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Would Love For Daniel To Discuss This Hand Of His


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#41 Phlatline

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Posted 30 March 2006 - 06:58 PM

View PostAndy Beal, on Thursday, March 23rd, 2006, 10:13 PM, said:

Well one might turn around and argue that variance is the friend of the fav since reducing variance just makes higher probability events more "likely" to occur on this small set of trials.On the other hand, one might very well argue that if you are the dog, then you NEED variance to help your low probability event "occur" and therefore should want to just run it once.
Above was your assertion in the direction I was talking about when I said "Andy, I think you have fallen into the gamble more to catch up when -EV and preserve bankroll mode when +EV mindset." Gamble more here means running once when -EV.Increased variance does not help a low probability event occur, it only increases the range of possible outcomes. Over a large sample the -EV will come out anyway.

View PostAndy Beal, on Thursday, March 30th, 2006, 4:45 PM, said:

On the other hand if the pot is $250K+ like on High Stakes Poker on GSN and Esfandiari turns over AA while Elezra turns over JJ, Elezra may very well choose to run it twice to give him a better chance at losing less. It just depends on the risk/reward characteristics of the individual players involved I think.
Becareful with your choice of words here, as Elezra does NOT have a better chance at losing less. He is simply limiting BOTH winning and losing potential, ie. variance. I agree that these considerations depend heavily on the risk tolerance of the individual player.Phlat__________ :club:

#42 Andy Beal

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Posted 30 March 2006 - 07:52 PM

View PostPhlatline, on Thursday, March 30th, 2006, 8:58 PM, said:

Above was your assertion in the direction I was talking about when I said "Andy, I think you have fallen into the gamble more to catch up when -EV and preserve bankroll mode when +EV mindset." Gamble more here means running once when -EV.Increased variance does not help a low probability event occur, it only increases the range of possible outcomes. Over a large sample the -EV will come out anyway.Becareful with your choice of words here, as Elezra does NOT have a better chance at losing less. He is simply limiting BOTH winning and losing potential, ie. variance. I agree that these considerations depend heavily on the risk tolerance of the individual player.Phlat__________ :club:
OH, again let me be clear. I wasn't saying that high variance "helps" a low probability event occur, I was saying one might try to argue that. Of course there is nothing one can do to hurt OR help a specific probablity event occur any more than one can affect the gravitational constant of the Earth.No he does have a better chance of losing less by running it 2x (of course, he has a worse chance of winning more too). It's basically the same caluclations that I already did for the Deeb-Negreanu hand here, post #7. If they run it once, DN has a 67% chance of losing $110k and 33% chance of winning $125.3k. If they run it twice, Daniel has a 44.4% chance of losing $110k, a 44.4% chance of chopping the pot (for a modest win of $7.65k due to the dead money in the pot), and only an 11.1% chance of winning $125.3k. So by running it twice, Daniel has a considerably better chance of NOT losing the max of $125.3k.
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#43 Shaffer

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Posted 05 April 2006 - 06:17 PM

In a situation where the EV is identical (as running it twice always is), I think the only variables in play are metagame effects of variance: (1) whether or not, if you bust the other player, they will get up and leave, and (2) whether or not, if you give them a beat, they will tilt.Say you've got a strong player sitting at the table that is unlikely to rebuy if they go bust, with a weaker player waiting in the wings. This pushes the advantage to running it once (whether dog or favorite), because of the chance to bust the player and lower the overall skill level of the table. Conversely, busting a weak player with a strong player waiting decreases your overall EV by making the table stronger. This situation gives value to running it twice, whether dog or favorite; keep the table weak, and increase your chances of getting your money in with better EV in future hands.On the psychological level, I can make an argument, I think, for running it only once as the dog in a situation where, if you lay a beat on a player for their whole stack, they will rebuy, go on tilt, and drastically increase your EV in the near future. In another hand, on an earlier night, DN was willing to call in what he knew was a very negative EV situation in order to put someone (I forget who off the top of my head) on tilt. While that's something I don't think I would ever do (and I don't agree with the play under the circumstances for sure), I see the logic behind it. In a situation that doesn't affect your EV, such as running it twice, I think you want to do whatever you can to bump up the variance if you feel you have a better psychological capacity for handling the swings than your opponent does.I think DN and FD's mutual decision to run it twice reflected their natural desire to lower variance as well as (possibly) a mutual respect insofar as neither one felt they would be likely to tilt the other. Maybe Freddy also felt like he had an advantage over DN so long as DN was playing hyper-aggressive, but felt that if he took 100k from him, Daniel would calm down, play tighter, and lower Freddy's EV - kind of an inverse tilt sort of thing.Who knows really? When EV is not affected, very small considerations can push the balance one way or another.

#44 Mriya

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Posted 07 April 2006 - 01:01 AM

first of all, I havent read all the replies, I just found 'Andy's' calculation is logically unjustified. Based on that, this can't be the Andy Beal who is a very good mathematician. Now, why did I say that?There is one fact, that 'Andy' ignored, and he called it 'subtlety', that there is only 52 card in a deck. Personally, I don't think this is subtle at all, nor it is a minor issue. In fact, this is the ONLY issue that matters. Ignoring this fact, which means the probability of drawing out as a underdog remains the same, the equity will OF COURSE, ALWAYS stays the same. You really dont need any calculation to draw that conclusion. Because all you did, is run the EXACTLY same hand a number of times, and each and everyone of them has the EXACTLY same probabilty of winning/losing.Now why isnt it good to run it more times when you are a FAVOURITE?NO, IT IS NOT THE VARIANCEThat is because, you are effectively giving more chances to get out DRAWN.REMEMBER there are only 52 cards in a deck, and are only a few cards that can help the DOG. As a favourite, you DONT WANT TO SEE THEM. If you ran it twice, you are given out more chances to SEE them, that is IT. The chances of hitting 3 cards when there is 35 cards left is much higher than there is 40 left, isn't that obvious?I think this is very simple to understand?

#45 Mriya

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Posted 07 April 2006 - 01:47 AM

Now because this is a 'mathematical' discussion, let's see some numbers. Lets use a rather simple example (coz I am lazy)AK vs KJ, and you are dead to a JACK on the flop: you have two chances to hit remaining 3 jacks in the deck, the estimated probability is: (considering only you and your opponent's hands)(3/45)*(39/44) = 5.9%-> one jack on the turn, a card that is NOT an ACE on the river (Situation A)(42/45)*(3/44) = 6.36%-> one card that is NOT an ace on the turn, and a jack on the river (Situation B)(3/45)*(2/44) = 0.3%-> hit runner runner Jacks (situation C)all toghether the estimated probability that KJ will draw out on AK is: 12.56% (might be a different from the simulation results, this is just an estimate anyway)So the probability of NOT drawn out is about 100% - 12.56% = 87.44% (Situation D)Now what if we run it twice? The first run obviously is identical to the abovenow the second run:provide that it is situation (D)we had a new set of probabilities, similar to the above caculation only now there are only 43 cards left: (3/43)*(37/42) + (40/43)*(3/42) + (3/43)*(2/42) = 13.12% (E)Notice 13.12% > 12.56%This is the probability that KJ drawn out on AK PROVIDE THAT it didnt drawn out in the first run, the overall probability for this to happen is: P(D) * P(E) = 87.44%*13.12% = 11.47%THIS IS A SPLIT POT, but this is only one situation.IN total we have the following situations: 1) when we draw out on AK at the one time, but didn't draw out at the other time (this has eight combinations, 1st turn, 1st river, 2nd turn, 2nd river, 1st turn and river and NO JACKS in 2nd, 1st turn and river and one ace one jack, 2nd turn and river and no jacks in 1st run, 2nd turn and river with one ace and one jack in 1st run) 2) when we draw out on AK at the first time AND we draw out on AK at the second time (this has eight combinations, 1st turn+2nd turn, 1st turn+2nd river, 1st river+ 2nd turn, 1st river + 2nd river, 1st turn and river + 2nd turn and no ace on the river, 1st turn and river + 2nd river and no ace on the turn, 1st turn and no ace on the river + 2nd turn and river, 1st river and no ace on the turn + 2nd turn and river) 3) we didnt draw out on neither runs: this is rather easy to calculate: 87.44%*(1-13.12%) = 75.98%Notice the drop in probability compare to situation Dwhich means what? your AK is that 87.44% - 75.98% = 11.46% less likely to hold for the whole pot!Since no one is paying me for doing this, Im not gonna bother about the details. But HOPEFULLY you can see why it is NOT GOOD to run more than once as a favourite.

#46 Shaffer

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Posted 07 April 2006 - 05:42 AM

I don't understand what all the math is supposed to be about here. The EV of running it twice is the same whether you run it once, twice, or eight times. The only thing you affect is the variance. In effect, you are putting yourself in an identical situation twice for half the stakes each time. Yes, if you are the favorite, you gain the additional chance of your opponent getting half the pot, but you also essentially kill your opponent's chance of scooping it with a bad beat. The only variable that is affected is your variance.

#47 Mriya

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Posted 07 April 2006 - 05:56 AM

READ my post carefully, your assumption is WRONGTHERE ARE ONLY 52 CARDS IN A DECK and it matters GREATLYthat is, if you didnt get the card you want from the 1st run, you have a HIGHER probability of hitting it at the 2nd run. If you run it a 3rd time, the probability is going to be even higher. Understood?if you cant understand the math, at least try to understand the logic

#48 dingas

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Posted 07 April 2006 - 11:33 PM

As I mentioned in the other thread, your logic is flawed. If you don't hit on the first run, you have a greater chance to hit on the second, but if you do hit on the first you have a lesser chance on the second. These two effects cancel each other out.The chance of only getting half is mitigated by the reduced chance of leaving with nothing, both for the favourite and for the underdog.
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#49 Shaffer

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Posted 08 April 2006 - 08:16 AM

Ah, I love being talked down to by someone that I (hope and pray) has never taken a statistics course. Mriya, you are simply wrong. Not "I disagree" wrong, but "2+2=5" wrong.Let's boil it down to the basics. Let's say you have your opponent drawing to 1 out with 1 card to come (under-set vs overset, for example). There's an even $1000 in the pot. With 4 cards on the board and 2 in each of your hands, that leaves 44 cards unaccounted for. You are a 43:1 favorite; your opponent has a 1 in 44 chance of sucking out.Therefore, running it once, your expected value is (43/44) * 1000 = $977.27.Running it twice isn't much more complicated. On the first card, your EV is calculated in exactly the same way: your chance of winning the first half of the pot is 43/44, just as it was running it once. Therefore, on the first run, your EV is (43/44) * 500 = $488.64On the second run, 1 out of every 44 times your opponent will have sucked out on the first run. When this happens, your opponent will have no outs left and be drawing dead, so your EV for this happening is (1/44)*(1/1)*500 = $11.36.The other 43/44 times, your opponent will not have sucked out on you and will be drawing live, with 1 out out of 43 cards. Therefore your EV in this instance is (43/44) * (42/43)*500 = $477.27.To get your total EV for running it twice you just add your three EV's together: $488.64 + $11.36 + $477.27 = (gasp!) $977.27, the exact EV of running it once.I've expanded this way more than it needs to be, with the hopes that you'll be able to follow and see why running it twice does not raise or lower your EV in any way at all. Your variance is reduced, but that's a complication more complex than you're probably ready for at this point.

#50 Vman96

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Posted 08 April 2006 - 01:41 PM

View PostShaffer, on Saturday, April 8th, 2006, 8:16 AM, said:

I've expanded this way more than it needs to be, with the hopes that you'll be able to follow and see why running it twice does not raise or lower your EV in any way at all. Your variance is reduced, but that's a complication more complex than you're probably ready for at this point.
Burnnnnnnnnned! You're the insult master! (Aqua Teen Hunger Force quote)And you're right...nice example too.

#51 Mriya

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Posted 08 April 2006 - 09:43 PM

I am glad that you have read the other post and was able to see my point. I have no objection to the conclusion that EV doesnt change when you consider the long run. My objection is to what Andy was saying:" run as many times as you like "That is just simply not correct. If anyone followed my calculation would understand this:If you are the favourite, your best chance to win the whole pot is to run it once. When you run multiple times, your chance of winning that proportion of the pot is going to be reduced. In your own example, this is to be shown by the reduced equity in the 2nd run: 1st run: $488.642nd run: $477.27Here the 2nd time your equity is reduced. That is for the same 500 dollars, your equity is higher in the first run. Sure it doesn't matter if you add them all up and talked about EV, by my point is very simple, if run it multiple times your are getting less in your chance of winning the whole 1000.Think it this way, if you deal every card in the deck out run by run, your opponent will almost always get a proportion of the pot. (apart from rare situations where your killer card presents itself in the same run with his) In this situation, he is getting a fair shot at 'his' equity. If now, everybody plays this way, all you have to do is to pay the exact money to get your equity and that would be a 'ultimate fair game' Because everyone makes the correct decisions and everyone gets his/her deserved equity (which is now the money they are paying), in a game like this everyone would go home with the money they brought in. But thats not poker. That would make the game pointless. Now, I think I have explained my point fairly simply, whether to appreciate it or not, it's up to you. As for your remarks on statistics and other things. Let me put it this way, I am NOT just a guy who had taken a statistic course in college.

#52 TheMathProf

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Posted 09 April 2006 - 09:32 AM

Allow me to make a couple of points:(1) The equity of the two runs are actually equal, unlike your claim. Your claim for the equity of the second run is $477.27, which actually only includes the times when you've won the first run, and also win the second. However, your second run also includes equity from the times where you've lost the first run (an equity of $11.36), which gives you a total second run equity of $488.63. You can't simply ignore this equity here.Now the equity of the first run is equal to $488.64 according to the calculations, which does seem to imply missing equity of a penny in the second run. This is just a particular nuance of the situation where we round to the nearest penny. As it turns out, the equity of the first run is exactly equal to 488 + 7/11, where 7/11 equals .63 repeating. Since the next digit is larger than 6, we round up, creating equity of $488.64. With the second run, the combined equity is 488 + 7/11 + 488 + 7/11 = 977 + 3/11, where 3/11 = .27 repeating, which rounds down, hence appearing to create the illusion of a missing penny.(2) The other problem with your arguments, however, seems to be your insistence on there only being value in winning the whole pot.In this situation, you can choose to win the whole pot 97.73% of the time and lose the whole pot 2.27% of the time.Or you can choose to half the pot 4.55% of the time and the whole pot 95.45% of the time.While some people find value in winning the whole pot the extra 2.27% of the time, other people find value in making sure that they NEVER walk away from the pot with nothing, which also happens 2.27% of the time.This is the same type of argument that I would expect from someone who insists on winning the most pots, regardless of how much he loses in the other pots, because the object of the game is to win pots.Except the problem is that it's not an argument of what's the way to win the most money in any given pot, but is what's the way to win the most money in the long term. As it turns out, it frankly doesn't matter, and it's only a question of variance.So if you're looking for the concession, of "Yes, it's more likely that you're going to win $1,000 in that pot by running it once", then yes, you're absolutely right. But you're wrong in your assertion that everybody would leave with the money they brought in with everybody getting "their fair share". People are still going to make decisions that have both positive and negative expected values to them, and there are going to be times in addition to those where the turn of the cards isn't going to resemble the long-term expected results. There's also the point to consider that not everybody makes the decisions that expected value would dictate.You could play this pot, exactly the way you've said, and walk home with nothing. The odds are slim, to be sure, but the odds aren't zero. If you end up with zero, you're certainly not heading home with what you came with.To further illustrate the point, imagine that the hand was run several times with one player ending up with a scant amount of cash based on actually hitting his hand, and the other player ending up with the majority of the 1000.Let's further suppose that on the very next hand, the exact same situation occurred, but with the players holding the opposite player's holdings. Are we to suggest now that the player should end up even in chips?The fact that the long run suggests that every player should be faced with situations from both sides doesn't mean that the opportunities will truly turn out to be equal, based on the players involved and based on the chip stack situations that they run into at the same time.

View PostMriya, on Saturday, April 8th, 2006, 10:43 PM, said:

I am glad that you have read the other post and was able to see my point. I have no objection to the conclusion that EV doesnt change when you consider the long run. My objection is to what Andy was saying:" run as many times as you like "That is just simply not correct. If anyone followed my calculation would understand this:If you are the favourite, your best chance to win the whole pot is to run it once. When you run multiple times, your chance of winning that proportion of the pot is going to be reduced. In your own example, this is to be shown by the reduced equity in the 2nd run: 1st run: $488.642nd run: $477.27Here the 2nd time your equity is reduced. That is for the same 500 dollars, your equity is higher in the first run. Sure it doesn't matter if you add them all up and talked about EV, by my point is very simple, if run it multiple times your are getting less in your chance of winning the whole 1000.Think it this way, if you deal every card in the deck out run by run, your opponent will almost always get a proportion of the pot. (apart from rare situations where your killer card presents itself in the same run with his) In this situation, he is getting a fair shot at 'his' equity. If now, everybody plays this way, all you have to do is to pay the exact money to get your equity and that would be a 'ultimate fair game' Because everyone makes the correct decisions and everyone gets his/her deserved equity (which is now the money they are paying), in a game like this everyone would go home with the money they brought in. But thats not poker. That would make the game pointless. Now, I think I have explained my point fairly simply, whether to appreciate it or not, it's up to you. As for your remarks on statistics and other things. Let me put it this way, I am NOT just a guy who had taken a statistic course in college.


#53 Barbarian

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Posted 21 April 2006 - 12:25 PM

*first post, high everyone :club: * Mriya seems to be arguing simply that running it once gives you the greatest chance of taking the whole pot. This is clearly true. Mriya also seems to be assuming taking the whole pot is 'best' or 'what one should want to do'. This is debatable. So Andy is not 'simply wrong' when he says he would like to run it as many times as possible when a favourite. It's his opinion, and Mriya doesn't seem to have offered any concrete argument against it. Mriya's argument essentially rests on the assumption that winning the whole pot has some kind of intrincic value; value other than the monetary value.

#54 cgrohman

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Posted 25 April 2006 - 02:44 AM

There might? also be a slight change in the math needed. When something is run twice after the flop, there are 2 independent turn and river cards. However, after the 1st run, those turn and river cards are NOT reshuffled back into the deck. Thus, if the person who is behind hits 1 (or 2) of his outs on the 1st run, he is less likely to hit them on the second and vice versa.

#55 topset

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Posted 20 September 2007 - 02:08 PM

Here it is very simply. We are playing with a deck of 16 cards containing:AAAA,KKKK,QQQQ,2222. No straights or flushes can come up. There are no burn cards.Player A holds: AAPlayer B holds: QQFlop/Turn is: A Q 2 2 8 cards remain in the deck. Only 1 of those improve Player B's hand and allow him to win--that card being the case Q. If any of the remaining 7 cards come Player A wins.If the players Agree to run the river 8 times, Player B will win only 1 out of those 8 times.Theory A: If the rule is that the who ever wins majority of the draws takes everything then Player A will always take everything. Player A should therefor elect to run it as many time as possible. In fact, in this particular contrived scenario, if he runs it at least three times, he is sure to win because he can lose only once out of those three times.Theory B: The pot is split up according to the number of times a player wins the redraws. This is the same as the two players deciding the odds and splitting up the pot right there. If player A is X% favorite and B is Y% favorite then A takes X% and B takes Y% out of the pot. This scenario does not change EV. It changes the variance only. It ensures that each player will only get what they are expected to win in the first place and takes the gamble out completely.Going back to theory A in which one player takes all, the favorite-to-win should always elect to run it as many times as possible and the underdog should clearly elect to run it as few times as possible. As I outlined earlier, the favorite-to-win will _always_ win if the hand is run only three times in that particular scenario.That should make it very clear to understand and help you determine what route to pursue next time you are in the situation--even though it seems counter-intuitive. If you are a favorite run it twice. If you're a dog, run it once.

#56 Naismith

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Posted 20 September 2007 - 02:15 PM

Who does everyone like at the final table of the main event? I think that Hachem guy might win even though he's the short stack.
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#57 Scott3705

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Posted 21 September 2007 - 06:03 AM

View PostNaismith, on Thursday, September 20th, 2007, 2:15 PM, said:

Who does everyone like at the final table of the main event? I think that Hachem guy might win even though he's the short stack.
Hachem won last year.... I've got my money on that Luckbox wearing the BoDog gear.

#58 Scott3705

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Posted 21 September 2007 - 06:04 AM

View Posttopset, on Thursday, September 20th, 2007, 2:08 PM, said:

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