Re: Expected Value problem

Hi terrorsquid,

Let's consider a single card and see if we can find the distribution of the number of times it is flipped-- say x. At each card flip, the probability that our special card is selected is 1/50. Since the flips are independent, the total number of times the card is flipped follows a Binomial distribution with n = 50 and p = 1/50. So the probability the card is flipped x times is

$\displaystyle P(x) = \binom{50}{x} (1/50)^x (49/50)^{50-x}$

Next, we would like to know the probability that x is even, i.e. P(0) + P(2) + P(4) + ... +P(50). To find this sum, expand (1/50 + 49/50)^50 by the binomial theorem:

$\displaystyle (1/50 + 49/50)^{50} = \sum_{x=0}^{50} \binom{50}{x}(1/50)^x (49/50)^{50-x}$

Then do the same for (1/50 - 49/50)^50:

$\displaystyle (1/50 - 49/50)^{50} = \sum_{x=0}^{50} (-1)^x \binom{50}{x}(1/50)^x (49/50)^{50-x}$

If we add these two equations, the odd-numbered terms cancel out and on the right-hand side we have twice the probability that x is even. So

$\displaystyle (1/50 + 49/50)^{50} + (1/50 - 49/50)^{50} = 2 P(\text{x is even})$

and

$\displaystyle P(\text{x is even}) = (1/2)\; [(1/50 + 49/50)^{50} + (1/50 - 49/50)^{50}]$

$\displaystyle = (1/2) \; (1 + (48/50)^{50})$

This is the probability that our special card is left with the red side up. There are 50 cards in all, so the expected number of cards left with the red side up is

$\displaystyle 50\; P(\text{x is even})= 25\; (1 + (48/50)^{50})$

and the expected number of cards left blues side up is just 50 minus the expected number left red side up, i.e.

$\displaystyle 50 - 25\; (1 + (48/50)^{50})$

which is about 21.75.