<p>There are several problems with your game that you are not considering, the first of which is that the risk of ruin is irrelevant.  The game is a race.  Your question is roughly equivalent to concerning yourself with your horse falling and dying in the race but with an objective to win the race.  The death of the horse is not the only way to lose.  Ruin is not the only way to lose.</p>
<p>In controlling the risk of ruin, you may guarantee that you lose the race.  You have two additional problems as well.</p>
<p>Because ruin is impossible in the real numbers, the game must have one unit as the smallest indivisible amount.  Now let us imagine that the payoff for some combination is 1.01:1, and you bet 1 unit.  You can only be paid 1 unit.</p>
<p>Your third problem is that this is a race being run by a random number generator.  If this were a game of skill instead of chance, then answers might be quite a bit different.</p>
<p>The winner will be the actor that succeeds only through randomness the most.  It is possible to maximize the return over infinite repetition of the game, but not in any one instance.</p>
<p>It is true that you could make a Kelly Bet for the first round, but what if one other actor doubles their money and you halve yours?  The two parties should even out as time goes to infinity, but this is a race to a finite goal.  This is a stopping time problem.  If the leader is four times closer than the last place person, does either change their strategy?  This is an all-or-nothing game.  It does not matter how close you got if you were not the first across the line.</p>
<p>Ruin, in this framework, is just one way not to cross the finish line.</p>
<p>If you were sincerely considering this game, then you would have a terrible decision tree because it needs to include the potential outcomes of every actor.  Your utility here is <span class="math-container">$c,c&gt;0$</span> if you win and <span class="math-container">$0$</span> if you lose.  This is not a game with a payoff for win, place or show.  There is no participation trophy either.</p>
<p>Nonetheless, ruin avoidance should only accidentally be a strategy because it would happen to be optimal to keep you from losing.</p>

CC BY-SA 4.0
2021-08-15T18:00:50.757Z
While ruin is categorically analogous to the horse dying, I think it differs enough by degree to be very relevant.  In practice, ruin will be the most likely terminal state if I wager too much, while being beaten to 100000 points will be the most likely terminal state if I wager too little.  All payouts are integer multiples of the wager size, so rounding isn't a problem (I will update the question with this information).  The third concern is correct, and I accept that solving the problem I framed does not solve the game, but it is a much more tractable problem, and should still be useful
user37672
CC BY-SA 4.0
2021-08-15T18:01:57.947Z
during game play at least some of the time (such as near the start of the game when no one has a commanding lead and point totals are low).
user37672
CC BY-SA 4.0
2021-08-15T04:51:48.717Z
2021-08-15T04:51:48.717Z
Dave Harris
<p>Suppose a number of players are given <span class="math-container">$100$</span> points each, and repeatedly engage in a gamble having positive expected value, with the goals of being the first player to reach <span class="math-container">$100000$</span> points.  Solving for the optimal bet size per round to balance risk of ruin and speed of point accrual is a very difficult game theory problem, so the reasonable approach seems to be to impose a heuristic constraint on the risk of ruin, and optimize speed relative to that constraint.  Let's say I decide I want no more than a <span class="math-container">$2$</span>% probability of going bankrupt before reaching the <span class="math-container">$100000$</span> point goal.  Now I can simulate the game a large number of times to find the bet size which reaches the goal in as few rounds as possible (probably the largest bet size) without exceeding the target risk of ruin.</p>
<p>However, suppose I am playing the game in practice, and have accrued a point total of <span class="math-container">$1000$</span>.  The remainder of the game is functionally equivalent to a new, identical game, except that in the new game, I start with <span class="math-container">$1000$</span> points instead of <span class="math-container">$100$</span> points.  In other words, the game I've been describing has an optimal substructure.  It seems to me that in the new game, my desired risk of ruin constraint should still be about the same, <span class="math-container">$2$</span>% in the example; there seems to be no fundamental reason (apart from assessing my position relative to other players, which is beyond the scope of this question) to suppose I would have a different risk tolerance in the race from <span class="math-container">$1000$</span> to <span class="math-container">$100000$</span> than in the race from <span class="math-container">$100$</span> to <span class="math-container">$100000$</span>.  Allowing the risk of ruin to decline too much as the game progresses is an unnecessary speed concession.</p>
<p>The problem is that changing my strategy as the game progresses to retarget a <span class="math-container">$2$</span>% risk of ruin means my original strategy actually had more than a <span class="math-container">$2$</span>% risk of ruin.  My present, <span class="math-container">$1000$</span> point self shouldn't care about this enough to dissuade me from retargeting (the past is in the past), but my past <span class="math-container">$100$</span> point self foreseeing this phenomenon cares a lot.  It seems that what's needed is a function which maps every possible point total between <span class="math-container">$1$</span> and <span class="math-container">$99999$</span> to a bet size, where the risk of ruin at every input is more or less constant (maybe lower at the very end of the game due to the discreteness of the problem), but it is unclear that such a function is possible, because attaining points would always reduce risk of ruin no matter what function is chosen.  Is is possible to construct such a function?  If not, how can I rationally play this game when I know my underlying risk of ruin preferences won't change much as the game progresses, and I will not stick to any possible strategy I create?</p>
<p>EDIT regarding nature of gamble, as per request in comments:  Suppose that each round consists of rolling fair dice, and the profit or loss a player attains for that round is computed based on the sum of the rolls (with a positive expected value).  Profit or loss are always an integer multiple of the bet size.  Each set of rolls is independent and identically distributed relative to their previous rolls and other players' rolls.</p>

CC BY-SA 4.0
2021-08-09T06:17:49.493Z
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.
kyle
CC BY-SA 4.0
2021-08-09T15:46:47.697Z
@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)."
user37672
CC BY-SA 4.0
2021-08-12T20:42:27.653Z
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.
Nuclear Hoagie
CC BY-SA 4.0
2021-08-12T21:42:17.733Z
@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.
user37672
CC BY-SA 4.0
2021-08-12T21:45:13.073Z
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.
user37672
CC BY-SA 4.0
2021-08-13T10:43:37.530Z
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%.
Criticize SE actions means ban
CC BY-SA 4.0
2021-08-13T13:27:09.303Z
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.
Nuclear Hoagie
CC BY-SA 4.0
2021-08-13T16:00:57.627Z
@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.
user37672
CC BY-SA 4.0
2021-08-13T16:05:18.490Z
@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"](https://www.lesswrong.com/posts/Kbm6QnJv9dgWsPHQP/schelling-fences-on-slippery-slopes)
Criticize SE actions means ban
CC BY-SA 4.0
2021-08-13T18:43:36.843Z
@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.
user37672
CC BY-SA 4.0
2021-08-13T18:46:24.457Z
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.
user37672
CC BY-SA 4.0
2021-08-14T20:05:59.337Z
Isn't it the optimal bet size *to maintain a certain chance of losing*? Not just the optimal bet size.
Criticize SE actions means ban
CC BY-SA 4.0
2021-08-14T20:17:14.070Z
@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.
user37672
CC BY-SA 4.0
2021-08-17T10:32:02.920Z
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?)
afreelunch
<decision-theory><expected-utility><rational-expectations><rationality><policy>
Analyzing a Gambling Race Paradox
user37672
# Analyzing a Gambling Race Paradox

(Asked by: username_0 on 2021-08-08)

Suppose a number of players are given $100$ points each, and repeatedly engage in a gamble having positive expected value, with the goals of being the first player to reach $100000$ points. Solving for the optimal bet size per round to balance risk of ruin and speed of point accrual is a very difficult game theory problem, so the reasonable approach seems to be to impose a heuristic constraint on the risk of ruin, and optimize speed relative to that constraint. Let's say I decide I want no more than a $2$% probability of going bankrupt before reaching the $100000$ point goal. Now I can simulate the game a large number of times to find the bet size which reaches the goal in as few rounds as possible (probably the largest bet size) without exceeding the target risk of ruin.

However, suppose I am playing the game in practice, and have accrued a point total of $1000$. The remainder of the game is functionally equivalent to a new, identical game, except that in the new game, I start with $1000$ points instead of $100$ points. In other words, the game I've been describing has an optimal substructure. It seems to me that in the new game, my desired risk of ruin constraint should still be about the same, $2$% in the example; there seems to be no fundamental reason (apart from assessing my position relative to other players, which is beyond the scope of this question) to suppose I would have a different risk tolerance in the race from $1000$ to $100000$ than in the race from $100$ to $100000$. Allowing the risk of ruin to decline too much as the game progresses is an unnecessary speed concession.

The problem is that changing my strategy as the game progresses to retarget a $2$% risk of ruin means my original strategy actually had more than a $2$% risk of ruin. My present, $1000$ point self shouldn't care about this enough to dissuade me from retargeting (the past is in the past), but my past $100$ point self foreseeing this phenomenon cares a lot. It seems that what's needed is a function which maps every possible point total between $1$ and $99999$ to a bet size, where the risk of ruin at every input is more or less constant (maybe lower at the very end of the game due to the discreteness of the problem), but it is unclear that such a function is possible, because attaining points would always reduce risk of ruin no matter what function is chosen. Is is possible to construct such a function? If not, how can I rationally play this game when I know my underlying risk of ruin preferences won't change much as the game progresses, and I will not stick to any possible strategy I create?

EDIT regarding nature of gamble, as per request in comments: Suppose that each round consists of rolling fair dice, and the profit or loss a player attains for that round is computed based on the sum of the rolls (with a positive expected value). Profit or loss are always an integer multiple of the bet size. Each set of rolls is independent and identically distributed relative to their previous rolls and other players' rolls.

**Tags:** decision-theory, expected-utility, rational-expectations, rationality, policy

> **Comments:**
> **username_1**: 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.  
> **username_0**: @username_1 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)."  
> **username_2**: 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.  
> **username_0**: @username_2 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.  
> **username_0**: 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.  
> **username_3**: 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%.  
> **username_2**: 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.  
> **username_0**: @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.  
> **username_3**: @username_0 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"](https://www.lesswrong.com/posts/Kbm6QnJv9dgWsPHQP/schelling-fences-on-slippery-slopes)  
> **username_0**: @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.  
> **username_0**: 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.  
> **username_3**: Isn't it the optimal bet size *to maintain a certain chance of losing*? Not just the optimal bet size.  
> **username_0**: @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.  
> **username_4**: 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?)  

---

## 1 answers

### Answer by username_5 on 2021-08-15. Score: 2

There are several problems with your game that you are not considering, the first of which is that the risk of ruin is irrelevant. The game is a race. Your question is roughly equivalent to concerning yourself with your horse falling and dying in the race but with an objective to win the race. The death of the horse is not the only way to lose. Ruin is not the only way to lose.

In controlling the risk of ruin, you may guarantee that you lose the race. You have two additional problems as well.

Because ruin is impossible in the real numbers, the game must have one unit as the smallest indivisible amount. Now let us imagine that the payoff for some combination is 1.01:1, and you bet 1 unit. You can only be paid 1 unit.

Your third problem is that this is a race being run by a random number generator. If this were a game of skill instead of chance, then answers might be quite a bit different.

The winner will be the actor that succeeds only through randomness the most. It is possible to maximize the return over infinite repetition of the game, but not in any one instance.

It is true that you could make a Kelly Bet for the first round, but what if one other actor doubles their money and you halve yours? The two parties should even out as time goes to infinity, but this is a race to a finite goal. This is a stopping time problem. If the leader is four times closer than the last place person, does either change their strategy? This is an all-or-nothing game. It does not matter how close you got if you were not the first across the line.

Ruin, in this framework, is just one way not to cross the finish line.

If you were sincerely considering this game, then you would have a terrible decision tree because it needs to include the potential outcomes of every actor. Your utility here is $c,c>0$ if you win and $0$ if you lose. This is not a game with a payoff for win, place or show. There is no participation trophy either.

Nonetheless, ruin avoidance should only accidentally be a strategy because it would happen to be optimal to keep you from losing.

> **Comments:**
> **username_0**: While ruin is categorically analogous to the horse dying, I think it differs enough by degree to be very relevant.  In practice, ruin will be the most likely terminal state if I wager too much, while being beaten to 100000 points will be the most likely terminal state if I wager too little.  All payouts are integer multiples of the wager size, so rounding isn't a problem (I will update the question with this information).  The third concern is correct, and I accept that solving the problem I framed does not solve the game, but it is a much more tractable problem, and should still be useful  
> **username_0**: during game play at least some of the time (such as near the start of the game when no one has a commanding lead and point totals are low).  

---

