>>13183551I think you phase this question in a needlessly complicated way, by using 10 & 100 instead of 1 & 10, by speaking of the "coin flip" as "game" itself and not reserve the world for the full scenario, and also by adding a needless "for the first time" that I didn't quite get the first time. But okay, so much for my autism.
What's standard theory is that after n random steps with equal chance for left and right, the expected distance away from the starting point is sqrt(n). So after 100 steps ("game wins", in your language), you're on average sqrt(100)=10 steps away from the start. (And one step is 10 points, in your language)
But steps away means either in the positive or negative direction.
Your question is more complicated than the above, since you only count outcomes at the upper end.
The general theory falls under
https://en.wikipedia.org/wiki/Hitting_timeIf you allow arbitrary negative states (e.g. player is at -24220 points but still keeps playing), then I think the hitting time to your 100 points might not be defined.
This is similar to how the Cauchy distribution
is a perfectly nice bump with
,
but interpreted as as probability distribution it has no variance, since
does not converge.
For a variation of your question - and maybe that's what you have in mind, if
S is the winning amount
k is the starting amount
0 is the losing amount (game ends)
and
p is the chance of going up (and not down) per step
and with
r=p/(1-p)
then
the chance of winning (ending up at S) is
or, in the limit p=50%
and the expected number of steps is
resp. for p=1/2
Here a script that confirms at least w numerically
https://gist.github.com/Nikolaj-K/72cf3acecd611f4b0091c236200baf2f>>131847214u
