What's the probability that the flop missed everyone?

[ctv1116]

OK, here's an interesting theoretical question. Let's say you hold 22 in the BB. It is folded to the button, who limps, and the SB completes, you check. Let's say the flop is K73r. What is the probablity that neither the Button nor the SB holds a pair (including hands like 44 or AA)?

EDIT: I guess the easiest way to do this is to write down every possible hand that beats 22, and calculate the probabilities.

EDIT2: Perhaps this can be extended to what if you held 44? 88? Perhaps this would be helpful in determining postflop decisions with low pairs.

[schubes]

Interesting question, I'll take a shot:

I'll find the probability that 2 hands do not have pairs given a flop that is not paired. I'll ignore the extra info you get from knowing your hand is 22.

prob 1st card of 1st hand doesn't pair flop: 40/49
prob 2nd card doesn't pair flop or 1st card: 36/48

prob 1st card of 2nd hand doesn't pair flop or either card of 1st hand: 32/47
given this, prob 2nd card doesn't pair first or flop: 34/36

prob 1st card doesn't pair flop but pairs one of the cards of 1st hand: 6/47
given this, prob 2nd card...: 35/46

((40/49)*(36/48)) * ((32/47)*(34/46) + (6/47)*(35/46))

= 36.76%

where did I screw up? /images/graemlins/tongue.gif

EDIT: I think the probability that they don't have a pair will be slightly less given that you have a pocket pair

[ctv1116]

I was thinking more along the lines of adding the probabilities of a player having AK, KK, KQ, KJ, KT, etc.

[schubes]

Turns out to be the same thing, only your way is a little more tedious /images/graemlins/wink.gif

Unless I made a mistake, I'm effectively adding up all possible hands that don't pair divided by all possible hands.

Now if you wanted to find out the probability that you were ahead when you had something like 88, the probability would be greater because you can include times your opponent makes a pair smaller than 8. Then maybe your approach would be needed. But in that case the answer would also depend on the specific flop.

For your original question (again, I may have made error) 22 will be beating 2 random hands a little over 1/3 of the time.

[schubes]

Playing poker all night, now I can't seem to sleep /images/graemlins/ooo.gif
I know I'm talking to myself here, but this is the exact answer to the problem using the fact that our hand is 22.

In situation A, hand 1 does not contain a 2, prob. is (36/47)*(34/46)
In B, hand 1 does contain a 2, prob. 2*(2/47)*(36/46)

a) If the 1st card of hand 2 doesn't pair flop or hand 1's cards, and isnt a 2:
For A, prob. is (28/45)*(32/44)
For B, prob. is (32/45)*(32/44)

b) 1st card pairs a card in hand 1, isn't a 2:
A, (6/45)*(33/44)
B, (3/45)*(33/44)

c) 1st card is a 2:
A, (2/45)*(34/44)
B, (1/45)*(35/44)

P =
P(A)*(P(a|A)+P(b|A)+P(c|A)) + P(B)*(P(a|B)+P(b|B)+P(c|B))
= (43!/47!)*( 36*34*(28*32+6*33+2*34) + 2*2*36*(32*32+3*33+35) )

=37.12%

Which is slightly more than using the easier way. I predicted it would be less: if you have a pocket pair shouldn't it be more likely that your opponents have pairs?

Anyone want to speculate why having a pair makes it slightly less likely that your opponents do? /images/graemlins/confused.gif

[catalinuzzu]

[topspinner]

Cat, that link goes to a foreign site, that I belive says this link doesn't exist.

[schubes]

[ QUOTE ]
The probability of not holding it is 1-(6/C(50,2)). The probability for 2 palyers not to hold a specific pair will be 2(1-(6/C(50,2)).

[/ QUOTE ]
I'm not sure why you think you can add here but it's not correct. Do you think its twice as likely that two players won't hold a specific pair? Shouldn't it be less likely?

1-(6/C(50,2)) is the probability that a player doesn't hold a specific pocket pair, the question is how often do two players not have any PP and not pair the flop.

Unless someone can show me a mistake, I'm pretty sure my way is correct and necessary. Look at the first way I did it if you want to see my method.

[BugsBunny]

I'm tired so I probably messed up somewhere but I make it ~36%.

This assumes that they play any 2 cards.
There are 47 unknown cards.

C(47,2) = 1081 (Possible hands)
C(38,2) = 703 (Possible hands that miss the flop (but may include a pp))

K, 7, and 3 are accounted for. 22 is in your hand. That leaves 9 other possible pocket pairs. 6 ways each = 54 total. Add 1 for the remaining 22 combo = 55 total pocket pair combos.

703-55 = 648 hands that miss the flop and aren't a pocket pair.

648/1081 = 0.5994449584 (chances that 1 opponent misses and has no pp)

0.5994449584^2 = 0.3593342581 (chances that both opponents miss and have no pp)

In reality opponents rarely play any 2 so the chances are probably better that they missed.

[Precision1C]

You hold 22 flop is K73. The odds of your opponent holding a pocket pair is 5.51% so the odds of neither holding a pocket pair is 89.3%. The odds of none of the 4 outstanding hole cards being a K73 are C(38,4) divided by C(47,4)= 41.4%. (note: slight inaccuracy since I am counting cases of pocket pairs in both the numerator and denominator) 89.3% times 58.6%=52.3%

So you are beat by a pocket pair 10.3% of the time, with a small chance of tying versus 22, and beat by a flopped pair 52.3% when there are no pocket pairs. You are holding the best hand 100-52.3-10.3=37.5% of the time versus 2 random hands. Note: Your actual odds might be improved since neither the button or small blind elected to raise preflop eliminating most pocket pairs and several combinations with kings.

[schubes]

I know there's no point in replying here, since we both agree on essentially the same number, and we both agree that the 2 opponents do not really have random hand distributions, but...

I like your way of figuring 648 hands that don't pair. Here's how I did it thinking about single cards in the deck instead of 2-card hand combinations:
[ QUOTE ]
In situation A, hand 1 does not contain a 2, prob. is (36/47)*(34/46)
In B, hand 1 does contain a 2, prob. 2*(2/47)*(36/46)

[/ QUOTE ]
Here I'm partitioning the hands of one opponent into 2 mutually exclusive cases (A and B). If you add them you should get the probability that one opponent missed the flop and doesn't have a pocket pair. I triple-checked everything and found an error: A should be (36/47)*(32/46)

Now P(A)+P(B)=(36*32 + 2*2*36)/(47*46)= 648/1081...the same number you got.

But here's my problem with your method:
[ QUOTE ]
0.5994449584^2 = 0.3593342581 (chances that both opponents miss and have no pp)


[/ QUOTE ]
You can't simply square this since the probability that one opponent doesn't have a pair isn't independent of another opponent missing. You have to multiply 648/1081 by the probability that the other player misses given that the first player misses. That's why I went through the more complicated P(A)*(P(a|A)+P(b|A)+P(c|A))+... thing.

Redoing it after I found my error, my number changes to 35.17% which confirms my intuition that having a pocket pair makes it slightly more likely others will have pairs.
(see me talking to myself in my earlier posts)

[FrankLu99]

50% either yes or no - well thats what "Enemy" tells me

[BugsBunny]

You're right - I said I was tired. (I knew I was missing something).

Chances of first player missing = 648/1081 = 0.5994449584

C(47,2) = total combinations possible for first player. We've removed 2 cards so the 2nd player has: C(45,2) = 990 possible combinations.

We've also removed 2 cards from the missed combinations possibilities. That leaves C(36,2) = 630 possibilities.

Instead of 9*6 + 1 possible pocket pairs we now have:
If no 2 in his hand:
7*6 + 2*3 + 1 = 49 pocket pairs
If he has a 2 in his hand:
8+6 + 1*3 + 0 = 51 pocket pairs.

He'll have a 2 (in a hand that missed):
2/47 * 36/37 + 36/47 * 2/46 = 0.0747056029

0.0747056029 * 51 = 3.8099857489
0.9252943971 * 49 = 45.3394254569

That gives us 45.3394254569 + 3.8099857489 = 49.1494112058 pp on average

630 - 49.1494112058 = 580.8505887942

580.8505887942/990 = 0.5867177665 (chance of second player missing if first player missed)

0.5867177665 * 0.5994449584 = 0.3517050071 or 35.17%

Which now matches your number, confirming the answer (using a slightly different method).

[BugsBunny]

The odds of your opponent holding a pocket pair is 5.51%

This number is wrong. The correct number is 5.09%. The chances of his holding a pair of K's, T's, 3's or 2's are reduced due to the cards already out. That changes the chances of neither holding a pp to 90.08%

Using groups of 4 also doesn't work, When counting groups of 4 a combination of AB CD is the same as AC BD, but this isn't accurate since in reality they're different sets of hands. It throws everything off.

Edit: The groups of 4 statement I made is only accuarte if you start including pairs in the hands. If you disregard pairs you can use groupings of 4.