Properly understood, we could say that the expected number of games to go bust is $\frac1{|1-2p|}.$ It doesn't make sense to say that it is $ \frac{1-p}{p(2p-1)}$ for $p > \frac12$ since this is decreasing as $p$ increases and at $p=\frac34$ it is $\frac23<1$ !

Actually, for $p=\frac34$ the probability of ever going bust is $\frac13.$ In $3N$ trials we would expect to go bust $N$ times and the expected number of games in one of those terminating trials is $\frac{2/3}{1/3}=2.$ If you do not go bust immediately, which happens $\frac34$ of the time, then the chance that you will ever go bust after that is $\frac19.$ We need to ask what is the probability $y=y(p)$ that you ever go bust. Certainly $1-p\leq y$ since that is the probability that you go bust on the first game. We would expect that $y=1$ for $0 \leq p<\frac12$ and is decreasing to $0$ on $[\frac12,1]$

The analysis is similar to that above for $x$ but with multiplication instead of addition. If $y$ is the probability of ever going bust from the position of $\\\$1$ then $y^2$ is the probability from the position of $\\\$2.$ So $$y=(1-p)+py^2$$ which has two solutions $y=1$ and $y=\frac{1-p}p.$

So $\frac{1-p}{p(2p-1)}$ is actually $xy=x\, \frac{1-p}p$ and $x=\frac1{2p-1}$ for $p>\frac12.$

Properly understood, we could say that the expected number of games to go bust is $\frac1{|1-2p|}.$ It doesn't make sense to say that it is $ \frac{1-p}{p(2p-1)}$ for $p > \frac12$ since this is decreasing as $p$ increases and at $p=\frac34$ it is $\frac23<1$ !

Actually, for $p=\frac34$ the probability of ever going bust is $\frac13.$ In $3N$ trials we would expect to go bust $N$ times and the expected number of games in one of those terminating trials is $\frac{2/3}{1/3}=2.$ If you do not go bust immediately, which happens $\frac34$ of the time, then the chance that you will ever go bust after that is $\frac19.$

We need to ask what is the probability $y=y(p)$ that you ever go bust. Certainly $1-p\leq y$ since that is the probability that you go bust on the first game. We would expect that $y=1$ for $0 \leq p<\frac12$ (and also $p=\frac12$) and that $y$ is decreasing to $0$ on $[\frac12,1]$

The analysis is similar to that above for $x$ but with multiplication instead of addition. If $y$ is the probability of ever going bust from the position of $\\\$1$ then $y^2$ is the probability from the position of $\\\$2.$ So $$y=(1-p)+py^2$$ which has two solutions $y=1$ and $y=\frac{1-p}p.$

So $\frac{1-p}{p(2p-1)}$ is actually $xy=x\, \frac{1-p}p$ and $x=\frac1{2p-1}$ for $p>\frac12.$

Properly understood, we could say that the expected number of games to go bust is $\frac1{|1-2p|}.$ It doesn't make sense to say that it is $ \frac{1-p}{p(2p-1)}$ for $p > \frac12$ since this is decreasing as $p$ increases and at $p=\frac34$ it is $\frac23<1$ !

Actually, for $p=\frac34$ the probability of ever going bust is $\frac13.$ In $3N$ trials we would expect to go bust $N$ times and the expected number of games in one of those terminating trials is $\frac{2/3}{1/3}=2.$

We need to ask what is the probability $y=y(p)$ that you ever go bust. Certainly $1-p\leq y$ since that is the probability that you go bust on the first game. We would expect that $y=1$ for $0 \leq p<\frac12$ and is decreasing to $0$ on $[\frac12,1]$

The analysis is similar to that above for $x$ but with multiplication instead of addition. If $y$ is the probability of ever going bust from the position of $\\\$1$ then $y^2$ is the probability from the position of $\\\$2.$ So $$y=(1-p)+py^2$$ which has two solutions $y=1$ and $y=\frac{1-p}p.$

So for $p=\frac34$ we have $y=\frac13$ and, given that you do go bust, the expected number of turns to do that is $2.$ If you do not go bust immediately (which happens $\frac34$ of the time, then the chance that you will ever go bust after that is $\frac19.$

Properly understood, we could say that the expected number of games to go bust is $\frac1{|1-2p|}.$ It doesn't make sense to say that it is $ \frac{1-p}{p(2p-1)}$ for $p > \frac12$ since this is decreasing as $p$ increases and at $p=\frac34$ it is $\frac23<1$ !

Actually, for $p=\frac34$ the probability of ever going bust is $\frac13.$ In $3N$ trials we would expect to go bust $N$ times and the expected number of games in one of those terminating trials is $\frac{2/3}{1/3}=2.$ If you do not go bust immediately, which happens $\frac34$ of the time, then the chance that you will ever go bust after that is $\frac19.$ We need to ask what is the probability $y=y(p)$ that you ever go bust. Certainly $1-p\leq y$ since that is the probability that you go bust on the first game. We would expect that $y=1$ for $0 \leq p<\frac12$ and is decreasing to $0$ on $[\frac12,1]$

The analysis is similar to that above for $x$ but with multiplication instead of addition. If $y$ is the probability of ever going bust from the position of $\\\$1$ then $y^2$ is the probability from the position of $\\\$2.$ So $$y=(1-p)+py^2$$ which has two solutions $y=1$ and $y=\frac{1-p}p.$

So $\frac{1-p}{p(2p-1)}$ is actually $xy=x\, \frac{1-p}p$ and $x=\frac1{2p-1}$ for $p>\frac12.$