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Gamblers Ruin - Markov Chain problem

  1. Oct 19, 2011 #1
    This is probably a noob question, background in probability theory isn't great but I was shown this problem in a lecture:

    "Suppose a gambler starts out with £n, and makes a series of £1 bets against the house.

    Let the probability of winning each bet be p, and of loosing be q = 1 − p. If the gambler’s capital ever reaches £0, he is ruined and stops playing; he remains at zero."

    Then this was stated:

    "In fact, he will reach zero with probability one: eventually he is bound to lose all his money, no matter how much he started with given any value of p between 0 ≤ p < 1."

    Apparently this is obvious, but I don't see why it is obvious. If p is large, say 0.99... why should he always, at some point reach the £0 boundary?
     
  2. jcsd
  3. Oct 19, 2011 #2
    I'm no expert, but logically, if he plays for long enough, he will encounter all possible combinations of results.
    Eventually he will have a string of losses long enough to bring him to 0, which forces him to stop. Having a high p will simply make this result more improbable, i.e. he will most likely play for longer. The final result is the same.
     
  4. Oct 19, 2011 #3
    Well I remember reading somewhere that in a simple one dimensional random walk, which I guess this problem comes under, given enough time every point will be reached. Thus at some time the £0 point must be reached, so he will always lose.

    Haven't seen a proof anywhere... for the moment, think I'll try running a few matlab scripts to test out the idea.
     
  5. Oct 19, 2011 #4
    Cool, I'll try to do the same in python.
     
  6. Oct 19, 2011 #5
    Right now my guy starts off with $100, he bets $1 each time, and he has a 50% chance of winning. He got to 0 on the 21 millionth iteration

    EDIT: Now I'm trying to see how many iterations it takes if he has an 80% chance of winning. (see you all tomorrow haha).

    EDIT2: I got impatient and I stopped my program. My guy has played almost 100 million times and he's banked 58 million dollars. :rofl:

    EDIT3: I think this logic makes sense if you really accept the idea of infinity. Question: how many times would I have to try flipping a coin 8 times to get 7 heads and 1 tails. If there is a finite answer (which of course there is), then there is no reason to think that this logic can't be extended. The answer to this type of question cannot be infinity, therefore the probability of any sequence appearing in an infinite amount of tries is 1.
     
    Last edited: Oct 19, 2011
  7. Oct 19, 2011 #6
    Well I did the same except with a £50 initial fund, 50% chance of winning... ran it a few times and got values between 100 and 500000 iterations.

    Put it up to 60%, started it 5 minutes ago... its still running :D
     
  8. Oct 20, 2011 #7

    chiro

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    Pick up an applied probability book for a more fleshed out exposition of this problem.

    One book that I'm familiar with is Introduction to Probability Models by Sheldon Ross.
     
  9. Oct 22, 2011 #8
    You didn't say how much he gets from each win, but assuming it's £1 on top of his original bet, the statement is definitely NOT true - for any p > 1/2 there's (theoretically) a finite chance that he can continue betting indefinitely. In fact, the exact probability can be expressed as a function of p and n, e.g. using branching processes and generating functions.
     
  10. Oct 22, 2011 #9

    phinds

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    It's been a VERY long time since I took Finite Math, but intuitively I'd say you HAVE to be right and if p is close to 1 then the chance of never losing should be quite high.
     
  11. Oct 22, 2011 #10
    The way you interpret the problem is probably different from the professors.

    If P = .99, then over time it WILL diverge into a .99 success rate. If you have a million euros to start with, then the chance of you going back to 0 is virtually 0. Definitely not guaranteed.

    I believe the professor only meant P = ≤.5. If that were the case then yes, almost surely, you would eventually go down to 0. But for anything higher than .5, no.
     
    Last edited: Oct 22, 2011
  12. Oct 22, 2011 #11

    lavinia

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    There are many ways to prove this... but why not just try counting the probabilities of reaching -n in a simple random walk?
     
  13. Oct 23, 2011 #12

    phinds

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    So, you disagree w/ all the rest of the answers in the post. Why is that? What is the math that says you are right?
     
  14. Oct 23, 2011 #13

    lavinia

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    Why not think the problem through for yourself? That's all I was saying. The simplest example is a fair coin. You should be able to generalize from that example. good luck.
     
  15. Oct 23, 2011 #14

    phinds

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    But I HAVE thought it through. I am convinced that if the probability of winning is .99, then the probability of ever getting to zero is negligible (not zero but infinitesimal)

    I do not understand why you think otherwise, and I do not see that a fair coin is evey remotely applicable to this situation because it has a probability of .5, not of .99

    IF the probability of winning is .5 then I would agree w/ you, but it is not so I do not.

    I ask again, What math makes you think that if the probability of winning is .99 you would have any significant change of getting to zero?
     
  16. Oct 23, 2011 #15
    After thinking a bit longer and doing some reading I'm in agreement with phinds, bpet and Firefight.

    I believe the Gamblers ruin only applies to problems whereby the expectation value of a bet is zero or less, so the chance of winning is .5 or less. For p > .5, the expectation value of a bet is positive and given a great number of bets the gambler will always make more money.

    The idea of betting against the house is that, given an opponent with limitless money, no matter what £n the gambler starts with, so long as his expectation value of winning is .5 or less he will always lose eventually. Apparently casinos make good use of this... they can make money off a fair game just by the fact they always have more money. Although casino games are rarely fair.

    Means I've probably misheard what the lecturer was saying, will ask him tomorrow exactly what he meant.
     
  17. Oct 23, 2011 #16

    phinds

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    Be open to the possiblity that he made a mistake and you didn't misunderstand him. Even profs make mistakes from time to time. ( Except Feynman. I really don't think he ever made mistakes. :smile: )
     
  18. Oct 23, 2011 #17
    If p(win)=0.99 and q(loss)=0.01, and each trial is independent for a fixed bet, then the probability of losing after n trials is [itex] 1 - 0.99^{n}[/itex]. It's clear that the probability of losing at least once becomes large for large n. However, the game is over if and only if the gambler losses all his money, and there is always a non-zero probability the gambler will lose for every trial.

    We have only specified an end point for losing, but not for winning. That's why the gambler's ruin rule holds in theory. Obviously 99 to 1 is pretty good odds, In the real world, you're lucky to get 50-50.
     
    Last edited: Oct 23, 2011
  19. Oct 23, 2011 #18

    phinds

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    I don't follow what this has to do with answering the original question. Of COURSE the odds of loosing once are significant. My point is that if you start of with much of anything your chances of ever getting down to zero are INsignificant.
     
  20. Oct 23, 2011 #19
    That depends on how much you bet on each trial as a percentage of your resources, and how long you play. The original question was a theoretical one.
     
    Last edited: Oct 23, 2011
  21. Oct 23, 2011 #20

    phinds

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    Fair enough. I was assuming you start with n dollars and bet one dollar per round. If n is at all significant, then my statement stands.
     
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