Probability of winning based on stop time 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.

• Well, if an event happens at exactly time $\tau$, then you want $N(\tau)-N(s)=1.$ – mjw Apr 30 at 20:22
• @mjw The way I read it we do not want any events in $(s, \tau]$ ? – Jac Frall Apr 30 at 20:24
• $(s,\tau)$ or $(s,\tau]$? – mjw Apr 30 at 20:26
• @mjw Please see the update, it is directly from the textbook. – Jac Frall Apr 30 at 20:32
• 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. – 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.
• Whoops, there is my problem, I assumed $s$ was the time that you stop, when actually you stop on the first event after $s$ – Jac Frall Apr 30 at 20:54