boltonguy
Flush
There was a discussion on the Hustler Live thread about running it once or twice and if there was any impact. Last night on the Hustler Live stream, Randall Emmett (who I just cant stand BTW) made a comment as he kept jamming flush draws on flop that he was expecting to regularly chop and running it once breaks his game a bit. So I wanted to dig into this. I'm trying to do math below and would appreciate any feedback and/or correction.
Let's look at the choice between running it once or twice in a classic "coin flip" scenario: AKs all-in pre versus 22 (not shareing any of the same suits as AKs). Equilab has this match at 50.08% for AKs and 49.92% for 22, so let's round and go with 50/50.
So in a HUP game with stacks of 100BB, player 1 dealt AKs and player 2 dealt 22, all the money gets in pre. Player 1 RFI, player 2 jams and player 1 calls. Is there an impact on the outcome to running it once or twice? Let's look at this from player 1's perspective with no blinds just to make the math easier:
In this scenario, player 1 has called 100BB to win 100BB. She has 50% probability of winning 100BB and 50% probability of losing 100BB. So Ev = Pwin*Value = Plose*Value = 100*0.5 - 100*0.5 = 0. So it looks like in a coin flip situation the EV is 0.
If we diagram it out using H(eads) and T(ails):
H = win 100BB, Ev = 50BB, p = 50%
T = lose 100BB, Ev = -50BB, p = 50%
Ev = 0
When we run it an even number of times, such as twice, we introduce the possibility that the players will tie and chop the pot. This doesnt change the expected value but it does change the probability distribution of the outcomes. If we diagram again using H9eads) and T(ails):
H = HH, Player 1 wins both boards = 100BB, p = 25%
H<
T = HT, players chop = 0BB, p = 25%
H = TH, players chop = 0BB, p= 25%
T<
T = TT, Player 1 loses both boards = -100BB, p = 25%
The Ev in this case is the sum of all the expected values * their probabilities and is still equal to zero.
However, we see that the probability of winning the entire pot has decreased from 50% when running it once to 25% when running it twice; conversely, the probability of losing one's entire stack has also decreased from 50% to 25%. This is because in a coin flip scenario the players will chop 50% of the time.
So while the EV doesnt change, because we introduce the possibility of chopping, the probability of the outcomes does change (no we chop half the time) and we see a game where players can call flips (which is probably exciting for the audience) with less risk of losing their stack.
So if you are ahead in a non-flip situation, while agreeing to run it twice wont change the odds of specific cards coming, you will introduce the possibility of chopping and will decrease the probability that you will drag the whole pot. Thoughts?
Let's look at the choice between running it once or twice in a classic "coin flip" scenario: AKs all-in pre versus 22 (not shareing any of the same suits as AKs). Equilab has this match at 50.08% for AKs and 49.92% for 22, so let's round and go with 50/50.
So in a HUP game with stacks of 100BB, player 1 dealt AKs and player 2 dealt 22, all the money gets in pre. Player 1 RFI, player 2 jams and player 1 calls. Is there an impact on the outcome to running it once or twice? Let's look at this from player 1's perspective with no blinds just to make the math easier:
In this scenario, player 1 has called 100BB to win 100BB. She has 50% probability of winning 100BB and 50% probability of losing 100BB. So Ev = Pwin*Value = Plose*Value = 100*0.5 - 100*0.5 = 0. So it looks like in a coin flip situation the EV is 0.
If we diagram it out using H(eads) and T(ails):
H = win 100BB, Ev = 50BB, p = 50%
T = lose 100BB, Ev = -50BB, p = 50%
Ev = 0
When we run it an even number of times, such as twice, we introduce the possibility that the players will tie and chop the pot. This doesnt change the expected value but it does change the probability distribution of the outcomes. If we diagram again using H9eads) and T(ails):
H = HH, Player 1 wins both boards = 100BB, p = 25%
H<
T = HT, players chop = 0BB, p = 25%
H = TH, players chop = 0BB, p= 25%
T<
T = TT, Player 1 loses both boards = -100BB, p = 25%
The Ev in this case is the sum of all the expected values * their probabilities and is still equal to zero.
However, we see that the probability of winning the entire pot has decreased from 50% when running it once to 25% when running it twice; conversely, the probability of losing one's entire stack has also decreased from 50% to 25%. This is because in a coin flip scenario the players will chop 50% of the time.
So while the EV doesnt change, because we introduce the possibility of chopping, the probability of the outcomes does change (no we chop half the time) and we see a game where players can call flips (which is probably exciting for the audience) with less risk of losing their stack.
So if you are ahead in a non-flip situation, while agreeing to run it twice wont change the odds of specific cards coming, you will introduce the possibility of chopping and will decrease the probability that you will drag the whole pot. Thoughts?