Kill game avg big bet

[Chizoad]

I'm trying to do some accounting for my spreadsheet, and was curious if anyone has tried to figure out the average size of a big bet in a kill game. This was a 3-6 dealer's choice game, with very little holdem played, mostly omaha and pineapple.

My working guess is 1.33*BB.

Thanks.

[IsaacW]

First you'd need to estimate the probability that a particular pot is killed, let's denote that P(kill).

Assuming that the bets double when the pot is killed and that a particular hand being killed does not affect the probability that the next hand will kill, then the average big bet size (in terms of the normal big bet size BB) is:

BB*(1 - P(kill)) + 2*BB*P(kill)

The second assumption is a bit tenuous, because in some kill games the kill is based on the size of the pot, and obviously the pots are more likely to be over the kill amount if the bets are double-sized. If the kill game was the type that killed after the same person won two pots in a row, then I imagine the second assumption would be OK.

[jar]

[ QUOTE ]

BB*(1 - P(kill)) + 2*BB*P(kill)


[/ QUOTE ]
Wouldn't this be clearer written as
BB + BB*P(kill)?

If the rule is 2 hands in a row is a kill, P(kill) is 1/n in an n handed game, if you assume that all players are equally likely to win a hand. How close that is to true depends greatly on the players in a game. I'd guess the probability is higher than 1/n in a game with some loose players and some tight players. The loose players are much more likely to win consecutive hands, since they play more hands.

[IsaacW]

You are right, I never said I was any good at simplifying... the algebra is what gets me, rarely the complex stuff [img]/images/graemlins/grin.gif[/img]

I think empirical evidence would be the way to go for estimating P(kill). Of course, SSTS would be a problem...

[IsaacW]

You could write it even more simply as BB*(1 + P(kill)). Only took me a week!