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Imagine a game of roulette in which the player has a chance of $\frac 1n$ to win. When he does win, he earns $k > n$ times his initial bet. In other words, it's a worthwhile game for the player, as its expected value is

$$\frac kn > 1$$

The player is allowed to play an infinite number of times, but needs to pick ahead of time a single amount of money $b$ that he must bet on every round. Moreover, he doesn't have an infinite amount of money to begin with, only \$$a$ (where $a \ge b$, obviously).

If the player wants a chance of exactly $2\%$ (or any other arbitrarily low number) of never losing all of his money, what portion of $a$ should he choose $b$ to be?

Rodrigo de Azevedo
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koorkevani
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  • Just to clarify, are $n$ and $k$ fixed beforehand (i.e. the player has no choice about them)? – Varun Vejalla Oct 14 '19 at 21:14
  • Also, are you saying that the player's money goes from $m$ to $m-b+bk$ if he wins and to $m-b$ if he loses? – Varun Vejalla Oct 14 '19 at 21:16
  • are you allowed to pick $b$ to be arbitrarily close to $0$? That will be slowest but safest – BGM Oct 14 '19 at 23:42
  • Is this homework? What have you tried? And how much do you know already about asymmetric random walks? – antkam Oct 14 '19 at 23:46
  • @automaticallyGenerated Everything you've said is correct. – koorkevani Oct 15 '19 at 12:01
  • @bgm Right, but I don't want the safest strategy. I want the strategy with exactly a change of 2% (or any other arbitrary number) to go bust. – koorkevani Oct 15 '19 at 12:01
  • @antcam It's not homework. To be honest, I'm trying to exploit an online game the most efficient way... I don't know much about statistics, except for basic combinatorics. – koorkevani Oct 15 '19 at 12:01
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    What you have is an asymmetric random walk with (1) unequal step sizes, and (2) unequal probabilities of going left or right. This MSE Q&A seems very relevant although there the steps are fixed at $+2, -1$ whereas yours are $+b(k-1), -b$. Some of the comments in that thread also seems relevant. – antkam Oct 15 '19 at 12:56
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    Also, do you know the Kelly criterion? It asks and answers a different question - what constant fraction of your bankroll should you bet (in contrast, you're asking what constant bet should you bet). So going bankrupt is not an issue in the Kelly problem (since you never bet 100% of your bankroll), but the goal is maximizing growth. Depending on your problem context it might be relevant. It is a very standard way to deal with how to bet in a favorable betting game. – antkam Oct 15 '19 at 12:59
  • @antkam I think that question addresses the probability of reaching exactly $0$. What this question is asking for is the probability of crossing $0$. – Varun Vejalla Oct 15 '19 at 23:13
  • @automaticallyGenerated - interesting point... but in that question, you cannot cross $0$ without first reaching $0$ due to the $(+2,-1)$ steps. so i thought it's "very relevant". however the recurrence method may not work here, if the steps $+b(k-1), -b$ are not integers (or at least, rationals). even if it works, the boundary conditions are different since you can jump across $0$... but can you "really"? you cannot jump to negative bankroll in "real life", right? so isnt this question a little underspecified, i.e. you cannot keep the constant bet $b$ if your bankroll $< b$? – antkam Oct 15 '19 at 23:55
  • I think the player is done if he can't pay the buy-in of $b$. – Varun Vejalla Oct 16 '19 at 22:55

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