Expected Value problem

**terrorsquid** · November 25th 2011, 08:53 PM

50 cards having one blue side and one red side are placed on a table all with the red face up. After 50 random card flips (a flipped card now displays its blue side, and if flipped again, it displays its red side etc.), what is the expected number of BLUE faces showing?

I haven't ever studied any statistics (I will be next semester), but I find it very interesting. I was wondering if anyone could show me what the standard way of solving this would be?

I tried to solve it just using logic, but came across some problems:

The total unconstrained number of possible combinations of red and blue cards (before the 50 random flips constraint) is 2^50. I can see that any combination that includes an odd number of blue faces (and therefore an odd number of red faces) cannot happen as 50 is an even number of flips, e.g., if you flipped 49 different cards, you would have 49 blue faces and 1 red face with one flip left over; regardless of which card is flipped in the 50th flip, there will be an even number of blue faces displayed.

Therefore, the number of possible outcomes is $2^{50} - ($ $50\choose{1}$ $+$ $50\choose{3}$ $+$ $50\choose{5}$ $+...+$ $50\choose{49}$ $) = \frac{2^{50}}{2}=2^{49}$

So, there are $50\choose{2}$ $+$ $50\choose{4}$ $+...+$ $50\choose{50}$ blue cards shown in the possible combinations, which is slightly less than half. So I guess slightly less than 25 blue cards on average?

Can anyone do this properly so I can see?

Thanks.

**awkward** · November 26th 2011, 06:27 AM

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

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:
$(1/50 + 49/50)^{50} = \sum_{x=0}^{50} \binom{50}{x}(1/50)^x (49/50)^{50-x}$
Then do the same for (1/50 - 49/50)^50:
$(1/50 - 49/50)^{50} = \sum_{x=0}^{50} (-1)^x \binom{50}{x}(1/50)^x (49/50)^{50-x}$

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
$(1/50 + 49/50)^{50} + (1/50 - 49/50)^{50} = 2 P(\text{x is even})$
and
$P(\text{x is even}) = (1/2)\; [(1/50 + 49/50)^{50} + (1/50 - 49/50)^{50}]$
$= (1/2) \; (1 + (48/50)^{50})$

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

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

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

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

which is about 21.75.

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