My Journey As A Professional Poker Player

Anthony Martino

Straight Flush
Joined
Sep 26, 2015
Messages
7,580
Reaction score
12,345
Location
Tampa, FL
Oh, last nights mixed game wasn't as bad as last times $500 loss, but I did take a $70 loss

Mostly thanks to starting with 3 card dugis multiple times and getting outdrawn by guys drawing 3, lol

Overall happy with how I played, we had two tables again plus some players who filled open seats

I was smarter this time, left at 130am so I didn't get stuck in the 2am closing line of cashouts
 

Legend5555

Flush
Joined
Jul 15, 2019
Messages
2,039
Reaction score
2,718
Location
Georgia
For those who don't speak Mathhead, in probability, binomial means "with replacement" (i.e., you're reshuffling the used cards back into the stub between runouts) and hypergeometric means "without replacement" (i.e., the way RIT works in real life). The EV for both players is unchanged in both cases.

Using my 36:8 river example from earlier, the way I analyzed it was the normal way, without replacement (hypergeoemtric). Running it exactly twice:

P(Hero scoops) =
P(Hero wins #1) × P(Hero wins #2 | Hero wins #1) =
36/44 × 35/43 ≈ 66.6%

P(Hero splits) =
P(Hero wins #1) × P(Hero loses #2 | Hero wins #1) + P(Hero loses #1) × P(Hero wins #2 | Hero loses #1) =
36/44 × 8/43 + 8/44 × 36/43 ≈ 30.4%

Equity ≈ 66.6% + (1/2)30.4% = 81.8%

For those who wanted some kind of mathematical evidence that running it twice is EV-neutral, this is one such data point. That 81.8% is the same approximate probability of Hero winning on a single runout (36/44 ≈ 81.8%), and it comes out exactly the same if you use the precise values instead of rounding. If anyone really wants me to strip out the specific values and algebraically prove the general case, I can do that too, but the explicit numbers are usually more helpful for non-math folks.

Now the binomial (with replacement) way:

P(Hero scoops) =
P(Hero wins #1) × P(Hero wins #2) =
36/44 × 36/44 ≈ 66.9%

P(Hero splits) =
P(Hero wins #1) × P(Hero loses #2) + P(Hero loses #1) × P(Hero wins #2) =
36/44 × 8/44 + 8/44 × 36/44 ≈ 29.8%

Equity ≈ 66.9% + (1/2)29.8% = 81.8%

Switching between the two models does not change equity (and thus EV) at all. The only real effect is that Hero has a slightly greater chance of scooping and a slightly lower chance of splitting. On the other side, Villain has a slightly greater chance of splitting and a slightly lower chance of scooping. The favorite is more likely to scoop and less likely to split, and the underdog is less likely to scoop and more likely to split, but it all comes out in the wash.

The only difference is in variance, which is greater when you run it out multiple times with replacement than without. To see why informally, look at the extremes. With replacement, it becomes possible, though unlikely, for either player to win 44 consecutive runouts. Without replacement, it's fully impossible for Villain to win more than 8 out of 44 runouts (or for Hero to win more than 36 out of 44).
I have a math degree. This is all good and everything. But why the hell are we getting into this in this thread? Is leaving it at "running it multiple times doesn't change the EV," not enough?
 

Bluegrass Poker

3 of a Kind
Joined
Dec 21, 2016
Messages
674
Reaction score
1,281
Location
Lexington, KY
For those who don't speak Mathhead, in probability, binomial means "with replacement" (i.e., you're reshuffling the used cards back into the stub between runouts) and hypergeometric means "without replacement" (i.e., the way RIT works in real life). The EV for both players is unchanged in both cases.

Using my 36:8 river example from earlier, the way I analyzed it was the normal way, without replacement (hypergeoemtric). Running it exactly twice:

P(Hero scoops) =
P(Hero wins #1) × P(Hero wins #2 | Hero wins #1) =
36/44 × 35/43 ≈ 66.6%

P(Hero splits) =
P(Hero wins #1) × P(Hero loses #2 | Hero wins #1) + P(Hero loses #1) × P(Hero wins #2 | Hero loses #1) =
36/44 × 8/43 + 8/44 × 36/43 ≈ 30.4%

Equity ≈ 66.6% + (1/2)30.4% = 81.8%

For those who wanted some kind of mathematical evidence that running it twice is EV-neutral, this is one such data point. That 81.8% is the same approximate probability of Hero winning on a single runout (36/44 ≈ 81.8%), and it comes out exactly the same if you use the precise values instead of rounding. If anyone really wants me to strip out the specific values and algebraically prove the general case, I can do that too, but the explicit numbers are usually more helpful for non-math folks.

Now the binomial (with replacement) way:

P(Hero scoops) =
P(Hero wins #1) × P(Hero wins #2) =
36/44 × 36/44 ≈ 66.9%

P(Hero splits) =
P(Hero wins #1) × P(Hero loses #2) + P(Hero loses #1) × P(Hero wins #2) =
36/44 × 8/44 + 8/44 × 36/44 ≈ 29.8%

Equity ≈ 66.9% + (1/2)29.8% = 81.8%

Switching between the two models does not change equity (and thus EV) at all. The only real effect is that Hero has a slightly greater chance of scooping and a slightly lower chance of splitting. On the other side, Villain has a slightly greater chance of splitting and a slightly lower chance of scooping. The favorite is more likely to scoop and less likely to split, and the underdog is less likely to scoop and more likely to split, but it all comes out in the wash.

The only difference is in variance, which is greater when you run it out multiple times with replacement than without. To see why informally, look at the extremes. With replacement, it becomes possible, though unlikely, for either player to win 44 consecutive runouts. Without replacement, it's fully impossible for Villain to win more than 8 out of 44 runouts (or for Hero to win more than 36 out of 44).
1613062644848.jpeg


I did enjoy the explanation and it made sense to this history degree guy. LOL.
 

Jimulacrum

Flush
Joined
Nov 16, 2014
Messages
1,898
Reaction score
2,495
Location
Nah Hampshah
I have a math degree. This is all good and everything. But why the hell are we getting into this in this thread? Is leaving it at "running it multiple times doesn't change the EV," not enough?
My last post was because @Rakrul had inquired about whether anything changes between the with/without replacement approaches. I had to indulge. Topics like this (i.e., analyzing optional bets that may or may not be fair to all participants) are interesting to me from a gamesmanship perspective. Hell, I published a whole book analyzing the effect of straddling on EV under different circumstances.

TLDR: Irrational compulsion.
 

Bluegrass Poker

3 of a Kind
Joined
Dec 21, 2016
Messages
674
Reaction score
1,281
Location
Lexington, KY
My last post was because @Rakrul had inquired about whether anything changes between the with/without replacement approaches. I had to indulge. Topics like this (i.e., analyzing optional bets that may or may not be fair to all participants) are interesting to me from a gamesmanship perspective. Hell, I published a whole book analyzing the effect of straddling on EV under different circumstances.

TLDR: Irrational compulsion.
What’s the book?
 

Anthony Martino

Straight Flush
Joined
Sep 26, 2015
Messages
7,580
Reaction score
12,345
Location
Tampa, FL
Lol, I have 8877 on a rainbow T76 flop and pot it, get called by J944 (hurr-durr) and he peels the 8 turn

I check behind but pay off the river bet when a 4 falls
 

Anthony Martino

Straight Flush
Joined
Sep 26, 2015
Messages
7,580
Reaction score
12,345
Location
Tampa, FL
Wish I had good news but stuck $1500 so far

Flopped quad kings but high hand was already a straight flush and 15 minutes left on clock

Next hand get it in with double suited AA64 and lose to JJTT when he flops a Jack

Shortly after have AAK2 double suited in a raised multiwaypot and see a J86 two of my suit flop, plus nut backdoor suit. Get it in 3 ways and whiff, up against JJxx again
 

Anthony Martino

Straight Flush
Joined
Sep 26, 2015
Messages
7,580
Reaction score
12,345
Location
Tampa, FL
Picked up suited aces again, got it in on flop behind 3 ways, turned nuts, river made me have to chop it but won some $$$ from 3rd guy in hand

Still stuck $1200
 
Top Bottom