The result of running it twice
I was playing in a 50/$1 cashgame at my buddy's house and this hand came up. I was in the SB my one friend on the button raised to $4.50 I had Aces (AS,AH) I called cuz all there was left was Button and the blinds, and could easily trap, my other Tight Aggresive friend makes it $11.50, the button folded I go all in over the top for my remaining $17 he calls and turns over QH,QD I suggest to run it twice cuz it was a big pot and I wanted to protect my stack (I only had 1 buyin that day), he agreed. 2 diamonds flopped (I forget the cards) the turn was a diamond and the river was the Ace of diamonds, ok fair enough, I'll chop then, my AA got cracked, but no.... the next board came K,A,3 turn T river J!!! That's right! My AA got cracked 2 consecutive times by QQ! Yup, I got unlucky.
Beats that are good like this one don't phase me, it only phases me when idiocy is in the mix, when people call for no reason even to them, or use luck to rationalize poor play.
Beats that are good like this one don't phase me, it only phases me when idiocy is in the mix, when people call for no reason even to them, or use luck to rationalize poor play.
Comments
I don't think either of you played it bad but sometimes cards do strange things.
/g2
Another /g2 instant classic....... well played.
He had one buy in and was playing with scared money.
AA over QQ is 81.5% favorite to win, 0.5% to chop. (for ease of math, I'm going to ignore the chop and I'll assume an 82% chance of winning
Running it once:
AA wins full pot: 82% of the time
AA loses full pot: 18% of the time
Running it twice:
AA wins full pot: 67% of the time
AA wins half pot: 30% of the time
AA loses full pot: 3% of the time
So, you're trading off some percentage of the full pot, to reduce the chances that you go broke.
Please run the analysis Zithal. Thanks.
Feel free to chat about it on MSN, Haddon.
Curious strategy...
The EV is no different either way, so what's the difference? It's a variance issue.
/g2
Huh? How is the EV the same?
/g2
That should go in the sticky. Interesting.
I will use this example to show that no matter how many times you run it, the EV does not change and only risk/variance is reduced. The pot is $47.50 [($4.5 + $17) * 2 + $4.50].
cards . win . %win . lose . %lose tie %tie EV
As Ah 1392614 81.33 312251 18.24 7439 0.43 0.815
Qd Qh 312251 18.24 1392614 81.33 7439 0.43 0.185
EV = 0.815 * $47.50 - 0.185 * ($47.50)
= $47.50 * (0.815 - 0.185) = $47.50 * 0.63
~ $30.
There is a 18.5% risk that Bobsi38127 will get a bad beat and bust out. By running it twice, he reduces his risk of losing ALL his money for the night to 3.4% (0.185 * 0.185). Running it twice does not change the probabilities and simply means splitting the pot in two.
EV = 1/2 * $47.50 * 0.63 + (1/2 * $47.50 * 0.63)
= $30.
Unfortunately for him in this example, the 3.4% chance of getting two bad beats happened and he busted out. I calculated the variance in the first case to be 1,361, then is reduced to 680 by running it twice.
Had he agreed to run it three times, the pot will be split in three but the probabilities and EV remains the same.
EV = 1/3 * $47.50 * 0.63 * 3
= $30.
However, he will have further reduced his risk of having an early night to only 0.6% (0.185 * 0.185 * 0.185).
Buddy doesn't let facts get in the way of a great story. The cards out don't make that big a difference in this example. Taking the Ad and 3 other diamonds out makes it approx 3.8% instead of Buddy's 3.4%. The AA is still a 80:20 favourite on the second run. Without knowing exactly the cards his approximation is good.
The monster pots are the rarities and no way am I passing up the best chance to scoop.
/g2
QQ will beat you with a flush (when it shares a suit with one of your Aces) only 1% of the time. QQ will beat you with a straight only 1.2% of the time.
AA hits a set and loses only 0.5% of the time whereas QQ hits a set and loses 2.9% of the time.
i still say your math is too simplistic, buddy. i don't have a 'run it x times' simulator but i sure wish i did!
each time you run it, chances are some of the outs for the underdog are used up (flush cards, straight cards) even if the underdog loses. for sure they are used up when he wins. and the rare occasion he hits and still loses also uses up his outs, too. i say his chances of winning the n+1 run, in general, are lower than his chances of winning the nth run due to these dead cards. more importantly, the odds of the AA hand busting out completely drop dramatically for n>3, even more than your numbers indicate.
so, given the option, i would always offer to run it at 4 or 5 times. frequently i'm giving up a small chunk of the pot but generally this guarantees me the the majority of the pot and protects my scaredy cat bankroll.
i want to run AA vs QQ 2,3,4 or more times for a million runs and see what the numbers are then. intuitively, i feel AA has a greater advantage the more times it is run since it wins 58% of the time unimproved and QQ fails to improve 75% of the time (with no dead cards). the more times you run it, the less chance QQ has to improve in each subsequent run the way i see it. tell me what i'm missing.
/g2
cards . win . %win . lose . %lose tie %tie EV
As Ah 1392614 81.33 312251 18.24 7439 0.43 0.815
Qd Qh 312251 18.24 1392614 81.33 7439 0.43 0.185
Approximately four out of five times, the player with AA will gain $100, while lose $100 the other times.
EV = 0.8133 * (+$100) + 0.1824 * (-$100) + 0.0043 * ($0)
= $81.33 - $18.24 + 0
= +$63.09
If you run it once, it will be impossible to gain the exact amount of $63. It will be an all-or-nothing run in which you either win $100 or lose $100, so the variance is large.
The more instances that you have AA against QQ, the closer you will get to your average expected profit of +$63 and the lower the variance will be. This is what happens when you run it more than once. The variance will keep getting lower, and your actual profit will converge closer and closer to your long-term EV of +$63.
You can do a mini-simulation by playing with yourself.
EV is calculated at a moment in time given the available information. If you peek at the flop, then you will have new information and there will of course be new probabilities and new EV. But pre-flop BEFORE any new information, the EV calculation is correct over the long run.
It seems that some readers are getting confused with running it more than once by peeking at what will happen in the first run. It does not change the fact that EV before any of the runs is the same and correct over the long run.
To paraphrase Led Zeppelin, "The EV remains the same"
/g2
If I ever play a cash game with anybody here, I will gladly accept running it x times whether I'm the underdog or favourite - since I know that EV will remain the same.
Simple explination:
Let's say that we ran every possibility of running it 3 times. Each player has their two cards, and then we deal out three sets of 5 cards:
A1 A2 A3 A4 A5
B1 B2 B3 B4 B5
C1 C2 C3 C4 C5
If we took into account every possibility for these 15 cards to be dealt, an individual set of cards will show up the same number of times in each of the 3 columns (A, B, or C). The EV on each row is the same no matter how many times it is run.
What possibly WILL change is the amount of variance involved. ie. Let's say that you were a 80% favorite with a hand preflop. If you ran it twice then you would be 64% to win the whole pot, 32% to split the pot, and 4% to lose the whole pot - but if you did not shuffle the deck, then there is a possibilty that even though the EV remains the same, the chances of splitting the pot may increase (since when you win one hand, some of your 'outs' for winning the next hand would be gone). I am in the process of running simulations on that right now.
I'll post my results here when they are complete.