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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.

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 $P_k = q^k-1p$ and

begineqnarray P &=& P_1+P_3+P_5+... \ \
&=&p+q^2p+q^4p+q^6p+....\ \
&=& pover 1-q^2\ \& =& 1over 1+q
endeqnarray

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 $$P= 2over 3$$

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.$$p=frac12+frac12(1-p)implies p=frac23$$

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.

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 $p=2/3$.

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)