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Doesn't it depend on the #/ distribution of coupons printed/toys manufactured though? To take an extreme example if there are 4 different toys and only 10 of ea

Doesn't it depend on the #/ distribution of coupons printed/toys manufactured though? To take an extreme example if there are 4 different toys and only 10 of each are manufactured, then the probability of getting 4 all the same is surely somewhat lower than it would be if you assume essentially an infinite toy supply where the chance of getting a particular toy is always 1/4.

mlyle 4y ago
We're effectively assuming that the number of toys are infinite and equally likely. Or, to put it slightly more formally, we're choosing each time from 4 equally likely possibilities, with replacement. Imagine you're drawing from a bag with 4 toys inside. How long until you've drawn each of the 4 toys, if you replace the toy in the bag after each choice?
- wizofaus 4y ago
  Yes I gathered, but in reality there's not an infinite number all produced in equal quantities. I'm curious how much it might change how many meals you'd need to buy if it was a limited production run, and even whether the total number of participants would make much difference (clearly it must make some - if only 1000 of each toy were produced and 1000 customers took part, surely very few would have been lucky enough to get all 4 toy types at all, as on average they could have only received 4).
  - mlyle 4y ago
    > Yes I gathered, but in reality there's not an infinite number all produced in equal quantities. Most of the McDonald's toys are made in equal quantities, and the quantity available dwarfs the quantity any one customer buys. > clearly it must make some - if only 1000 of each toy were produced and 1000 customers took part, surely very few would have been lucky enough to get all 4 toy types at all I don't think the odds change much in this case. 4 provides you with a very small chance from a bottomless pit -- 3/4 * 1/2 * 1/4 = .09375 chance of getting all 4 with with purchased. It's slightly better if there's a small number, say, 4 of each, because you remove the ones you don't want from contention. You have a 12/15 chance of the second not matching (3 toys remain that you don't want; and 4 toys each remain of each of the 3 kinds you want); then a 8/14 chance of the third not matching, and then a 4/13 chance of the fourth one not matching-- or a ~14% chance. So even when we say there's only 4 of each toy, the odds don't change -that- much. Once we get to 1000 of each, the odds basically are the same as the infinite case. 3000/3999, 2000/3998, 1000/3997 are practically 3/4, 1/2, 1/4 -- ~.09389. > as on average they could have only received 4 Well, we already know from the infinite case you need to buy more than 4 to have a good shot.

mlyle 4y ago

We're effectively assuming that the number of toys are infinite and equally likely. Or, to put it slightly more formally, we're choosing each time from 4 equally likely possibilities, with replacement. Imagine you're drawing from a bag with 4 toys inside. How long until you've drawn each of the 4 toys, if you replace the toy in the bag after each choice?

wizofaus 4y ago
Yes I gathered, but in reality there's not an infinite number all produced in equal quantities. I'm curious how much it might change how many meals you'd need to buy if it was a limited production run, and even whether the total number of participants would make much difference (clearly it must make some - if only 1000 of each toy were produced and 1000 customers took part, surely very few would have been lucky enough to get all 4 toy types at all, as on average they could have only received 4).
- mlyle 4y ago
  > Yes I gathered, but in reality there's not an infinite number all produced in equal quantities. Most of the McDonald's toys are made in equal quantities, and the quantity available dwarfs the quantity any one customer buys. > clearly it must make some - if only 1000 of each toy were produced and 1000 customers took part, surely very few would have been lucky enough to get all 4 toy types at all I don't think the odds change much in this case. 4 provides you with a very small chance from a bottomless pit -- 3/4 * 1/2 * 1/4 = .09375 chance of getting all 4 with with purchased. It's slightly better if there's a small number, say, 4 of each, because you remove the ones you don't want from contention. You have a 12/15 chance of the second not matching (3 toys remain that you don't want; and 4 toys each remain of each of the 3 kinds you want); then a 8/14 chance of the third not matching, and then a 4/13 chance of the fourth one not matching-- or a ~14% chance. So even when we say there's only 4 of each toy, the odds don't change -that- much. Once we get to 1000 of each, the odds basically are the same as the infinite case. 3000/3999, 2000/3998, 1000/3997 are practically 3/4, 1/2, 1/4 -- ~.09389. > as on average they could have only received 4 Well, we already know from the infinite case you need to buy more than 4 to have a good shot.