# Probability, coin flip

2 people each toss coin n times. what would be the probability that they will toss the same number of head??

--my try--

I think the probability of toss head is $1/2 = 0.5$

there are two people. so $(1/2)^2$.

tossing $n$ times, so the probability would be $(1/2)^{2n}$

but I'm not quite sure..

• Do you know how to calculate, for example, the probability that they both get exactly three heads? Do you know how to calculate the probability that, say, the 1st person gets exactly three heads? – Gerry Myerson Jan 23 '13 at 3:45
• @GerryMyerson Hi. I'm assuming that the probability that they both get exactly three heads would be (1/4)^3. and the 1st person gets exactly three heads would be (1/2)^3 – hibc Jan 23 '13 at 3:50
• Why are you assuming something, when it is possible to calculate it? The probability the 1st person gets exactly three heads is not $(1/2)^3$. This is where you have to start. – Gerry Myerson Jan 23 '13 at 4:04
• ok so 1st person gets exactly three heads from n toss would be [(factorial n)/ (factorial 3)(factorial n-3) ] * (1/2)^n – hibc Jan 23 '13 at 4:31
• Right. Now can you go on to do the original question? – Gerry Myerson Jan 23 '13 at 5:16

The chance you quote of $(\frac 12)^2$ is indeed the chance that the two people will both toss heads on one throw each. Then $(\frac 12)^{2n}$ is the chance that they will both throw heads on all $n$ tosses. The problem seems to ask something different. If $n=5$, A throws HHTTT, and B throws HTTHT, they have both thrown $2$ heads, though not at the same time. As I read the question, this should count as success. If so, you need to calculate the sum of probabilities (A gets no heads)(B gets no heads)+(A gets 1 head)(B gets 1 head)+ ... (A gets 5 heads)(B gets 5 heads). You have already done the last, and the first is the same, so you just have two more to do.