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@Matthew: Here is how I calculated it:

You want a specific set of 20 stickers. Imagine you have already \(n\) of these. The probability that the next sticker you buy is one that you want is
$$\frac{20-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not wanted})\times\mathbb{P}(\text{sticker after next is wanted})$$
$$=\frac{662+n}{682}\times\frac{20-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next wanted sticker is
$$\left(\frac{662+n}{682}\right)^{i-1}\times\frac{20-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find the next wanted one:
$$\sum_{i=1}^{\infty}i \left(\frac{20-n}{682}\right) \left(\frac{662+n}{682}\right)^{i-1} = \frac{682}{20-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{19}\frac{682}{20-n} = 2453 \text{ stickers}.$$
Matthew
on /blog/56
               
@Matthew: Thank you for the calculations. Good job I ordered the stickers I wanted #IRN. 2453 stickers - that's more than the number you bought (1781) to collect all stickers!
Milad
on /blog/56
               
@Matthew: Here is how I calculated it:

You want a specific set of 20 stickers. Imagine you have already \(n\) of these. The probability that the next sticker you buy is one that you want is
$$\frac{20-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not wanted})\times\mathbb{P}(\text{sticker after next is wanted})$$
$$=\frac{662+n}{682}\times\frac{20-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next wanted sticker is
$$\left(\frac{662+n}{682}\right)^{i-1}\times\frac{20-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find the next wanted one:
$$\sum_{i=1}^{\infty}i \left(\frac{20-n}{682}\right) \left(\frac{662+n}{682}\right)^{i-1} = \frac{682}{20-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{19}\frac{682}{20-n} = 2453 \text{ stickers}.$$
Matthew
on /blog/56
               
@Matthew: Thank you for the calculations. Good job I ordered the stickers I wanted #IRN. 2453 stickers - that's more than the number you bought (1781) to collect all stickers!
Milad
on /blog/56
               
@Matthew: Here is how I calculated it:

You want a specific set of 20 stickers. Imagine you have already \(n\) of these. The probability that the next sticker you buy is one that you want is
$$\frac{20-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not wanted})\times\mathbb{P}(\text{sticker after next is wanted})$$
$$=\frac{662+n}{682}\times\frac{20-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next wanted sticker is
$$\left(\frac{662+n}{682}\right)^{i-1}\times\frac{20-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find the next wanted one:
$$\sum_{i=1}^{\infty}i \left(\frac{20-n}{682}\right) \left(\frac{662+n}{682}\right)^{i-1} = \frac{682}{20-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{19}\frac{682}{20-n} = 2453 \text{ stickers}.$$
Matthew
on /blog/56
               

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