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Posers and Puzzles

Posers and Puzzles

  1. 05 Jun '08 22:49
    Suppose there is a betting game at the local fairgrounds. In each round of this game, a mouse is freed at the beginning of a circuit and then runs into either hole 1, hole 2, or hole 3 -- each hole with equal chance. Each round before the mouse is freed, you can bet on which hole the mouse will run into. If you are correct, then you get back twice the money you bet. If you are wrong, you lose the money you bet. The minimum bet is 1 dollar. There is no maximum bet. You may play as many rounds as you like, provided of course that you have money to meet the minimum bet for each round.

    Now, suppose you have 100 dollars on you, and you go to play this game with the sole objective of leaving with more money than you came with. What is your optimal playing strategy and what can we say about the associated probability of success (again, success is your leaving the game with some amount > 100 dollars)?

    I think I have an optimal strategy, but I would like to see if anyone else can come up with something better.
  2. 06 Jun '08 01:45
    Bet the minimum on the first bet. If you lose, double your previous bet. If you lose again, double the last bet once again.

    eg: bet $1...lose, bet $2...lose, bet $4...lose, bet $8...and so on. Probabilities are (1 in 3) that you will win at least once before you lose your $100.
  3. Standard member TheMaster37
    Kupikupopo!
    06 Jun '08 10:24
    Originally posted by fratt12
    Bet the minimum on the first bet. If you lose, double your previous bet. If you lose again, double the last bet once again.

    eg: bet $1...lose, bet $2...lose, bet $4...lose, bet $8...and so on. Probabilities are (1 in 3) that you will win at least once before you lose your $100.
    But is this the best strategy?
  4. Standard member wolfgang59
    Infidel
    06 Jun '08 11:36
    Originally posted by fratt12
    Bet the minimum on the first bet. If you lose, double your previous bet. If you lose again, double the last bet once again.

    eg: bet $1...lose, bet $2...lose, bet $4...lose, bet $8...and so on. Probabilities are (1 in 3) that you will win at least once before you lose your $100.
    This is certanly the best strategy for $63 or $127 but what is your strategy for $100?

    When you are $63 down and have $37 left your last bet was $32 so you cannot double up!
  5. 06 Jun '08 12:00
    I would go to the shop, spend $2 on a piece of cheese. I would then place the cheese in hole Nº1, and bet $98 for hole Nº1
  6. 06 Jun '08 13:45
    Originally posted by wolfgang59
    This is certanly the best strategy for $63 or $127 but what is your strategy for $100?

    When you are $63 down and have $37 left your last bet was $32 so you cannot double up!
    The idea is good. However, you don't have to double exactly.
    What will happen if you use the serie $1, $2, $ 5, $ 10, $ 20, $ 50, $100 instead of just double up at every loss?
  7. Standard member PBE6
    Bananarama
    06 Jun '08 13:54
    Originally posted by Guanche
    I would go to the shop, spend $2 on a piece of cheese. I would then place the cheese in hole Nº1, and bet $98 for hole Nº1
    Gouda choice!
  8. Standard member forkedknight
    Defend the Universe
    06 Jun '08 14:48 / 1 edit
    Originally posted by FabianFnas
    The idea is good. However, you don't have to double exactly.
    What will happen if you use the serie $1, $2, $ 5, $ 10, $ 20, $ 50, $100 instead of just double up at every loss?
    maybe 1, 2, 5, 10, 30, 52,

    because once you lose the $30 bet, you will have $52 dollars left.
  9. Standard member wolfgang59
    Infidel
    06 Jun '08 14:57
    Originally posted by FabianFnas
    The idea is good. However, you don't have to double exactly.
    What will happen if you use the serie $1, $2, $ 5, $ 10, $ 20, $ 50, $100 instead of just double up at every loss?
    More than doubling previous bet is inefficient. Remember the goal is to come out winning! Winning by $1 is just as good as winning by $20 (in the problem)

    Your method will actually decrease the chances of winning since after a losing streak of 6 the doubling system still allows you $37 to play with.

    My hunch is to use these amounts
    1, 2, 4, 8, 16, 32 , 14,

    If my $14 bet wins I put all my $51 on next bet.

    If my $14 bet loses I bet all my remaining $13, then my $26, then $51

    With the odds against me I want to minimise the number of bets so hoarding money is irrational - I need to get to my goal asap.

    maybe I'll do the probabilities later ........
  10. Standard member forkedknight
    Defend the Universe
    06 Jun '08 15:18
    Originally posted by forkedknight
    maybe 1, 2, 5, 10, 30, 52,

    because once you lose the $30 bet, you will have $52 dollars left.
    if you want to _continue_ betting and have the best chance of winning, you would need to adjust your betting scheme every time you win to utilize your entire bankroll and the maximum number of bets.
  11. Standard member forkedknight
    Defend the Universe
    06 Jun '08 15:22
    Originally posted by wolfgang59
    More than doubling previous bet is inefficient. Remember the goal is to come out winning! Winning by $1 is just as good as winning by $20 (in the problem)

    Your method will actually decrease the chances of winning since after a losing streak of 6 the doubling system still allows you $37 to play with.

    My hunch is to use these amounts
    1, 2, 4, 8, 16, ...[text shortened]... s irrational - I need to get to my goal asap.

    maybe I'll do the probabilities later ........
    that may be true if winning _at all_ is your only goal, but in real life, you want to maximize your profits while giving yourself the maximum number of bets.

    If the chances of losing your entire bankroll are small enough, you can win enough early on to gain another bet, thereby decreasing your chances of losing even more.
  12. 06 Jun '08 18:26
    Originally posted by fratt12
    Bet the minimum on the first bet. If you lose, double your previous bet. If you lose again, double the last bet once again.

    eg: bet $1...lose, bet $2...lose, bet $4...lose, bet $8...and so on. Probabilities are (1 in 3) that you will win at least once before you lose your $100.
    This works great for the first 6 potential rounds (betting 1, 2, 4, 8, 16, 32). Any first win as you move through those rounds will constitute success. But the probability that you lose all 6 rounds should be (2/3)^6, which is around 0.09. In that case, what do we do with the remaining 37 dollars?
  13. 06 Jun '08 18:29 / 1 edit
    Originally posted by wolfgang59
    More than doubling previous bet is inefficient. Remember the goal is to come out winning! Winning by $1 is just as good as winning by $20 (in the problem)

    Your method will actually decrease the chances of winning since after a losing streak of 6 the doubling system still allows you $37 to play with.

    My hunch is to use these amounts
    1, 2, 4, 8, 16, ...[text shortened]... s irrational - I need to get to my goal asap.

    maybe I'll do the probabilities later ........
    This is exactly what I was thinking. I'm struggling with whether or not it's optimal, though.

    EDIT: I mean that I was thinking about 14 next. But if you lose the 14, you're left with 23, not 13. So there's that problem with your analysis.
  14. Standard member uzless
    The So Fist
    06 Jun '08 20:26
    Originally posted by wolfgang59
    With the odds against me I want to minimise the number of bets so hoarding money is irrational - I need to get to my goal asap.

    maybe I'll do the probabilities later ........
    Quite the opposite...you want to maximize the number of bets in order to give yourself more chances to win.
  15. Standard member wolfgang59
    Infidel
    07 Jun '08 09:12
    Originally posted by LemonJello
    This is exactly what I was thinking. I'm struggling with whether or not it's optimal, though.

    EDIT: I mean that I was thinking about 14 next. But if you lose the 14, you're left with 23, not 13. So there's that problem with your analysis.
    13 + 14 isnt 37? Damn.

    OK with $23 left lets go for a 3$ bet. If that wins bet $26, if it loses and you are left with $20 bet $6 then if that wins bet the whole $26. If the $6 bet lost it would be $12 next ... etc.

    The idea being to get to a {100/(2^n) +1} amount so that you can double up to get over $100 asap.