5/5 NL...Doing Business????

[lapoker17]

do it sometimes, don't do it others. just don't become known as the guy who always runs it twice.

[TheWorstPlayer]

I know what you're saying, but this doesn't sound like a game full of guys who are going to be taking shots at you because you run it twice. It sounds like a nice soft home game where everyone runs everything twice. So he should just always run it twice here.

[cbloom]

(I wrote this a while ago, HTML)

In cash games it's common to "run it twice". What this means is when the players are all-in before the river, they will
draw the remaining cards twice. If one player wins both times, he scoops the pot. If each player wins once, the pot is
split. The funny thing about how this is usually done is that the drawn cards are not shuffled back into the deck, you
simply draw once, leave those cards in the muck, then draw again.

<P>

This is very bad for the EV of someone with few outs. Consider the case that you only have one out - you can only possibly
win once, because if you win one you have no more cards to hit for the other draw! Let's look at the math in one specific
case, Hold'em, with two players cards up, and just one more card to come, so 8 cards are known and 44 remain unknown. After
one draw there are 43 unknown. The player who's behind has N outs.

<P>

If you just draw once, your EV in units of pots is N/44

<P>

If you draw twice you can either win both : N/44 * (N-1)/43 , or more likely you hit the first and miss the second or
miss the first and hit the second, and if those happen you get half EV :

<P>
<PRE>

1/2 * (N/44 * (1 - (N-1)/43)+ (1 - N/44) * (N-1)/43)

=
[ N * (N-1) + 1/2 * (N * (43 - (N-1))+ (44 - N) * (N-1)) ] / (44*43)

=
[ N * (N-1) + 1/2 * (N * 43 - 2N*(N-1) + 44 * N - 44) ] / (44*43)

=
[ 1/2 * (N * 87 - 44) ] / (44*43)
=
[ N * 87/2 - 22 ] / (44*43)

=
N/44 * 87/86 - 22 * (44*43)

=
(N/44) * 87/86 - 1/86

</PRE>

Well, this is always less than the EV of just drawing once. The difference is ((N/44) - 1)/86. If you somehow had > 44 outs
it would be +EV to draw twice (silly), and the less outs you have the worse it is. In the extreme case of having only 1 out,
you lose 1.1% of the pot in EV by drawing twice !!

<P>

So, the moral of the story is : if you ever get in a big cash pot and someone offers to draw twice - you should refuse
if you're drawing and accept if you have the better made hand. (note that the player drawing may actually be "ahead" in
the sense that if they have more than 22 outs they are favored to win, but it's still a bad move for them to accept the
double run - drawing players should only run once). People often justify running it twice as a way to reduce variance.
It sure is, but if you're playing hold'em your variance is already huge and you should just worry about EV.

[cbloom]

Actually that math is messed up. Stupid algebra! Anyhoo, I think the correct result when you correctly add the odds of hitting both is something like (N - 1)/(44*43*2) , which is very small so pretty irrelevant unless the pot is huge.

[scdavis0]

cbloom,

It's amazing to me that you could spend so much time working out the math on this and still come up with the wrong answer. I hope for your sake that your post is some sort of joke. I'll let you spot the error in your logic.

Running it once, twice, thrice, the whole deck -- it does nothing to EV.

[scdavis0]

Here's a reasonable formulation in the case of ONE out.

Let S = size of the pot

If you run it once your EV = S/44

If you run it twice, one of 3 things will happen

1. You'll miss twice
2. You'll hit the first board but miss the second
3. You'll miss the first board but hit the second

To get our overall EV, we simply sum the product of each probability by the percentage of the pot we will receive should they occur.

Scenario 1 -- miss twice:

We will miss on the first card 43/44. Given that we missed the first card, we will miss the second card 42/43. The product of these probablities is 21/22. We will miss both boards 21/22 and receive zero.

Scenario 2 -- hit 1st and miss 2nd

We will hit on the first board 1/44. Given that we hit the first board, we will miss the second board 43/43. The probability is therefore the product -- 1/44. In this case we get half of the pot.

Scenario 3 -- miss the first but hit the second

We will miss the first board 43/44. Given that we missed the first, we will hit the second 1/43. The product of these is again 1/44.

Our overall EV is therefore:

EV = (S * 21/22 * 0) + (S * 1/44 * 1/2) + (S * 1/44 * 1/2)
= S/44

Thus running it twice with one out does NOTHING to EV. For fun, solve it generally for any number of outs!

(If anyone cares to see the general solution I'll be happy to provide)

[pokerjoker]

Allin on turn with villian having 1 out and the cards are resuffled inbetween tries.

Heres an easy equation w/o algebra.

1/44 x 1/44 =villians chance of winning

43/44x43/44 =your chance of winning

chance of tie (1/44x43/44)x2

if you don't do it twice then

43/44 u win
1/44 villian wins


I dont feel like mathing this out 'cause I have no calculator here.

[PokerSlut]

If you want to lock in the best deal for yourself, in that situation your opponent has 2 outs, so offer to run it 3 times.

[scdavis0]

run the whole deck out.. now your EV is infinity!

[cbloom]

Hmm.. of course you're right, apparently I suck at math.

BTW I see term 3 as

(1 - N/44) * N/43

which of course is indeed the same as term 2 :

N/44 * (1 - (N-1)/43)

though it's not obvious to me until you reduce them and see they are indeed equal.