A Slightly Less Patient Coin Flipper

This problem can be solved by using the formula of the expected value of the binomial distribution:

$\displaystyle E(x)=np$

Where the variable $x$ is the number of heads shown in the top face. For this problem $E(x) =5$ and $p=0.5$ , so we can solve for $n$ (which is the number of tosses):

$\displaystyle n=\frac{E(x)}{p}$

$\displaystyle n=\frac{5}{0.5}$

$\displaystyle \boxed{n=10}$

Suppose Timmy had arrayed in front of him, before he started flipping, his five favorite coins, each with a different number on its face. Starting with the first coin, coin 1, he flips until he attains a head. He then switches to coin 2, and so on until he has flipped a head with all 5 coins.

It follows that the expected number of flips necessary for Timmy to receive one head on each of his five special coins is the sum of the expected number of flips necessary to receive a head for each coin.

Since the coins in the problem are identical, we seek 5 * (Expected # of flips to flip a head on a fair coin).

= 5 x ( (1)( 1/2) + (2)(1/4) + (3)(1/8) + ... (n)(1/(2^n)...)

This sum can be rearranged into:

(1/2 + 1/4 + 1/8 ...)(1 + 1/2 + 1/4 + 1/8...) = 2. (Try it!)

So! The answer is: $\boxed{10}$

En = 1 + 0.5En + 0.5E(n-1)...where E1 = 2