# Problem 73 Solution

First I'll show that if the probability of an event happening is p then the mean number of trials to obtain a sucess is 1/p.

```Number   Probability
of         of
Trials     Success
------   -----------
1           p
2           pq
3           pq2
4           pq3
.           .
.           .
.           .
```
Since there must eventually be a sucess the sum of probabilities is:
p + pq + pq2 + pq3 + ... = 1.

The mean number of trials (m) is: m = p + 2pq + 3pq2 + 4pq3 + ...

mq = pq + 2pq2 + 3pq3 + ...

m-qm= p + pq + pq2 + pq3 + ...

m(1-q) = 1.

m = 1/(1-q) = 1/p.

Second, the answer to the problem can be express as the sum of the following:

• Number of trials to get a first toy
• Number of trials to get a second toy once you have one toy
• Number of trials to get a third toy once you have two toys
• Number of trials to get the final toy once you have three toys
The number of trials to get one toy is obviously one.

Once you have one the probability of getting a different toy in the next box is 3/4, thus the expected number of trials is 1/(3/4) = 4/3 to get the second toy.

By the same logic the number of trials to get the third is 1/(1/2)=2 and 1/(1/4)=4 for the final toy.

Summing these yields 1 + 4/3 + 2 + 4 = 25/3.

I'd like to thank Michael Brasher for providing this solution. Below is my old solution which is much less elegant, I recommend you ignore it.

The probability of it taking i boxes to get all four prizes is the probability that the first (i-1) will contain three different kinds of the toy, times (1/4).

The probability that in j boxes there will be three or less kinds of toys is 4*(3/4)j where j>3.

The probability that in j boxes there will be exactly three kinds of toys is 4*(3/4)j - 12*(1/2)j + 12*(1/4)j. This is rather hard to explain why but basically there are 4 different sets of 3 toys, 6 different sets of 2 toys, and 4 different sets of 1 toy, use this information to not double count combinations of 2 prizes or 1 prize.

Thus the answer is 1/4*sum for j=3 to infinity of j*(4*(3/4)j - 12*(1/2)j + 12*(1/4)j).

Theorem: The sum for j=1 to infinity of i*(1/n)i = n/(n-1)2.

From there is is just simple math to find that the answer is 25/3.

Michael Shackleford, A.S.A.