I've strongly know this piece of information for years, but I feel like it's such a useful thing to know yet commonly not grasped (especially as more and more of our lives are taken up by video games with dice rolls) and this is well presented to such an audience.
I think you're not explaining the entire situation. 36.6% will not get a lucky pull, but 26.5% will get more than one lucky pull. That's how it averages out to 1% drop rate, the chance of multiple hits vs the chance of zero.
What if I stop buying when I get the legendary loot box? I only need one. You've calculated the odds of getting at least one legendary. I'm not even sure what to ask. Perhaps a calculation that includes money spent. Maybe: how much money would I need to spend to have an X chance of getting at least one legendary. Thank you for this video. You've expanded my understanding.
Let's say the precise question you are asking is: I would like to buy x lootboxes such that I have a p chance at getting one legendary. Assume n is the odds of pulling a legendary. In other words, you have a 1 in n chance at getting a legendary. Hint: Let's define a lootcrate as n lootboxes. The probability of getting one legendary from a lootcrate is 1-1/e, as we explained in the video. The number of lootcrates we buy is x/n. So all we need to do is multiply our probability that many times and solve. (1-1/e)^(x/n) = p As usual, remember that the 1-1/e probability isn't exact, but it is a very close estimate for large n values. For most applications, this is good enough, and if it isn't, it's not too difficult to replace it with a more precise value (as shown in my video).
@@zhulimath Too much words; I do fewer: "x" = lootboxes "p" = desired chance "n" = odds per lootbox n^x = p Solve for lootboxes: *x = ln( p ) / ln( n )* ("ln" is the natural logarithm)
Huh? intuitively, I buy a box at 1% success rate. how come it means i don't always get one for each hundred? seems to be a contradiction inherent to probability..
It's possible that you could miss on all 100 boxes, and so the probability can't be 100%. The reason why it still "balances out" is because it's possible to succeed more than once.
An easier way to picture it is assume you have 1000 pulls. The legendary item should occur once every 100 pulls. Hope I don't fry my brain trying to make this work. The idea is 1 in 100. But not one in every hundred you set as a reference - which would become the pity system. You can get two limited at 56 and 57, and go all the way to 250 without getting another. On paper, its cheating, but well within the bounds of the random system i.e 2 pulls in less than 300 pulls. If you get another legendary at 280 - it balances out. So there is a risk of clamping the successes the higher you go, since the sample rate isn't per 100, a specific metric you choose to start counting at, but every hundred existing up to infinity. Oh well, some people reach the clamping section earlier, others later, but its always unique as elaborated in this video.
if pitty at 100 and you only pull 100, it would be 1 - (0.99^99*0.5) = 81.51 %. Just take the complement of all the failure multiply together, all the failure are 99 percent and pitty has 50 percent failure.
I also thought about a couple of other systems: - The probabilities don't change, but there's a way of turning unwanted items into other items (like "dust" in Hearthstone). So if you don't get a legendary, but you get a lot of rares, you might still be able to make a legendary out of the items you got. - The probabilities change with every pull. I think this is more often used for attacks that have a chance of an extra effect -- the chance might be 8% if it happened last time, plus another 8% for each attack since it last happened (so it'd be guaranteed to happen after 12 failures, since the probability would be 104%), and the stated chance would be what this averages out to (25%, IIRC). But I don't think there needs to be a guarantee -- the probability could get closer and closer to 100% without ever reaching it. For example, it could be as if the items were pulled from a machine that initially contains 70 commons, 24 uncommons, 5 rares, and 1 legendary, and for each item pulled, another set of items were added. Using the first pulls shown in the video, and assuming a legendary is pulled on the 5th try, that would result in the following odds: 1. Common pulled. New odds: 139/199 common, 48/199 uncommon, 10/199 rare, 2/199 legendary. 2. Uncommon pulled. New odds: 209/298 common, 71/298 uncommon, 20/298 rare, 3/298 legendary. 3. Rare pulled. New odds: 279/397 common, 95/397 uncommon, 29/397 rare, 4/397 legendary. 4. Common pulled. New odds: 348/496 common, 119/496 uncommon, 39/496 rare, 5/496 legendary. 5. Legendary pulled. New odds: 418/595 common, 143/595 uncommon, 49/595 rare, 5/595 legendary...
I thought he was going to be talking about the actual drop rates in particular games. 1% for the legendary is pretty generous, I think in reality it's more like 0.0001%.
![This Counting Problem Changed My Life. Can You Solve It? [Gridwalking]](http://i.ytimg.com/vi/3B-D3w292TI/mqdefault.jpg)
![This Counting Problem Changed My Life. Can You Solve It? [Gridwalking]](/img/tr.png)
I've strongly know this piece of information for years, but I feel like it's such a useful thing to know yet commonly not grasped (especially as more and more of our lives are taken up by video games with dice rolls) and this is well presented to such an audience.
This is fantastic. Thanks for breaking this down in such a clear way.
Now that I've actually watched this is a pretty good video. Good job man
intro is fire.
This is super useful and very well explained, thank you.
man, I should have really paid more attention in the discrete mathematics classes.
underated
0:06 was also an overwatch lootbox
I think you're not explaining the entire situation. 36.6% will not get a lucky pull, but 26.5% will get more than one lucky pull. That's how it averages out to 1% drop rate, the chance of multiple hits vs the chance of zero.
1:05 The video clearly states that the process will stop after pulling one lucky poll
@@glumbortango7182 Sounds like an unfair election process.
What if I stop buying when I get the legendary loot box? I only need one. You've calculated the odds of getting at least one legendary. I'm not even sure what to ask. Perhaps a calculation that includes money spent. Maybe: how much money would I need to spend to have an X chance of getting at least one legendary. Thank you for this video. You've expanded my understanding.
Let's say the precise question you are asking is:
I would like to buy x lootboxes such that I have a p chance at getting one legendary.
Assume n is the odds of pulling a legendary. In other words, you have a 1 in n chance at getting a legendary.
Hint: Let's define a lootcrate as n lootboxes. The probability of getting one legendary from a lootcrate is 1-1/e, as we explained in the video.
The number of lootcrates we buy is x/n. So all we need to do is multiply our probability that many times and solve.
(1-1/e)^(x/n) = p
As usual, remember that the 1-1/e probability isn't exact, but it is a very close estimate for large n values. For most applications, this is good enough, and if it isn't, it's not too difficult to replace it with a more precise value (as shown in my video).
@@zhulimath Thank you!
@@zhulimath Too much words; I do fewer:
"x" = lootboxes
"p" = desired chance
"n" = odds per lootbox
n^x = p
Solve for lootboxes:
*x = ln( p ) / ln( n )*
("ln" is the natural logarithm)
Huh? intuitively, I buy a box at 1% success rate. how come it means i don't always get one for each hundred? seems to be a contradiction inherent to probability..
It's possible that you could miss on all 100 boxes, and so the probability can't be 100%. The reason why it still "balances out" is because it's possible to succeed more than once.
If you got exactly 1 legendary every 100 rolls, it couldn't be random because they would have to be spaced out evenly.
An easier way to picture it is assume you have 1000 pulls. The legendary item should occur once every 100 pulls. Hope I don't fry my brain trying to make this work. The idea is 1 in 100. But not one in every hundred you set as a reference - which would become the pity system. You can get two limited at 56 and 57, and go all the way to 250 without getting another. On paper, its cheating, but well within the bounds of the random system i.e 2 pulls in less than 300 pulls. If you get another legendary at 280 - it balances out. So there is a risk of clamping the successes the higher you go, since the sample rate isn't per 100, a specific metric you choose to start counting at, but every hundred existing up to infinity. Oh well, some people reach the clamping section earlier, others later, but its always unique as elaborated in this video.
Tl:dr. 1% of a 1000 is 10. 10 pulls can be spread through 1000 without existing at 100 intervals. The higher it goes the worse it gets
The chance you get no legendary sis balanced out by the chance you get multiple legendary
But, what about if the probability of that to get a legendary decreses, example from 5% to .25% by a constant decrease of .25%
What about if the pity system isnt 100% for the legendary but 50% how could you calculate that?
if pitty at 100 and you only pull 100, it would be 1 - (0.99^99*0.5) = 81.51 %. Just take the complement of all the failure multiply together, all the failure are 99 percent and pitty has 50 percent failure.
I also thought about a couple of other systems:
- The probabilities don't change, but there's a way of turning unwanted items into other items (like "dust" in Hearthstone). So if you don't get a legendary, but you get a lot of rares, you might still be able to make a legendary out of the items you got.
- The probabilities change with every pull. I think this is more often used for attacks that have a chance of an extra effect -- the chance might be 8% if it happened last time, plus another 8% for each attack since it last happened (so it'd be guaranteed to happen after 12 failures, since the probability would be 104%), and the stated chance would be what this averages out to (25%, IIRC). But I don't think there needs to be a guarantee -- the probability could get closer and closer to 100% without ever reaching it. For example, it could be as if the items were pulled from a machine that initially contains 70 commons, 24 uncommons, 5 rares, and 1 legendary, and for each item pulled, another set of items were added. Using the first pulls shown in the video, and assuming a legendary is pulled on the 5th try, that would result in the following odds:
1. Common pulled. New odds: 139/199 common, 48/199 uncommon, 10/199 rare, 2/199 legendary.
2. Uncommon pulled. New odds: 209/298 common, 71/298 uncommon, 20/298 rare, 3/298 legendary.
3. Rare pulled. New odds: 279/397 common, 95/397 uncommon, 29/397 rare, 4/397 legendary.
4. Common pulled. New odds: 348/496 common, 119/496 uncommon, 39/496 rare, 5/496 legendary.
5. Legendary pulled. New odds: 418/595 common, 143/595 uncommon, 49/595 rare, 5/595 legendary...
Good command of the manim, i place an announcement in the comments of 3blue1brown, I hope you will have several views soon, you are very talented.
I thought he was going to be talking about the actual drop rates in particular games. 1% for the legendary is pretty generous, I think in reality it's more like 0.0001%.
huh, higher than i initially thought.
I want to believe that you are secretly MusicalBasics..! Your voices are so similar
So this is why my RNG is so bad
is that an apex lootbox
I think calling e the natural number is a bit weird. It is not a natural number after all. I get why you say it though.
This is fantastic. A truly great mind of our age.
Bro can you friend me on apex, also amazing video