[bryanlarsen]

If there's a highway made of pennies that is being maintained by a steamroller, you can make a very consistent but small profit by picking up pennies every day. Except for that one day that you get run over by a steamroller.

A better example is imagine that every day you bet on coin flips. Every day you go home after you're up $1. You start off with a $1 bet, and double the bet every time you lose. So for example one day, you might lose $1, then lose $2, then win $4 for a total profit of $1.

It's a foolproof strategy! You win $1 every day and you can't lose. But of course it assumes that both you and the house have an infinite stake. But if not, one day the house flips 16 heads in a row and you don't have $65536 to bet a 17th time so you go home down $65535. Kind of makes your $1 daily wins look pretty stupid now, doesn't it?

There are tons of options plays available on the stock market that have a risk profile similar to the coin flip example.

[slv77]

This is called a martingale bet (https://en.m.wikipedia.org/wiki/Martingale_(betting_system)). Doubling down on losses would be an example of a martingale bet.

[jiscariot]

I think this is also a reason why tables generally have set limits.

[confidantlake]

I don't think so. If you as the player make bets a series of bets, each with a negative expected value then your total expected value will also be negative. It doesn't matter if you double after every loss.

The limits are mostly because the casino can't afford to take on a 20 billion dollar bet from someone like Bezos. Even if it has a positive expected value, they will still go broke the 49% of the time they lose it.

[j7ake]

The martingale strategy works because this theoretical gambler has infinitely deep pockets to withstand the losses.

Setting bet limits can reduce the effectiveness of the martingale strategy.

[confidantlake]

I don't think it matters, even with infinite pockets. Let's say 3 is the max number of losses we will accept. 50-50 coin flip, start of betting a dollar. 7 out of 8 times we win a dollar. 1 out of 8 time we lose a dollar, double lose 2 dollars, double, lose 4 dollars quit.

(7/8)(1) + (1/8)(-7) = 0

Generally: n = number of losses before quit.

E(x) = (1-.5^n)(1) + (.5^n)(-2^n+1)

     = (1) - (1/2^n) - (2^n-1)/(2^n)
     

     = (2^n)/(2^n) - 1/(2^n) - (2^n+1) / (2^n)
     
     = (2^n -1 - 2^n + 1) / (2^n)
     
     = 0

So with a 50-50 our expected value is 0 even with an infinite bankroll. Which makes sense, there is no way to transform a series of neutral or negative expected value bets into a positive expected bet by combining them.

[slv77]

With an infinite bankroll there is no reason to stop at three losses. In an infinite series of fair coin flips there will be deviations in the distribution around the mean. A gambler can simply choose to stop at any point of excess “wins” as long as he isn’t stopped out due to exceeding his bankroll.

[j7ake]

The crazy thing about infinite pockets is that the strategy will favour the gambler even if the odds are 1-99 (or any nonzero probability of winning).

Using expected value for assess winning strategies don’t work with infinite bankroll.