Which situation has better odds

While I was analyzing the hand pairs, I came across a interesting situation. I pose it as a quiz:

In heads up following are the four situations:

Situation Hand1 Hand2
1 AA KK
2 AA 44
3 KK QQ
4 44 22

In which situation hand 1 has better odds.

In all of them, hand 1 has better odds. What’s the trick?

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Hand one has better odds in each situation.

I would say the best odds would be in hand #4, because 4 is a higher pair and has some blocking effect in preventing the 2s from making any straight. In the other three hands, the lower pair has a blocking effect on some or all the straights that can be made by the overpair and the lower pair can make some straights that are not blocked by the upper pair, for example KQJT9 is not blocked by AA.

The 2s can block the 4s from making the straights A2345, 23456, but cannot block 34567, 45678. However, the 2s cannot make any straight without a 4. If a 4 does fall on the board giving the 2s a straight, it is still not over as the 4s might make a full house if the fifth card pairs any of the other four cards in the straight.

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I didn’t make it clear, In all situations hand 1 has better odds. Among these odds which one is better.

Answer and the interesting thing to note is in all situations hand 1’s odds are just about the same, around 82/18%, irrespective of whether KK or 44 is called against AA. Similarly AA vs 44 or 44 vs 22 the odds are about the same. Except a small variation (about two percent) in straight draws and blockers as @MekonKing pointed out.

This is because only way hand 2 can win is by drawing a set, set then full house or four of a kind. That is two outs of the same card in all cases.

No, this is not correct either, because the lower pair can hit a straight or a flush that the upper pair does not participate in. If the suits of the upper pair are both the same as the suits of the lower pair, then that increases the likelihood of the upper pair winning. If they are both different, then that throws another lifeline to the lower pair. If the pairs have one card of the same suit, that slightly reduces the chances of the lower pair.

Of course we are assuming (although it is not stated in the quiz) that these hands are simply put all-in against each other preflop and see five cards on the table, and leaving out considerations of position, and whether 22 might fold to a three bet preflop or whether other pairs might fold at the flop. With QQ unimproved on the flop, a player might well fold if the flop came AKx, or in fact with any A.

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By and large its true, straight and flush draws gives about 2% difference against a 82% odd. Its just about the odds, not considering the play which will be difficult to put as a question.

Whole point is to highlight a not so obvious situation that a 22 nearly has the same odds of winning against AA or 33.

Here is another one, easy though:

Game: Royal
Number of players: 6

Situation 1:
Open: SA CA HQ DQ HJ 
Hand: SQ CQ

Situation 2:
Open: SA CA HQ DQ HJ (same as above)
Hand: DA CQ

Which hand has better odds.

I don’t understand the question. The hands in Situation 1 and Situation 2 cannot coexist on the same board, since both share the queen of clubs.

Situation 1 is the second nuts, and loses only to the sole combination of AdAh. I’d expect to get value from someone with only one ace, since from their perspective they would lose to only your hand, and split the pot with the other ace. There are going to be far more times when you’re up against players with just one ace than when your opponent stacks you.

Situation 2 is the nuts, because you block AA and QQ. However, I wouldn’t expect a naked queen to call you, because there would be too many aces out there. As a result, the only action you’re likely to get is from aces that split with you.

Personally, I’d prefer to be in Scenario 1, despite the risk of being up against the nuts. Long-term, there’s more value to be had.

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Situation 2 has better odds. Way that hand can be defeated is by someone drawing AA or QQ, both are blocked by the hand by holding AQ. Its a top hand with 100% probability (including splits). In situation 2 hand can be defeated by AA, 6.4% probability. Hand has the odds of 93.6%.

I am not talking about play here, though I agree with your argument of playing. This is just a simple of question of hand odds. To point out a situation where a full house has better odds than quads.

I would play both hands the same way ignoring the 6.4% odd that QQ can lose, unless there is some strong indication that another player holds AA.

Okay, now I REALLY don’t understand the question.

In Situation 2, given a 20-card deck with 7 known cards, there are 78 potential competitor hands. You lose to only one of them. 1/78 is about 1.3%. How do you get to 6.4%?

This is six players table. There are five other players. 1.3% is the probability that one player draws AA. With five players it will be 1.3 (1.28 to be more accurate) x 5 which is 6.4%.

I would say A AA KK

Yes, AA vs KK and KK vs QQ has better odds for hand 1 if we consider the straight draws, otherwise they all have the same odds of winning as explained here.