I came across this problem in Scientific American (vol. 321.5, Nov 2019, p. 73), and it was troubling.

Say your very generous dad gives you $100 to go to a casino that offers a game where it pays 20 percent if you flip heads and takes away 17 percent if you flip tails. By one equation, the equity-value positive.

½ x $20 + ½ x -$17 = EV + $1.50

But the casino isn’t as harebrained as you might think. The rule is that you have to bet everything you have over ten coin tosses. You cannot pocket your winnings until the tenth flip, and you have to bet everything you have each time, whether you are winning or losing. In other words, your ante is multiplied by 1.2 for heads up and 0.83 for tails over ten trials. Assuming that you flip tails five times and heads five times, your equity value calculation is

1.2 x 1.2 x 1.2 x 1.2 x 1.2 x 0.83 x 0.83 x 0.83 x 0.83 x 0.83 x $100 = 98.02, or EV - $1.98

It doesn’t matter which order you flip heads or tails. The result is the same. If you flip the coin 100 times, your EV is about minus $33!

The article in Scientific American claimed that the second equation explains why, mathematically, rich people with more capital come out ahead over poor people, even when the odds are stacked against the rich people.

The result was troubling to me. My friend who is a professional poker player was flabbergasted as well. Any ideas about how these equations apply to poker?

The reason is well known to investors. If you lose 20% of your money, you must then gain 25% just to be even. More specifically after losing 17%, you then need to gain 20.5% to get even. Not just 20%. So this is not a winning game.

Yes, the article said that investors are aware of this math. What are the implication for poker players?

For tournament play, does it explain why the biggest stack on the table has an advantage beyond just the luxury of staying alive after losing to a smaller stack all in?

For cash games, does it mean that you always have to replenish your stack when you start dropping below a certain spr in many pots? Do you need a bankroll not just so that you don't go bust but also because the player with more capital has an advantage beyond just regular EV probabilities?

It doesn't apply. In poker, you aren't required to go all in every hand.

Adding to what Didace said, you aren't required to play at stakes that are too high for your bankroll. For this reason, a poker player can keep their risk low and their growth factor larger than 1.

You’re getting different answers because the second equation doesn’t do an EV calculation because it doesn’t consider all outcomes. Only ones where number of heads equal number of tails.

A sequence of +EV bets can’t be -EV. You correctly calculated a single wager as a factor 1.015 increase in bankroll, so 10 wagers would be a (1.015)^10 = 1.16 factor increase.

You can alternatively calculate the EV doing a weighted average of all outcomes, which is more work than I’m willing to do on a phone calculator, but it’s easy to verify for two wagers.

.25*(1.2^2)+.25*(.83^2)+.5*(1.2*.83) = 1.030225= 1.015^2 as expected.

The stuff about rich vs poor people seems like nonsense, although there are other reasons having wealth makes it easier to obtain more.

Edit: I haven’t read the article. Just going off of what you posted. Are you sure the article isn’t talking about expected median bankroll growth rather EV? It’s well know that betting too large a fraction of your bankroll will lead to median bankroll decline. Look up the Kelley Criterion.

It's not 10 separate wagers. It's one wager.

The math doesn’t care about semantics.

You are, of course, correct. The point I was trying to make - which I did a poor job of with my quick reply - is that I think it would help the op if he thought about the scenario as only one bet.


Also, This is very important. I think the OP thinks that 5 heads and 5 tails is the average outcome. It would be a good exercise for the OP to figure out the result if there were 10 tails in a row and compare it to 10 heads in a row. He would never go completely broke, but the upside is pretty high. Just a small section of the possible results, but helps to get him thinking why 5 and 5 does not provide the average net gain/loss.

Okay. I hope I didn't seem too snarky. I think both ways of thinking have advantages for sure. You may be right that OP would be helped more by thinking of it as a single bet.

Quite a good discussion about this (called "volatility drag") in Red-Blooded Risk: The Secret History of Wall Street (Aaron Brown used to post here and has a couple of good posts about this too if you search his name [or look at the links here]).

Juk

This is so sneaky. The expected value is positive, but we're forced to bet just over twice the Kelly Criterion. In other words, the bet itself is +EV, however we're forced to "overbet our bankroll", which is losing long term.

The Expected Value is actually +16.05% overall. (For 10 flips)

The Kelly Criterion says we should risk 8%, but we're forced to risk 17%.
It's a well known fact that if you bet twice the Kelly amount, on average the geometric size of your bankroll will not grow at all, and anything larger than that will on average cause it to shrink.

https://www.lesswrong.com/posts/BZ6X...elly-criterion

Lastly, here's a spreadsheet that calculates the expected value across all outcomes for 10 flips.

https://docs.google.com/spreadsheets...t?usp=drivesdk

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How does this relate to poker?

This is actually important. It teaches us that overbetting our bankroll is losing long-term, even if we have an edge over our competition.

Yes but the simple reason is called Risk Of Ruin. It is a factor in all betting systems.

If you lost a million flips in a row playing this game, your bankroll will still be positive (assuming granular bets). So risk of ruin may be the wrong term here.

I think it's more related to volatility drag as the other guy pointed out.

Actually I defined volatility drag in the first response in this thread, using gambling terminology instead of investment terminology. But your statement above ignores the rules of the game. The volatility drag and the large RoR are both a result of the betting requirements. Proper betting, if allowed, would eliminate both.

Risk of Ruin usually means you can no longer continue to bet. That never happens in this game. Even if you lose all 10 flips you would still walk away with $15.50.

This is semantics though. The point is that overbetting your bankroll is a losing proposition even if you're making +EV bets.

Agreed, and it's possible I misunderstood the rules.

If we have to bet our entire net worth every time we can’t expect to make a profit from this game. That makes it a bad bet in a theoretical setting with very strict rules but that doesn’t mean it’s a bad bet for most people in practice due to the fact that the lost money is easily replenished if we don’t run above average. Our net worth isn’t effectively $100 if it can increase through other means.

I got a really good answer to this question on stackexchange that deserves to be shared here.

https://math.stackexchange.com/a/3523621/617737

I think what they are saying is when the number of heads exactly equals the number of tails , the player loses money. But if you remove that restriction, the player has +EV since the amount of money he makes when the number of H>T is much greater than the money he loses when the number of H=<T. For example, when he hits 10 heads wins in row he reaches $619, a large gain of $519, and if he loses 10 tails in a row he still will have $15.52, a minimal loss of $85. 6 Hs of 10 tosses gets him to $142, gain of $42, and 4 Hs of 10 tosses gets him to $68 , a loss of only $32. Since the casino cant really restrict the number of Heads and Tails the casino will lose in the long run.

What does it mean for poker players? I would speculate that it means that successful players make their profit by winning big pots when running hot and losing smaller amounts when on a losing hands streak or on a 50% hand winning session.