4 ms·

Classic recursive or iterative problem that an undergraduate Statistics major needs to be able to derive directly not using formulas. Its wild how the chance of

Classic recursive or iterative problem that an undergraduate Statistics major needs to be able to derive directly not using formulas. Its wild how the chance of going broke is your money in bet units above both sides money at 50/50 and the odds go dramatically worse with any house edge.

andirk 4y ago
Is it because one can get ruined by hitting 0 but a "win" is subjective because there's no absolute upper bound? If one could go in to negative at 50/50 odds, then the game will dance around that $1 starting point forever?
- hackerlight 4y ago
  If there's no absorbing states and it's the cumulative sum of infinite coin flip (-1, 1), it'll visit every integer an infinite number of times with probability 1 a.s..
  - andirk 4y ago
    So in real-world, the gambler would owe all the money in the world at some point. But also win all the money in the world at another point! And then cash out and he wins? Sounds beatable.
    - hackerlight 4y ago
      The gambler will owe all the money (and then some) with probability 1 a.s., but it's not guaranteed to happen because a.s. implies it's not a certainty.
im3w1l 4y ago
> Its wild how the chance of going broke is your money in bet units above both sides money at 50/50 It's actually quite simple. In a fair game you expect to leave with as much money as you started with. initial money = expected final money = p * all money
- krackers 4y ago
  To be clear, the reason why expected final money is `p * all money` is because the only "absorbing states" are the win state and lose all money state, so it's `p * max money + (1 - p) * 0`, right? Also, it's not quite clear to me why in this case initial money = expected final money. For the related game where you go double-or-nothing on each bet, it's evident this holds since the expected value of your next turn is the same as your current (I believe it's called martingale property), but that doesn't hold for the game in the article where you win +1 or -1 unit each bet, since if you e.g. have 4 units and the win state is 5 units, you're more likely to win than lose. Edit: nevermind, I think it is a martingale since if you're at 4 units the expected value of the next state is 0.5*5 + 0.5*3 = 4. I guess my confusion stemmed from the fact that even though the long-term expected value is unchanged, you can still have good odds of winning (which is balanced by a small chance of a catastrophic loss), e.g. 4 = 0 * 1/5 + 5 * 4/5.

andirk 4y ago

Is it because one can get ruined by hitting 0 but a "win" is subjective because there's no absolute upper bound? If one could go in to negative at 50/50 odds, then the game will dance around that $1 starting point forever?

hackerlight 4y ago
If there's no absorbing states and it's the cumulative sum of infinite coin flip (-1, 1), it'll visit every integer an infinite number of times with probability 1 a.s..
- andirk 4y ago
  So in real-world, the gambler would owe all the money in the world at some point. But also win all the money in the world at another point! And then cash out and he wins? Sounds beatable.
  - hackerlight 4y ago
    The gambler will owe all the money (and then some) with probability 1 a.s., but it's not guaranteed to happen because a.s. implies it's not a certainty.

im3w1l 4y ago

> Its wild how the chance of going broke is your money in bet units above both sides money at 50/50 It's actually quite simple. In a fair game you expect to leave with as much money as you started with. initial money = expected final money = p * all money

krackers 4y ago
To be clear, the reason why expected final money is `p * all money` is because the only "absorbing states" are the win state and lose all money state, so it's `p * max money + (1 - p) * 0`, right? Also, it's not quite clear to me why in this case initial money = expected final money. For the related game where you go double-or-nothing on each bet, it's evident this holds since the expected value of your next turn is the same as your current (I believe it's called martingale property), but that doesn't hold for the game in the article where you win +1 or -1 unit each bet, since if you e.g. have 4 units and the win state is 5 units, you're more likely to win than lose. Edit: nevermind, I think it is a martingale since if you're at 4 units the expected value of the next state is 0.5*5 + 0.5*3 = 4. I guess my confusion stemmed from the fact that even though the long-term expected value is unchanged, you can still have good odds of winning (which is balanced by a small chance of a catastrophic loss), e.g. 4 = 0 * 1/5 + 5 * 4/5.