One more probability..

[Lestat]

We both have fair coins.

We are both going to flip them 3 times.

If we both get the same number of heads I'll pay you 2 dollars, if we get different number of heads you pay me a dollar.

Will you play this game?

Brian

 

[aalox]

For problems like this, which involve an either-or situation which is repeatedly tested, use the binomial distribution.

P(k out of N) =
N!
---------- x (p^k)(q^(N-k))
k!(N-k)!

where:
N = the number of opportunities for event x to occur;
k = the number of times that event x occurs or is stipulated to occur;
p = the probability that event x will occur on any particular occasion; and
q = the probability that event x will not occur on any particular occasion.

In this case, N = 3, k ranges from 0 to 3, and p and q are both 0.5. For the three tosses done by each person, the probabilities are
# of heads(k) P
---------------------
0 1/8
1 3/8
2 3/8
3 1/8

These are the probabilities for each person. To find the probabilities from the two of you, multiply the probabilities and make a matrix of the different possibilities.
\ you 0 1 2 3
friend
0 1/64 3/64 3/64 1/64
1 3/64 9/64 9/64 3/64
2 3/64 9/64 9/64 3/64
3 1/64 3/64 3/64 1/64

The total probability that you and your friend will get the same number of heads is therefore (1/64 + 9/64 + 9/64 + 1/64) = 20/64 = 5/16 = 31.25%
To determine if this is a good game or not, determine the expectation value of the money you get by multiplying the probability of each outcome by the money you get for the outcome.

Same # of heads = 31.25% * $2 = 62.5 cents per game
Diff. # of heads = 68.75% * -$1 = -68.75 cents per game

So if you keep playing this game, you will tend to lose money over time.
Don't play!