Why Your f is optimal

From Stanford Wong's BJ21

Posted by MathProf on 15 Dec 1997, at 4:47 a.m., in response to Re: Utility functions and optimal f , posted by ML on 14 Dec 1997, at 1:39 p.m.

Mike:
In you little hypo, you assummed that there would be the expeceted number of Wins and Losses. For example, with p=.505 and q=.495, you assummed there would be 505 wins and 495 losses. Under these circumstances, the Kelly fraction will actually optimize the bankroll. (This is an easy Calculus exerice. It happens that, in this case, the quantity that you want to amxiimize is essentially your Expected Logarithm). However, you will will not always have the expected number of wins and losses.

However, if it is the case that (as assummed above) if we have the expeceted number of each outcome, then the kelly fraction will optimize the Final bankroll.

Let us suppose that, in your hypo, we ran into a round of good luck and won 51% of our games, lossing 49% of them. Then, had we been with a Fraction of 2, we would have the greatest final bank. Conversly, if we had run a string of bad luck and won only 50% of the time, then the fraction of "0" would give us our greatest exnding bankroll.

This can be generalized. Imagine 3 people betting on the same sequence of biased-coin tosses (or other game of chance). One is named K for Kelly, A for "aggresive" and T for timid. K bets strict Kelly, A bets with a higher fraction and T bets with a lower fraction.

Now if they run into the expected amount of luck, K will have a larger Bankroll than the otehrs at then end of the day. However if they were lucky enough, then A would outperform K. K will outpeform T. However this will happen less than 50% of the time (How much less depends on how aggressive A is annd on how manyt rials we have. If we have very many trials, the prob will be low that A will beat K.)

Now suppose that they run into a string of "Bad Luck". Then T will actually outperform K, who certainly outperforms A. Again this ahppens less than 50% of the time. So we have three cases: 


Case 1:  Good Luck:    A  beats K who beats T (<50% prob)

Case 2:  Normal Luck:  K  beats A and T

Case 3:  Bad Luck:     T  beats K who beats A (<50% prob.)

Note that K beats A in cases 2 and 3 (more than 50% of the time) and beats T in cases 1 and 2 (more than 50% of the time.)

So for any other Betting Fraction, K will outperform it more than 50% of the time.

Now this "more than 50%" is an understatement! If they are playing long enough, then Cases 1 and 3 become very unlikely, and K will outperform "most of the Time". As N increases to infinity, the Probability that K will approaches 1.

This is the basis of the statement (which is in Epstein, maybe due to Thorp?) that the Kelly fraction will outperform any other fraction most of the time.

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