Two of 'em

Originally posted by forkedknight
Intuitively, it should be 41 cents. You're taking half the coins, so on average, you should have half your pennies, half your nickles, and half your dimes.

Can you prove it?

I can. And it is indeed 41.

First way:
There are 924 combinations. In each one there are 6 coins, so if you list all the combinations that's a total 5,544 coins. Each of the 12 coins appears in those as many times, which is 5,544 / 12 = 462 times. Therefore, the total value of coins showing in those 924 combinations is

462 x (2 x 1 + 4 x 5 + 6 x 10) = 37,844 cents.

Divide that by 924, and you get 41.

Second way:
When all else fails, brute force sometimes works. List all the possible combinations of taking 6 out of 12, count how many permutations there are, multiple by the value of the coins in that set, sum up, and divide by 924. As there are only three kinds of coins, it's simple enough. Listing dimes first, then nickles, then pennies, looks like this. There are only 15 possibilities. The whole thing adds up to 37,884 cents, as above, with an average of 41p.
6-0-0 x1 x60p
5-1-0 x24 x55p
5-0-1 x12 x51p
...
1-3-2 x24 x27p
0-4-2 x1 x22p

nice one, yes, your first way is a good proof and shows that for any combination of coins in your pocket, if you pick N of them out, on average you will get the average_coin_value*N