0
$\begingroup$

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..

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Gerry Myerson Jan 23 '13 at 3:45
  • $\begingroup$ @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 $\endgroup$ – hibc Jan 23 '13 at 3:50
  • $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jan 23 '13 at 4:04
  • $\begingroup$ ok so 1st person gets exactly three heads from n toss would be [(factorial n)/ (factorial 3)(factorial n-3) ] * (1/2)^n $\endgroup$ – hibc Jan 23 '13 at 4:31
  • $\begingroup$ Right. Now can you go on to do the original question? $\endgroup$ – Gerry Myerson Jan 23 '13 at 5:16
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you @Ross Millikan but I'm still confused. Do you mean I should use summation notation ?? $\endgroup$ – hibc Jan 23 '13 at 3:40
  • 2
    $\begingroup$ hibc, the notation you use is not as important as the concepts you understand. Ross is telling you that you have to compute a bunch of probabilities, multiply them in pairs, and then add up the products. Do you understand each step of that? If not, ask Ross to clarify whichever step you don't get. $\endgroup$ – Gerry Myerson Jan 23 '13 at 3:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.