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I understand the process of solving this problem but there is one thing that I am confused about.

Why, in order to win, do we need $N(\tau)-N(s)=1$? I thought that the number of events in $(s, \tau]$ is given by $N(\tau)-N(s)$ so it would make sense that we want this to to be $0$, in order to win.

How can I get the probability of winning in this stopping game until nth event happens? Link to question asked before but this specific problem I am having was not addressed.

  • $\begingroup$ Well, if an event happens at exactly time $\tau$, then you want $N(\tau)-N(s)=1.$ $\endgroup$ – mjw Apr 30 at 20:22
  • $\begingroup$ @mjw The way I read it we do not want any events in $(s, \tau]$ ? $\endgroup$ – Jac Frall Apr 30 at 20:24
  • $\begingroup$ $(s,\tau)$ or $(s,\tau]$? $\endgroup$ – mjw Apr 30 at 20:26
  • $\begingroup$ @mjw Please see the update, it is directly from the textbook. $\endgroup$ – Jac Frall Apr 30 at 20:32
  • $\begingroup$ Okay, got it, thank you! I guess that in the light of the update, these comments are unnecessary. Glad you got a clearer picture and an answer that makes sense. $\endgroup$ – mjw May 1 at 19:31

If there are zero events in the interval $(s,\tau)$, then you lose because you never stop.

If there are two or more events in $(s,\tau)$, then you lose because you stop on the first one, and then a second one occurs before $\tau$, so you did not stop on the last event before $\tau$.

Only when there is exactly one event in $(s,t)$ do you win, because then that event you stop on is the last event to occur.

  • $\begingroup$ Whoops, there is my problem, I assumed $s$ was the time that you stop, when actually you stop on the first event after $s$ $\endgroup$ – Jac Frall Apr 30 at 20:54

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