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A gambler plays a fair game where he can win or lose $\$1$ in each round. His initial stock is $\$200$. He decides a priori to stop gambling at the moment when he either has $\$500$ or $\$0$ in his stock. Time is counted by the number of rounds played.

  1. Show that the probability that he will never stop gambling is zero.

  2. Compute the probability that at the time when he stops gambling he has $\$500$ and the probability that he has $\$0$.

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    $\begingroup$ Look up gambler's ruin. The game is fair, so the expectation at the end must be his starting capital. $\endgroup$ – Ross Millikan Jun 17 '18 at 17:55
  • $\begingroup$ this is a random walk with absorbing barries [math.stackexchange.com/questions/373634/… ]. For the shortcut read Ross. $\endgroup$ – Boyku Jun 18 '18 at 16:46
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i) I do not know a great way to do, but I'll try to make a proof. He will eventually get to $0$ or $500$ since he moves up one dollar or down one dollar without a pattern that restricts the $0$ or $500$ option.

ii) The probability is $\boxed{\frac25}.$ The only thing I can think of is states. First, try $0$ or $400$ and they go to either one with probability $\frac12.$ Then when at $400,$ it can go to $300$ or $500$ with equal probability too. At $300,$ it can go to $100$ or $500.$ At $100,$ it can go to $0$ or $200.$ This has five variables and is much simpler to solve. Specifically, find $b$ in the system of equations $b=0.5d,d=0.5+0.5c,c=0.5a+0.5,a=0.5b.$ Solving this gives us $b=\frac25.$

To clarify, in each multiple on $100,$ we try either a) The amount of money needed to reach $500$, if possible, or b) If a isn't possible, then double or nothing. Since it has equal probability in each step, the states in the better idea are possible.

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  • $\begingroup$ You're welcome. $\endgroup$ – Jason Kim Jun 17 '18 at 18:27
  • $\begingroup$ Also, I made a simpler way... $\endgroup$ – Jason Kim Jun 17 '18 at 18:32
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    $\begingroup$ You can use the fact that the expectation is constant in a fair game, so it has to be $200$ at the end. That gives the odds of stopping at $0$ or $500$. $\endgroup$ – Ross Millikan Jun 17 '18 at 18:36
  • $\begingroup$ What i dont understand is this part "at 400, it can go to 300 or 500. At 300, it can go to 100 or 500. At 100, it can go to 0 or 200" $\endgroup$ – Mark Jun 24 '18 at 12:38
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    $\begingroup$ Done.Thanks a lot $\endgroup$ – Mark Jun 24 '18 at 15:52
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This is a standard gambling problem.

To prove that the game stops, you could note that it stops when a sequence of 300 wins occur. Now, you could try to link this action to a geometric distribution and you find that it will stop eventually.

To calculate the probability, Ross Milikan made a nice comment. Note that, we know the expectation of his end capital and note that his end capital can either be $0$ or $500$. The calculation of the probability should now be easy.

If you have any questions, feel free to ask them.

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  • $\begingroup$ Hello. P(n)=p(1-p)^n its the geometric distrubition formua. our p is 0.5 and n is 300 but i cant get 0 can you pls helo $\endgroup$ – Mark Jun 24 '18 at 14:22
  • $\begingroup$ Probability of a sequence of 300 wins in a row is equal to: $0.5^{300}$. If you look at the points $1,301,601,\dots$, you can check whether the sequence of $300$ wins in a row occurred. The first time that this happens is distributed according to the geometric distribution. The thing you want here is that the first time this happens is finite. And this is indeed the case for a geometric distribution with any $p>0$. $\endgroup$ – Stan Tendijck Jun 24 '18 at 17:38

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