Kelly criterion for 3 outcomes I have been exploring the Kelly criterion for optimizing the bet size for a two outcome bet situation. I'm having trouble applying this to a three outcome bet.
I may refer to this excellent thread: Kelly criterion with more than two outcomes, answered by David Speyer. The thing that troubles me is that the result of his answer is a single number for all three bets combined. In my mind, there should be three different "optimal" bet sizes, depending which outcome you bet on.
Say the odds for a match is the following:
Team1 wins, 1.54 - probability 65% 
Team2 wins, 4.00 - probability 25% 
Draw, 10.00 - probability 10%
You can only bet on one outcome, and there is only one of the outcomes that happens.
Explanation much apreciated. Thanks!
 A: You are correct that there are three different optimal bet sizes. This is due to the requirement that the gambler commits to a specific outcome of the team game.
Three are three outcomes to the team game:


*

*Team1 wins

*Team2 wins

*Draw


However, for any of those outcomes, there are only two outcomes to the gambler's bankroll:


*

*The gambler bet on the correct outcome of the team game


*

*The gambler's bankroll increases by the net odds received on the wager


*The gambler bet on the wrong outcome of the team game


*

*The gambler's bankroll decreases by the amount wagered



The Kelly criterion is only concerned with the outcomes to the gambler's bankroll. Since the gambler is required to commit to a choice on the outcome of the team game, and this choice affects the outcomes to the gambler's bankroll, this choice must be determined before the evaluation of the Kelly criterion.
There are thus three optimal bet sizes, each dependent on the gambler's respective commitment to the outcome of the team game:


*

*Team1 wins


*

*Net odds received on wager: $0.54$

*Probability of winning: $0.65$

*Probability of losing: $0.25+0.10 = 0.35$

*Fraction of the current bankroll to wager:
$$f^* = \frac{(0.54)(0.65)-(0.35)}{(0.54)} \approx 0.00185$$


*Team2 wins


*

*Net odds received on wager: $3.00$

*Probability of winning: $0.25$

*Probability of losing: $0.65+0.10 = 0.75$

*Fraction of the current bankroll to wager:
$$f^* = \frac{(3.00)(0.25)-(0.75)}{(3.00)} = 0.00$$


*Draw


*

*Net odds received on wager: $9.00$

*Probability of winning: $0.10$

*Probability of losing: $0.65+0.25 = 0.90$

*Fraction of the current bankroll to wager:
$$f^* = \frac{(9.00)(0.10)-(0.90)}{(9.00)} = 0.00$$




In the question "Kelly criterion with more than two outcomes" (where a colored jelly bean is grabbed at random from a bag of 10 colored jelly beans), there are three possible outcomes in the gamble:


*

*Black Jelly Bean: no payout (i.e. simply lose wagered amount)

*Blue Jelly Bean: net odds received on the wager = $10$

*Red Jelly Bean: net odds received on the wager = $30$


Each of these outcomes affect the gambler's bankroll in a distinct way from the other outcomes. However, since no choice is involved in this gamble, there is only a single optimal bet size, that is determined by the three possible outcomes to the gambler's bankroll.
This is the difference between your team game example and the jelly bean bag example: choice. A choice must be committed before the Kelly criterion can be determined.
