We did an example in class where we had to figure out what the expected value was to get all 6 prizes in the cereal boxes. In other words, how many boxes would you expect to buy in order to get all the 6 different prizes. We did random trials and the average came out to be 14.7, which is what it should be. He said that it was right because

6/1 + 6/2 + 6/3 + 6/4 + 6/5 + 6/6 = 14.7

He said that he got those numbers because the probability of getting the first prize it 6/6 (any prize), the next one is 5/6 (you can't get the same prize, they need to be all 6 different prizes), the next 4/6 and so on. All you do is take the reciprocal.

The problem that he gave us is now there are 3 prizes. We have to do the same thing like the example. However, all the prizes are not = this time. If they were, it would be
3/3 + 3/2 + 3/1 = 5.5

Prize A = 1/2
B = 1/4
C = 1/4

that is the probability of each of the 3 prizes coming up, because they make more of prize A.

If any one could figure this out it would help a lot!!
Thanks!

I don't think this is so simple...

This is a variation on the "cereal box problem" - see Cereal Box Problem: Theoretical Answer (http://www.mste.uiuc.edu/reese/cereal/theor.html) for a discussion on the original problem. However, the bit about making one of the prizes a higher probability than the others is an interesting twist. I think I know how to find the answer, but to understand it I think you first need to understand why it is that the expected value in the original problem is 6/6 + 6/5 + 6/4 +... +6/1. One explanation is that the expected value of the number of turns it takes for a probability p event to occur is 1/p. So, the expected number of turns it takes to get the first prize is simply 1. For the second prize the probability of getting something other than the first prize is p=5/6, so it takes on average 6/5 turns to get the second prize. Then to get the third prize the probability of selecting something other than the first or second prize is 4/6, so you add 6/4, and so on. Hopefully this is a reasonably clear explanation.

Now, for your problem where the probabilities vary: the probability of the 2nd selection not being the same as the first depends on what came up on the first selection. For example, if the 1st prize selected is the high probability one, then the expected number of turns to next select a lower probability prize is 4/2, and the third low probability prize is 4/1. That's one possible path - there are 2 others - take the weighted average of the 3 possible paths to find the total expected value.

I apologize if this isn't too clear - let me know if you need further explanation.