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Question:

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?

My attempt:

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 $X-1$

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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.

Doubt:

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?

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  • $\begingroup$ Notice that the question is NOT "from the time you enter the room until his next win" but "from his last win to his next win". Since he loses the game he is playing when you enter the room, those ate not the same. $\endgroup$ – user247327 Jan 21 at 12:53
  • $\begingroup$ X is after the time you enter to his first win. Y is after his last win to the time you enter. so $X+Y-1$ is after his last win to just before his current win. correct? $\endgroup$ – abhishek Jan 21 at 12:58
  • $\begingroup$ Am I the only one who thinks the wording is ambiguous? If we interpret "at midnight" to be a specific midnight, e.g. the midnight at the end of Halloween 2019-10-31, then the stmt just means conditioning on that game being lost, which is independent of numbers of games from that game to next win and back to most recent win. So both are geometric. However, if we interpret "at midnight" to mean some "random" moment in time, then this becomes a steady-state problem, and that "midnight" moment is more likely to be in a longer losing streak. Perhaps I'm confused or over-thinking this? $\endgroup$ – antkam Jan 21 at 18:42
  • $\begingroup$ this problem is in the 6th chapter. and the text does not even use the word "steady" until the 7th chapter. So, maybe "at midnight" should be interpreted as "a specific midnight"? by the way, according to you, how does $Y$ become a geometric r.v ? $\endgroup$ – abhishek Jan 22 at 8:58
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If we interpret "at midnight" to be a specific midnight, say the midnight at the end of Halloween 2019-10-31, then the problem stmt just means conditioning on that game being lost, which is independent of both

  • $X =$ no. of consecutive further losses (excluding current game) until next future win, and

  • $Y =$ no. of consecutive previous losses (excluding current game) since most recent win.

So the r.v. you seek is simply $Z= X+Y+1$. $X$ is clearly geometric, but then so is $Y$ if you look back in time.

BTW note that I defined $X,Y$ differently from you, and according to wikipedia there are two different but both common definitions of geometric variable, one including the first success, and one excluding. My variables are the latter version with mean $E[X] = E[Y] = {1-p \over p}$.

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