Odds of catching 2 outer?

[ImNew]

I just busted out in a 22$ MTT, I had KQ and my opponent had K5, I check raised him all in on a board of KQ6, no flush draw was possible, he caught two running fives on turn and river to beat me, what is the odds?

Plesae help by showing how you work out the exact %, thanks a lot, I'm still tilting.

[Acehawk74]

I *Think* it would be:
(3/45) * (2/44) = 11.21%

With 45 unknown cards in the deck, there are 3 fives left. With 2 cards to come, the chance of the first one coming is 3 out of 45 possible cards. Once the 2nd one hits, the chance the next 5 comes on the river is 2 out of 44 remaining cards. Someone please correct me if I am wrong, but I think this is right.

Acehawk

[LetYouDown]

[ QUOTE ]
I *Think* it would be:
(3/45) * (2/44) = 11.21%

[/ QUOTE ]
Left side is right...no idea how you came up with the right side.

He could also catch running 6's for a split, but for the 5-5:

C(3,2)/C(45,2) = 3/990 = .303...%

[ImNew]

I just ran the cards on a Texas Hold'em Odds Calculator:
http://www.cardplayer.com/poker_odds/texas_holdem/index.php

It said that I was a 99.1% favourite, so the odds of two running fives isnt 9% I assume?

[Acehawk74]

whoops, that 11% figure is the two of those added.. not multiplied. Sorry bout that, *hit + instead of * on the calc*

Can you explain that Combination formula? I remember it vaguely, but my understanding of it is weak.

[Acehawk74]

no, its not. Letyoudown was right, the odds of two 5's hitting in your scenario is .303%

odds of catching running 5's is 1 in 341... if you watch WPT @ all, last year Helmouth was a victim of this(he had xJ vs someone's J7) flop came KJ5 (no flush draw) turn 7, river 7, they showed the odds on the commercial as one of the Poker Quiz qustions... What were the odds of running 7's... answer was 1 in 341

[LetYouDown]

[ QUOTE ]
Can you explain that Combination formula? I remember it vaguely, but my understanding of it is weak.

[/ QUOTE ]
C(3,2)/C(45,2) = 3/990 = .303...%

C(3,2) is the number of ways to choose two remaining 5's from the three remaining 5's...regardless of order. C(45,2) is the number of possible turns and rivers with 45 cards remaining in the deck.

I don't know where the 1/341 came from...unless they made not of cards that were folded.

[kiddj]

[ QUOTE ]
I just ran the cards on a Texas Hold'em Odds Calculator:
http://www.cardplayer.com/poker_odds/texas_holdem/index.php

It said that I was a 99.1% favourite, so the odds of two running fives isnt 9% I assume?

[/ QUOTE ]
No.

First of all: 100% - 99.1% = 0.9% (not 9%)

This 99.1% takes into account the fact that running sixes (or running aces) would be a split pot.

[LetYouDown]

[ QUOTE ]
This 99.1% takes into account the fact that running sixes (or running aces) would be a split pot.

[/ QUOTE ]
And running A-A.

[kiddj]

Yeah, i fixed it, but I guess i wasn't quick enough. /images/graemlins/wink.gif