B
BenAWhile
Rising Star
Bronze Level
What size bankroll do you need to play a break-even gambling game with no rake?
Let's play a simple game. I'll flip a (fair) coin and you guess if it comes up heads or tails. If you're right, you win a dollar. If you're wrong, you lose a dollar.
The expected value for this game is zero. In the long run, you will break even...
...but...
...is breaking even the most likely outcome?
No. I'm about to show you that the more times you play this game, the LESS likely it is that you will break even.
Let's say you play the game twice. Here are the four possible outcomes:
1) You win, then win again. (Up $2)
2) You win, then lose. (Break even)
3) You lose, then win. (Break even)
4) You lose, then lose again. (Down $2)
If you play twice, there is a 50% chance you'll break even (two out of four outcomes). So what happens if you play four times?
I won't list all 16 possible outcomes, but of those 16, only SIX of them consist of two wins and two losses (the break-even results). Therefore, you break even only 6 of 16 times, or 37.5% of the time.
Here's how the math works:
Play 2 times, 50% chance to break even (1 win, 1 loss).
Play 4 times, 37.5% chance of 2 wins, 2 losses.
Play 8 times, approx. 27.3% chance of 4 wins, 4 losses.
Play 16 times, approx. 19.6% chance of 8 wins, 8 losses.
Play 32 times, approx. 14.0% chance of 16 wins, 16 losses.
Play 100 times, approx. 8.0% chance of 50 wins, 50 losses.
Play 1,000 times, approx. 2.5% chance of 500 wins, 500 losses.
You can see the pattern here. Your chance of getting exactly as many wins as losses, which is the requirement to break even, goes down as the number of times you play goes up. Achieving the expected value of zero becomes LESS likely the MORE you play the game because it's not the expected value that determines your results here, it's the variance (or its square root, the standard deviation).
So... what's the most you could reasonably expect to ever be up (or down) in this game?
That math puzzle is much harder to solve, and I didn't solve it. Comtet and Majumdar did in 2005 (https://arxiv.org/pdf/cond-mat/0506195.pdf) based on the work of many people before them. Thanks to their work, here's the approximation formula for any large (at least 50) number of times you play this exact game:
N = Number of times you play the coin-flip game
sqrt = square root of
Approximate expected value of the most you can be up or down in N plays is...
sqrt(2 * N / pi) - 0.5 + (1 / sqrt(N))
(Warning: the "-0.5" part is what makes the math difficult, because that number only fits cases where you win $1 or lose $1 exactly 50% of the time. These two mathematicians proved that the middle term, the "-0.5" part, needs to be recalculated for every different variation of this game, including if you change the odds of winning/losing or the amount you win/lose.)
The standard deviation, however, is always sqrt(N), which means that 95% of the time, the most you will ever be ahead or behind in this game will be the expected value described above plus roughly twice the square root of the number of times you play it.
Real examples:
If you play 100 times, it is 95% likely you will be up or down by as much as $28 at some point.
If you play 1,000 times, it is 95% likely you will be up or down by as much as $88 at some point.
If you play 10,000 times... $280.
If you play 100,000 times... $884.
If you play 1,000,000 times... $2,797.
So if you play the coin-flip game a million times, you'd better have a bankroll of at least $2,800 to reduce the risk of going broke to less than 5%.
And that's if you play a break-even game that has minimal variance and no rake! Crazy, huh?
My name is Tarl Kudrick, and I hope you found this post interesting.
Let's play a simple game. I'll flip a (fair) coin and you guess if it comes up heads or tails. If you're right, you win a dollar. If you're wrong, you lose a dollar.
The expected value for this game is zero. In the long run, you will break even...
...but...
...is breaking even the most likely outcome?
No. I'm about to show you that the more times you play this game, the LESS likely it is that you will break even.
Let's say you play the game twice. Here are the four possible outcomes:
1) You win, then win again. (Up $2)
2) You win, then lose. (Break even)
3) You lose, then win. (Break even)
4) You lose, then lose again. (Down $2)
If you play twice, there is a 50% chance you'll break even (two out of four outcomes). So what happens if you play four times?
I won't list all 16 possible outcomes, but of those 16, only SIX of them consist of two wins and two losses (the break-even results). Therefore, you break even only 6 of 16 times, or 37.5% of the time.
Here's how the math works:
Play 2 times, 50% chance to break even (1 win, 1 loss).
Play 4 times, 37.5% chance of 2 wins, 2 losses.
Play 8 times, approx. 27.3% chance of 4 wins, 4 losses.
Play 16 times, approx. 19.6% chance of 8 wins, 8 losses.
Play 32 times, approx. 14.0% chance of 16 wins, 16 losses.
Play 100 times, approx. 8.0% chance of 50 wins, 50 losses.
Play 1,000 times, approx. 2.5% chance of 500 wins, 500 losses.
You can see the pattern here. Your chance of getting exactly as many wins as losses, which is the requirement to break even, goes down as the number of times you play goes up. Achieving the expected value of zero becomes LESS likely the MORE you play the game because it's not the expected value that determines your results here, it's the variance (or its square root, the standard deviation).
So... what's the most you could reasonably expect to ever be up (or down) in this game?
That math puzzle is much harder to solve, and I didn't solve it. Comtet and Majumdar did in 2005 (https://arxiv.org/pdf/cond-mat/0506195.pdf) based on the work of many people before them. Thanks to their work, here's the approximation formula for any large (at least 50) number of times you play this exact game:
N = Number of times you play the coin-flip game
sqrt = square root of
Approximate expected value of the most you can be up or down in N plays is...
sqrt(2 * N / pi) - 0.5 + (1 / sqrt(N))
(Warning: the "-0.5" part is what makes the math difficult, because that number only fits cases where you win $1 or lose $1 exactly 50% of the time. These two mathematicians proved that the middle term, the "-0.5" part, needs to be recalculated for every different variation of this game, including if you change the odds of winning/losing or the amount you win/lose.)
The standard deviation, however, is always sqrt(N), which means that 95% of the time, the most you will ever be ahead or behind in this game will be the expected value described above plus roughly twice the square root of the number of times you play it.
Real examples:
If you play 100 times, it is 95% likely you will be up or down by as much as $28 at some point.
If you play 1,000 times, it is 95% likely you will be up or down by as much as $88 at some point.
If you play 10,000 times... $280.
If you play 100,000 times... $884.
If you play 1,000,000 times... $2,797.
So if you play the coin-flip game a million times, you'd better have a bankroll of at least $2,800 to reduce the risk of going broke to less than 5%.
And that's if you play a break-even game that has minimal variance and no rake! Crazy, huh?
My name is Tarl Kudrick, and I hope you found this post interesting.
