|
|
#1
06-29-2005, 05:47 AM
|
|||
|
|||
Odds vs. Bankroll.
Okay, I can't remember where it was from, but somewhere I remember hearing something like "By giving up a 1% edge, even the richest man in the world will go broke."
This, however, must be assuming that "the richest man in the world" and his opponent are playing off identical bankrolls, which by the virtue of the richest man in the world being the richest friggin' man in the world, can't be. If this were true, then Doyle and Co. would've had no objections to play Andy Beal for whatever stakes he chose. Surely, the pros would have at least a 1% advantage over Beal, but by exposing themselves to limits above their bankroll, they exposed themselves to a higher risk of ruin. I'm sure all you guys understand this better than I do, but it leads me to my question. At what point does your bankroll become more important than your edge? For example, lets say two guys are playing a series of heads up matches. However, there are restrictions. Player 1 has 600 in chips. Player 2 has 2400 in chips. Betting is not allowed. Both players have to post the 60 chip ante and then turn over their hands and have it played out face up. The other stipulation is that both players are dealt the same hands every time. Player 1 is always dealt 7c7h (56%)_ Player 2 is always dealt Ac9h (44%). The players play until one player has all the chips. What percentage of these matches will Player 2 win? What will the standard deviation be per 100 games? If you can please show the calculation involved in solving this kind of problem, I will be very grateful as I really want to know how to do this. 
|
|
#2
06-29-2005, 09:03 AM
|
|||
|
|||
|
Re: Odds vs. Bankroll.
When a constant amount is wagered at each step, and you have x units and your opponent has y units, and you win with probability p, your probability of winning the freezeout is
1-c^x --------- 1-c^(x+y) where c = (1-p)/p. When p=1/2, this degenerates, but the limit is the correct value x/(x+y). [ QUOTE ] Player 1 has 600 in chips. Player 2 has 2400 in chips. Betting is not allowed. Both players have to post the 60 chip ante and then turn over their hands and have it played out face up. The other stipulation is that both players are dealt the same hands every time. Player 1 is always dealt 7c7h (56%)_ Player 2 is always dealt Ac9h (44%). [/ QUOTE ] x=10. y=40. c=749550/955129=.784763 (throwing out the ties). 1-c^10 ------ = .911415 1-c^50 That's the probability player 1 wins, despite starting with only 20% of the chips. Here are the probabilities of winning with other numbers of chips out of 50: 0: 0 1: .215 2: .384 3: .517 4: .621 5: .702 10: .911 15: .974 20: .992 25: .997669 30: .999310 40: .999944 49: .99999850 50: 1
|
|
#3
06-29-2005, 02:21 PM
|
|||
|
|||
|
Re: Odds vs. Bankroll.
Here is a link to the derivation of the formula that pzhon is using. This is the same formula in a different form. I agree with his result.
[ QUOTE ] What will the standard deviation be per 100 games? [/ QUOTE ] Since his probability of winning a match is 91.1%, his standard deviation for 100 matches is sqrt[100*0.911*(1-0.911)] =~ 2.8 matches.
|
|
#5
06-29-2005, 05:21 PM
|
|||
|
|||
|
Re: Odds vs. Bankroll.
[ QUOTE ]
I missed something. Where did you get c=749550/955129 ? I thought it would have been .56/.44 or something like that. Am I blind, or just stupid? ;-) [/ QUOTE ] He's analyzing Ac 9h vs. 7c 7h. If he won exactly 56% and lost exactly 44%, then it would be .44/.56. Note that c is only (1-p)/p if there are no ties. In general, it is P(loss)/P(win). http://twodimes.net/h/?z=1057453 pokenum -h ac 9h - 7c 7h Holdem Hi: 1712304 enumerated boards cards win %win lose %lose tie %tie EV Ac 9h 749550 43.77 955129 55.78 7625 0.45 0.440 7c 7h 955129 55.78 749550 43.77 7625 0.45 0.560
|
|
#7
06-29-2005, 06:43 PM
|
|||
|
|||
|
Re: Odds vs. Bankroll.
If you use Enumerate All, PokerStove does not estimate. It calculates the exact equity. In fact, PokerStove gives the exact same numbers at twodimes.
- Andrew www.pokerstove.com
|
 |
|
|
Linear Mode
