Razz - paired cards

pococurante · #1 11-13-2007, 01:33 AM

Hmm.... possibly the best way is just to count all the possible combinations including suits.

We know he holds the 2 of diamonds. The other potential cards he could hold: three 5s, four 4s, three 3s, two As.

A3 x6 - no
A4 x8 - no
A5 x6 - yes
34 x12 - no
35 x9 - yes
45 x12 - yes

That's 27 hands where the 5 pairs him, and 26 hands where it doesn't. 27/53 = there's a 50.94% chance he got paired on 4th street.

BruceZ · #2 11-13-2007, 03:02 AM

[ QUOTE ]
Hmm.... possibly the best way is just to count all the possible combinations including suits.

We know he holds the 2 of diamonds. The other potential cards he could hold: three 5s, four 4s, three 3s, two As.

A3 x6 - no
A4 x8 - no
A5 x6 - yes
34 x12 - no
35 x9 - yes
45 x12 - yes

That's 27 hands where the 5 pairs him, and 26 hands where it doesn't. 27/53 = there's a 50.94% chance he got paired on 4th street.

[/ QUOTE ]

Right except that there is only 1 ace unaccounted for. The possibilities then become:

A3 x3 - no
A4 x4 - no
A5 x3 - yes
34 x12 - no
35 x9 - yes
45 x12 - yes

This gives a probability of pairing the five of 24/43 =~ 55.8%.

*TT* · #3 11-13-2007, 03:53 AM

Bruce - pococurante got it right, there are two A's unaccounted for. By using combinations the probability of the villain pairing his 5 is 50.9%.

But now comes the second part of my question, is there a shortcut for calculating this probability on the fly rather than typing out the combinations by hand?

BruceZ · #4 11-13-2007, 04:08 AM

[ QUOTE ]
Bruce - pococurante got it right, there are two A's unaccounted for. By using combinations the probability of the villain pairing his 5 is 50.9%.

[/ QUOTE ]

Look at the post you mentioned on the Stud forum. Player 1 has the A [img]/images/graemlins/spade.gif[/img], player 6 has the A [img]/images/graemlins/heart.gif[/img], and hero has the A [img]/images/graemlins/diamond.gif[/img]. Only the A [img]/images/graemlins/club.gif[/img] is unaccounted for.

Your statement that there are 12 bicycle cards unaccounted for is incorrect. There are 6 exposed, and since we are told that opponent's first 3 cards are different, we know that he doesn't hold another 2, which leaves 20 - 6 - 3 = 11 bicycle cards.

[ QUOTE ]
But now comes the second part of my question, is there a shortcut for calculating this probability on the fly rather than typing out the combinations by hand?

[/ QUOTE ]

Not if you want to take into account the exact cards that are unaccounted for.

pococurante · #5 11-13-2007, 04:48 AM

[ QUOTE ]
Only the A [img]/images/graemlins/club.gif[/img] is unaccounted for.

[/ QUOTE ]

ahh... good eye there

*TT* · #6 11-13-2007, 05:19 AM

Thanks Bruce, your right there is only one A unaccounted for, I must have made an typo when I tallied up the cards which I carried over here.

so using combinations I have learned that the villain pairs the board 56% of the time assuming he holds any 3-card bike. If we remove the roughest bike hand of (45)2 from his capping range to more accurately represent his likely holdings if he is slightly loose then there is a 39% chance he paired. In the event we open the villain's range to include all but the roughest sixes then there is a 28% chance the villain paired the 5.

thanks a bunch for your help!

Lottery Larry · #7 11-13-2007, 10:03 AM

Does it make sense to call with bike cards WIHTOUT the Ace, if we see 2 Aces?

I'll have to go read that hand in the Stud forum

Lottery Larry · #8 11-13-2007, 11:12 AM

Make that, THREE aces are known, with one folded and two raising.

What bike hands do we fold here, having completed in early position on 3rd?

SGspecial · #9 11-13-2007, 12:07 PM

[ QUOTE ]
Does it make sense to call with bike cards WIHTOUT the Ace, if we see 2 [sic - actually 3] Aces?

[/ QUOTE ]
No. But assuming that most razz players will act sensibly is a bad assumption, and will usually get you into trouble. Given this particular villain, any 3-card bike or 6 (no matter how rough) would be worth an UTG completion and then a cap.