Running it twice

Sometimes when two people get all-in during a hand, they choose to "run it twice." Normally when two people are all-in, the remaining community cards are dealt, and the winner wins the entire pot. When players run it twice, the remaining community cards are dealt twice, and a player wins the pot only if he wins both times (and the pot is split if each player wins once). For example, suppose the players are all-in preflop with AA against KK and decide to run it twice. If the first set of community cards is 2389J, and second set is 3489K, then the players chop the pot because the AA player won the first hand and the KK player won the second hand (assume no one hit a flush).

Players seem to choose to run it twice fairly often for big pots on televised high stakes cash games (I guess the casino allows it). Pokerstars doesn't have the option to run it twice, but I think Full Tilt or another major site has the option if both players agree. I've even seen players run it three times once on tv.

Why would players choose to run it twice? To reduce variance. Suppose both players are all-in in a 50/50. For this example suppose that the community cards are dealt with replacement (though in reality they are dealt without replacement). If they just ran it once, then each player would win the pot P with probability 1/2, and win 0 with probability 1/2. So the expected payoff is
(1/2) P + (1/2) 0 = P/2
and the variance is (P/2)^2 = P^2/4.
If they ran it twice, then both players win the pot with probability 1/4, and with probability 1/2 they chop. So the expected payoff is still P/2, but the variance is now (1/2)*(P/2)^2 = P^2/8. So the variance has halved, while the expected payoff remains the same. Thus if both players are risk-averse, they should choose to run it twice in this setting.

Risk-averse players would actually prefer to run it 3 times than to run it twice, assuming replacement. In fact, the optimal way to allocate the pot would be to not even run it at all, and just give each player his expected payoff. For example, if player 1 wins with probability q and there is no replacement, then you should just give him q*P and give player 2 (1-q)*P, and not even deal out the cards. This would give both players their expected payouts with zero variance, which is clearly desirable for risk-averse players. This would also help make games run faster, since no more cards need to be dealt once people are all-in. But I have a feeling that casinos won't be using this rule anytime soon ...

The previous analysis holds for cash games; however, in big tournaments there are actually benefits of taking a high-variance strategy if the payoff structure is very top-heavy. I'm pretty sure no sites or casinos allow players to run it twice in tournaments though.

In the previous examples we assumed the cards were dealt with replacement; but in reality they are not. Does this change anything? It isn't obvious at all, but it turns out that your expected payoff is the same in both settings.

Let P be the size of the pot.
Let p be the probability P1 wins the first hand.
Let w be the probability P1 wins the second hand given he wins the first hand.
Let L be the probability P1 wins the second hand given he loses the first hand.

P1's expected payoff running it once is p*P
P1's expected payoff running it twice (without replacement) is
pwP + p(1-w)P/2 + (1-p)LP/2
= (P/2)(pw + p + L - pL)

Now, note that the probability that P1 wins the second hand is the same as the probability that he wins the first hand. This is because for community card sequences c1 and c2, the probability c1 is the first sequence and c2 is the second sequence is the same as the probability that c2 is the first sequence and c1 is the second sequence.

So Prob(P1 wins 2nd hand) = p.
But we know (Prob P1 wins 2nd hand) = Prob(P1 wins 2nd hand | P1 wins 1st hand) * Prob (P1 wins 1st hand) + Prob(P1 wins 2nd hand | P1 loses 1st hand) * Prob(P1 loses 1st hand) = wp + L(1-p)

So we have p = wp + L(1-p)
So substitution into the expression above yields:
P1's expected payoff running it twice (without replacement) is
(P/2)(2p) = p*P

So P1's expected payoff is the same running it once and twice.