Timeline for Analyzing a Gambling Race Paradox
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Aug 17, 2021 at 10:32 | comment | added | afreelunch | If gambles are binary, shouldn't everyone just choose a bet size such that, if they win the bet, they will hit the target? (Or if the maximum bet is constrained to prevent this, won't they just bet the maximum allowable amount?) | |
| Aug 15, 2021 at 18:03 | history | edited | user37672 | CC BY-SA 4.0 |
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| Aug 15, 2021 at 4:51 | answer | added | Dave Harris | timeline score: 2 | |
| Aug 14, 2021 at 20:17 | comment | added | user37672 | @user253751 Yeah, optimal would be fleshed out as reaching the target in as few rounds (on average) as possible, without ever exceeding a maximum chance of losing, but possibly setting some edge cases aside, this will entail always maintaining that maximum as closely as possible. | |
| Aug 14, 2021 at 20:05 | comment | added | Criticizing Israel not allowed | Isn't it the optimal bet size to maintain a certain chance of losing? Not just the optimal bet size. | |
| Aug 13, 2021 at 18:46 | comment | added | user37672 | My player does not need to figure out when to stop playing the game, but to find the optimal bet size as a function of current point total. Ceasing play (the strategy which bets 0 at every point total) seems to be an obviously suboptimal strategy. | |
| Aug 13, 2021 at 18:43 | comment | added | user37672 | @user253751 That is an interesting article. I think what he describes is ultimately a Sorites Paradox; i.e., it is very unlikely that adding any one additional minute of game play will cause the transition from sufficient sleep to insufficient sleep, but he knows that when many such minutes are added, an unidentifiable one of those minutes will constitute that critical minute. I'm not confident that his "Schelling Fence" solution is really the best he can do under this uncertainty, but at any rate, I'm afraid I do not currently see how this applies to my problem. | |
| Aug 13, 2021 at 16:05 | comment | added | Criticizing Israel not allowed | @user37672 Well, then if you have used up your 2% allowed probability of ruin at the first time step, you must play the remainder of the game with a 0% allowed probability of ruin. If I understand your question you are asking about the philosophical justification for why you wouldn't bump your chance back up to 2%? It reminds me of the "1%-more-murderous Gandhi problem" | |
| Aug 13, 2021 at 16:00 | comment | added | user37672 | @user253751 The probability of ruin at the initial point total (initial time step or later return to that total) is necessarily the probability of ruin overall. Also, the probability of ruin at another point total is necessarily the probability of ruin overall in a new game in which the player begins with that point total. You're right about the next thing you say, which is a big part of what constitutes the paradox. | |
| Aug 13, 2021 at 13:27 | comment | added | Nuclear Hoagie | I don't understand why you presuppose there must be a strategy that allows you to hit your risk target either overall or at each time point - you say the notion that all strategies have too-high or too-low risk must be false. But here's a simple counterexample: flip a coin, on heads you win double your bet and on tails you get nothing, and you win when you get to 100 points. Victory is guaranteed regardless of strategy - it is impossible to have a 2% chance of failure. If you make the losses very small but non-zero, your chance of failure is still very low no matter what you do. | |
| Aug 13, 2021 at 10:43 | comment | added | Criticizing Israel not allowed | Do you want your probability of losing at all to be 2%, or your probability of losing at each given time step to be 2%? Consider the case where I keep rebalancing my strategy to have a 2% chance of losing in the next time step and 0% thereafter, but I rebalance it after every time step. I run this for 100 time steps. My total chance of losing is actually quite high, much higher than 2%. | |
| Aug 12, 2021 at 21:45 | comment | added | user37672 | I used the model of "changing strategy" merely for clarity in understanding the paradox, which is that a player with constant risk preferences seemingly cannot rationally commit to a strategy in which risk monotonically decreases as their point total increases. At some point (possibly at the start), risk will either be too high or too low, and the player will scrap the strategy. But if all possible strategies have this feature, then it seems the player cannot rationally play the game according to any strategy, which is clearly false. | |
| Aug 12, 2021 at 21:42 | comment | added | user37672 | @NuclearHoagie I am using "strategy" in the same way you have defined it, which I suspect may point to the heart of the paradox. To "change strategy" would be to have a complete algorithm prescribing an action from every game state (call this Strategy A), but to set it aside and create a new complete algorithm prescribing an action from every game state (Strategy B). I think your point is that this behavior is functionally equivalent to a single strategy whose actions are identical to A when the player would follow A, and B when the player would follow B, which I fully appreciate is true. | |
| Aug 12, 2021 at 20:42 | comment | added | Nuclear Hoagie | A strategy is a complete algorithm for decision making at any possible game state. One cannot "change strategy" partway through a game - that is the strategy itself. Your strategy must dictate what to do at every possible game state, otherwise it's not a strategy. For arbitrary games, I don't see any expectation that risk of ruin should be able to remain constant - one can imagine a game where victory is inevitable after a certain point total, in which risk of ruin goes to 0 after some point. | |
| Aug 12, 2021 at 20:24 | history | edited | user37672 | CC BY-SA 4.0 |
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| Aug 9, 2021 at 15:46 | comment | added | user37672 | @kyle Indeed, I have considered this necessary edge case deviation. That is what I meant by "more or less constant (maybe lower at the very end of the game due to the discreteness of the problem)." | |
| Aug 9, 2021 at 6:17 | comment | added | kyle | It is a little tough to answer without a few more details (e.g., what is the nature of the gamble?), but I suspect the answer is either, "it cannot be done," or otherwise unsatisfying. For example, take the case of 99999 coins: if I lose, I'll have less than 99999 coins and a subsequent 2% chance of bankruptcy; if I win, I win the game and do not go bankrupt. This is only consistent with a 2% chance of bankruptcy today if I gamble all 99999 coins AND losing everything is possible AND has a specific probability. | |
| Aug 8, 2021 at 18:42 | review | First posts | |||
| Aug 12, 2021 at 18:54 | |||||
| Aug 8, 2021 at 18:41 | history | asked | user37672 | CC BY-SA 4.0 |