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Recently, I read this article on expected value. The author cites the example of a poker player. If he has a hand that gives him a 1 in 5 chance of winning the pot, would you take the chance? The answer of the majority of people would be to say ''Are you crazy!!? I would never do that!” The answer should be “It depends!?”. For example, if I have a 1 in 5 chance of winning the $500 pot or a 4 in 5 chance of losing my 100 dollars bet, then the answer would be different if you know a bit about probability theory. In this specific case, I would win an average of 20 dollars on each hand. In my perspective, I think I would take the bet, but I am still confused! Does it depend of a certain variance? If I bet everything I have on a certain hand even though on average I am positive, there's still a risk I would lose everything. How should I demystify my questions?

J.Doe
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  • The $500 pot includes the $100 that you're betting. In other words, you have a 1 in 5 chance of gaining $400, which evens out with the 4 in 5 chance of losing $100. – Théophile Dec 31 '22 at 04:26
  • @Théophile You know what I am trying to say ... Let's say the pot is 1000 dollars instead – J.Doe Dec 31 '22 at 04:31
  • In fact, I don't know what you're trying to say. You mentioned an article about a poker game where the expected value is $0$, and then you said that you would win an average of 20 dollars per hand, and that you were confused about this. I was showing where your calculation was wrong, but perhaps I misunderstood the question? – Théophile Dec 31 '22 at 04:36
  • Expectations are particularly useful when you have independent repeated experiments: The law of large numbers says that the time averages will get close to the one-shot expectation. So if you get to play the game many times, and if the expectation is in your favor, it may be good to play. – Michael Dec 31 '22 at 04:41
  • There is massive theory behind this, so I am sure somebody more familiar with stochastic processes will eventually be able to help. To me, it looks like a generalisation of random walk where the steps are: $a=500$ (with probability $p=20%$) and $b=-100$ (with probability $1-p=80%$), the starting value is $x_0$ (e.g. $\ge 100$) and you can ask many questions, such as "what is the probability that I lose all the money?", i.e. land on a state $x_n\lt 100$. –  Dec 31 '22 at 05:56

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Working with expected value is sufficient when the value is small compared to your economy or you can play enough times for the law of large numbers to kick in. It fails when enough is at stake that losing is worse than winning even if the expected value is positive. Expected value assumes that every dollar is the same value to you. At large stakes that fails and generally you should be more conservative.

In this answer I talk about an extreme case-you are offered very nice odds, so by expected value you should bet all your money each time. The chance you wind up broke is very high, but if you don't you are so rich that the expected value is also very high.

Ross Millikan
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  • I've been thinking about this recently, and I don't think that repeatability is sufficient to ensure that +EV decisions are good ones. – Zubin Mukerjee May 02 '23 at 14:33
  • In the toy game I'm thinking of, you pay $x$ to play, with payoffs of $1.6x$ and $0.6x$, each with $50%$ probability. The expected payoff is $1.1x > x$, and it's infinitely repeatable. It's known that 1D random walks return to the origin infinitely often, so by continuously playing this game we will inevitably reach points with exactly $n$ wins and $n$ losses. These correspond to an overall payoff of $0.96^n$, and this will occur for arbitrarily large $n$. So if you play this +EV repeatable game your overall payoff will inevitably go to $0$?! – Zubin Mukerjee May 02 '23 at 14:44
  • @ZubinMukerjee: the random walk returning does not matter. It represents having won and lost the same number of times. At that point you are ahead because you win more than you lose. After a reasonable number of throws the $-3 \sigma$ case is positive and you are almost sure (in the normal sense, not the mathematical one of probability $1$) to be ahead. – Ross Millikan May 02 '23 at 15:24
  • I did not specify the game properly. I am thinking of playing with all your money each game. For example WLWL would correspond to $$1 \to 1.6 \to 0.96 \to 1.536 \to 0.9216$$ In this way, the points of random-walk-returning correspond to multiplying your initial amount by $0.96^n$ (with $n$ wins and $n$ losses) despite every single individual decision being a positive EV decision. I think that is interesting – Zubin Mukerjee May 02 '23 at 16:23
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    It is no different than the case I linked to. You are correct that an equal number of wins and losses when betting all your money will lead to a loss. The EV is still positive because cases with more wins than losses make you very rich. This is a softened version as the one I linked to has one loss leaving you broke. The failure of EV is not mathematical, it is that we don't value every dollar equally. Especially large gains are not valued as high as EV says by most people – Ross Millikan May 02 '23 at 19:56
  • Thank you. I think your last point relates to economists' idea of decreasing marginal utility of money – Zubin Mukerjee May 02 '23 at 22:35
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    Exactly. One heuristic is to take the log of the amount of money you have and maximize the expected value of that. It makes going broke have value $-\infty$ which is not a bad idea. This is the basis of the Kelly criterion for sizing bets. – Ross Millikan May 03 '23 at 00:06