Kelly % = W – [(1 – W) / R]
Where:
W = Winning probability
R = Win/loss ratio
So if W = 0.505, R should be 101:99 ~= 1.02, and plugging the numbers in we get:
Kelly % = 0.505 – [(1 – 0.505) / 1.02] ~= 0.0197
i.e., the optimal amount to risk is about 2% of the bankroll. You are at 1%, or a half-Kelly. You are trading some growth for less volatility.
If someone thinks I've got this wrong I welcome the correction.
I think you've got this wrong...
You have to look at it from the house point of view.
Consider the 49.5% game. The house has a 50.5% chance of winning, or W = 0.505.
R is the amount the house wins when it wins (1 unit) divided by the amount it loses when it loses (1 unit), so R = 1/1 = 1
Kelly = W - (1-W)/R = 0.505 - (1 - 0.505)/1 = 0.505 - 0.495 = 0.01 = 1%.
Notice it's the same for all bets (funny that - evidence for intelligent design??):
>>> w = 0.02 (98% game)
>>> r = 1 / 0.01020408
>>> w - (1-w)/r
0.01
>>> w = 0.25 (75% game)
>>> r = 1 / 0.32
>>> w - (1-w)/r
0.01
>>> w = 0.5 (50% game)
>>> r = 1/0.98
>>> w - (1-w)/r
0.01
>>> w = 0.505 (49.5% game)
>>> r = 1/1
>>> w - (1-w)/r
0.01
>>> w = 0.75 (25% game)
>>> r = 1 / 2.96
>>> w - (1-w)/r
0.01
>>> w = 0.999999 (0.0001% game)
>>> r = 1 / 989999.0
>>> w - (1-w)/r
0.01
More generally, payout * chance = 1 - edge,
(1)W = house_chance = 1 - chance
(2)R = 1 / (payout-1)
(3)Kelly = W - (1-W)/R
= (1 - chance) - chance*(payout-1)
(using 2 and 3)= 1 - chance - payout*chance + chance
(expanding)= 1 - payout*chance
(cancel chance)= edge
(using 1)So the Kelly factor is the same as the house edge, whatever the edge.
I never worked that out like that before, but it's a nice result.
