If you bet more than Kelly, you’ll experience lower average returns and higher variance.
No. As they discovered in the dialog, average returns is maximized by going all-in on every bet with positive EV. It is typical returns that will be lower if you don’t bet Kelly.
The reason I brought this up, which may have seemed nitpicky, is that I think this undercuts your argument for sub-Kelly betting. When people say that variance is bad, they mean that because of diminishing marginal returns, lower variance is better when the mean stays the same. Geometric mean is already the expectation of a function that gets diminishing marginal returns, and when it’s geometric mean that stays fixed, lower variance is better if your marginal returns diminish even more than that. Do they? Perhaps, but it’s not obvious. And if your marginal returns diminish but less than for log, then higher variance is better. I don’t think any of median, mode, or looking at which thing more often gets a higher value are the sorts of things that it makes sense to talk about trading off against lowering variance either. You really want mean for that.
The reason why variance matters is that high variance increases your odds of going broke. In reality, gamblers don’t simply get to reinvest all of their money. They have to take money out for expenses. That process means that you can go broke in the short run, despite having a great long-term strategy.
Therefore instead of just looking at long-term returns you should also look at things like, “What are my returns after 100 trials if I’m unlucky enough to be at the 20th percentile?” There are a number of ways to calculate that. The simplest is to say that if p is your probability of winning, the expected number of times you’ll win is 100p. The variance in a single trial is p(1-p). And therefore the variance of 100 trials is 100p(1-p). Your standard deviation in wins is the square root, or 10sqrt(p(1-p)). From the central limit theorem, at the 20th percentile you’ll therefore win roughly 100p − 8.5sqrt(p(1-p)) times. Divide this by 100 to get the proportion q that you won. Your ideal strategy on this metric will be Kelly with p replaced by that q. This will always be less than Kelly. Then you can apply that to figure out what rate of return you’d be worrying about if you were that unlucky.
Any individual gambler should play around with these numbers. Base it on your bankroll, what you’re comfortable with losing, how frequent and risky your bets are, and so on. It takes work to figure out your risk profile. Most will decide on something less than Kelly.
Of course if your risk profile is dominated by the pleasure of the adrenaline from knowing that you could go broke, then you might think differently. But professional gamblers who think that way generally don’t remain professional gamblers over the long haul.
(Variance is “expected squared difference between observation and its prior expected value”, i.e. variance as a concept is closely linked to the mean and not so closely linked to the median or mode. So if you’re talking about “average” and “variance” and the average you’re talking about isn’t the mean, I think at best you’re being very confusing, and possibly you’re doing something mathematically wrong.)
I’m sorry that you are confused. I promise that I really do understand the math.
In repeated addition of random variables, all of these have a close relationship. The sum is approximately normal. The normal distribution has identical mean, median, and mode. Therefore all three are the same.
What makes Kelly tick is that the log of net worth gives you repeated addition. So with high likelihood the log of your net worth is near the mean of an approximately normal distribution, and both median and mode are very close to that. But your net worth is the exponent of the log. That creates an asymmetry that moves the mean away from the median and mode. With high probability, you will do worse than the mean.
The comment about variance is separate. You actually have to work out the distribution of returns after, say 100 trials. And then calculate a variance from that. And it turns out that for any finite n, variance monotonically increases as you increase the proportion that you bet. With the least variance being 0 if you bet nothing, to being dominated by the small chance of winning all of them if you bet everything.
I think the disagreement here is on what “average” means. All-in maximises the arithmetic average return. Kelly maximises the geometric average. Which average is more relevant is equivalent to the Kelly debate though, so hard to say much more
No. As they discovered in the dialog, average returns is maximized by going all-in on every bet with positive EV. It is typical returns that will be lower if you don’t bet Kelly.
Dang it. I meant to write that as,
That said, both median and mode are valid averages, and Kelly wins both.
The reason I brought this up, which may have seemed nitpicky, is that I think this undercuts your argument for sub-Kelly betting. When people say that variance is bad, they mean that because of diminishing marginal returns, lower variance is better when the mean stays the same. Geometric mean is already the expectation of a function that gets diminishing marginal returns, and when it’s geometric mean that stays fixed, lower variance is better if your marginal returns diminish even more than that. Do they? Perhaps, but it’s not obvious. And if your marginal returns diminish but less than for log, then higher variance is better. I don’t think any of median, mode, or looking at which thing more often gets a higher value are the sorts of things that it makes sense to talk about trading off against lowering variance either. You really want mean for that.
The reason why variance matters is that high variance increases your odds of going broke. In reality, gamblers don’t simply get to reinvest all of their money. They have to take money out for expenses. That process means that you can go broke in the short run, despite having a great long-term strategy.
Therefore instead of just looking at long-term returns you should also look at things like, “What are my returns after 100 trials if I’m unlucky enough to be at the 20th percentile?” There are a number of ways to calculate that. The simplest is to say that if p is your probability of winning, the expected number of times you’ll win is 100p. The variance in a single trial is p(1-p). And therefore the variance of 100 trials is 100p(1-p). Your standard deviation in wins is the square root, or 10sqrt(p(1-p)). From the central limit theorem, at the 20th percentile you’ll therefore win roughly 100p − 8.5sqrt(p(1-p)) times. Divide this by 100 to get the proportion q that you won. Your ideal strategy on this metric will be Kelly with p replaced by that q. This will always be less than Kelly. Then you can apply that to figure out what rate of return you’d be worrying about if you were that unlucky.
Any individual gambler should play around with these numbers. Base it on your bankroll, what you’re comfortable with losing, how frequent and risky your bets are, and so on. It takes work to figure out your risk profile. Most will decide on something less than Kelly.
Of course if your risk profile is dominated by the pleasure of the adrenaline from knowing that you could go broke, then you might think differently. But professional gamblers who think that way generally don’t remain professional gamblers over the long haul.
(Variance is “expected squared difference between observation and its prior expected value”, i.e. variance as a concept is closely linked to the mean and not so closely linked to the median or mode. So if you’re talking about “average” and “variance” and the average you’re talking about isn’t the mean, I think at best you’re being very confusing, and possibly you’re doing something mathematically wrong.)
I’m sorry that you are confused. I promise that I really do understand the math.
In repeated addition of random variables, all of these have a close relationship. The sum is approximately normal. The normal distribution has identical mean, median, and mode. Therefore all three are the same.
What makes Kelly tick is that the log of net worth gives you repeated addition. So with high likelihood the log of your net worth is near the mean of an approximately normal distribution, and both median and mode are very close to that. But your net worth is the exponent of the log. That creates an asymmetry that moves the mean away from the median and mode. With high probability, you will do worse than the mean.
The comment about variance is separate. You actually have to work out the distribution of returns after, say 100 trials. And then calculate a variance from that. And it turns out that for any finite n, variance monotonically increases as you increase the proportion that you bet. With the least variance being 0 if you bet nothing, to being dominated by the small chance of winning all of them if you bet everything.
I think the disagreement here is on what “average” means. All-in maximises the arithmetic average return. Kelly maximises the geometric average. Which average is more relevant is equivalent to the Kelly debate though, so hard to say much more