Run it twice, calculating equity % for both runs.

You can use Equilab for this and you remove the cards of the first showdown like this:
Dead cards: A♠T♣2♥4♠2♣
       Equity   Win   Tie
SB     88.40%  88.15%   0.25% { QhQs }
BB     11.60%  11.35%   0.25% { ThTs }

That means you have 88% equity in run 2 to win the 2nd showdown.

Beware that you can's run Equilab if you have your PokerStars/FTP client open.

Cheers,
SDK1987

Great, so in the underdog equity vision :
- 18,43% of win flop1
- 11,35% of win flop2 (after winning First flop; otherwise he runs again at 20%-80% QQ)

For BOTH events happening and TT win it all = 2,09% ?
Is this correct ?

So in this example "running it twice" preflop grants me (QQ):
- ??? % of winning both
- ??? % of winning one pot
- 2,09% of losing both

Could you please help me figure out these % ?
Just a ballpark % as there will always a minor split %.

Why exactly are you asking this? Your EV is the same if you run it just once or if you run it 10 times.

Originally posted by la55i
Why exactly are you asking this? Your EV is the same if you run it just once or if you run it 10 times.

I agree that long term EV is the same, no problem with that.

I'm considering "this event" outcome only. Say it's the final table live heads'up for 1 Million USD, only the outcome of THIS event would matter, after this, you retire from poker.

If I run it once, I have 80-20% for QQ. - 1 out of 5 times I will be busted and villain wins it all. Correct.

If we run it twice, the chance of TT winning both boards wouldn't be much much lower than 20% ? (Say 18,4% * 11,35% = 2% only). - Is this correct ?

So in this singular event, if running twice, how many times QQ will win only half the pot ?
That's the question, not worrying about long term EV.

I want to know what's the % that QQ wins it all, % that splits the pot and keep going on the heads'up and the chance that TT win's it all by busting you on both runs.

Yes if you run it twice, villain has just about 2% chance to win both runs.

On the first run you have something like 81% equity. If you then add 5 dead cards that are not T (so villain has 2 outs still on the second run) then your equity drops to something like 80%. Of course it could decrease even more if you hit a set on the first run, because then if villain hits a set on the second run you have less outs to improve your hand. So to be fair and take some kind of average value, let's say your equity on the second run is 79%?

You will win .81*.79 = 64% of the time.

So if you win both runs 64% of the time and villain wins both 2% of the time, then you would split the pot 34% of the time?

That's exactly what I'm trying to figure it out (for a singular event).

If this is correct and we round up to:
64% QQ wins it all
34% each player wins half pot (you get your $ back minus rake)
2% busted

Would it effectively means giving away "16%" of my 80%-20% all or nothing, to be 98% sure I win or remain on the game. Is this calculation right?

My brain is a bit fuzzy now, but yes that seems about right. If you run it twice, it has the same expectation on the long run but it will cut down the variance. You are more likely to stay in the game.

However, I'm not sure if that is always the best thing to do. I'm not a tourney player and don't know a lot of those but I would guess that at some stages of the game you might want to take a bigger risk and double up because it could help you later on the tournament. For example if you are on a final table and you and your villain have 2 of the biggest stacks in the table. There are still one table full of players. Do you want to make sure you can keep on playing or do you want to take slightly bigger risk and bust the villain and get a huge chip lead?

I don't know what would be the correct thing to do there but that just came to my mind.