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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

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    $\begingroup$ Welcome to mathematics stackexchange. What have you tried? $\endgroup$ – Alain Remillard Oct 31 '19 at 18:27
  • $\begingroup$ 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. $\endgroup$ – InterstellarProbe Oct 31 '19 at 18:40
  • $\begingroup$ 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? $\endgroup$ – antkam Oct 31 '19 at 18:57
  • $\begingroup$ Why would "The loser will continue to flip until he gets a Head." What is the point of that? $\endgroup$ – David G. Stork Oct 31 '19 at 19:09
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    $\begingroup$ 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 $4$? $\endgroup$ – antkam Oct 31 '19 at 19:26
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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 $P(W \neq L)$?

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