To say you would bet the whole $1k is absurd.
Agreed.We can argue about what constitutes risk and how to measure it, but it looks undeniable that the "optimal strategy" is a function of the risk tolerance.
To the chap who is asking if Kelly applies to a one spin scenario, the answer is yes.
Red-16 1-spin scenario has the odds stacked against it, 89.2% chances of losing your bet. You have to believe in Santa to make that *one* bet.
Kelly prescribes betting 8% of the bankroll on Red-16. 8% of $1000 is $80.
With that single $80 bet, you have an 89.2% probability of losing $80, and a 10.8% probability of making $2800 (80 * 35). Therefore, the expectancy of that single bet is:
0.892 * (-80) + 0.108 * (+2800) = $231.
QED.
close, reduces variance by 1 divided by root 2 = 71%![]()
Please, prove it.
Red-16 1-spin scenario has the odds stacked against it, 89.2% chances of losing your bet. You have to believe in Santa to make that *one* bet.
But if you do, then just bet whatever you are comfortable losing. This not a matter of long-term profit maximization.
One last comment as I have other things to do this week-end: whet makes you think that the odds of hitting Red-16 are best on the 1st spin than on the 2nd? Assuming no win, your 1st bet is $80, the second one $73, the 3rd one $67, etc. - this makes absolutely no sense, since all these spins have the same win probability.
Stop the dogma, and start thinking.
Hey, Visaria, can you point me to the way of calculating the relationship between Kelly and the standard deviation? What I am looking for is a way to maximize the (growthRate/stdev ) ratio. That is, the question is, by how much do I need to reduce Kelly so that I can maximize the quantity which I call the "risk-adjusted Kelly"? Thanks.