Is Running it Twice a Sucker Bet? (Long)

[TrvChBoy]

Comments please...

Is Running it Twice a Sucker Bet?

The newest poker television shows including GSN’s High Stakes Poker and UPN’s Cash Poker introduced us to the concept of Running it Twice. To Run it Twice, generally the following situation is present

The game is a cash game, not a tournament
The pot is heads-up (just two players)
One player is all-in
There is at least one more card to come

If both players agree to Run it Twice, the dealer deals the remaining cards twice. For example, if there is only one card left to come, the dealer would deal two separate river cards.

If one player wins with both river cards, that player wins the entire pot
If one player wins with one river card and the other player wins with the other river card, the pot is split.

The question I had when seeing this is: What is the advantage or disadvantage of this bet? I searched on the Internet and read a lot of opinions that there is no real advantage or disadvantage; it only reduces variance by increasing the chances of a split. I also wondered if there was more of an advantage if one player was a clear favorite. Of course, being from the Show-Me-State, I was frustrated with reading other people’s opinions on the subject, so I decided to calculate the Expected Value (EV) of running it twice.

EV is the correct way to calculate whether a bet is a sucker bet or not and it is as simple as multiplying the amount you expect to win by the probability of winning, then subtracting the amount you expect to lose multiplied by the probability of losing.

For example, the EV of betting $1 on red on an American roulette wheel is

Win: Red: +$1 x 18/38 = + $0.47
Lose: Black -$1 x 18/38 = - $0.47
Lose: Green -$1 x 2/38 = - $0.05
Total: = - $0.05

Since there are 18 reds, 18 blacks, and 2 greens out of 38 possible outcomes on the American roulette wheel.

So the EV of betting red on an American roulette wheel is -$0.05, and over a long period of spins, I expect to lose an average of $0.05 per spin. Since the EV is negative, this qualifies as a sucker bet. I may win any given bet, or any given night, or any given week, but over thousands of bets, I will lose, and at the end of thousands of bets I will have lost about $0.05 per spin.

EV is the correct way to evaluate any bet, and it is the correct tool to see if Running it Twice is a sucker bet or not.

For this investigation, I calculated the EV of six scenarios:

Running it Once as a Long Shot
Running it Twice as a Long Shot
Running it Once as Even Money
Running it Twice as Even Money
Running it Once as the Favorite
Running it Twice as the Favorite

For this test, I selected a heads-up Hold-Em game with $100 from each player in the pot and one card to come. In this situation, there are 2 cards in Player A’s hand, 2 cards in Player B’s hand, and 4 cards on the board for a total of 8 seen cards and 44 unseen cards.

For the long shot scenarios, I gave our underdog 4 outs to a win, consistent with an inside straight draw.

For the even-money situation, I gave each player 22 outs to a win, which might be impossible in Hold’Em, but is a good check on my math, since the EV must be exactly $0.00 for this perfect coin flip situation.

For the favorite scenarios, I gave our favorite player 40 outs to a win, which is consistent with a made hand threatened only by an inside straight draw. Notice that this is the exact opposite of the long-shot scenario, so it is another good check on my math; the EV for the favorite scenario must be the same number as the EV of the long-shot scenario, but the EV for the favorite scenario will have a positive sign.

I will leave the calculations in the files section for those who wish to review it. Here are my findings

Running it Once as a Long Shot (4 outs): EV -$81.82
Running it Twice as a Long Shot (4 outs): EV -$81.82

Running it Once as Even Money (22 outs): EV $0.00
Running it Twice as Even Money (22 outs): EV $0.00

Running it Once as the Favorite (40 outs): EV +$81.82
Running it Twice as the Favorite (40 outs): EV +$81.82

From this experiment, I conclude that there is no long-term financial advantage or disadvantage to Running it Twice. The EVs are the same whether you are way behind, looking at a coin flip, or way ahead.

One thing Running it Twice does do: It increases your chances of splitting the pot while at the same time decreasing your chances of winning or losing the entire pot.

Running it Once as a Long Shot (4 outs)

Win the pot: 9.1%
Split the pot: 0.0%
Lose the pot: 90.9%

Running it Twice as a Long Shot (4 outs)

Win the Pot: 0.6%
Split the Pot: 16.9%
Lose the Pot: 82.5%

To check my math, see the Excel spreadsheet at
http://www.saltlakepoker.com/poker/runningittwice.xls

So when running it twice as a long shot, you get the advantage of splitting more often, but you give up a considerable amount of your probability of winning the entire pot. Since the EV of Running it Once is the same as the EV of Running it Twice, there is no financial advantage or disadvantage in the long run, but you may appreciate the reduction in variance. When running it twice, your short-term fluctuations will be lower, but your long-term overall win or loss will be unchanged. It is desirable to reduce your short-term fluctuations when playing high for your bankroll, or when playing against opponents who are less skilled than you.

When you get the chance to run it twice, decide whether or not you want to decrease your chances of winning and increase your chances of splitting. If you do, then go ahead and run it twice…whether you are way behind, even money, or way ahead, I assure you it’s not a sucker bet.

Organizer Eric (Full Tilt: TrvChBoy)
http://poker.meetup.com/939/
nelsonericrobert@yahoo.com

[Ncoe]

Isn't this intuitively obvious for you?

[OrangeKing]

Yeah, I'm surprised it required this long an explanation. It's well known that running it twice (or any number of times) simply reduces variance without changing EV. It's simply like being in that situation twice - you're helping yourself get to the long run faster.

[Vetgirig]

Runnings it twice is done for one and only one reason - to lower the variance! It doesn't change the EV but it does make it so that one are not fair behind och far ahead of once expected value in the long run from this kind of situations.

Running it twice takes away some of the luck moment and makes it more a skill game.

[TrvChBoy]

[ QUOTE ]
Isn't this intuitively obvious for you?

[/ QUOTE ]
Yep, but I was frustrated with only reading other people’s opinions on the subject with no supporting documentation. There are people that will argue it is not neutral EV, and this is a better defense than just quoting Internet opinions.

[DarkMagus]

There's nothing wrong with verifying a concept for yourself just to be sure. If nothing else, it will increase your understanding of the concept more than just reading and accepting it will.

[tarheeljks]

use the search function; this question gets posted a lot

[robsmith82]

While your analysis is completely correct if the 2 runs were independent events, did you take into account the card removal effect of running it twice? I'm not proficient enough to model this mathematically, but I know enough statistics to see that card removal will play a (albeit very small) part in skewing the EV's. If someone could investigate this for me or point me to a link where this has been looked at that would be great.

[OrangeKing]

[ QUOTE ]
While your analysis is completely correct if the 2 runs were independent events, did you take into account the card removal effect of running it twice? I'm not proficient enough to model this mathematically, but I know enough statistics to see that card removal will play a (albeit very small) part in skewing the EV's. If someone could investigate this for me or point me to a link where this has been looked at that would be great.

[/ QUOTE ]

I don't have a link, but this doesn't skew the EV at all. This part isn't intuitively obvious, because after you take the first card off, the EV of the second run is different. However, this is only because the first run already was a win/loss for one player, thus taking one of their cards out of the deck. The EV of both runs is still equal to running it once.

Here's an example:

I have a one outer on the river, and we decide to run it twice. There are 44 cards left in the deck. The pot is 1. My EV is 1/44 (.023) of the pot, while yours is 43/44 (.977).

We run it once. 43/44 of the time, you win the first half pot (.5). 1/44 times, I win the pot (.5).

After I win, you always win the second run. So 1/44 times, we split the pot; .5 for me, .5 for you.

43/44 times, you've won .5. I will now win the other half 1/43 times; you will win that half 42/43. The overall EV of this situation is still 1/44 for me:

(1/44)(.5) + (1/43)(43/44)(.5)
(.023)(.5) + (.023)(.5)
.023 (or 1/44)

[TrvChBoy]

[ QUOTE ]
While your analysis is completely correct if the 2 runs were independent events, did you take into account the card removal effect of running it twice? I'm not proficient enough to model this mathematically, but I know enough statistics to see that card removal will play a (albeit very small) part in skewing the EV's. If someone could investigate this for me or point me to a link where this has been looked at that would be great.

[/ QUOTE ]

I did consider the removal, and the EV's were identical, even when I showed 6 decimal places.

Here is the link to my spreadsheet:
Running It Twice EV Spreadsheet