# Coin Flip: Expected number of flips

Two Players are playing a coin flips game. The game ends when at least one player has a Head. Whoever gets the Head first would be the winner, and the other would be the loser. The loser will continue to flip until he gets a Head. What's the expected number of flips for the winner? What about the loser? the probability of ending a game is 3/4 since the only case of continuing is both tails, which has a probability of 1/4; so the expected number of flips for the winner would be 1/(3/4) = 4/3. I'm wondering if the expected number of flips for the loser is (4/3) + 2

• Welcome to mathematics stackexchange. What have you tried? – Alain Remillard Oct 31 '19 at 18:27
• Hint for expected number of flips for the winner: Suppose the game has no winner on turn $n-1$. The game has a winner on turn $n$ if either player flips a head. The only way the game has no winner on turn $n$ is if both players flip tails. – InterstellarProbe Oct 31 '19 at 18:40
• Can you verify my understanding? (1) The players take turns alternately, with player A going on turns $1, 3, 5, \dots$ and the other player B going on turns $2,4,6\dots$. (2) The event "winner has 5 flips" is equiv. to the first Head appearing in turn $9$ (A is winner and she has 5 flips) or turn $10$ (B is winner and he has 5 flips). Am I right? – antkam Oct 31 '19 at 18:57
• Why would "The loser will continue to flip until he gets a Head." What is the point of that? – David G. Stork Oct 31 '19 at 19:09
• If both players flip TTTH individually, the winner's number of flips is $4$, right? But what is the loser's number of flips? Also $4$? – antkam Oct 31 '19 at 19:26

HINT

This answer interprets the OP question this way:

• $$W=$$ no. of flips by the winner $$= \min(T_1, T_2)$$,

• $$L=$$ no. of flips by the loser $$= \max(T_1, T_2)$$,

• where player $$i$$ first flips Heads at turn $$T_i$$.

In particular, if $$T_1 = T_2$$ (i.e. they get their first Heads at the same turn) then $$W=L=T_1=T_2$$.

As many have pointed out, $$W$$ is Geometric with success prob $$3/4$$ and $$E[W] = 4/3$$.

$$L$$ however is not Geometric. You can explicitly find $$P(L=l)$$ and then $$E[L]$$, but there is a faster way:

• Let $$X=L-W =$$ the no. of additional flips by the loser.

• If $$W=L$$, then $$X=0$$.

• If $$W \neq L$$, then the loser needs to keep flipping. Conditioned on $$W \neq L$$, you are right that $$X$$ is Geometric with success prob $$1/2$$ and $$E[X \mid W \neq L] = 2$$.

• So $$E[L] = E[W] + E[X] = E[W] + P(W=L)\times 0 + P(W \neq L)\times 2$$

• Can you find the value of $$P(W \neq L)$$?