London Colin said:
So my description of Kelly as EV/Payoff would only hold if we were talking about the EV in those terms. i.e. ignore pushes.
Does that all make sense?
Hi Colin,
It makes perfect sense, you can ignore pushes if you rescale the probabilities correspondingly. I.e. p(Win) -> p(Win) / (1 - p(Push)), and p(Loss) -> p(Loss) / (1 - p(Push)).
Then you indeed have a Kelly bet for the simplest 3-outcome bet (Win/push/loss).
The reason is, Kelly betting maximizes log-utility bankroll estimate (that is best estimated bankroll growth).
For a 2-outcome bet it is
maximize p*log(1 + f*b) + q * log(1 - f) for f
with the Kelly result
f = (b p - q)/b (with p+q=1)
If there is a push, and p is the probability to win, q to lose (and p+q<1), the Kelly criterion will be (see WolframAlpha result)
f = (b p - q)/(b p + b q)
which is elimininating the push by conditional (no push) probabilities p -> p/(p+q) and q -> q/(p+q).
Hence, the generalized Kelly criterion for a 3-outcome (again pushes) bet is simple, where p1,p2,p3 are the probabilities for result 1,2,3. and b1,b2,b3 are the payouts (p1+p2+p3=1)
maximize 1*log(1 + f*b1) + p2*log(1 + f*b2) +p3*log(1 + f*b3) for f
The Kelly-equivalent result will be
f = (p1 b1 + p2 b2 + p3 b3 - p2 b1 - 1) / (p1 b2 b3 + b1 p2 b3 + b1 b2 p3 - 1)
For an N-outcome bet, we could then assume it is
f = (p1 b1 + p2 b2 + ... + pN bN - 1) / (b1*b2*...*bN * (p1/b1 + p2/b2 + ... + pN/bN) - 1)
(without proof)