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Two players, $A$ and $B$, alternately and independently flip a coin and the first player to obtain a head wins. Player $A$ flips first. What is the probability that $A$ wins?

Official answer: $2/3$, but I cannot arrive at it.

Thought process: Find the probability that a head comes on the $n$th trial. That's easy to do, it's just a Geometric random variable. Then find the probability that $n$th turn is player's A turn. Finally, multiply both probabilities.

When I came up with each probability, both of them depended on the amount of trials $n$, so my answer was a non-constant function of $n$.

However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.

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marked as duplicate by Key Flex, Theo Bendit, José Carlos Santos, Kemono Chen, mrtaurho Feb 10 at 10:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that. $\endgroup$ – Pakk Feb 8 at 21:41
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    $\begingroup$ I have understood the flaw in my method by understanding the correct answer, but thank you anyways $\endgroup$ – Victor S. Feb 8 at 21:56
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So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and

\begin{eqnarray} P &=& P_1+P_3+P_5+... \\ \\ &=&p+q^2p+q^4p+q^6p+....\\ \\ &=& {p\over 1-q^2}\\ \\& =& {1\over 1+q} \end{eqnarray}

where $p$ is probability that head comes in one toss and $q=1-p$.

So if the coin is fair, then $p=1/2=q$, so $$P= {2\over 3}$$

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Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=\frac12+\frac12(1-p)\implies p=\frac23$$

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  • $\begingroup$ Is this way doable even if the coin is not fair? $\endgroup$ – Aqua Feb 9 at 7:43
  • $\begingroup$ @greedoid Sure. If the coin lands heads with probability $h,$ then $p=h+(1-h)(1-p)$ $\endgroup$ – saulspatz Feb 9 at 14:19
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Here's another approach.

Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.

In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).

But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.

This gives A a $\frac23$ chance of winning to $\frac13$ for B.

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Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So $$p = .5+ .25p$$ which yields $p=2/3$.

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P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)

A loses first round, B loses second round, A wins the third round (0.5)^3

And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2

P(A winning) = 0.5/(1-0.25)

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