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Thread: Anti-DN/Ivey pool?

  1. #161
    I could be wrong (this type of math is not in my wheelhouse) but I think your mistake is looking at them singularly in each event. They can't bust at the same point so they are never x% to win, it changes when one of them busts.

    For simplicity's sake suppose we have a 10 player tournament and Ivey and Negreanu are obviously more likely to win than the other players:

    Players 1-8: 9% to win
    Ivey:14% to win
    Negreanu: 14% to win

    you're looking at it as if they have a 28% chance to win, which they do at the outset. But with 5 players left suppose Ivey is eliminated, Negreanu is no longer 14% to win, his equity jumps up to about 19%, so even when one is eliminated the other gains, and in 100-150 player fields the added win-equity could be pretty substantial, maybe as high as 1-2% depending on when the other player busts, and even higher at a final table. Over the course of 50 events these .5%'s add up.

    You have to calculate it as if they are 28% to win at the start only, but when one is eliminated you have to factor in the other's new equity which will always be + (how much depends on entries, current chip-stacks and how far along the event is). The way you calculated your odds doesn't account for this readjustment when one of them busts and the other is still in it.
    I write things about poker at my Poker Blog and elsewhere on the Internets

  2. #162
    Owner Dan Druff's Avatar
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    Quote Originally Posted by Steve-O View Post
    I could be wrong (this type of math is not in my wheelhouse) but I think your mistake is looking at them singularly in each event. They can't bust at the same point so they are never x% to win, it changes when one of them busts.

    For simplicity's sake suppose we have a 10 player tournament and Ivey and Negreanu are obviously more likely to win than the other players:

    Players 1-8: 9% to win
    Ivey:14% to win
    Negreanu: 14% to win

    you're looking at it as if they have a 28% chance to win, which they do at the outset. But with 5 players left suppose Ivey is eliminated, Negreanu is no longer 14% to win, his equity jumps up to about 19%, so even when one is eliminated the other gains, and in 100-150 player fields the added win-equity could be pretty substantial, maybe as high as 1-2% depending on when the other player busts, and even higher at a final table. Over the course of 50 events these .5%'s add up.

    You have to calculate it as if they are 28% to win at the start only, but when one is eliminated you have to factor in the other's new equity which will always be + (how much depends on entries, current chip-stacks and how far along the event is). The way you calculated your odds doesn't account for this readjustment when one of them busts and the other is still in it.
    I understand what you are saying mathematically, and in theory you are correct.

    However, I don't believe that this really will play much into the situation, because one player busting out of a large/medium field event is negligible (no matter how good), and there will be so many good opponents (especially late) in the small-field events that I don't think a Negreanu or Ivey bustout changes very much as far as percentage for the other to win.

  3. #163
    Quote Originally Posted by Dan Druff View Post
    Quote Originally Posted by Steve-O View Post
    I could be wrong (this type of math is not in my wheelhouse) but I think your mistake is looking at them singularly in each event. They can't bust at the same point so they are never x% to win, it changes when one of them busts.

    For simplicity's sake suppose we have a 10 player tournament and Ivey and Negreanu are obviously more likely to win than the other players:

    Players 1-8: 9% to win
    Ivey:14% to win
    Negreanu: 14% to win

    you're looking at it as if they have a 28% chance to win, which they do at the outset. But with 5 players left suppose Ivey is eliminated, Negreanu is no longer 14% to win, his equity jumps up to about 19%, so even when one is eliminated the other gains, and in 100-150 player fields the added win-equity could be pretty substantial, maybe as high as 1-2% depending on when the other player busts, and even higher at a final table. Over the course of 50 events these .5%'s add up.

    You have to calculate it as if they are 28% to win at the start only, but when one is eliminated you have to factor in the other's new equity which will always be + (how much depends on entries, current chip-stacks and how far along the event is). The way you calculated your odds doesn't account for this readjustment when one of them busts and the other is still in it.
    I understand what you are saying mathematically, and in theory you are correct.

    However, I don't believe that this really will play much into the situation, because one player busting out of a large/medium field event is negligible (no matter how good), and there will be so many good opponents (especially late) in the small-field events that I don't think a Negreanu or Ivey bustout changes very much as far as percentage for the other to win.
    Certainly, but the difference when they both are in say the final 50 (or even 100) isn't negligible imo. Even if you consider it an average of .25% added to their win rate, over 50 events it adds up. Even if they don't gain added equity (i.e. the remaining field is just as good) they still gain equity along with the rest of the field. The deeper both make it the more this comes into play.

    I wonder what the odds of both players making final 20 or so would bed for each event, as this would drastically change their odds when on is eliminated adding at least x% to their win rate (I'm too lazy to calculate it out)?

    *EDIT* need someone good at math to chime in here right about now
    I write things about poker at my Poker Blog and elsewhere on the Internets

  4. #164
    Gold 4BET's Avatar
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    Fucking Ivey giving us a early unwanted sweat, Phil Ivey US 440,000 85,000
    Michael Mizrachi US Busted
    -Allergic to the struggle

  5. #165
    Really good article on WSOP ROI here:

    http://t.co/jzrULTnBWv

    The event that Ivey is in now is probably his 4th best chance to win, right after the other PLO events. Not really worried about DN at all.

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    Last edited by Shizzmoney; 05-28-2014 at 07:52 PM.

  6. #166
    Quote Originally Posted by Steve-O View Post
    I could be wrong (this type of math is not in my wheelhouse) but I think your mistake is looking at them singularly in each event. They can't bust at the same point so they are never x% to win, it changes when one of them busts.
    In any reasonable size field this effect is extremely small/negligible.
    Quote Originally Posted by abrown83
    I'm going to come across as a bit of a douche but I really know more about this then anyone on this board by miles.

    ...if Trump is nominee he wins Presidency easily. Angry Blue Collar Whites will have record turnout.

  7. #167
    Owner Dan Druff's Avatar
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    JC Tran flopped a small flush while Ivey flopped a straight with no flush draw.



    Onto the next one.

  8. #168
    Quote Originally Posted by Corrigan View Post
    Quote Originally Posted by Steve-O View Post
    I could be wrong (this type of math is not in my wheelhouse) but I think your mistake is looking at them singularly in each event. They can't bust at the same point so they are never x% to win, it changes when one of them busts.
    In any reasonable size field this effect is extremely small/negligible.
    I definitely agree with this, but these 2 will be playing in quite a few events where they are likely to be in the final 100 together. It stands to reason that they will be in the final >50 together multiple times as well, which is the point I don't think it's negligible anymore, and if they both make the same final table (or even top 20) it's certainly significant as we could be talking a full % point or more added to their win rate.

    If I was going to bet against DN/PI I'd have someone good with the math run the numbers looking at their WSOP cashes and WSOP final tables to determine the frequency they would both make deep runs in the same events, especially the 2-7 and One Drop type tourneys.
    I write things about poker at my Poker Blog and elsewhere on the Internets

  9. #169
    Gold 4BET's Avatar
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    Quote Originally Posted by Dan Druff View Post
    JC Tran flopped a small flush while Ivey flopped a straight with no flush draw.



    Onto the next one.
    This was a nice hand as Ivey got it all in drawing almost stone dead, Only outs being runner runner chop, We have faded the first of many sweats, 2/7 low ball and one drop I suspect will be the biggest sweats
    -Allergic to the struggle

  10. #170
    They both played the $1k PLO tournament (Event #3). Negreanu after he busted the Mixed Max, and Ivey before he busted the Mixed Max.
    Although they are both listed on the wsop.com chip counts as having survived Day 1, they are not listed in the official WSOP end of the day report. The event got 1,128 runners, and according to WSOP.com, 106 remain. The chipleader has 133k. Wsop.com's end of the night update (5/29/2014 1:29:35 AM PST) also states "Players busting today included Phil Ivey, Daniel Negreanu..." (They really should fix their chip counts with 106 players left!)

    Anyway, later today is the $10k 2-7 Limit Triple Draw Lowball event (#5), which is one of the high buy-in small field events that Negreanu and Ivey have their best shots in. (My calculations in post #75 above gave them each a 60:1 shot to win Event #5).
    Last edited by Sandwich; 05-29-2014 at 04:38 AM. Reason: Edited to reflect info in WSOP Official report.

  11. #171
    Quote Originally Posted by Dan Druff View Post
    Blue: 0.67% * 8 = 5.33%
    Green: 0.40% * 3 = 1.20%
    Red: 0.25% * 9 = 2.25%
    Orange: 0.17% * 5 = 0.85%
    TOTAL = 9.63%
    I do not believe your math is correct. Let's take a heads-up event with only 2 players. Player A has a 50% chance of winning the event. Player A plays two of these events. He does not have a 50% + 50% (100%!) chance of winning at least one of them! Instead we must take the probability that A wins neither, i.e. losing both in a row which equals (1/2)*(1/2) or 1/4 or 25%. This is the same as the probability of flipping tails twice in a row. The probability then of this *not* happening (i.e. Player A does *not* lose both heads up matches) is 1 - 1/4 which equals 3/4 or 75%.

    Revising the above, we then say that the likelihood of, say, DN losing all of the events is:

    P(LOSE) = (99.33%^8) * (99.66%^3) * (99.75%^9) * (99.83%^5) = 90.93%.

    Therefore the probability of winning one of them is

    P(WIN AT LEAST ONE) = 1 - P(LOSE) = 9.07%.

    Close to what you had but a bit more accurate.

    If we double all the amounts since DN and PI each play then we have a probability of them losing all of the events as:

    P(LOSE) = (99.33%^16) * (99.66%^6) * (99.75%^18) * (99.83%^10) = 82.69%

    and therefore:

    P(WIN AT LEAST ONE) = 1 - P(LOSE) = 17.31%

    Or am I mistaken?

     
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  12. #172
    Quote Originally Posted by system.out.println View Post
    Quote Originally Posted by Dan Druff View Post
    Blue: 0.67% * 8 = 5.33%
    Green: 0.40% * 3 = 1.20%
    Red: 0.25% * 9 = 2.25%
    Orange: 0.17% * 5 = 0.85%
    TOTAL = 9.63%
    I do not believe your math is correct. Let's take a heads-up event with only 2 players. Player A has a 50% chance of winning the event. Player A plays two of these events. He does not have a 50% + 50% (100%!) chance of winning at least one of them! Instead we must take the probability that A wins neither, i.e. losing both in a row which equals (1/2)*(1/2) or 1/4 or 25%. This is the same as the probability of flipping tails twice in a row. The probability then of this *not* happening (i.e. Player A does *not* lose both heads up matches) is 1 - 1/4 which equals 3/4 or 75%.

    Revising the above, we then say that the likelihood of, say, DN losing all of the events is:

    P(LOSE) = (99.33%^8) * (99.66%^3) * (99.75%^9) * (99.83%^5) = 90.93%.

    Therefore the probability of winning one of them is

    P(WIN AT LEAST ONE) = 1 - P(LOSE) = 9.07%.

    Close to what you had but a bit more accurate.

    If we double all the amounts since DN and PI each play then we have a probability of them losing all of the events as:

    P(LOSE) = (99.33%^16) * (99.66%^6) * (99.75%^18) * (99.83%^10) = 82.69%

    and therefore:

    P(WIN AT LEAST ONE) = 1 - P(LOSE) = 17.31%

    Or am I mistaken?

    you are correct
    Quote Originally Posted by abrown83
    I'm going to come across as a bit of a douche but I really know more about this then anyone on this board by miles.

    ...if Trump is nominee he wins Presidency easily. Angry Blue Collar Whites will have record turnout.

  13. #173
    Quote Originally Posted by Steve-O View Post
    Quote Originally Posted by Corrigan View Post

    In any reasonable size field this effect is extremely small/negligible.
    I definitely agree with this, but these 2 will be playing in quite a few events where they are likely to be in the final 100 together. It stands to reason that they will be in the final >50 together multiple times as well, which is the point I don't think it's negligible anymore, and if they both make the same final table (or even top 20) it's certainly significant as we could be talking a full % point or more added to their win rate.

    you are conditioning on the fact that they both make it deep, but we don't care what the probability is that they win given (conditioned on) that they make it deep. We only care about the probability that they win a tournament given a field size. In any decent size field (i.e. not a 2 table SNG) their probabilities of winning or more or less independent events.

    I get what you are saying, but that is a different probabilistic event. Like the chance they both make it deep (lets say final 6-7% of a field) even with a skill edge (lets say 10% chance of doing it) is very small 10% * 10% = 1%. Now you have to multiply the effect you are talking about ("a full % point") by this 1% because it is a parlay of events, and you can see it should be negligible on overall calculations.
    Quote Originally Posted by abrown83
    I'm going to come across as a bit of a douche but I really know more about this then anyone on this board by miles.

    ...if Trump is nominee he wins Presidency easily. Angry Blue Collar Whites will have record turnout.

  14. #174
    Owner Dan Druff's Avatar
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    Quote Originally Posted by 4BET View Post
    Quote Originally Posted by Dan Druff View Post
    JC Tran flopped a small flush while Ivey flopped a straight with no flush draw.



    Onto the next one.
    This was a nice hand as Ivey got it all in drawing almost stone dead, Only outs being runner runner chop, We have faded the first of many sweats, 2/7 low ball and one drop I suspect will be the biggest sweats
    To be fair, he repeatedly got lucky to survive as long as he did, including a 2-outer.

  15. #175
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    Quote Originally Posted by system.out.println View Post
    Quote Originally Posted by Dan Druff View Post
    Blue: 0.67% * 8 = 5.33%
    Green: 0.40% * 3 = 1.20%
    Red: 0.25% * 9 = 2.25%
    Orange: 0.17% * 5 = 0.85%
    TOTAL = 9.63%
    I do not believe your math is correct. Let's take a heads-up event with only 2 players. Player A has a 50% chance of winning the event. Player A plays two of these events. He does not have a 50% + 50% (100%!) chance of winning at least one of them! Instead we must take the probability that A wins neither, i.e. losing both in a row which equals (1/2)*(1/2) or 1/4 or 25%. This is the same as the probability of flipping tails twice in a row. The probability then of this *not* happening (i.e. Player A does *not* lose both heads up matches) is 1 - 1/4 which equals 3/4 or 75%.

    Revising the above, we then say that the likelihood of, say, DN losing all of the events is:

    P(LOSE) = (99.33%^8) * (99.66%^3) * (99.75%^9) * (99.83%^5) = 90.93%.

    Therefore the probability of winning one of them is

    P(WIN AT LEAST ONE) = 1 - P(LOSE) = 9.07%.

    Close to what you had but a bit more accurate.

    If we double all the amounts since DN and PI each play then we have a probability of them losing all of the events as:

    P(LOSE) = (99.33%^16) * (99.66%^6) * (99.75%^18) * (99.83%^10) = 82.69%

    and therefore:

    P(WIN AT LEAST ONE) = 1 - P(LOSE) = 17.31%

    Or am I mistaken?
    Yes, you're right.

    Sadly this actually came to mind but I was too rushed and lazy to really think about it, so I just did it the "easy way" which was also somewhat inaccurate. My stats professor from 1991 would be really disappointed with me.

    Adding the purple events:

    P(LOSE) = (99.33%^16) * (99.66%^6) * (99.75%^18) * (99.83%^10) * (97.50% ^ 2) * (98.89% ^ 2) = 76.87%

    1 - 76.87 = 23.13%

    That makes our bet look slightly better.


  16. #176
    I bet 2500 also but someone I know that's pretty smart made their side about -220 - obviously it all hinges on how much better they are than the competition. 2014 is not 2008 and the proliferation of mixed games online has blunted their edge in those events, but for the 1-drop i think their edge is way higher. Ivey might have the added benefit of playing with some of the "rich businessmen" in cash games and he isn't going to "dog it" for the money once they get deep in this event.

    The other consideration in determining their edge is structure. The more ghetto the structure, the more it benefits us. Did the 1-drop final table play really short 2 years ago? I watched Druff's final table last year in the limit hold-em and thought it was pretty damn crap-shooty and it was the determining factor in why I bet 2500.

    Someone's edge over the field can be significantly negated by the structure. I hope they fucked it up really bad for our sake.

  17. #177
    It just happened Todd I posted PI at 2/1. If some of the others think DN has a worse chance of winning a bracelet than him,you've got better than evens? I dig holes for a living so what do I know.......I will buy a $100 right now of the DN/PI bet 10 mins.......
    cmoney :It would be nice if Mexico could simply get human feces out of its drinking water

  18. #178
    Owner Dan Druff's Avatar
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    Quote Originally Posted by YUUP View Post
    It just happened Todd I posted PI at 2/1. If some of the others think DN has a worse chance of winning a bracelet than him,you've got better than evens? I dig holes for a living so what do I know.......I will buy a $100 right now of the DN/PI bet 10 mins.......
    Not sure what you are trying to say here.

    The KPR/Ivey bet at 2:1 suggests that there is a 1-in-3 chance of Ivey winning a bracelet in 2014.

    It is generally believed that Ivey has a better chance at winning one than Negreanu does.

    If Ivey really has a 1-in-3 chance. and Negreanu has a 1-in-4 chance, then it's a completely even bet on our side.

    If they have a worse chance than that -- even slightly -- the bet is +EV.

    I still think the best way to calculate it is using the per-event calculations based upon expected fields.

    Here is an interesting calculator that someone made regarding this bet: http://shiffman.net/wsop/

  19. #179
    What are the odds of Vanessa Selbst winning 2 WSOP bracelets this year?

    Phil and Daniel look a little too distracted this year.
    Ivey is under pressure to make money, he will be more busy playing cash games at Aria or Bellagio.
    Negreanu will take time off to give seminars at the Choice Center.
    Both have entourages that want them to stay out late partying and drinking.

    Vanessa Selbst should have the gold bracelet she wins today melted down and wear it around her neck, she should go around the Rio giving everyone the finger.



  20. #180
    Quote Originally Posted by Dan Druff View Post

    Here is an interesting calculator that someone made regarding this bet: http://shiffman.net/wsop/
    The calculator now reads JSON data to seed it. You can submit a pull request on github and I'll adjust accordingly.

    https://github.com/shiffman/wsop/blo...ages/events.js

    (You have to sign up for an account and then you can click 'edit').

    Am going to try to make it so you can set your own parameters and then tweet out a link.
    Last edited by system.out.println; 05-30-2014 at 08:49 AM.

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