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Suppose there's a casino that has a game where you sample from a Cauchy (100, 1) distribution (mode is 100). If the sample is positive, then the casino pays you that amount, otherwise you'd have to pay the casino.

My question is, what would be a value for which it's "rational" to play this game? Standard ways, like having a value less than the expected value, doesn't work since the Cauchy distribution has no mean.

I know my question is kind of vague, because I'm not sure how to define rational. On one hand, there definitely should be a way to value this game. For example if it's free, I'm sure most people would agree to play this game. Would a fair value be at the mode at 100?

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    $\begingroup$ OP writes: "If the sample is positive, then the casino pays you that amount, otherwise you'd have to pay the casino". Pay the casino what? $\endgroup$
    – wolfies
    Apr 22, 2023 at 12:02
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    $\begingroup$ Sorry if that was confusing. I meant if the sample is 10, you are paid \$10 from the casino. If the sample is -10, then you pay the casino \$10. $\endgroup$ Apr 22, 2023 at 18:20

2 Answers 2

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The Cauchy distribution has an infinite range. It is difficult to imagine how you would consider payouts of large sizes that make either the player or the casino go bankrupt.

The game has some decent odds for small values of pays. E.g. the odds of you paying 100 versus the casino paying 100 are 1:40001. So it seems a good bet.

However the game has a devilish drawback, which is the risk that you have the pay all the wealth that you possess and even more, getting into debts that you will never be able to pay off.

To estimate whether this game makes sense to play (for free) one could compute the probability of a gambler's ruin. One could describe the random walk of the gains and profits after $n$ steps as a sum of Cauchy distributed variables $Y(n) = \sum_{i}^n X_i$ and have an absorbing boundary at bankruptcy of the player and of the casino.

In the case of a random walk with steps of only +1 and -1, then you can regard this as a Martingale and the ratio of the probability of bankruptcy of for player and casino are equal to the amounts of money that they have at the start. E.g.

$$\frac{P(\text{player bankrupt})}{P(\text{casino bankrupt})} = \frac{\text{starting money casino}}{\text{starting money player}}$$

One can imagine a similar ratio for the case of steps according to a Cauchy distribution. But, each step the odds of winning a are larger than the odds of loosing, because the Cauchy distribution has a location of 100. This makes that it is much more likely that the casino goes bankrupt, than the player going bankrupt in comparison to the simple random walk.

Most bets will make the profit increase by 100 and this continues untill the casino goes bankrupt. The number of steps before the casino goes bankrupt will be roughly around $M_{casino}/100$ where $M_{casino}$ is the total money of the casino. The danger is is when during the $M_{casino}/100$ steps the player will go bankrupt.

We can approximate this by multiplying the probabilities of bankruptcy by the player during each step while assuming that the steps are 100 each time.

$$P(\text{player bankrupt}) \approx 1-\prod_{i=0}^{M_{casino}/100-1}1-F(-M_{player}-100i;100,1)$$

where $F$ is the CDF of the Cauchy distribution.

If $M_{player} = 100000$ and $M_{casino} = 1000000$ then this probability is 0.76%.

A simulation of 200 paths could look like:

simulation

Here 1 out of 200 players went bankrupt. Do you want to risk that?

sample = function(gambler = 100000, casino = 1000000) {
   X = c(0) 
   while((X[1] > -gambler) * (X[1]< casino)) {
     X = c(X[1]+rcauchy(1,100,1),X)
   }
   X
}

set.seed(1)      
plot(-100,-100, type = "l", ylim = c(-10^5,10^6), xlab = "number of gamble's", ylab = "accumulated profit/loss", xlim = c(0,10000*1.1))

pb = 0
for (i in 1:200) {
   Y = sample()
   lines(rev(Y), col = rgb(0,0,0,0.1))
   if (Y[1]<0) {pb = pb+1} 
}
pb

### estimate computation 
x = seq(0,1000000/100-1)
1-prod(1-pcauchy(-100000-x*100,100,1))
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    $\begingroup$ Thanks, your answer makes a lot of sense to me. This was on my mind for a while. $\endgroup$ Apr 22, 2023 at 18:28
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I think we should always keep in mind the distinction between formal modelling and reality. This affects at least two aspects of the problem given here.

  1. Does the Cauchy model appropriately model the real situation? Chances are it doesn't, as both the casino and the player will have upper bounds on the amounts they can pay out, so in fact both players may find out that they cannot do what is required in case of certain results. The problem would then change into a problem with a truncated Cauchy, in which obviously expected values are well defined if you know where the truncation is.

  2. How do we model "rationality"? The expected value is a popular candidate of course, but one can well consider other candidates. For example, one could well argue (ignoring problems with item 1) that the problem is symmetric around 100, so the fair price would be 100 because this makes the payout distribution identical for both players. However, considering potential consequences (which would involve reality again, i.e., what exactly would happen if the payout would be very high; also how is utility related to amount of money for both involved), a player may not consider rational playing at all due to a nonzero bankrupt probability and maybe also considering the utility of unbounded gain in terms of money in fact bounded. So one would arguably need to translate money payouts into actual utilities, and these may not be symmetric, and may again depend on the specific situation of everyone involved.

On one hand, there definitely should be a way to value this game.

Wishful thinking, maybe? You may define one, see above, however this means that you enforce your definition of rationality on the situation at hand. There is no "meta-objective/meta-rational" way to ultimately argue in what unique way rationality should be formally defined.

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