How many times should I run it? (1 Viewer)

[timinater]

The breakdown of how often each scenario occurs within each option I guess would involve computing some kind of parlay of my ~2/3 chance of winning on each board. This is where my Statistics 101 knowledge breaks down... Someone? Anyone?

Your expected value is always the same the moment you are all-in, no matter how many times you run it. You're always a 65% favourite (roughly). When you run it more than once what changes is how likely you both are to realize your true equity in the hand.

Disclaimer: I'm no stats expert but I did pay attention to the Grade 12 math unit that focused on gambling

You are looking for the conditional probability - the chance of a certain number of successes in a sequence of dependent events.

Conditional probability is more work than I feel like doing - so how about the binomial probability? This isn't perfect because the trials aren't independent, we're of course not shuffling the turn and river cards back in. As the run outs come your equity in subsequent run outs would differ slightly, but at the time of the decision, none of the run outs have happened so for comparison sake I think it is close enough. I've used the raw equity and ignored ties.

If anything, the "Lose All" for running 2x+ is a touch low, given that if you lose one run out, you would be less likely to lose another once since the villain would have less outs on subsequent run outs.

Look at the last row - "Lose All" see how it is lower the more times you run it? That's the real key here. Running it 2x+ is just to reduce that risk of ruin (variance) that comes with poker. Hopefully this helps illustrate it for you.

Result	Run it 1x	Run it 2x	Run it 3x	Run it 4x
Win 1	65%	45%	24%	11%
Win 2	-	42%	44%	31%
Win 3	-	-	27%	38%
Win 4	-	-	-	17%
Lose All	35%	12%	4%	1.5%

I used this calculator if you want to fiddle with the numbers: https://www.omnicalculator.com/statistics/binomial-distribution

Edit: so If the question is how many times should I run it to approximate mine and the villains true equity depends on the equity.

1. If you are racing, run it twice
2. if you are 2:1 run it three times
3. If you are 3:1 + run it 4x, but better yet don't run it multiple times at all and scoop the whole pot

 

[Moxie Mike]

I'll always run it as many times as the other players wants. Depends on the texture of the game and less on the individual situations. I was in a game last week where pretty much every all in confrontation with cards to come was ran 3x. Just keep in mind it creates more work for the dealer so tip generously regardless of the runout.

As to your equity question, I've seen guys insist on seeing their opponents' holdings before deciding, which I feel is bad etiquette. The point in running it 3x with the nuts is to prevent getting scooped, which is very unlikely with 3 run-outs as a 2-1 fav. Conversely, when you're behind or suspect you're behind, it really comes down to how much you feel like gambling. Personally if I've got the worst of it in a bad spot (pair vs pair AIPF for example), I'd just as soon run it once and hope to suck out since there's virtually no way I can scoop all 3 run-outs.

So IMO, there's really no equity threshold where it becomes 'correct' to run it more than once.

 

[kirchhausen]

Your expected value is always the same the moment you are all-in, no matter how many times you run it. You're always a 65% favourite (roughly). When you run it more than once what changes is how likely you both are to realize your true equity in the hand.

Disclaimer: I'm no stats expert but I did pay attention to the Grade 12 math unit that focused on gambling

You are looking for the conditional probability - the chance of a certain number of successes in a sequence of dependent events.

Conditional probability is more work than I feel like doing - so how about the binomial probability? This isn't perfect because the trials aren't independent, we're of course not shuffling the turn and river cards back in. As the run outs come your equity in subsequent run outs would differ slightly, but at the time of the decision, none of the run outs have happened so for comparison sake I think it is close enough. I've used the raw equity and ignored ties.

If anything, the "Lose All" for running 2x+ is a touch low, given that if you lose one run out, you would be less likely to lose another once since the villain would have less outs on subsequent run outs.

Look at the last row - "Lose All" see how it is lower the more times you run it? That's the real key here. Running it 2x+ is just to reduce that risk of ruin (variance) that comes with poker. Hopefully this helps illustrate it for you.

Result	Run it 1x	Run it 2x	Run it 3x	Run it 4x
Win 1	65%	45%	24%	11%
Win 2	-	42%	44%	31%
Win 3	-	-	27%	38%
Win 4	-	-	-	17%
Lose All	35%	12%	4%	1.5%

I used this calculator if you want to fiddle with the numbers: https://www.omnicalculator.com/statistics/binomial-distribution

Edit: so If the question is how many times should I run it to approximate mine and the villains true equity depends on the equity.

1. If you are racing, run it twice
2. if you are 2:1 run it three times
3. If you are 3:1 + run it 4x, but better yet don't run it multiple times at all and scoop the whole pot

If these numbers are true, it would appear run it once is your best option.

 

[timinater]

@kirchhausen I rounded some of the numbers so could be +/- .5% or so.

 

[Taghkanic]

ResultRun it 1xRun it 2xRun it 3xRun it 4xWin 165%45%24%11%Win 2-42%44%31%Win 3--27%38%Win 4---17%Lose All35%12%4%1.5%

At last! This table is what I was looking for.

Would like to understand the math behind the stats, but this will help me verify what I think it may be.

 

[ruskba]

It’s pretty simple...35% to lose ^ how many chances to lose

35% * 35% for two boards
35% * 35% * 35% for three
...etc

 

[Taghkanic]

It’s pretty simple...35% to lose ^ how many chances to lose

35% * 35% for two boards
35% * 35% * 35% for three
...etc

I understand how to compute the frequency of best case (win all) and worst case (lose all) scenarios. What I’m less sure of is the mixed outcomes: e.g., if you run it three times, how often do you win one runout and lose two?

 

[ruskba]

I think just replace the percentages with what outcome you want?

Win 2/3, 35% * 65% * 65%

Edit: I think you need to multiply that outcome by how many possibilities there are, so win lose lose, lose win lose, lose lose win would mean * 3? Maybe I’m just making stuff up now.

 

[Frogzilla]

It’s binomial probabilities and you can run it by hand. The math assumes the events are independent (as if you could have same river on both runouts) which is not true, but close enough to give you a really good idea and makes the math . In either case, the equity is absolutely 100% identical.

Prob of winning is p, prob of losing (1-p).

1 runout
Win once: 1*p
Win none: 1*(1-p)

2 runouts:
Win twice: 1*p*p
Win once: 2*p*(1-p)
Win none: (1-p)*(1-p)

3 runouts:
Win all three: 1*p*p*p
Win twice: 3*p*p*(1-p)
Win once: 3*p*(1-p)*(1-p)
Win none: (1-p)*(1-p)*(1-p)

 

[WedgeRock]

You're always a 65% favourite (roughly).

Disagree.

He has 7 outs on the turn and if he doesn't hit them, 4 more outs (11 total) on the river.

If he hits just one of those outs on the first board, he is less likely to improve on later boards. Assume you run it three times, and JA hit turn/river on board one. On board two, he only has 5 outs on the turn. Of course, it works in reverse too. If the runout is clean for the hero on board one, villain's equity improves on subsequent boards.

It may not be much, but it's not 65/35 for all boards.

 

[ando5k]

Disagree.

He has 7 outs on the turn and if he doesn't hit them, 4 more outs (11 total) on the river.

If he hits just one of those outs on the first board, he is less likely to improve on later boards. Assume you run it three times, and JA hit turn/river on board one. On board two, he only has 5 outs on the turn. Of course, it works in reverse too. If the runout is clean for the hero on board one, villain's equity improves on subsequent boards.

It may not be much, but it's not 65/35 for all boards.

It's not enough to matter in the decision making for sure. We are talking 2 boards he loses 4.4 to the turn and 2.2% on the river with any cards he hits on board one. Thats not enough to change the decision to run once or twice or three time or 10 times. What it comes down to is the decision being a loss averse decision or not. Are you afraid to lose a big pot and that money by playing the odds and game the correct way and getting your money in good? If the answer is yes, switch to limit games where you can flat twice with your straight and not have to deal with large pot decision making.

 

[Taghkanic]

It’s binomial probabilities and you can run it by hand. The math assumes the events are independent (as if you could have same river on both runouts) which is not true, but close enough to give you a really good idea and makes the math . In either case, the equity is absolutely 100% identical.

Prob of winning is p, prob of losing (1-p).

1 runout
Win once: 1*p
Win none: 1*(1-p)

2 runouts:
Win twice: 1*p*p
Win once: 2*p*(1-p)
Win none: (1-p)*(1-p)

3 runouts:
Win all three: 1*p*p*p
Win twice: 3*p*p*(1-p)
Win once: 3*p*(1-p)*(1-p)
Win none: (1-p)*(1-p)*(1-p)

Perfect. Now I understand. Thanks.

 

[timinater]

He has 7 outs on the turn and if he doesn't hit them, 4 more outs (11 total) on the river.

100% correct, of course equities can change when the turn comes, but that wouldn't change the EV at the decision point which was what I'm referring to if that wasn't clear.

It may not be much, but it's not 65/35 for all boards.

Agreed.

I get what you're saying about the first runout affecting subsequent runouts (I touched on it in the post you are referencing). I'm sure it could be calculated using a conditional probability, but I'll leave it to an actual math-er.

 

[MatB]

Run it twice for good karma, If you're ahead you're ahead both times. IF they get there for half. so be it.

Mathematically running it once may be optimal, but this is where chance/ variance comes in. And being ahead 65% doesn't stop you losing 100% of the time playing out the same scenario 50 or 60 hands in a row. Mathematically you shouldn't lose, but we all know you can. And will. We just don't know if it will be this hand.

Also i do like the 3 times as was said before, SOMEONE must feel some pain !

EDIT to add, i know OP looking for more mathematical solutions, this just my 2 cents, number wizardry be damned!

 

[Frogzilla]

It may not be much, but it's not 65/35 for all boards.

It is the same equity though. You’re right about winning one board decreasing your chance of winning the 2nd. You’re right again about not winning 1st board increasing chance of winning the 2nd. However, both of these effects provably offset each other exactly, and your equity in 2nd board is exactly the same as the 1st (unsurprising, as there is nothing special about the 4th card from the top vs the 2nd card from the top).

 

[WedgeRock]

You’re right about winning one board decreasing your chance of winning the 2nd. You’re right again about not winning 1st board increasing chance of winning the 2nd. However, both of these effects provably offset each other exactly

So I didn't think they would offset each other exactly. Working the simple math on that, it looks like my assumptions were wrong. If you win the first board, the chance of winng the second board increases! If you lose the first board, your chances of winning the second board increase even more!

You have to fade 7 outs on the turn + 10 outs on the river. Assume you win on the first board, you'll have to fade the same 7 and 10 outs.

1st board: (38/45*34/44)=63.3%

Now, 2nd board: (36/43*32/42)=63.8%

Your chance of winning the second board increases by 0.5%.

Now assume you lose the first board. You'll only have to fade 6 turn outs and 9 river outs in the second board.

1st board is the same: (38/45*34/44)=63.3%

2nd board now is: 1-(37/43*33/42) -- assuming the turn card was not a turn/river on the first board -- = 67.6%

Losing the first board increases your chance of winning by 4.3%

 

[msuroo]

Generally my non-analytical approach has been to ask to run it more times when I’m (likely) behind, and fewer when I’m (likely) behind.

There’s been a lot of good discussion in the thread, but this stuck out to me - I feel exactly the opposite. Take the example to the extreme - either a 99% favorite/dog, and you can run the board either 1 time or 100 (I know there’s not enough cards for that, just bear with me for the hypothetical). If I were the favorite, I would absolutely want to run that board 100 times. - all but assured of realizing full equity. If I was the dog, I would much prefer to run it once. No chance of coming away a winner on 100 boards, but can always spike variance on an isolated event.

It’s not an apples to apples comparison, bc in most sports wins/losses are binary results whereas poker results are measured in dollars (so the EV is the same no matter how many times you run it), but there’s a reason underdog sports teams employ high variance strategies.

 

[Frogzilla]

So I didn't think they would offset each other exactly. Working the simple math on that, it looks like my assumptions were wrong. If you win the first board, the chance of winng the second board increases! If you lose the first board, your chances of winning the second board increase even more!

You have to fade 7 outs on the turn + 10 outs on the river. Assume you win on the first board, you'll have to fade the same 7 and 10 outs.

1st board: (38/45*34/44)=63.3%

Now, 2nd board: (36/43*32/42)=63.8%

Your chance of winning the second board increases by 0.5%.

Now assume you lose the first board. You'll only have to fade 6 turn outs and 9 river outs in the second board.

1st board is the same: (38/45*34/44)=63.3%

2nd board now is: 1-(37/43*33/42) -- assuming the turn card was not a turn/river on the first board -- = 67.6%

Losing the first board increases your chance of winning by 4.3%

38/45*34/44 is 65.2%

And if we do conditional probs...you’re 65.2% to be in the situation #1 with 63.8% equity for board #2, and 34.8% to have 67.6% equity for board #2. 65.2%*63.8%+34.8%*67.6%= 65.2%

You really don’t have to worry about math though it’s always always always the same EV. 3-way all-in? Same EV. Split pot game? Same EV. Preflop with lots of different ways for JTs to crack KK? Same EV.

 

[Hornet]

I hate to add another tangent to this thread, but the math is simple. As others have explained, mathematically, there is no difference in equity between running it once, twice, thrice, or 40 times.

What is more interesting to me is how fold equity changes based on how many times I am willing to run it. There are definitely people in my home game who are more likely to fold big draws to me knowing I will only run it once. They don’t want to get felted, and that gives me an advantage, so long as everyone knows up front I normally only run it once (the exception being monster pots).

 

[Kyle]

I always like running it twice no matter what and try and let the entire table aware of this. I never expose my cards until they agree one way or the other. Their decision must be made before they see my cards. My only rule that I have for myself.

I also expect the other player to agree to run in twice when I call their all in for consistency. (These are againest players that I regularly play with for the most part at any stakes.) It also keeps them game a bit more friendly when felting someone in a big pot.

I figure most of the time I know I will have the best hand percentage wise and want the other players to make poor decisions based on that. Also good for cooler situations such as your example.

For example hey he will run it twice so I will call his all in. More often then not that is a winning scenario in the long run. It also allows them to gamble more vs me. And some insurance should they suckout on the 1st board. Sometimes they win both but that is part of it. It's poker.

In your example you should win both boards most of the time.

I think running it 2x is good for the game especially in home games.

 

[Taghkanic]

As others have explained, mathematically, there is no difference in equity between running it once, twice, thrice, or 40 times.

I think everyone understands that now, if they didn’t already. In the very long run, if you are a favorite to win 2 out of 3 runouts, if a similar situation comes up enough times, your results are eventually going to be 2 out of 3.

Eventually.

Obviously, that word is why people propose running a single hand more than once, to give a slightly better chance of realizing your “true” equity.

If everyone took a reductive approach (“ah, it doesn’t matter, in the long run it all evens out”) then no one would ever propose running it added times... except I guess for fun.

Even if you play poker for decades, certain situations are not going to come up enough times to truly smooth down or up to that “true” 65%. (I think it was PokerNews which recently posted an article about how—even though we all get dealt random cards and boards from the same decks—there actually are going to be some “luckier” and “unluckier” players, to the extent that few people play anything close to enough live hands to completely erase variance. You can be a very good player and not get past the bubble in 25 tournaments in a row. It just happens.

So the question was never about what equity one will eventually realize over a million iterations of a similar situation. The question was: What are the variety of outcomes possible when running it 2+ times, how often will each outcome occur, and (given, here, 65% equity) what is the short-term impact of such a decision, in relation to what one is hoping to accomplish. Take the whole pot down? Try to realize your (lesser but more reliable) equity? Be gambly? Be friendly?

If I have my life savings in the pot, and I am going to win 2/3 of the time running it once, but go bankrupt 1/3 of the time, I am going to ask to run it as many times as the opponent will agree to. If the pot is not more than I can earn in a few hours, or I’m feeling flush and loose, I may say screw it, just run it once. And if you lose, hope you wind up with another 60-80% equity situation like that again very soon.

The more interesting question (and the one which I was trying to get to the math of) was, how to compute how often each scenario comes up (win the whole pot, win 2/3rds of the pot, chop, lose 2/3rds, lose it all, etc.). I appreciate that several people actually provided the data and methodology for doing so. It helps to understand statistics better in general, and provides a guide for understanding when and to what extent to run out extra boards or streets, depending on those goals.

 

[DrStrange]

where to begin . . . . . . ?

All the posts saying the expected value is the same however you run it out are correct.

What changes with running it more than once is the variance. The more times you run the hand, the lower the variance. Taken to the extreme, the players could just chop the pot based on their equity. < assuming they could agree on how much equity they have. >

Reducing variance might be good for the game, or it might not. Depends on the players involved and the circumstance.

Good for the game does not mean good for Hero. Hero might enjoy variance or might need to "get lucky" to get back to even for the night. Hero might have meta game purposes that benefit from situations unhealthy for the game as a whole.

While it is true that actual results for the population as a whole converges on the expected value, there will still be significant variation centered around the expected value. Someone is going to be luckier than their skill earned and someone is going to have poorer results than expected. That is how variance works and why we "try our luck" in casinos. If you were 100% certain to lose a tenth of your money to the house every time you visited Vegas, you wouldn't play. While luck in general will average out - - - - YOUR luck will run good or bad over your life. It is quite unlikely for an individual's luck to average out to zero at the end of their life.

This discussion breaks down if the size of the money at risk is large, whatever large is to you. We could spend a long time discussing "utility theory" and how it relates to uber, ultra large stakes poker. But this sort of thing almost never happens. Utility theory posits that people value money more as they get closer to being broke AND ultra wealthy people value money more as they get wealthier. [ that is one reason why billionaires engage in risky or even illegal behavior to get richer, as if an additional ten billion dollars would make a difference to someone so rich. ] We almost all know not to sit down in a $25,000/$50,000 blind game because we can't afford to lose. { noting that there are a few people who could easily absorb million dollar swings, but such people aren't likely playing $2/$5 big bet poker. }

Playing modest stakes poker, run it how many times you want. If the stakes are so high that the variation becomes stressful, perhaps the game is too rich. Thus the question shouldn't be about reducing variation it might be better to focus on not playing too high.

DrStrange

 

[ReallyGoodUsername]

Apparently, 4x is the sweet spot if you're an 85:15 dog.

Love math so love the chat and input but I still can’t get over how much I loved watching Phil lose in this case (ftr I also love watching him win)

 

[abby99]

Love math so love the chat and input but I still can’t get over how much I loved watching Phil lose in this case (ftr I also love watching him win)

View attachment 302111View attachment 302112
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Link to full video in post #29. To his credit, Hellmuth's suggesting running it 4x to give Wiggins a better shot at winning at least part of the pot. I don't think this is what he had in mind.

 

[Taghkanic]

1) Great post, Dr. Strange

2) Here’s the article I mentioned above. It wasnt in PokerNews:

https://www.pokerlistings.com/strategy/the-mathematical-truth-some-players-have-to-run-bad

“Everyone will be dealt hands from this distribution each time they sit down and, in theory, they will all be dealt the "same" hands. In reality, of course, this sameness is only reached when an infinite number of hands have been dealt. Frankly, I don't have time to wait for this and neither do you.”

“Gus Hansen was once asked by a reporter what role luck played in poker. He responded that in any given session it probably accounted for about 90% of his outcomes. Over a month, he guessed it was about 10 or 15%; over a year it was down to around 2-5%.”

 

[Emlemleml]

If you can benefit by playing vs possible tilted opponents, and at the same time withstand from tilting yourself.
Then I would definitely only run it once.