Quote:
Originally Posted by whosnext
Since I have a program to handle simple cases like this, here are the Expected Values and Median Values under the assumptions of Mr. Vince's example.
Suppose you are a gambler with a starting bankroll of 1. Suppose you are faced with a series of 10 independent binary bets each of which pays 2:1 and each of which wins 50% (and loses 50%). Suppose further that you must wager a fixed fraction of your then-current bankroll on each of the ten bets. Call this fixed fraction F.
As shown above, the Kelly fraction given by the formula (bp-q)/b becomes (2*.5-.5)/2 = .25 = 25%.
Here is the table of how the Expected Value and Median Value vary depending upon the Fixed Fraction chosen.
| Fixed Fraction | Expected Value | Median Value |
|---|
| 0% | 1.00 | 1.00 |
| 10% | 1.63 | 1.47 |
| 20% | 2.59 | 1.76 |
| 25% | 3.25 | 1.80 |
| 30% | 4.05 | 1.76 |
| 40% | 6.19 | 1.47 |
| 50% | 9.31 | 1.00 |
| 60% | 13.79 | 0.53 |
| 70% | 20.11 | 0.19 |
| 80% | 28.93 | 0.04 |
| 90% | 41.08 | 0.01 |
| 100% | 57.67 | 0.00 |
As in the previous simple numerical example posted above, EV is maximized at F=100% while Median is maximized at F=Kelly=25%.
Why does kelly not maximise median here?
Roll =R0*(1+k*2)^(n/2)*(1-k)^(n/2)
For any n (possibly only even), using calculus kelly will airways be the k to maximise growth of the median punter.
Or is this the wrong way to look at it for small n, because the cases where n is small and odd you wouldn't really get the median pinters return.
