TITLE: Playing the St. Petersburg Lottery until I lose everything
QUESTION [13 upvotes]: This question continues the following question: Calculating the probability of winning at least $128$ dollars in a lottery St. Petersburg Paradox
Here is a lottery: A fair coin is flipped repeatedly until it produces "heads." If the first occurrence of heads is on the $n$th toss, you are paid $2^{n−1}$. So for instance, if heads appears on the first toss, you are paid 1 dollar; if heads appears for the first time on the second toss, you are paid 2 dollars, and so on.
Say the cost of entering the lottery is 10 dollars and I start with 100 dollars. I'll play the lottery again and again until I have more than 10 dollars to play the game.
Now my question is, what is the probability that I can go on playing the game?
Example: Say during the first game, I get a heads on the first toss then I'll be left with 91(=100-10+1) dollars. I'll play the game again by paying 10 dollars and this time I may get a heads on the 7 toss and now I'll have 145(=91-10+64) dollars and so on.
Edit: Or Is it possible to find the probability of surviving the nth round given that I have survived all the previous rounds??

REPLY [2 votes]: Let $P_n$ be the probability of losing (not indefinitely having enough money to play) if you start with $n$ dollars. Now, we can observe that the probabilities don't change after one step, except it's now distributed over the possible values of $n$. As such, we can express that, for $n\geq 10$,
$$
P_n = \frac{P_{n-9}}2 + \frac{P_{n-8}}4 + \frac{P_{n-6}}8 + \cdots + \frac{P_{n-10+2^k}}{2^{k+1}} + \cdots
$$
which states our requirement that the limit to infinity doesn't change. We also have that $P_n=1$ for $n<10$.
Investigations of this system using Octave, by constructing the matrix system for this, while taking $P_n=0$ for $n>n_{max}$ (this creates an $(n_{max}-9)\times(n_{max}-9)$ matrix system), produce an interesting result:
As we increase $n_{max}$, $P_n$ gets larger for each $n$... in a way that suggests that they are tending to 1. That is, it is looking as though, for any finite $n$, the final result will actually be $P_n=1$, meaning that, given enough time, the player will run out of money at some point. For instance, for  $n_{max}=5009$, we get $P_{100}\approx 0.9960939$. For $n_{max}=10009$, we get $P_{100}\approx 0.9977098$. Here are the probability curves that result for $n_{max}$ = 2509, 5009, and 10009 (note: x coordinate is shifted by 9, so the first curve runs from 1 to 2500 rather than 10 to 2509):

Note that these effectively show the probability of losing if the player has $n$ dollars at the start and the house has $n_{max}-n$ dollars that they can give - that is, if the player wins enough to reach $n_{max}$, then they have cleaned out the house's bank, and have "won", while if the player drops below 10 dollars, they have "lost". Reminder: the $P_n$ shown in the graph shows the probability of losing.