Question about "running it twice."

[chesspain]

I noticed from watching an episode of "High Stakes Poker" that when a hand was run twice, the first set of board cards was not reshuffled back into the deck before the next board was run. I would have thought that each board would be an independent trial, but obviously that isn't how it is done.

Can anyone shed some light on the this?

[Iron Tigran]

It doesn't affect the odds to just run it twice w/out reshuffling. Just reducing variance.

[dogdrool]

[ QUOTE ]
I noticed from watching an episode of "High Stakes Poker" that when a hand was run twice, the first set of board cards was not reshuffled back into the deck before the next board was run. I would have thought that each board would be an independent trial, but obviously that isn't how it is done.

Can anyone shed some light on the this?

[/ QUOTE ]

You might want to go ask over in the Theory or Probability forum.

But replacing and reshuffling does not effect the odds. I once had to put a lot of thought into this to prove it to myself. My main counter point was this:

Suppose that one of the players has a one-outer. If he spikes that one out on the first run, and the cards aren't replaced, then he has no chance of hitting it on the second one. So at best, over two runs, he can win once. He cannot win both runs.

But, with replacement, he could get super lucky and spike his one out on both runs and scoop the whole thing.

Even so, the two methods are equivalent in terms of EV. If you work out the math, the reason is this:

Without replacement, consider the case where the player does not hit is one out on the first run. Then the odds of him spiking the one out on the 2nd run is increased.

Extended further, if the board was run 9 times without replacement, then the player would be guaranteed (almost) to spike the out on exactly one of the runs. But without replacement, it's very possible that he would fail to spike the out on any of the runs.

[pudley4]

Simplify it down to two cards. Player A wins if card 1 hits, Player B wins if card 2 hits. Obviously the EV for each player is .5

If there's no replacement and it's run twice, each player will win once and lose once. EV is (1*1 + 0*1)/2 = .5 for each player.

If there is replacement, then Player A will win it twice (.5*.5)=25%, he'll win it once (.5*.5 + .5*.5)=50% and he'll win it zero times (.5*.5)=25%. This also gives him an EV of (2*.25 + 1*.5 + 0*.25)/2 = .5