You don’t get exactly 1⁄3 of a win with no variance in either case. You get exactly 1 win, 1⁄3 of the time, and no win 2⁄3 of the time.
As an example when betting on green, suppose there’s a 1⁄3 chance of 30 blue and 30 green balls, 1⁄3 chance of 60 green, 1⁄3 chance of 60 blue. And there’s always 30 red balls.
There is a 1⁄3 of 1⁄3 chance that there are 30 green balls and you pick one. There is a 2⁄3 of 1⁄3 chance that there are 60 green balls and you pick one. There is no chance that there are no green balls and you still pick one. Therre is no other way to get a green ball. The total chance of picking a green ball is therefore 1⁄3, that is, 1⁄3 of 1⁄3 plus 2⁄3 of 1⁄3. That means that 1⁄3 of the time you win and 2⁄3 of the time you lose, just as in the case of betting on the red ball.
A distribution of 1 one third of the time and 0 two thirds of the time has some computable variance. Whatever it is, that’s the variance in your number of wins when you bet on green, and it’s also the variance in you number of wins when you bet red.
Like I said below, write out the actual random variables you use as a Bayesian: they have identical distributions if the mean of your green:blue prior is 30 to 30.
There is literally no sane justification for the “paradox” other than updating on the problem statement to have an unbalanced posterior estimate of green vs. blue.
Bayesian reasoning is for maximizing the probability of being right.
Kelly´s criterion is for maximizing aggregated value.
And yet again, the distributions of the probabilities are different, because they have different variance, and difference in variance give different aggregated value, which is what people tend to try to optimize.
Aggregating value in this case is to get more pies, and fewer boots to the head. Pies are of no value to you when you are dead from boots to the head, and this is the root cause for preferring lower variance.
This isn´t much of a discussion when you just ignore and deny my argument instead of trying to understand it.
If I decide whether you win or lose by drawing a random number from 1 to 60 in a symmetric fashion, then rolling a 60-sided die and comparing the result to the number I drew, this is the same random variable as a single fair coinflip. Unless you are playing multiple times (in which case you’ll experience higher variance from the correlation) or you have a reason to suspect an asymmetric probability distribution of green vs. blue, the two gambles will have the exact same effect in your utility function.
The above paragraph is mathematically rigorous. You should not disagree unless you find a mathematical error.
And yet again I am reminded why I do not frequent this supposedly rational forum more.
Rationality swishes by over most peoples head here, except for a few really smart ones.
You people make it too complicated. You write too much. Lots of these supposedly deep intellectual problems have quite simple answers, such as this Ellsberg paradox. You just have to look and think a little outside their boxes to solve them, or see that they are unsolvable, or that they are wrong questions.
I will yet again go away, to solve more useful and interesting problems on my own.
Oh, and Orthonormal, here is my correct final answer to you: You do not understand me, and this is your fault.
No, because expected value is not the same thing as variance.
Betting on red gives 1⁄3 winnings, exactly.
Betting on green gives 1⁄3 +/- x winnings, and this is a variance, which is bad.
You don’t get exactly 1⁄3 of a win with no variance in either case. You get exactly 1 win, 1⁄3 of the time, and no win 2⁄3 of the time.
As an example when betting on green, suppose there’s a 1⁄3 chance of 30 blue and 30 green balls, 1⁄3 chance of 60 green, 1⁄3 chance of 60 blue. And there’s always 30 red balls.
There is a 1⁄3 of 1⁄3 chance that there are 30 green balls and you pick one. There is a 2⁄3 of 1⁄3 chance that there are 60 green balls and you pick one. There is no chance that there are no green balls and you still pick one. Therre is no other way to get a green ball. The total chance of picking a green ball is therefore 1⁄3, that is, 1⁄3 of 1⁄3 plus 2⁄3 of 1⁄3. That means that 1⁄3 of the time you win and 2⁄3 of the time you lose, just as in the case of betting on the red ball.
A distribution of 1 one third of the time and 0 two thirds of the time has some computable variance. Whatever it is, that’s the variance in your number of wins when you bet on green, and it’s also the variance in you number of wins when you bet red.
Like I said below, write out the actual random variables you use as a Bayesian: they have identical distributions if the mean of your green:blue prior is 30 to 30.
There is literally no sane justification for the “paradox” other than updating on the problem statement to have an unbalanced posterior estimate of green vs. blue.
Bayesian reasoning is for maximizing the probability of being right. Kelly´s criterion is for maximizing aggregated value.
And yet again, the distributions of the probabilities are different, because they have different variance, and difference in variance give different aggregated value, which is what people tend to try to optimize.
Aggregating value in this case is to get more pies, and fewer boots to the head. Pies are of no value to you when you are dead from boots to the head, and this is the root cause for preferring lower variance.
This isn´t much of a discussion when you just ignore and deny my argument instead of trying to understand it.
If I decide whether you win or lose by drawing a random number from 1 to 60 in a symmetric fashion, then rolling a 60-sided die and comparing the result to the number I drew, this is the same random variable as a single fair coinflip. Unless you are playing multiple times (in which case you’ll experience higher variance from the correlation) or you have a reason to suspect an asymmetric probability distribution of green vs. blue, the two gambles will have the exact same effect in your utility function.
The above paragraph is mathematically rigorous. You should not disagree unless you find a mathematical error.
And yet again I am reminded why I do not frequent this supposedly rational forum more. Rationality swishes by over most peoples head here, except for a few really smart ones. You people make it too complicated. You write too much. Lots of these supposedly deep intellectual problems have quite simple answers, such as this Ellsberg paradox. You just have to look and think a little outside their boxes to solve them, or see that they are unsolvable, or that they are wrong questions.
I will yet again go away, to solve more useful and interesting problems on my own.
Oh, and Orthonormal, here is my correct final answer to you: You do not understand me, and this is your fault.