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Let's say I can play a game by paying a ticket:

  • I have $1$ ticket and the game is:
    • I have $1/2$ chance to get nothing for my ticket
    • and $1/2$ chance to get $3$ tickets.
  • I play the game until i don't have any tickets.
  • Intuitively I would think the game is infinite with probability $ > 0$.
  • I tried to calculate the probability of it being finite, this way: If $F$ is the event "the game I play with $1$ ticket is finite", then with a probability tree I can see that $$\mathbb{P}\left(F\right) = 0.5 + 0.5\,\mathbb{P}(F)^{3}. $$ According to $\tt WolframAlpha$, there is $3$ solutions, a negative, something between $0$ and $1$ and $1$. I'm pretty sure $1$ is not $\mathbb{P}(F)$ but I don't know why,

Can someone explain it to me $?$. I'm looking for an argument to rule out the solution $1$ specifically, if there is one.

Felix Marin
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Maxime Weill
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    Let $0<p<1$ be the root of $x=0.5x^3+0.5$, and $X_n$ be the number of tickets after playing $n$ times. It’s easy to see that $\mathbb{E}[p^{X_n}]=p$ by induction. Thus $P(X_n = 0)=P(p^{X_n} \geq 1) \leq \frac{\mathbb{E}[p^{X_n}]}{1}=p$. Now $F$ is the increasing reunion of the events $X_n=0$ so $P(F) \leq p < 1$. – Aphelli Mar 29 '21 at 22:03
  • There are many bounds derived for these sort of games, unfortunately I can't recall the names. – Dole Mar 29 '21 at 22:10
  • Related https://math.stackexchange.com/questions/1252066/probability-of-asymmetric-random-walk-returning-to-the-origin – leonbloy Mar 30 '21 at 01:45

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