Two players, Jim and Tom, each are going to flip 3 fair coins. What is the probability that they will get the same number of heads?
I am reviewing for an exam and cannot figure out how to get to this answer. The key says 5/16
P(both 3 heads) = $\displaystyle (\frac{1}{2})^3 \times (\frac{1}{2})^3$
P(both 2 heads) = $\displaystyle 3(\frac{1}{2})^3 \times 3(\frac{1}{2})^3$
I used 3 because you can have HHT, HTH or THH
P(both 1 head) = $\displaystyle 3(\frac{1}{2})^3 \times 3(\frac{1}{2})^3$
Same thing here, TTH, THT, HTT
P(both 0 head) = $\displaystyle (\frac{1}{2})^3 \times (\frac{1}{2})^3$
Then add all of them to give 5/16
Let $\displaystyle A $ be the event that they both flip the same number of heads.
Condition on the number of heads Jim flips.
$\displaystyle P(A) = P(A|0)P(0) + P(A|1)P(1) + P(A|2)P(2) + P(A|3)P(3) $
$\displaystyle = \binom{3}{0} (1/2)^{0} (1/2)^{3}*\binom{3}{0} (1/2)^{0} (1/2)^{3} + \binom{3}{1} (1/2)^{1} (1/2)^{2}*\binom{3}{1} (1/2)^{1} (1/2)^{2} $ $\displaystyle + \binom{3}{2} (1/2)^{2} (1/2)^{1}*\binom{3}{2} (1/2)^{2} (1/2)^{1} + \binom{3}{3} (1/2)^{3} (1/2)^{0}*\binom{3}{3} (1/2)^{3} (1/2)^{0} $
$\displaystyle = \frac{1}{8}*\frac{1}{8} + \frac{3}{8}*\frac{3}{8} + \frac{3}{8}*\frac{3}{8} + \frac{1}{8}*\frac{1}{8} = \frac{5}{16}$