Running It Twice and Insurance
 

I know what it mean and how to run it twice but what effect does it have on the outcome and why do players do it and how did it get started
InsuranceI have no clue what this mean, when to player are all in whatdoes it mean when one player offer another Insurance in poker
I can figger that he could be like BJ insurance but how do you fig out the odds each hand is diffrent

any help will be great

Re: Running It Twice and Insurance
 

running it twice reduces variance. The odds are the same each time, you just smooth out the results a little bit. It is most desirable if you are playing over your head and the suck out would ruin you.

Expand the concept, say you can run it a million times. With pocket kings against AQ, you're going to lose about 28% of the time. If you run it once, you'll either lose the whole pot or win the whole pot. If you run it a million times, you'll win ~72% of the pot, period. Running it twice doesn't get you this close to the average, but it will smooth it out some.

Insurance is a similar concept where you get $X to end the hand right there. In the above example, you know you should average 72% of the pot, but on this particular hand, you could lose the whole thing. Instead, the house pays you 65% of the pot hand and it is over. You are happy because you didn't bust out and got some profit that is close to what you deserved in the hand, whereas the house is happy because they gain 7% equity right there, and can take a high volume of these bets to guarantee themselves profit.

Re: Running It Twice and Insurance
 

As for running twice, If you ask to run it twice, make sure you return the favor later in the session. Of course that should go without saying. Once you start running it twice or three times, you leave yourself vulnerable to future multi-runnings.

I personally do not subscribe to running the board twice.

Re: Running It Twice and Insurance
 

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you leave yourself vulnerable to future multi-runnings.

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What would be the problem with that?

Re: Running It Twice and Insurance
 

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you leave yourself vulnerable to future multi-runnings.

[/ QUOTE ]

What would be the problem with that?

[/ QUOTE ]

Isn't that blatantly obvious?

Re: Running It Twice and Insurance
 

It doesn't affect your EV at all, and it lowers your variance. To what are you vulnerable?

So no, its not obvious.

Re: Running It Twice and Insurance
 

As posted before, running it twice only makes the chance on a split pot bigger. EV remains unchanged. In the long run it doesn't influence anything.

Insurance is letting someone off for a bit worse price than they get in the long run. Say we go allin and you have 28% equity.... I let you keep 25% of the pot without showdown. You are glad since you are going to bust 72% of the time and now you have some chips left. I am happy, because I just made a little more than I should have in the long run and didn't give you a chance to suck out.

It's mathematically ok to give insurance, but never to take it (unless you have no bankroll left if you lose and have to eat, but in that case you did something else wrong). Most pokerplayers don't do insurance either way, I believe.

Re: Running It Twice and Insurance
 

It is absolutely incorrect to say that Running more then once is the same EV. That is myth it is not. Its obvious, AA vs. 33. AA are 80/20. Well if we hit the 3 on the first run it DOES NOT GO BACK IN THE DECK! Therefore to hit it a second time you only have one out. Therefore, running it twice gives you 6 to 1 instead of 4 to 1. THis is the most extreme example but it is the same wiith flush draws and straight draws if you hit you are missing an out. I play in a NYC club and have thought about this a lot. The only time you should do business(run it more then once) is when you are AHEAD not behind. Can anyone argue with this logic?

Re: Running It Twice and Insurance
 

You don't even need to know math to see running it twice doesn't effect your results.

In the AA vs 33 example you know it's about 80/20. Just look at it this way. 4 cards are dead so there's 48 left. If you run the whole board you can do it about 10 times (assume no burn). In those 10 times you'll hit both 3's, usually on seperate hands, and usually win those hands. So the majority of the time you'll win 2 of 10. Occasionaly you'll make quads and just win that one. Occasionaly you'll win with a straight or, if you have different suits, a flush. Just the rough look at cards in the deck and what makes your hand shows you your equity is around 20% no matter how many times you run it.

Re: Running It Twice and Insurance
 

You are incorrect WildThought. Even though the cards remain out of the deck EV stays exactly the same...

Look at this simplified example:

We have a 2 street, one card per street game with a 12 card deck (four each of Aces, Kings and queens)...

I have KQ. You have AQ. The flop is K. I go allin. You can only win by spiking an A on the second and last street. So you have 3 outs out of 7 remaining cards. Or 3/7 equity... Let's say the the blinds were high enough to make this a call for you.

Now lets see what happens if we run it twice.

You win by winning it twice. The chance for this to happen is: 3/7 * 2/6 = 6/42

You lose by losing it twice. The chance for this to happen is: 4/7 * 3/6 = 12/42

The other 24/42 we have a split pot. This is correct, because: 3/7 * 4/6 * 2 = 24/42...

So you expect to win (6 - 12) / 42 = -6/42 = -1/7 when running it twice.

And you expect to win (3 - 4) / 7 = -1/7 when running it once.

EV has not changed, the amount of losses per wins has become greater (2:1 instead of 4:3), but in return you get a lot more split pots when running it twice.

The same stuff applies to full scale holdem. I can give you proof of this, but that would get rather complicated. And I hope you intuitivey see this is so.

Hope this explains.

Re: Running It Twice and Insurance
 

Wildthought, MVD gives you a good example of why you are wrong, but here's the intution:

EV stands for "Expected Value," that is, how often you EXPECT to win before you see any more cards. In your example, your EV of the 2nd running has fallen after hero spikes his card, but that changed only AFTER you saw that he spiked his card. But, the EV of the second running only changes AFTER the first running, and so your example CANNOT disprove anything about the expected value BEFORE both runnings.

Further, your EV represents what your actual equity would converge to if you ran the possible board out an infinite number of times. Although the cards you deal on turn 1 will necessarily not constitute what appears on turn 2, both runs are possible if you run the hand only once. Thus, if you ran out 2 boards an infinite number of times, and the single board an infinite number of times, the various card combinations would occur with identical frequencies. As a result, EV is not affected.

Re: Running It Twice and Insurance
 

Yes... well explained ev_slave (with that name you should be able to [img]/images/graemlins/smile.gif[/img])

It's the same error in logic as this riddle induces... I'll post it here and give the answer with explanation later. Have fun.

You are in a quiz. There are three safes. One of them contains a million dollars. The other two are empty.

You get to pick one safe, A, B or C.

The quizmaster opens one of the other two safes and shows you it's empty.

You get the choice: opening the safe that you first chose or switch to the other one.

Should you stay with your safe, switch, or doesn't it matter?

Re: Running It Twice and Insurance
 

Running it twice is bad if you have an edge in an uncapped game and aren't above your roll, unless the guy is likely to leave as soon as he beats you, but would play on after a split pot.

Re: Running It Twice and Insurance
 

I must be stupid. But, If I am open neded with 8 outs. I spike a card. Now, I have 7 outs. Don't I have less a chance of winning the bot pots since one of my outs is gone?

If I run it once, I have roughly 32% chance of spiking. If I run it the second time I only have a 28% chance. No one can argue that I have the same chance the second time as the first. What am I missing?

Re: Running It Twice and Insurance
 

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Running it twice is bad if you have an edge in an uncapped game and aren't above your roll, unless the guy is likely to leave as soon as he beats you, but would play on after a split pot.

[/ QUOTE ]

Yeah, that's really the only reason that you would not want to run it twice. If you have the max buyin, and some bad players have more than the max buyin, you'd rather not run it twice so that you give yourself a bigger chance of doubling up and being able to play deep stacked against the bad players.

Once you've already built a stack up and have the bad players covered (or close to it), then you would definitely want to run it twice so you can have a better chance to keep playing deep stacked.

Re: Running It Twice and Insurance
 

[ QUOTE ]
I must be stupid. But, If I am open neded with 8 outs. I spike a card. Now, I have 7 outs. Don't I have less a chance of winning the bot pots since one of my outs is gone?

If I run it once, I have roughly 32% chance of spiking. If I run it the second time I only have a 28% chance. No one can argue that I have the same chance the second time as the first. What am I missing?

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Do an EV calc.

I'll simplify the math a little. Suppose there's $100 in the pot, one card to come on the river and you have some sort of draw with 4 outs, with 44 cards left in the deck (4 board cards and both players' cards gone) so you're 9.1% to hit.

If you run it once the EV is simply (0.091)*($100) = an EV of $9.10.

For running it twice:

The 9.1% you hit the first time, you win $50 of the pot but you're down to 3 outs out of 43 cards, which is 7.0%.
The 90.9% you don't hit the first time, you don't win the first pot but then you're up to 4 outs out of 43 cards, which is 9.3%.

So your EV for running it twice is:

(0.091)*[ $50 + (0.070)*($50) ] + (0.909)*[ $0 + (0.093)*($50) ] = $9.10

Exactly the same EV as if you'd run it once.

Moral: You're correct that hitting the first time reduces your odds of hitting the second time. But remember that missing the first time increases your odds of hitting the second time. These differences cancel each other out.

Re: Running It Twice and Insurance
 

[ QUOTE ]
I must be stupid. But, If I am open neded with 8 outs. I spike a card. Now, I have 7 outs. Don't I have less a chance of winning the bot pots since one of my outs is gone?

If I run it once, I have roughly 32% chance of spiking. If I run it the second time I only have a 28% chance. No one can argue that I have the same chance the second time as the first. What am I missing?

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U are not stupid... It's a very human flaw in reasoning. Look at my simplified example. The card IS left out after the first win/lose.

What you are missing is that you evaluate the situation AFTER you have more information. Before you do so you do not know what is going to happen the first time. Yes, your chances of winning are far lower, but so are your chances of losing.

You have to win BOTH the first AND the second time to win.
You have to lose BOTH the first AND the second time to lose.

After one lose or one win the first one you can calculate your odds (like you do), but that wouldn't be fair. When you make the decision to run it twice you are facing both deals.

Re: Running It Twice and Insurance
 

[ QUOTE ]
I must be stupid. But, If I am open neded with 8 outs. I spike a card. Now, I have 7 outs. Don't I have less a chance of winning the bot pots since one of my outs is gone?

If I run it once, I have roughly 32% chance of spiking. If I run it the second time I only have a 28% chance. No one can argue that I have the same chance the second time as the first. What am I missing?

[/ QUOTE ]

I think you are missing the following:

You are correct, that IF you hit one of your outs the first time, you will have a lower chance to hit one in the second run.

BUT: If you dont hit your outs the in first run, you will have a HIGHER chance to hit in the second run. This balances out with the first effect, such that the overall expectation remains neutral.

eg: You have 8 outs and run the river twice. After missing the first run, you will have 8/42 probability to hit in the second run - instead of 8/44.


I think the above might be more intuitive for you, but its basically overly complicated thinking. A better way to look at it imo:

If you are playing brick & mortar, the cards are already shuffled when you decide to run it twice. That is, the arrangement of cards will not change any more.

The probability of the first card being an out of yours is obv. 8/44. The probability of the second card being an out is also 8/44, obv (as long as we dont know the first card). So - at the moment you decide to run it twice - you are basically taking two smaller 8/44 gambles for half the pot each, instead of one big 8/44 gamble for the whole pot. It should be clear that this is EV-neutral and variance reducing.

Note to self: Read the last posts of a thread before replying. Yes, i basically just re-stated what is said in the posts above [img]/images/graemlins/wink.gif[/img] Meh, heh.

Re: Running It Twice and Insurance
 

Wow,

I thought I was smart about this and was readt to write an essay about the correct way of doing business. Thanks for the schooling, very instructive.

Andy

Re: Running It Twice and Insurance
 

[ QUOTE ]
Yes... well explained ev_slave (with that name you should be able to [img]/images/graemlins/smile.gif[/img])

It's the same error in logic as this riddle induces... I'll post it here and give the answer with explanation later. Have fun.

You are in a quiz. There are three safes. One of them contains a million dollars. The other two are empty.

You get to pick one safe, A, B or C.

The quizmaster opens one of the other two safes and shows you it's empty.

You get the choice: opening the safe that you first chose or switch to the other one.

Should you stay with your safe, switch, or doesn't it matter?


[/ QUOTE ]


No responses... I'll give the answer anyways:
You should switch.

When you first choose a safe you have 1/3 chance of winning and 2/3 chance that it's in one of the others.

If you had chosen the correct safe already the quizmaster could open one of both safes.

If you had chosen an incorrect safe the quizmaster could only have opened the one empty safe that remains. This information increases the chance that the other safe has the goods.

In other words:

When you don't switch: 1/3 of the time you chose the correct safe in the first place and win.

When you do switch: 1/3 of the time you chose the correct safe the first time and now lose. But 2/3 of the time you didn't and the only alternative safe left (the quizmuaster just took away one choice) must be the correct one.

Re: Running It Twice and Insurance
 

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Insurance is letting someone off for a bit worse price than they get in the long run. Say we go allin and you have 28% equity.... I let you keep 25% of the pot without showdown. You are glad since you are going to bust 72% of the time and now you have some chips left. I am happy, because I just made a little more than I should have in the long run and didn't give you a chance to suck out.

It's mathematically ok to give insurance, but never to take it (unless you have no bankroll left if you lose and have to eat, but in that case you did something else wrong). Most pokerplayers don't do insurance either way, I believe.

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Either I am confused by your post or you are. The person in the lead pays a premium to insure his hand. If your foe has 8 outs on the river, he has ~16% equity, but it costs you 3:1 to insure. In other words, for every dollar you insure, whoever accepts the insurance (the house or the player who is drawing) gets 25% equity instead of 16%.

So, when you want to avoid a suckout when in the lead, it is going to cost you plenty to insure.

Re: Running It Twice and Insurance
 

[ QUOTE ]
It is absolutely incorrect to say that Running more then once is the same EV. That is myth it is not. Its obvious, AA vs. 33. AA are 80/20. Well if we hit the 3 on the first run it DOES NOT GO BACK IN THE DECK! Therefore to hit it a second time you only have one out.

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You are right but wrong. No one is saying running it twice is the same EV on each run as running it once.

However, if you consider both runs together vs. running it once, it is the exact same EV.

Re: Running It Twice and Insurance
 

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As for running twice, If you ask to run it twice, make sure you return the favor later in the session.

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?? The person in the lead always has the choice. Once, twice (if the foe agrees), or insurance (assuming the house will lay it).

So, just because you offer to run it twice on a given hand when you are in the lead does not mean you are obligated to offer it every time you are in the lead. Nor does it obligate you to accept someone else's offer of running it twice later in the session.

Re: Running It Twice and Insurance
 

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Insurance - I have no clue what this mean, when to player are all in what does it mean when one player offer another Insurance in poker.

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Typical holdem insurance chart is:

1 out 20:1
2 outs 12:1
3 outs 8:1
4 outs 6:1
5 outs 5:1
6 outs 4:1
7 outs 3.5:1
8 outs 3:1
9 outs 2.75:1
10 outs 2.5:1
11 outs 2.25:1
12 outs 2:1

& so on until 24 outs is even money

If someone has 2 outs twice on the flop, it is counted as 4 outs on the insurance board. You double the outs if you get all in on the flop and want to insure.

1 out is 2.3% true equity, and if you are insuring against a 1 outer, you give up 1/21 or 4.8%. So, you give up 2.5% of your edge. In other words, the player in the lead will be paying "premium" and giving up some of his edge to insure against a suckout.

Here is how much of your edge that you give up when you insure:

1 out - 2.3% premium

2 outs 3.1-3.3% premium

3 outs 4.3% premium

4-5 outs 5.2-5.5% premium

6-12 outs 5.8-8.1% premium

Usually, you pay more premium when you insure on the flop than on the turn. In other words, 4 outs twice on the flop = 8 outs and you give up 7.8%. But if your foe has 8 outs with 1 card to come, you are paying 6.8% premium.

For me, I rarely insure a hand unless:

1. The game is good and I don't have access to more cash if they suckout, or

2. The have 1, 2 or 3 outs, because these are the outs with the least premium.

Another mistake people make is insuring the whole pot to the felt. The more you insure, the more premium you pay. A good rule of thumb is to only insure what you put into the pot.

So, if I insure half the pot against a 2 outer, I am theoretically only giving up 1.6% of my edge. If I insured the full put, I would be giving up the full 3.1-3.3% of my edge.

If I want to reduce my variance and they have more than 3 outs, I will usually run it twice.

If I am behind and have more than 3 outs, I always take the insurance if offered. It's the only opportunity I know in poker to make money when you have the worst of it. If I am behind with 1-3 outs, I might take a shot at sucking out because the "premium" is small.

Re: Running It Twice and Insurance
 

The easiest way to explain why multiple runnings have the same EV is to take any hold'em example in which there is 1 card to come and show the effect of running it 44 times.

Re: Running It Twice and Insurance
 

[ QUOTE ]
[ QUOTE ]
Yes... well explained ev_slave (with that name you should be able to [img]/images/graemlins/smile.gif[/img])

It's the same error in logic as this riddle induces... I'll post it here and give the answer with explanation later. Have fun.

You are in a quiz. There are three safes. One of them contains a million dollars. The other two are empty.

You get to pick one safe, A, B or C.

The quizmaster opens one of the other two safes and shows you it's empty.

You get the choice: opening the safe that you first chose or switch to the other one.

Should you stay with your safe, switch, or doesn't it matter?


[/ QUOTE ]


No responses... I'll give the answer anyways:
You should switch.

When you first choose a safe you have 1/3 chance of winning and 2/3 chance that it's in one of the others.

If you had chosen the correct safe already the quizmaster could open one of both safes.

If you had chosen an incorrect safe the quizmaster could only have opened the one empty safe that remains. This information increases the chance that the other safe has the goods.

In other words:

When you don't switch: 1/3 of the time you chose the correct safe in the first place and win.

When you do switch: 1/3 of the time you chose the correct safe the first time and now lose. But 2/3 of the time you didn't and the only alternative safe left (the quizmuaster just took away one choice) must be the correct one.

[/ QUOTE ]

Yeah, I play this one on my friends. I believe it's known as the "Montey Hall Game" in the literature. It really tests people's intuition. Often they'll argue with oyu even after you explain it to them.

Re: Running It Twice and Insurance
 

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I must be stupid. But, If I am open neded with 8 outs. I spike a card. Now, I have 7 outs. Don't I have less a chance of winning the bot pots since one of my outs is gone?

If I run it once, I have roughly 32% chance of spiking. If I run it the second time I only have a 28% chance. No one can argue that I have the same chance the second time as the first. What am I missing?

[/ QUOTE ]

Do an EV calc.

I'll simplify the math a little. Suppose there's $100 in the pot, one card to come on the river and you have some sort of draw with 4 outs, with 44 cards left in the deck (4 board cards and both players' cards gone) so you're 9.1% to hit.

If you run it once the EV is simply (0.091)*($100) = an EV of $9.10.

For running it twice:

The 9.1% you hit the first time, you win $50 of the pot but you're down to 3 outs out of 43 cards, which is 7.0%.
The 90.9% you don't hit the first time, you don't win the first pot but then you're up to 4 outs out of 43 cards, which is 9.3%.

So your EV for running it twice is:

(0.091)*[ $50 + (0.070)*($50) ] + (0.909)*[ $0 + (0.093)*($50) ] = $9.10

Exactly the same EV as if you'd run it once.

Moral: You're correct that hitting the first time reduces your odds of hitting the second time. But remember that missing the first time increases your odds of hitting the second time. These differences cancel each other out.

[/ QUOTE ]

But are you in essence cutting your EV in half?
I completely understand your mathmatical example of the EV being the same for both runs, but you are giving your opponet two times to crack your hand.

Re: Running It Twice and Insurance
 

[ QUOTE ]
[ QUOTE ]
As for running twice, If you ask to run it twice, make sure you return the favor later in the session.

[/ QUOTE ]

?? The person in the lead always has the choice. Once, twice (if the foe agrees), or insurance (assuming the house will lay it).

So, just because you offer to run it twice on a given hand when you are in the lead does not mean you are obligated to offer it every time you are in the lead. Nor does it obligate you to accept someone else's offer of running it twice later in the session.

[/ QUOTE ]

I fully understand there is there is no moral obligation to offer or accept future offers; but some people do not understand that and they get pissy if you do not offer it again in the future. I prefer to keep it straight on the board. Prevents the possibility of unwanted bulljive.

Re: Running It Twice and Insurance
 

Soju,

You clearly don't understand the math.

Re: Running It Twice and Insurance
 

Soju - do you think that burning three or four cards instead of one before showing the river will change the EV?

Re: Running It Twice and Insurance
 

[ QUOTE ]
But are you in essence cutting your EV in half?
I completely understand your mathematical example of the EV being the same for both runs, but you are giving your opponent two times to crack your hand.

[/ QUOTE ]
Think of it this way, if you get your money in as a 2:1 favorite, you HAVE to lose 1/3 of the time. Essentially you should only be winning 2/3 of that pot. The more you run it the closer to that expectation you'll be. The only question is do you want the spikes, aka variance, (as you win the whole pot 2/3 of the time or lose the whole pot 1/3 of the time)? In the long run you still win as much or lose as much as you're supposed to in that situation.

Re: Running It Twice and Insurance
 

[ QUOTE ]

[ QUOTE ]

Yes... well explained ev_slave (with that name you should be able to )

It's the same error in logic as this riddle induces... I'll post it here and give the answer with explanation later. Have fun.

You are in a quiz. There are three safes. One of them contains a million dollars. The other two are empty.

You get to pick one safe, A, B or C.

The quizmaster opens one of the other two safes and shows you it's empty.

You get the choice: opening the safe that you first chose or switch to the other one.

Should you stay with your safe, switch, or doesn't it matter?

[/ QUOTE ]

No responses... I'll give the answer anyways:
You should switch.

When you first choose a safe you have 1/3 chance of winning and 2/3 chance that it's in one of the others.

If you had chosen the correct safe already the quizmaster could open one of both safes.

If you had chosen an incorrect safe the quizmaster could only have opened the one empty safe that remains. This information increases the chance that the other safe has the goods.

In other words:

When you don't switch: 1/3 of the time you chose the correct safe in the first place and win.

When you do switch: 1/3 of the time you chose the correct safe the first time and now lose. But 2/3 of the time you didn't and the only alternative safe left (the quizmuaster just took away one choice) must be the correct one.

[/ QUOTE ]

wtf does dis have 2 do w/ the op's question? noone answered b/c it isn't related by any means 2 the original post or any of the replies. by the way, this is the worst explanation i've ever read/heard for the monty hall problem. vos savant should kick ur ass. and u prolly suck @ poker.

Re: Running It Twice and Insurance
 

[ QUOTE ]
[ QUOTE ]

[ QUOTE ]

Yes... well explained ev_slave (with that name you should be able to )

It's the same error in logic as this riddle induces... I'll post it here and give the answer with explanation later. Have fun.

You are in a quiz. There are three safes. One of them contains a million dollars. The other two are empty.

You get to pick one safe, A, B or C.

The quizmaster opens one of the other two safes and shows you it's empty.

You get the choice: opening the safe that you first chose or switch to the other one.

Should you stay with your safe, switch, or doesn't it matter?

[/ QUOTE ]

No responses... I'll give the answer anyways:
You should switch.

When you first choose a safe you have 1/3 chance of winning and 2/3 chance that it's in one of the others.

If you had chosen the correct safe already the quizmaster could open one of both safes.

If you had chosen an incorrect safe the quizmaster could only have opened the one empty safe that remains. This information increases the chance that the other safe has the goods.

In other words:

When you don't switch: 1/3 of the time you chose the correct safe in the first place and win.

When you do switch: 1/3 of the time you chose the correct safe the first time and now lose. But 2/3 of the time you didn't and the only alternative safe left (the quizmuaster just took away one choice) must be the correct one.

[/ QUOTE ]

wtf does dis have 2 do w/ the op's question? noone answered b/c it isn't related by any means 2 the original post or any of the replies. by the way, this is the worst explanation i've ever read/heard for the monty hall problem. vos savant should kick ur ass. and u prolly suck @ poker.

[/ QUOTE ]

Lol... quite an angry little fellow are you?

I'm sorry that I didn't explain the problem or the solution to your standards [img]/images/graemlins/tongue.gif[/img]. I'm also sorry that I haven't made the connection to the running it twice stuff clear enough for you so I'll do my best to explain.

The relation is that the odds change only after new information has become available. Some of the replyers analysed the problem by assuming they knew the outcome of the first run. In that case more information about the cards is available and the odds are different, just like the safes where the information became available only after the safe was opened. This is what I though caused the misunderstandings.

Re: Running It Twice and Insurance
 

What if you only got 1 out? If you run it thrice it's impossible to win.

Re: Running It Twice and Insurance
 

[ QUOTE ]
What if you only got 1 out? If you run it thrice it's impossible to win.

[/ QUOTE ]

1 out has 1/44 equity (2.27%) with 1 to come running it once.

Twice, it has 0.5 * 1/44 + 0.5 * (43/44 * 1/43 + 1/44 * 0/43) = 2.27%

Same 2.27% for thrice, foursies, or however much you want to run it.

You can't win the whole pot running it twice with 1 out. You can only win half the pot. You can only win 1/3 pot running it thrice with 1 out.

Re: Running It Twice and Insurance
 

Let's say you hold: 55
Villain holds: 44

Board: 4459A

If you run it once you can still catch the case 5 and scoop the whole pot. But if you run it thrice you will lose no matter what.

Re: Running It Twice and Insurance
 

[ QUOTE ]
Let's say you hold: 55
Villain holds: 44

Board: 4459A

If you run it once you can still catch the case 5 and scoop the whole pot. But if you run it thrice you will lose no matter what.

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Again, I say:

"You can't win the whole pot running it twice with 1 out. You can only win half the pot. You can only win 1/3 pot running it thrice with 1 out."

Over the long run, however, your average will be 2.27% of the pot no matter if you run it once 1000 times or if you run it thrice 1000 times.

Re: Running It Twice and Insurance
 

"You can only win 1/3 pot running it thrice with 1 out."

Could you elaborate on this? How can you win 1/3 pot? You either win the pot, lose the pot or split the pot. If you run it thrice with 1 out you always lose the pot because in order to win the pot you need to win at least 2 of the 3 runs. This is impossible as we only got 1 out.

Edit: Nevermind... I just learned that the pot will be split in thirds when running it thrice