# Running it twice, odds calculation

I'm looking at a setting in a No limit Texas Hold'em hand. We start the hand with 6 people, only two will go to the flop. Ok, I'll admit this is a very unlikely situation:

Player 1 starts the hand with: T♥ T⋄ Player 2 starts the hand with: A♠ Q♠

Flop comes: T♠ T♣ Q⋄

Player 2 bets on the overpair with backdoor straight draw and backdoor flush draw. Betting goes round, player 2 ends up going all in. Player 1 obviously makes the call. Players agree to run it twice.

To win player 2 needs to hit either the Q⋄ Q♣ or J♠ K♠ for the higher quads or the royal flush.

Calculating the odds of player 2 winning both pots, would I be correct in saying:

``````   2/1081 will give me the odds for the first run
1/990 will give me the odds for the second run
``````

Somehow it does not feel right to multiply these two with one another to get the odds of player 2 winning the entire pot. It seems like I treat the two draws as independent, while in fact the results of the two will influence one another. If the first comes Qd Ks the person is drawing dead on the second. (I realise that on such ridiculously small odds this is marginal, but I am curious how to deal with it)

How should I adjust for this ?

Welcome to Poker.SE!

Your instinct to multiply is not incorrect, but I think there are some subtleties to the calculation which you might have missed.

There are 990 (`COMBIN(45,2)`) possible turn/river combinations (ignoring the order in which the cards come) on the first run and 2 of them are winners for Player 2.

Therefore, their chance of winning on the first run is `2 / 990 = 0.002 (0.2%)`.

After the first run, there are now 903 (`COMBIN(43,2)`, as 2 more cards are now known) possible turn/river combinations and only one of them is a winner for Player 2 (as they cannot hit the same cards which already won them the first run). Therefore, the chance of Player 2 winning the second run after winning the first is `1 / 903 = 0.001 (0.1%)`.

Putting this all together, the chances of Player 2 winning both runs would be:

`(2 / 990) * (1 / 903) = 0.000002237...`

or roughly 0.000224%, or 1 in 446,985.

As always, my math might be wrong, but this is my understanding! :)

• Ah, yes of course. I was calculating it as if i was one of the people in the hand with only the knowledge of that player. My mistake. Feb 2, 2017 at 18:53
• Would it make sense to multiply it with the 4 hands it is not allowed to be ? Qd Ks, Qd Js, Qc Ks, Qc Js. i.e. multiply the above by (990-4)/990 to rule this out ? I'm not sure if that solves the lack of independence in the draws Feb 2, 2017 at 18:54
• You probably could arrive at the same result by a different method (explicitly accounting for hitting individual cards out of the combos which win for P2), but it shouldn't affect the final result as to the chances of P2 winning both runs. Feb 2, 2017 at 19:06