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Posers and Puzzles

Posers and Puzzles

  1. Standard member PBE6
    Bananarama
    21 Nov '06 19:26
    I was just thinking about gambling and calculating the risk of ruin. For simplicity's sake, let's consider a game with only two outcomes (win or lose), a probability of winning Pw = 1 - Pl, and a given number of betting units B to gamble with. What is the probability that you will lose all your betting units at some point in the game?

    (I know approximate formulas exist for this, but can anyone come up with an analytical solution? It's more complicated than you'd think...)
  2. 21 Nov '06 23:18
    Originally posted by PBE6
    I was just thinking about gambling and calculating the risk of ruin. For simplicity's sake, let's consider a game with only two outcomes (win or lose), a probability of winning Pw = 1 - Pl, and a given number of betting units B to gamble with. What is the probability that you will lose all your betting units at some point in the game?

    (I know approximate f ...[text shortened]... can anyone come up with an analytical solution? It's more complicated than you'd think...)
    If the game goes on infinitely long, isn't the answer 1? (unless P_w = 1). It's basically a random walk, which will reach zero at some point.

    If it's not inifinitely long, what's the stopping criterion?

    If that's not it, I think you need to be a bit more specific with the problem.
  3. Subscriber AThousandYoung
    It's about respect
    21 Nov '06 23:27
    Originally posted by PBE6
    I was just thinking about gambling and calculating the risk of ruin. For simplicity's sake, let's consider a game with only two outcomes (win or lose), a probability of winning Pw = 1 - Pl, and a given number of betting units B to gamble with. What is the probability that you will lose all your betting units at some point in the game?

    (I know approximate f ...[text shortened]... can anyone come up with an analytical solution? It's more complicated than you'd think...)
    How much can one bet at a time?
  4. Standard member PBE6
    Bananarama
    22 Nov '06 14:53
    Originally posted by mtthw
    If the game goes on infinitely long, isn't the answer 1? (unless P_w = 1). It's basically a random walk, which will reach zero at some point.

    If it's not inifinitely long, what's the stopping criterion?

    If that's not it, I think you need to be a bit more specific with the problem.
    Yes, you're right. I should have said "what's the risk of ruin after playing 'n' hands?"
  5. Standard member PBE6
    Bananarama
    22 Nov '06 14:54
    Originally posted by AThousandYoung
    How much can one bet at a time?
    For simplicity's sake let's say you only bet one unit at a time, and you either win or lose a unit as the case may be.
  6. Standard member uzless
    The So Fist
    22 Nov '06 15:45
    Originally posted by PBE6
    For simplicity's sake let's say you only bet one unit at a time, and you either win or lose a unit as the case may be.
    50% you will lose all your units over N hands
  7. 22 Nov '06 16:06 / 2 edits
    Originally posted by mtthw
    If the game goes on infinitely long, isn't the answer 1? (unless P_w = 1). It's basically a random walk, which will reach zero at some point.

    If it's not inifinitely long, what's the stopping criterion?

    If that's not it, I think you need to be a bit more specific with the problem.
    this is simply not true. This is only true if your expected value is negatif, thus Pw is smaller than Pl
  8. 22 Nov '06 16:25 / 1 edit
    Originally posted by Darrie
    this is simply not true. This is only true if your expected value is negatif, thus Pw is smaller than Pl
    I withdraw the previous statement here. While doing the calculation I realised you're quite right (though the answer is 1 if Pw = Pl = 1/2 as well). Sorry!
  9. Standard member uzless
    The So Fist
    22 Nov '06 16:27
    Originally posted by mtthw
    Afraid not. A one-dimensional random walk (which this is, until you hit zero) given infinite time will hit all points an infinite number of times "almost surely" (i.e. with probability 1). As long as P_w is less than one you're stuffed in the long run. Infinity is a long time!
    I would argue that for every bet, not only does it create an infinite amount of lines that end in bankruptcy, it also creates and infinite amount of lines that continue forever.

    That's why I reason it to be 50% over infinite hands.
  10. 22 Nov '06 16:42 / 1 edit
    Originally posted by uzless
    I would argue that for every bet, not only does it create an infinite amount of lines that end in bankruptcy, it also creates and infinite amount of lines that continue forever.

    That's why I reason it to be 50% over infinite hands.
    You've got to be careful with inifinities

    Look at it this way. Let's say B = 1, and P_w = 0.1. The risk of ruin is at least 0.9, as there's a 0.9 chance of losing everything on the first turn. So it can't be 50%. It either has to be 1, or a function of B and P_w.
  11. Standard member PBE6
    Bananarama
    22 Nov '06 17:19
    Originally posted by mtthw
    If the game goes on infinitely long, isn't the answer 1? (unless P_w = 1). It's basically a random walk, which will reach zero at some point.

    If it's not inifinitely long, what's the stopping criterion?

    If that's not it, I think you need to be a bit more specific with the problem.
    Actually, the answer may not be 1 even for an infinite number of hands. Here's a draft formula I'm working out. The probability of winning "m" bets and losing B bets is:

    P = A(m)*(Pw^m)*(Pl^(B+m))

    where A(m) is the number of valid ways to arrange the wins and losses. This is a little bit complicated, because not all arrangements are valid (you can't have more than B-1 losses in a row before getting a few wins), but as an upper estimate for A(m) we can say it's the total number of ways to arrange "m" wins in B+m trials, which is equal to (B+m)choose(m). The total risk of ruin is just the sum of these terms from m=0 to m=b as b --} infinity. This sum converges for 0 < Pw < 1.
  12. 22 Nov '06 17:31
    Originally posted by PBE6
    Actually, the answer may not be 1 even for an infinite number of hands. Here's a draft formula I'm working out. The probability of winning "m" bets and losing B bets is:

    P = A(m)*(Pw^m)*(Pl^(B+m))

    where A(m) is the number of valid ways to arrange the wins and losses. This is a little bit complicated, because not all arrangements are valid (you can't ha ...[text shortened]... um of these terms from m=0 to m=b as b --} infinity. This sum converges for 0 < Pw < 1.
    As I belatedly realised, and posted above, you're right.

    For the infinite case, you can use a difference equation. If R(B) is the probability of ruin if you start at B, then:

    R(B) = Pw R(B+1) + Pl R(B - 1)

    You can find more details at: http://www.fooledbyrandomness.com/gamblersruin.pdf (saves me typing out the argument), but for Pw > 1/2 you get:

    R(B) = (1/Pw - 1)^B
  13. Standard member uzless
    The So Fist
    22 Nov '06 17:32 / 1 edit
    Originally posted by mtthw
    You've got to be careful with inifinities

    Look at it this way. Let's say B = 1, and P_w = 0.1. The risk of ruin is at least 0.9, as there's a 0.9 chance of losing everything on the first turn. So it can't be 50%. It either has to be 1, or a function of B and P_w.
    You can't lose everything on the first turn if you can only bet one at a time....unless the question assumes you start with only one piece to bet with
  14. 22 Nov '06 17:36
    Originally posted by uzless
    You can't lose everything on the first turn if you can only bet one at a time....unless the question assumes you start with only one piece to bet with
    As I said, let B = 1. I took one specific case and showed the answer couldn't be 50%. Therefore it can't be 50% in all cases.
  15. Standard member uzless
    The So Fist
    22 Nov '06 17:40 / 1 edit
    Originally posted by mtthw
    As I belatedly realised, and posted above, you're right.

    For the infinite case, you can use a difference equation. If R(B) is the probability of ruin if you start at B, then:

    R(B) = Pw R(B+1) + Pl R(B - 1)

    You can find more details at: http://www.fooledbyrandomness.com/gamblersruin.pdf (saves me typing out the argument), but for Pw > 1/2 you get:

    R(B) = (1/Pw - 1)^B