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
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 $P(W \neq L)$?