# Fair coin tosses

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### Fair coin tosses

Every person In a group of n has a fair coin, which he flips until he gets head. At the end of the game, a total of 2k+1 flips (odd number) have been made. Find the probability that there were more heads than tails.

Are we going to get the binomial (2k+1 select k+1)(1/2)^(k+1)(1/2)^k multiplied by n? Or the sum of all the ways to select >k+1 from 2k+1?

I would appreciate your assistance!

Alex.vollenga

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Joined: Wed May 17, 2017 10:04 am
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### Re: Fair coin tosses

I am thinking that in all sequences of outcomes of n people, the last outcome is always H. We have for example TTTH or TTH or H.
For the number of H to be > number of T, we must have the following;
We must have a good number of H in the first toss (probability 1/2) and for the rest of sequences, the total number of T must be less than the total number of H at the first toss. For example, H, H, TH, TTH, H, H, TH, TH, TTTH.
The total probability must be 1/$2^{p1}$+1/$2^{p2}$+...+1/$2^{pn}$ where p1+p2+...+pn=2k+1 (the total number of tosses) and there must be P(total probability)>1/2.
We also have that the total number of HEADS is n (since each player stops tossing once he gets a H), so it must be n>(2k+1)/2.

How do we calculate the required probability?

Alex.vollenga

Posts: 7
Joined: Wed May 17, 2017 10:04 am
Reputation: 2

### Re: Fair coin tosses

By intuition only, I suspect it converges to zero (as a resemblance to a harmonic series for a very big n), but I can't prove it.

Is it correct that I take the products of probabilities of individual series of tosses of each person? Or should I take the sum?
(1/2+1/4+1/2+1/8+...)?

Alex.vollenga

Posts: 7
Joined: Wed May 17, 2017 10:04 am
Reputation: 2

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