Posted by Dr Fro 10:48 AM

I love High Stakes Poker. I watch every episode. I wanted to puke on behalf of Kid Poker when his boat (66655) lost to Gus Hansen's quads (5555x). Gus hit a one-outer on the turn. Brutal!

One thing I like about the program is that it shows a lot of the shenanigans that go on ancillary to the actual poker (e.g. the prop bets on the composition of the flop). It also routinely shows players deciding to "run it twice", a practice that was common in the $2-$5 PLHE game I played in Houston but is not common anywhere I currently play.

I have often wondered (but never actually calculated) if this is inherently beneficial to either player. Intuitively, I figured that if it were done "with replacement", the effect on either player's EV would be nil. However, the cards dealt on the first run are not replaced in the deck. I had always assumed that the "without replacement" must affect the math, but I had never sat down to actually prove this. Now I have.

Guess what I learned. Do you think that "running it twice" systematically favors the EV of the favourite, the underdog or neither?

2 Comment(s):

Posted by  Padilla, at 12:34 PM, August 16, 2006  

We're talking clean outs here...

Intuitively:
You already know the EV of the first 'run'. The favorite is ahead by a percentage. If you assume those percentages hold up, then the 2nd 'run' would depend on which side of those percentages the first 'run' fell into.

If there was a 4-1 favorite in run #1, and the favorite lost, he's now an even greater favorite in run #2, as the dog's outs are even less.
If there was a 4-1 favorite in run #1, and the favorite wins, you'd think that the dog would pick up some odds...and he may, but he may also still be a sizeable dog in run #2.
My guess is that when it's closer to 50/50 going into the deal, you may switch sides between favorite and dog, but as always, you want to get your money in with the best hand.

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Posted by  Junelli, at 1:07 PM, August 17, 2006  

I don't think it would change at all.

The number of outs you have is the same in either instance, because you decide to run it twice prior to seeing either "run".

Therefore, if you have 9 outs, that's exactly what you have: 9 outs. You have it twice, but only for half the pot. Therefore, I would assume that the odds remain the same.

Your chances of winning the entire pot stay the same. 9 outs turns into 18 outs, but you have to win it twice.

Your chances of winning half the pot double as your outs double. But then again you're only winning half the pot.

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