# Nontransitive dice – how to win every time

You can still lose. I remember seeing a proof back in college that if your opponent has an unlimited bankroll, they will eventually bankrupt you - *even if the odds are in your favor.* Eventually you will hit a losing streak long enough to wipe you out.

I think our professor called it “the Gambler’s Nightmare.”

I’m guessing bankroll is limited by the available money in the world economy. Any mathematicians want to figure out how big of a starting pool you’d need to reliably scam the entire planet?

That’s the one. Way counterintuitive!

EDIT: To have 100% certainty, the casino needs an infinite bankroll - but as an approximation, very large numbers, such as Las Vegas, will bankrupt small numbers, such as me.

Yeah, “double your last bet” is a system that always seems enticing to people when they first think or it or hear of it, but it is literally no better than any other system, so long as you’re stuck with boring-old finite amounts of money, as we unfortunately are in the physical universe.

The usual system (there are plenty of variations, but they’re all the same under the hood) is simply to double your last bet if you lose. So you lose a dollar, bet two. You lose that, you’ve lost three total, so bet four. You lose that so bet eight. Now if you finally win, you’ve won a dollar.

The belief of that gambler is that they *must* win eventually, their streak of losing can only go on so long, so by sticking to this system they are guaranteed to always win.

The thing is, no matter how large your initial bankroll, you will *always* just break even in a 50-50 game, and lose when the odds are not in your favor. If your initial bankroll is a million dollars, then (on average) one in a million times you will lose 20 coin flips in a row, and you’ll lose every dollar you won on the other 999,999 flips. Furthermore, your odds are exactly the same that the 20-in-a-row losses will happen before you’ve won a million dollars (potentially landing you hundreds of thousands of dollars in debt) as afterwards (landing you with a profit).

Your Mean Expected Value of playing the game is exactly the same whether or not you decide to use the system.

(The other thing to think about is that the amount you’re betting each time goes up *really fast*, and yet you’re still betting to win that one dollar each time. So pretty soon you’re laying down thousand-dollar bets just for the chance to gain a dollar.)

Interestingly enough, even if your opponent can forbid you from choosing the die immediately to the left, for three of your opponent’s choices you can at least break even. If your opponent selects die 1, 2, or 4 and forbids you from choosing die 4, 1, or 3 then you should choose die 3, 4, or 2 respectively.

Choosing die 3 versus die 1 will make you win 5/9 of the time (unless die 1 rolls a 4 and die 3 rolls a 2) while the die 4 versus die 2 match is even odds (die 4 wins if it rolls a 5 and loses if it rolls a 1.)

If your opponent chooses die 3 and forbids you from choosing die 2, the best you can do is to choose die 1 and win 4/9 of the time instead of choosing die 4 and winning just 1/3 of the time. [Or if you’re just trying to avoid losing, choose a second copy of die 3. If you can do that you tie 5/9 of the time and win 2/9 of the time, meaning you lose only 2/9 of the time.]

I believe that experiment is under way. With sypathetic “investigators” and “enforcers”, you can keep gambling with other people’s money until there is no more money in the system. From what I hear, the experiment runs until a 6 degree Celsius world average temperature increase occurs, at which point the winners sink everything into oxygen futures, and the losers learn to hold their breath.

https://mathsgear.co.uk/products/non-transitive-grime-dice has a cute video with James Grime and Matt Parker, and you can find out about what happens if you have 2 of each colour

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