Your cousin has been playing the same video game from time immemorial. Assume that he wins each game with probability $p$. independent of the outcomes of other games.
At midnight,you enter his room and witness his losing the current game. What is the PMF of the number of lost games between his most recent win and his first future win?
what is happening inside the room is basically a bernoulli process.
So, when i enter the room and start observing, the process starts fresh.
So, the number of games until(and including) the first future win after the current lost game, is a geometric r.v $X$,with parameter $p$. So, number of games lost after the current lost game is $X-1$
i still am unable to figure out the geometric distribution of r.v. $Y$, which represents the number of losses after the most recent win, until(and including) the current lost game.
the answer is $X+Y-1$, where $X,Y $~$ Geom(p)$.
And this made me think that a geometric r.v. is defined as "no. of independent trials until the first success".
Hence the trials can be in the future or in the past.
And if we take the above statement to be correct, then $Y$ is a geometric r.v representing the number of trials from(and including) the most recent win, until(and not including) the current lost game. So then, add 1 (for the current lost game) and subtact 1 (for the most recent win), i.e
$Y+1-1=$number of losses after(and not including) the most recent win, until(and including) the current lost game.
is this reasoning correct?