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This is my second question following this post.

Three players are playing a game. They all have small amounts of money, let say: player 1 has $\$a$, player 2 has $\$b$, and player 3 has $\$c$, where $a<b<c$. The probability of each player wins each turn of the game is $p$ for player 1, $q$ for player 2, $r$ for player 3, and $s$ for having draw, where $p+q+r+s=1$. The losers will transfer a dollar ($\$1$) to the winner for each turn. The game ends until one player has all the money. What is the probability of each player going bankrupt? What is the expected number of turns so that only one player left as the winner?

Suppose that they play blackjack, if player 1 gets 20 points, player 2 gets 19 points, and player 3 gets 18 points, then the winner of that turn is player 1, so the other two players must pay a dollar to the player 1. If there are two players get, for example, 19 points and the another player gets 18 points, then that turn is considered draw. If they all get 19 points, this is also considered draw. If one player loses all the money, then he will stop playing and only two player will continue the game with probability of winning for each player is $x$ and $y$, also the probability of draw is $z$. Each turn will be repeated until one player has all the money.

To be honest, I can't answer this question and I really don't get it. I left my answer sheet totally empty for this one. (─‿‿─)

Please help me to answer this question and provide a simple explanation about the answer you submit. Every answer would be greatly appreciated.

Community
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Anastasiya-Romanova 秀
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  • How much money is wagered each turn? – RandomUser Apr 28 '14 at 17:12
  • @RandomUser Sorry, I'm sleepy. I've edited my question. Thanks for the correction. – Anastasiya-Romanova 秀 Apr 28 '14 at 17:23
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    You also did not specify what happens when one player looses all the money. How the probability of winning changes for the remaining 2? – nsg Apr 28 '14 at 17:42
  • @nsg If one player loses all the money, then he will stop playing and only two player will continue the game. – Anastasiya-Romanova 秀 Apr 28 '14 at 17:46
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    @V-Moy equally you don't specify what happen with the probabilities when one of the players loses. – rlartiga Apr 28 '14 at 18:36
  • @V-Moy I would try to answer this question but I need to know what happens to the probabilities $p, q, r, s$ when one of the players drops out. – Caleb Stanford Apr 29 '14 at 02:03
  • @rlartiga I don't get this question and I've edited the question according to my thought. I hope this already states all the rules. – Anastasiya-Romanova 秀 Apr 29 '14 at 13:48
  • @Goos I don't get this question and I've edited the question according to my thought. I hope this already states all the rules of games. – Anastasiya-Romanova 秀 Apr 29 '14 at 13:50
  • @V-Moy The answer is going to become really ugly with you introducing so many different variables $p, q, r, s, x, y, z$. Are you asking about the game blackjack specifically? And another thing, with $x, y, z$ you need to specify which probability applies to which player. – Caleb Stanford Apr 29 '14 at 13:52
  • In particular there is not going to be any simple answer, let alone a "simple explanation" if you have so many variables we have to deal with. – Caleb Stanford Apr 29 '14 at 13:53
  • @Goos It's OK, I'll try to deal with it. ಥ_ಥ – Anastasiya-Romanova 秀 Apr 29 '14 at 13:55
  • @V-Moy The set-up seems like an extension of the typical Gambler's Ruin Problem but with three players and ties allowed. For two players there are nice closed-form formulas, but as you get 3 players it seems the solutions are approximated by two-dimensional Brownian motion. Putting ties into pictures complicates the situation further. Though I would suggest familiarizing yourself with the "regular" Gambler's Ruin problem, then do an extension to 3 players, and then allow ties in 2 person case and finally in 3 person case. – Grid Apr 29 '14 at 19:18
  • Also there seem to be several research articles on the extensions of the Gambler's Ruin Problem (e.g. http://www.sciencedirect.com/science/article/pii/S0893965908001572#). Hmm actually I reread your question and you said it was an exam question. It seems I misinterpreted your question then. If I had to guess there were probably a lot of simplifying assumptions in the exam question which made the question much easier to analyze, but perhaps you're unable to reproduce the exact statement of the question from memory. – Grid Apr 29 '14 at 19:19
  • @Grid My brother also said about gambler's ruin problem but he didn't explain that (as usual), when I read by myself about that I didn't understand about Martingale & Markov's stuff. This was my essay exam but the committee restricted the students to take the exam papers. – Anastasiya-Romanova 秀 Apr 30 '14 at 16:00
  • @V-Moy Yes, it certainly does seem like an extension of Gambler's ruin and to understand it properly, you would need to read up on Markov chains. If I had to guess the committee just wanted go gauge how you would approach the problem, actually solving the problem to completion is a nontrivial task. To best understand your problem it's easier to start with the general (basic problem) with two players and no ties. Then in each step either the player wins a point or loses a point (see: http://www.columbia.edu/~ks20/FE-Notes/4700-07-Notes-GR.pdf) for a reference. – Grid Apr 30 '14 at 17:10
  • @Grid I can find the expected number of turns for 2 players if the probability of winning for each player is $\frac{1}{3}$ and the probability of draw is $\frac{1}{3}$. Also for 3 players if the probability of winning for each player is $\frac{1}{4}$ and the probability of draw is $\frac{1}{4}$. But I can't find for the general problem of GR for 2 and 3 players as the question stated. I think the committee does only want to mess up with us. Ugh!? :@ – Anastasiya-Romanova 秀 Apr 30 '14 at 17:55
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    You can ignore the ties and scale up the probabilities $p,q,r$ appropriately. Just pretend the ties don't happen and scale the expected number of turns up. You have to specify what the probabilities when the game drops to 2 players – Ross Millikan Apr 30 '14 at 18:10
  • Mr. @RossMillikan, how do I scale up them? I don't get it. Would you post your answer to this question, please? This question is really difficult for me but I really want to know how to answer it because I hate know nothing & I really wonder how this problem can be answered. Thank you Sir. – Anastasiya-Romanova 秀 May 01 '14 at 07:29
  • I don't have an answer, or I would post it. My point was that you can ignore the ties and consider $a$ to win $\frac p{1-s}$ of the time, $b$ to win $\frac q{1-s}$ of the time, etc, when you think about who wins. Then multiply the expected number of turns you get by $\frac 1{1-s}$ to get the duration. – Ross Millikan May 01 '14 at 13:34
  • Mr. @RossMillikan: My brother also suggested about weighted the probability. He suggested player 1 and 2 had a big chance to win because they had much more money than player 3, so he said the new probability after 3 lost was $\frac{p}{p+q+s}$ and $\frac{q}{p+q+s}$ but I don't know to use it. – Anastasiya-Romanova 秀 May 01 '14 at 15:21
  • After 3 loses it might be $\frac p{p+q}$ and $\frac q{p+q}$. The probabilities have to add to $1$. – Ross Millikan May 01 '14 at 15:27
  • Mr. @RossMillikan: The probability of draw is also weighted, $\frac{s}{p+q+s}$. – Anastasiya-Romanova 秀 May 01 '14 at 16:16

2 Answers2

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For fixed, $p,q,r,s$, let $f(a,b,c)$ denote the probability that player 1 is the first to go bankrupt, say. Then we have the recursion $$f\tag1(a,b,c)=pf(a+2,b-1,c-1)+qf(a-1,b+2,c-1)+\\+rf(a-1,b-1,c+2)+sf(a,b,c) $$ and boundary conditions $f(0,b,c)=1$, $f(a,0,c)=f(a,b,0)=0$ for $a,b,c>0$. We must also add $f(a,0,0)=0$, $f(0,b,0)=f(0,0,c)=1$ to account for the possibility that two players get bankrupt together (whereas $f(0,0,0)$ is undefined). Note that the same question for player 2 and player 3 results in the same recurson $(1)$, but different boundary conditions. Also note that $(1) $ can be transformed to $$\tag2f(a,b,c)=p'f(a+2,b-1,c-1)+q'f(a-1,b+2,c-1)+r'f(a-1,b-1,c+2) $$ with $p'=\frac p{1-s}$ etc. (as was suggested in the comments). For concrete values, it is enough to use $(2)$ to climb from the boundary cases to the specific values, for example $f(1,1,1)=q'+r'$ and $f(a,1,1)=0$ for $a>1$. For the general (i.e. abstract) solution, it woul dreally be helpful to understand Markov chains, eigenvalues and stuff.

Hagen von Eitzen
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    I'm sorry but I'm not familiar with your notation Mr. @Eitzen. Would you elaborate a bit more, please? It's okay if you use Markov chains & stuffs. I'll try to learn & understand all of them. シ – Anastasiya-Romanova 秀 May 06 '14 at 13:36
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For starters, you need to be more specific about that probability of a draw. It's possible that the top two players tie and the bottom-ranked player loses, and it's also possible that all $3$ players tie. Your draw probability doesn't distinguish these different cases, and yet these different cases will affect the result. You need to say how the money is divided when two or more top players tie, and also specify what is the probability that an arbitrary pair of two players tie for the top, and what is the probability that all three players tie for the top. Otherwise the problem is ill-posed.

user2566092
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