coin flip

**ihavvaquestion** · June 1st 2010, 10:19 AM

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

**Plato** · June 1st 2010, 11:06 AM

$\frac{1} {{64}} + \frac{9} {{64}} + \frac{9} {{64}} + \frac{1} {{64}} = \frac{5} {{16}}$

**Unknown008** · June 1st 2010, 11:26 AM

P(both 3 heads) = $(\frac{1}{2})^3 \times (\frac{1}{2})^3$

P(both 2 heads) = $3(\frac{1}{2})^3 \times 3(\frac{1}{2})^3$

I used 3 because you can have HHT, HTH or THH

P(both 1 head) = $3(\frac{1}{2})^3 \times 3(\frac{1}{2})^3$

Same thing here, TTH, THT, HTT

P(both 0 head) = $(\frac{1}{2})^3 \times (\frac{1}{2})^3$

Then add all of them to give 5/16

**Random Variable** · June 1st 2010, 11:33 AM

Let $A$ be the event that they both flip the same number of heads.

Condition on the number of heads Jim flips.

$P(A) = P(A|0)P(0) + P(A|1)P(1) + P(A|2)P(2) + P(A|3)P(3)$

$= \binom{3}{0} (1/2)^{0} (1/2)^{3}*\binom{3}{0} (1/2)^{0} (1/2)^{3} + \binom{3}{1} (1/2)^{1} (1/2)^{2}*\binom{3}{1} (1/2)^{1} (1/2)^{2}$ $+ \binom{3}{2} (1/2)^{2} (1/2)^{1}*\binom{3}{2} (1/2)^{2} (1/2)^{1} + \binom{3}{3} (1/2)^{3} (1/2)^{0}*\binom{3}{3} (1/2)^{3} (1/2)^{0}$

$= \frac{1}{8}*\frac{1}{8} + \frac{3}{8}*\frac{3}{8} + \frac{3}{8}*\frac{3}{8} + \frac{1}{8}*\frac{1}{8} = \frac{5}{16}$

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