two remaining options, but there is still only a 33% chance that you actually PICKED the door that the car is behind. So it is in your favor to switch.
Because some people refuse to believe that it is beneficial to switch and require this experiment. Some people are so stubborn that absolutely nothing you say would convince them, even after experimenting on simulations.
if I try out for an NFL football team, there are two outcomes: I make the team or I don't. By your logic, I have a 1/2 of making the team, which is obviously not true. Having two options means the probabilities of the two options must sum to 1; it does not mean they must be equal.
"the host doesn't give you another opportunity to choose a door again"
You originally said:
"if the host didn't open a door at all, but gave you an option to switch from your one door, to the two remaining doors."
Assuming you mean "switch from your one door [to choose one of the] two remaining doors" then you are left with the original choice you started with, which is choosing 1 out of the three options. Of course, that assuming that's what you meant to say, rather than what you wrote...
Don't think that this is a simple thing. For many years a lot of mathematicians thought that in the second instance there was 50% of chances. And many of the common people that act like they "get it" and "everyone who doesn't get it is a stupid" are, in my opinion, not really getting it, but only accepting the "truth" that is being said to them. It took me like months to accept that there was really 2/3 chances of winning with the switch.
1=goat
2=goat
3=car.
I pick number 1, 2 is revealed. i switch to 3 to get car.
I pick number 2, 1 is revealed. i switch to 3 to get car.
I pick 3, the number is revealed, i switch to another number I get goat.
therefore switching gives you 2/3 chance of winning the car.
Regarding the coin toss comparison - fine, if you don't like that, then how about this:
There are two jars filled with with black and white marbles. One jar has a 50-50 ratio of black and white, and the other has a 30-70 black-white ratio. Now a game show host picks one marble from each jar, blindly - neither you nor the host sees what colors were picked. He then, without looking, places the marbles in two boxes, and the boxes are closed. You know only which box has a marble from which of...
damn you're smart.
...still would have ended up in a situation where you have one door that might hold the price. The timing of the reveal of one of the false doors does not matter - mathematically, the situation of either revealing one of the false doors and then letting you switch, is equivalent to the situation where you are allowed to switch to both of the remaining doors. That's why there's the 2/3 chance to win with switching - there's no real "paradox" there, just the counterintuitive feel to the solution.
My analysis already acknowledges that it's most likely better to swap, because the probability that your first choice was correct is only 1/3. So I don't see why you're telling me something that I've already acknowledged...Though, my first comment doesn't acknowledge the switch, as it was merely focused on explaining that the player is always left with 2 remaining options. If you would have bothered reading my comments you would have seen I've already acknowledged that it's better to swap.
@bucko386 "If he offered you the swap without opening any doors you would say YES" your comment is incorrect, without opening the door there is no variable change, therefore no point taking the swap, change your mind as many times as you want , with 3 doors its 33.3/33.3/33.3 chance on all doors. when monty shows you 1 incorrect door that brings in variable change, now the chance of the door you originally picked stays at 33% but the other remaining door is now 66%
Acknowledging that it is better to switch only makes you half right. The problem is that you have also asserted that a probability of 1/2 is also correct, when it is not. The "evidence" or "logic" you seem to be resting on is that if you have two possible outcomes, then the chance that one outcome is correct is 1/2. This is only true if the events are equiprobable. Let me give you an example,
...you swap to the other TWO doors instead of revealing where the goat is behind one of those two doors. The ONLY difference there is that in this latter scenario, the answer would be intuitive and obvious, that you should switch - two doors is better than one!
But that's exactly what you get if you switch AFTER the one false door out of the two remaining is revealed. In the altered scenario that door would just have been opened, to reveal a goat, after you made the choise to switch - you....
I did address your marble example, inadvertently. It's essentially your 'swap opportunity of choosing 2 doors simultaneously' example. My analysis already concludes it's better to swap, because the odds that your 1st choice was correct is only 1/3. However, you're still left with 2 options - 1 goat, 1 car - & you're only aloud to pick 1 of the 2 remaining doors, thus giving a 1/2 probability, by default. Nevertheless, my analysis still concordantly acknowledges the likelihood for swapping.
It has been three years and I do not know if you keep thinking the same thing, but just in case.
Think of the odds as the approximate fraction of times an event will be fulfilled after numerous repetitions. For example, a coin has probability 1/2 to fall heads or tails, so after 1000 attempts it should have left about 500 times heads, and about 500 times tails. The margin of error must be small.
When we say that the probability of winning in the Monty Hall problem is 1/3, we refer when your choice NEVER changes, so after many attempts you would have won about a third of the time maintaining that same strategy. You will always keep the door of the initial election, you will not make again a random selection among the remaining two doors, and I think that is the greatest confusion.
The probability of being successful at first was 1/3. If anyway you will not change the choice, in the end you have to win the prize many times as if the presenter had never opened the other door. That action does not affect the result. It's like if you have a box with 99 white balls and only 1 black; grab one without seeing it and keep covered in hand. The probability of having caught the black is 1/100. If someone removes all other balls in the box leaving only one, but you will not have the opportunity to change which is already in your hand, then the probability of having caught the black will not increase to 1/2. In the Monty Hall problem is the same: if you will never change, the proportion of times you win must be the same as if they never give you the opportunity to change.
If this is not clear, another way to look at it is thinking that the probability of an event is equal to the number of favorable cases divided by the number of possible cases, only if the cases are equaly likely. For example, if a dice was not balanced, the probability to be out wouldn't be 1/6 for each number, despite they are still six possible outcomes. In this case, the fact that gives greater probability weight to a gate than the other is that yours was completely randomly selected from three possible doors, while the other was selected with the obligation that the car must remain covered. Think about this: when opening the other, does it give you additional information about your door? The answer is no, you could have chosen one with a goat ore one with a car, and anyway it was going to be an additional goat, so seeing that in another door there is a goat does not make yours more likely to have the car. Instead, you effectively have new information about the other door closed. Suppose you had chosen the door 1 and the presenter opened the 3. You know he is forced to leave the car covered. If he did not open 2, it is probably because the car was behind door 2 and not behind 3, and so he was forced to open 3. Or maybe the election between 2 and 3 was irrelevant because both had goats, but this is more difficult, because that implies you chose the correct door at first, which has probability 1/3.
Confusion usually comes because some people see the two stages as independent games, when in fact they are not. If at first you had chosen the car (1 case), changing will make you lose. If at first you had chosen a goat (2 cases), changing will make you win. As seen, if changing makes you win or not depends on your first election.
If in the second stage the choice was randomly, then the probability of winning would be 1/2. To understand why this doesn't contradict previous conclusion is that it is good to think the odds as a fraction of times you succeed after numerous attempts. Making randomly, about half of the time you pick the same door you chose at first, and about half of the time you pick the door the presenter left closed. This is because randomly there is nothing that forces to pick one door more times than the other. In the half you pick the same door you chose at first, 1/3 you should win, but that represents 1/2 * 1/3 = 1/6 of the total, and you should lose 2/3 of the time (2/6 of the total). In the half you chose the presenter's door, you should win 2/3 (2/6 of total) and lose 1/3 (1/6 of the total).
C: Changing
N: No changing
N C
Win---> 1/6 + 2/6 = 3/6 = 1/2
Lose--> 2/6 + 1/6 = 3/6 = 1/2
Wondering your thoughts on this? I think why people get confused is that after they pick the 1 of 3 cards, and then one goat is revealed, what they don't get is that they have already made the 1 in 3 choice, and that revealing the other goat does not change things as is. Just seeing one card that is not the desired one does not change the odds of what has already happened. But now, they are given the choice of choosing between the two cards not yet revealed thinking they have a 50/50 chance?
The only thing that matters is your initial choice.
Two out of 3 times, you will pick a goat. Obviously, staying with your first choice will result in you getting a goat 66.666...% of the time, on average.
You will only select the car 1 out of 3 times. Staying with your initial pick will result in you getting the car only 33.333...% of the time.
The strategy is obvious - switch doors and you should win 2 out of 3 times.
I really don't see why people have such a hard time understanding that.
incorrect. i still CHOOSE to stay, which makes it 50/50
@paddygilbert
I made this video in response to those who *swear* up and down, on all that is holy, that the final probability of switching is 50/50 and not the correct 33/66. There is just nothing you can say to these people to convince them otherwise. I got so frustrated with one of these people that I made this video. If you ever get into this argument, send that person here and tell them to do this experiment.
No, you did not do any valid ananylsis, it never was valid and it isn't now.
Two options means there is a 50/50 chance- unless you can explain how one option is more or less likely to occur than the other. And we have done so here.
Two choices means 50-50 doesn't always apply, and this is one of those cases. You used a false statement to do your "analysis" so it was never valid at all.
Now admit you were wrong.
*SIGH*
2 out of 3 times you will choose a goat initially, right? That means that if you stay with your initial choice, 2 out of 3 times you will get a goat, right?
On the other hand, if you trade doors after Monty shows you a goat, you will win the car 2 out of 3 times, right?
It doesn't matter if you play once, a hundred times, or a gazillion times - the odds stay the same and you will win about 66.6% of the time if you swap doors.
But I'm not sure if you get this or not by looking at one of your other comments. I won't be bothered with reading the book-worthy list of replies, so if you you get this, just ignore my comments and leave it for someone else to read in the hopes that it helps them understand; I rather not to go back and forth on this with you (or anyone else for that matter) unless you are confused/want clarification.
BTW, 5Cats, you are so prolific here, the system is marking you as spam. I am trying to catch your comments and clear them for flight.
Oh! I missed that part! That's kind of funny, eh?
Just came to look at that recent comment. It still "bends my brain" a little but it's still 1/3 and 2/3 after the 'extra' card is removed. It was 1/3 for your card in BOTH rounds? Perhaps it's easier to understand it like that?
If it's 1/3 with 2 other cards, then each card is 1/3.
If it's still 1/3 with ONE other card? That card other must be at 2/3 odds.
That makes sense? I hope it can help someone grasp this slippery problem.
5Cats I think you have it. The mistake that I and most others make is to "remove" the first door when it is opened and somehow think that now only two doors are involved when we make our choice. Three doors are always involved.
What if there were TWO contestants and each of these two contestants were ardent mathematicians, both with an innate understanding of the principle behind the Hall paradox. After each choose a door, and the third unchosen door was opened to reveal a zonk, would each of the two players, if given the chance, leap at the opportunity to accept the other fellow's door? (And in this scenario the 2 scholars are rivals who would not agree to sell the car and split the proceeds) Hmm?
Johnny Radionic ™
You missread, or misunderstood my comment regarding the altered scenario for the monty hall problem.
The point was that the host doesn't give you another opportunity to choose a door again - he gives you an opportunity to choose between the door you originally chose, or the TWO doors that you didn't choose.
This is a binary choise: stick with your door, or choose BOTH the other doors. No third option there.
And that is exactly mathematically equivalent to how the game actually goes.
I knew before you reveal the card which one was the "car" so to mean this aint proven
This will make it click: The host is not able to open the car door.
@paddygilbert, I didn't make this for you, you get it. You need to hook up with one of these knuckleheads who swear the answer is 50/50. *Then* you'll see why I made it.
Relax; don't let that temper boil. Maybe you'll be able to comprehend these probabilities one of these day...
The fact that there WAS ONCE 2/3 doors that are goats becomes irrelevant as soon as 1 of the those 2 doors are revealed, and hence eliminated from the option, because then, there are only 2 options left - 1 goat, 1 car - to choose from during the swap opportunity. It doesn't matter if there were 3 goats and 1 car, if 2 goats are revealed before the opportunity to swap, because either way, you're still left with only 2 options; 1 car, 1 goat; it's always 50/50
....And are you saying that since this is a binary choise between two options, and one choise gives you better odds than the other, this should somehow be considered paradoxical?
Of course there are no paradoxes involved. No more than there are in the monty hall problem - the probability calculations are straight forward, and there's no requirement AT ALL that having a blinded choise between two options must imply 50-50 chances.
U know U can even blame the weather for that.
...And even then he'll probably just tell you that it doesn't solve the paradox that supposely exists. I don't know if anything we say will ever convince him that there is no paradox. Two options doesn't make it 50/50. Well, sometimes it does, and if you can logically demonstrate why one is more likely to occur than the other (as done here), then it's NOT 50/50. And that's exactly what happened here.
"This is only true if the events are equiprobable."
False. Equiprobability would only determine the relevancy of the probabilities; not whether or not they're true. As I've already denoted previously, you could argue that the 1/3 probability is more relevant than the 1/2 probability, as the 1/3 probability information can produce better tangible results than the 1/2 probability information, but that doesn't make the other probability untrue; it's entirely erroneous to assume it does.
"No, if you are only left choose 1 out of 2 options, then you will have a 1/2 probability of choosing the correct option... So both probabiliteis are correct."
WRONG.
..the original jars, and you know the ratios of black-and-white marbles in those jars.
Let's call the 50-50 jar "Jar A", and the box with the marble out of that jar "Box A", and the other jar and box "Jar B and "Box B" respectively.
Now you are given a choise between the boxes - if the box you pick has a white marble in it, you win a million dollars, if it has a black marble in it, you win nothing.
Are you now saying that there somehow should be a 50/50 chance of winning with either box?...
"No actually it's wrong"
No, it's right. In fact I have logically demonstrated my analysis through deductive reasoning; so yeah, it's reasonable. The onus is now for you to disprove me; this has yet to occur.
"a 50/50 chance you're going to die tommorow[sic]"
False comparison, unless you're talking about suicide/euthanasia, because a suicide/euthanasia are choices to make, like in the monty hall choices; yes or no; 1/2...
Either something happens or it doesn't; that's the basis of reality.
@Solsane Nice. I like it.
what do you think "If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal?"
Incorrect. If you stay you have a 1 in 3 chance of winning.
Last time I watched gameshow, there were no cards and no dice
lol
You are a complete idiot.
Sorry, got the wrong impression from the argument you're having with WhyTauIsRight AndPiisWrong - if you're clear on the issue of the probabilities, I don't know what the argument is about.
No, it's right.
"The percent can't be added"
Yes they can.
"If the guest show the car you still have the 33%?"
No.
"And 30/70(any side) is the normal result of the 50/50 odds."
Wrong.
Yes it is 50% after you are left with two choices and I'll prove it quite simply. If half the time you switch and half the time you stay you will win 50% of the time. LOL I am so damn smart. Who is smarter than me? No one. Of course if you switch 100% of the time you will win a lot more cars. So people are basically arguing two different points. Some people say that after one choice is eliminated you have a 50/50 chance of winning. They are right. Other people say that if you switch every time you will win 2/3 of the time. They are right too.
CacacacaYEAH - I recommend searching RUclips for "The Monty Hall Problem Explained" - watch the video by axmurderer27. That's perhaps the clearest explanation of the problem, and why switching does double your chances from aprox. 33% to aprox. 66%. If that's not enough, look up the Wikipedia article on "Monty Hall Problem"; it explains it in various ways, including a mathematical proof. The article also gives an interesting history of the problem, and how it's fooled even mathematicians.
@TheWatchOrgie -- Funny, but a coincidence. Wish I'd planned it that way.
No, if you are only left choose 1 out of 2 options, then you will have a 1/2 probability of choosing the correct option. But simultaneously, the probability that your first choice was correct is only 1/3, which means you should most likely swap. So both probabilities are correct. Just for fun, here's another probability that's also true. The probability that all your choices are correct is 2/5. 5 is the total number of options you've had throughout, and 2 is the number of possible right choices.
Relevancy? A probability is only relevant if it is true. 1/3 is the more relevant probability here because it is the mathematically derivable and empirically provable true probability of the event. 1/2 is just something you made up with zero mathematical or empirical evidence. It is not an assumption to say that 1/2 is untrue; it is a provable fact.
Why can't you just admit you're wrong?
Ahh, so that's what the argument is about. Of course there isn't any real paradox - there's only the conflict between "common sense" expectation, and the math. The probabilities aren't in any sense "paradoxical".
Yes, you would have an advantage. But your friend is still acting as the game show host. He is still looking behind the doors the way the game show host would.
"if you don't like that"
It simply wasn't comparable to the original premise. It's apparent to me now that you meant what you wrote before, so I will only reiterate that I understand what you're saying, but it's still a false comparison. There's simply nothing in the original monty hall premise which causes the doors to have an irregularity. You don't get "2 doors for one choice", there's no '1 transparent door' (or anything else that would cause an irregularity in the doors) ...etc
Having two choises does NOT mean that the probability is automatically 50-50. This is your basic assumption, and it is simply false. A fair coin flip gives a 50-50 probability of heads or tails, but an irregular coin may produce, say, a 55-45 outcome probability.
If there were just two doors originally, a goat behind one, and a car behind one, placed randomly, then the chance of picking a car WOULD be 50-50. But you cannot ignore the relevant details of the problem, which scew....