19

I was watching a World Series of Poker video, and noticed a strange thing with the percentages that each player had for their chance to win the hand.

Before the flop, after all but 3 players had folded, the percentages were like this:

Player A: 9♥9♣ - 35%

Player B: A♦Q♦ - 40%

Player C: J♥10♦ - 25%

Player C folded, and immediately player A went to 56%, while player B went to 44%. Player B was the favorite based on his cards, but after player C folded, player A became the favorite. How is this possible?

Here is the video in question, at the relevant timestamp:

https://www.youtube.com/watch?v=oZP3OrtcO-Q&feature=youtu.be&t=1m22s

Improve this question
11
  • 2
    Shouldn't this be asked on/migrated to Poker SE?
    – Glorfindel ♦
    Sep 30, 2016 at 17:04
  • 4
    Alice has $35, Bob has $40 and Carol has $25. Carol dies and leaves $21 to Alice in her will and $4 to Bob. Before Carol's death, Bob was the richest of the three of them, but after her death, now Alice is the richest. How is this possible? Now -- how is the "puzzle" I have presented any different from the situation you have presented? I do not see why you think this is an impossible situation; can you explain why you believe it is impossible?
    – Eric Lippert
    Sep 30, 2016 at 17:34
  • 2
    Note that these percentages take into account the removal from the possible community cards of all cards which were known to be in the hands of players who had already folded. This is what accounts for the resulting percentages being somewhat different than an analysis would show when only considering the cards shown.
    – Makyen
    Sep 30, 2016 at 17:38
  • 1
    @Glorfindel We do get poker questions here all the time; I'm not sure what the criteria is for which site is better.
    – GendoIkari
    Sep 30, 2016 at 17:40
  • @EricLippert That's a pretty good analogy. What I was missing in the first place was that the cards that would have originally helped player C win would be better for player A than player B.
    – GendoIkari
    Sep 30, 2016 at 17:41

4 Answers 4

Reset to default
25

Suppose the five community cards wind up being 9♦, 10♥, 10♣, J♠, and 3♠. Then C would win with a better full house than A, but A's full house is better than B's pair of 10's. Before C folds, this would count towards his 25% win chance; after he folds it counts for A's 56%. When C folds, we're essentially taking his 25% chance to win, and splitting it up between the two remaining players. The issue then becomes that the cards that help C out also help A more than they help B, so most of C's 25% goes to A rather than B after C folds.

Improve this answer
5
  • Ah, I must've thought the Jack was another 9 when I came up with the hand. I'll fix the example.
    – Michael Snook
    Sep 29, 2016 at 20:26
  • Example is spot on, but you could do more about explaining why A enjoys a greater gain from C's fold than B. While B is favored 40% before the fold, that still means he loses to 60% of the possible community card permutations.
    – Drunk Cynic
    Sep 29, 2016 at 21:10
  • 2
    A much more extreme, but simplified sample. A is winning 60%, B is 0, C is 40%. B is 0 because every combination that B wins, A will win even more. Now A folds. All of this 60% went to B, so be now becomes 60% to win, and C stays at 40%.
    – Nelson
    Sep 30, 2016 at 6:08
  • @Nelson is such a hand possible; that every possible card which would help player B actually gives A then win?
    – GendoIkari
    Sep 30, 2016 at 17:42
  • Never mind, I guess with enough card eliminated from options by being in other players' hands it is. B has KK , and A has AA of the same 2 suits as B. At least 3 other players already folded KKQQQQ.
    – GendoIkari
    Sep 30, 2016 at 17:47
14

While I don't have the exact math to prove it, it certainly seems reasonable. The question to ask is, in the 25% of cases where C would've won, who would've had the second best hand? If the answer is almost always A, then you'll get the results you saw.

Here the thing that really boosts A's chance are having a 9 on the table. But when C is in play, that also leads to a lot of straight possibilities that would give the hand to C. With C out of the picture, those hands strongly favor A.

Note that B's chances to win have NOT gone down. They've actually increased from 40% to 44%. It is just that now all the remaining hands go to a single player, so the comparison looks different. B was always more likely to lose than win. He was just the individual most likely to win.

Improve this answer
1
  • It's almost the Spoiler effect in action.
    – Nate Diamond
    Sep 29, 2016 at 23:45
2

When player C folded, both player's chances went up. B's chances rose from 40% to 44%, while A's chances rose from 35% to 56%.

What happened was that A captured "most" of C's chances. The reason is that both players would benefit from a 9 that would make a straight for A, and a "set" or full house for C.

B benefited from capturing C's chances of making a second pair behind his "set" (or pair) (not a lot). Meanwhile, A benefited more from C's chances of getting the 9.

So B is no "less" probable a winner when C folded, but s/he went from first of three to second of two in terms of "relative" probability.

Improve this answer
1

B did not go from a probable winner to a probable looser
The odds went up (slightly)
Your odds would never go down as a result of a fold

The odds were

Player A: 9♥9♣ - 31%

Player B: A♦Q♦ - 43%

Player C: J♥10♦ - 26%

After C folded

Player A: 9♥9♣ - 56%

Player B: A♦Q♦ - 44%

Consider the 99
As a pair it is currently ahead
But any of 4 cards will beat it - AQJT
There are also straight and flush draws

After the fold the 99 is still ahead
But now now only 2 cards will beat it AQ (and flush and straight)

AQ only went up slightly as it still has to improve to win
If AQ improves to like a pair then JT would need 2 pair or a set and that is just not likely

It was the correct for B to call even though he was behind. He was getting correct pot odds to call.

Jarvis was down to 23 big blinds so he felt like it was time to make move. If he could fold one player out and only facing two overs then a good play. If he cannot fold a player out or facing an over pair then a bad play. That is poker.

I watched the hand again. When B only called you could put him on AQ or lower. AA, KK, JJ, TT, and AK would have raised (against Duhamal a chip leader playing a lot of pots). A was counting on folding out B and C not having a monster. As is turned out C was raising light. When it got to B he should mathematically call unless he is facing AA or QQ and that is not likely. The right mathematical call.

Improve this answer

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .