Probability question/riddle

    • epaga
      epaga
      Bronze
      Joined: 05.03.2005 Posts: 89
      Just played a 6-player SNG and won it.

      In the SNG I encountered two coinflip situations (e.g. AK vs QQ) where I would have busted before the bubble if I had lost, and (obviously) won both of them.

      Now I got to thinking: doesn't this mean this SNG was basically -EV for me? 3 out of 4 times I would have had $0, only the 1 time out of 4 did I still have a chance (and not even a sure chance at that!) to make any money.

      Now obviously coinflip situations are a very common occurrence but good players still play +EV SNGs...so where is the error in my thinking?
  • 8 replies
    • EagleStar88
      EagleStar88
      Bronze
      Joined: 06.10.2008 Posts: 7,359
      I'm not really an expert on such things to be honest epaga, but there are some great articles on the subject of EV etc in the strategy guides (both NL & SnG's).

      I'm sure a more experienced/better qualified member will come up with some calcs/advice/experience shortly though.

      Congrats on your win :s_thumbsup:

      Bart
    • LgWz
      LgWz
      Black
      Joined: 26.05.2007 Posts: 7,641
      You should calculate the equity you had against the opponent's range, and not vs the particular hand he had... let's say you get QQ vs AK all in preflop, it's a flip... but he might go all the way with hands that you have crushed such as AQ or JJ, etc.
    • andreibalint
      andreibalint
      Bronze
      Joined: 11.04.2009 Posts: 872
      Not a big player nor a math geek but I know this: Different hands in poker are independent. That means (logically) that they don't have any connection between them. And one of the few useful things I learned in my universty is how to calculate odds of this kind of events...
      I don't know how to incorporate this in SnG's but in a purely statistical manner you were 50% * 50% = 25% favourite to win both hands. You just multiply the probabilities.
      This means that if you go all in 3 times with AA in a tournament you were actually a coinflip away of being eliminated: 80%*80%=64%*80%= somewhere less then 50% I'm after some beer. Kind of logical anyway since you know that you are 80% to win each time, so you win 4 times and lose one time. But you already won 3 times, what do you think is next? Win the last time or lose ~ 50% each. Don't ask me which one of them but Phill Gordon or Greenstein said this also (I tend to say Greenstein on Ace on the river). Hope it helps.
    • DevilChess
      DevilChess
      Global
      Joined: 21.03.2009 Posts: 488
      The reason I think your reasoning is wrong is this:
      First:
      You cannot consider past events to determine if a SINGLE decision was +EV or -EV (unless that event affects the outcome of your decision, like for example your table image resulting of playing aggressively), since past events DO NOT affect future outcome of cards. (if you had AA the past hand, you've got the same chance of getting AA again)

      Second:
      Considering the COMPLETE tournament, you must also add-in the likelyhood of ALL your opponents losing that tournament in their all-in moves, since you may have had 25% chance of winning those two hands, but if the sum of odds of winning that tournament of your 9 opponents was 75%/9 (100% minus 25%) of winning , then you're getting 25% odds to win 1st Place is very good:
      1 to 3 odds, of 1 to 5 (SnG 50% Prize 1st place payout) giving you 66% (2/3) ROI (not counting rake), which is VERY GOOD.

      In general, counting ALL of your moves, you should be getting better %odds than the average share of odd% of getting ITM, and as long as you play good poker, that will be true.

      But a SINGLE move can only be made determining PRESENT information, without saying: "Oh, I have gone all-in 4 times with AA, I have AA again, I must fold." That is obviously wrong.
    • viewer88
      viewer88
      Bronze
      Joined: 19.04.2008 Posts: 5,545
      coin flips are unavoidable sometimes, and while the coinflip might be -$EV the action leading to the coinflip (pushing) can still be +EV

      Say you are playing a 6max sng with top 2 payout and you are 3handed (bubble): BU with ~same stacksize shoves; SB (huge chipleader) folds and you have about the same amount of chips as BU.

      You are holding AK. You know BU pushes something like A5+ K8+ QJ+ 22+ in this spot. You call and villain turns over QQ; it sucks and you lose money (because you lose 50% of the time and get 0$). Calling is still a good move though because you are only flipping against the 22+ part of his range (and you're fucked against AA and KK). The call agains this pushrange is +$EV.
      You don't want to be results oriented here, think in ranges!


      Now lets go to another situation: you are playing in an extreme 6handed WINNER TAKES ALL sng with 2 extreme donks and 4 perfect players (you are one of the perfect players).
      The 2 donks get killed early and you play against the 3 other sharks.
      You have no edge here, everyone has the same chance to win (you won't always have the same amount of chips, sometimes you will have more sometimes less; in average you will have the same amount of chips so it evens out over a large sample).

      Now you are basicly coinflipping, you have only 1/4 chance to make money. This is NOT -$EV because 1 out of 4 times you will win 6 buy-ins, so you earn 6/4 = 3/2 buyin in average.

      Coinflipping is a necesity, buth you will still make money because the 2 fish donated there money to the 4 sharks at the start of the sng.


      I hope this made any sence, the 2nd situation was an "ideal" situation to answer your questing with a simple example.
    • andreibalint
      andreibalint
      Bronze
      Joined: 11.04.2009 Posts: 872
      But a SINGLE move can only be made determining PRESENT information, without saying: "Oh, I have gone all-in 4 times with AA, I have AA again, I must fold." That is obviously wrong.

      Haha, of course I'm not saying that. And I told you I'm not talking about EV here, I'm talking about the chances to win the pot. And you can determine your overall EV.

      Aces three times, again. Statistically a coinflip. I'm really right here :D
      Now, how to calculate EV? Well I'll do it like in trading. Should be the same idea. Expectancy=%Won*How much you won - %Lost*How much you lost.
      If you play a cash table for 1$ you get 0.80$-0.20% = 0.6$, so you expect to win 60 cents everytime you go all in with aces on average.
      Hmmm, in a tournament it gets tricky. On the bubble like you said. The only way I can figure it out is to suppose that if you win you will surely get ITM.
      So EV=80% * at least 2 times your investment (3rd place) - 20%*your investment (entry fee). Let's say you play for 1$. That is: 0.8*2$-0.2*1$=1.4$ at least!!! (if you finish 3rd). So it's of course +EV. Every time you push with aces you can expect to make at least 1.4$ in this spot. (10 fish SnG payout 20%-30%-50%). Happy now? :)
    • andreibalint
      andreibalint
      Bronze
      Joined: 11.04.2009 Posts: 872
      Forgot about your first post. You just change the numbers. In my opinion going for a flip on the bubble it's slightly +EV: -EV if you get third because of the comission and +EV if you actually finish higher, and you'll finish often enough.
    • viewer88
      viewer88
      Bronze
      Joined: 19.04.2008 Posts: 5,545
      payout in regular 6max sng is pretty top heavy (80-20? 70-30?) so you can take more risks on the bubble because first plays pays alot more.