How often does 68o outflop AJo?

[deepdowntruth]

How often will 68o outflop AJo (ignoring the effect on the hand's outcome of the turn and river)? And how precisely does one figure this? I want to assume that the four hole cards are known.

Also, is there a utility available that will calculate hand matchups assuming only three community cards (rather than five) are dealt?

[BruceZ]

[ QUOTE ]
How often will 68o outflop AJo (ignoring the effect on the hand's outcome of the turn and river)? And how precisely does one figure this?

[/ QUOTE ]

Flop 1,2, or 3 from {6,8} with no A or J: 6*C(36,2) + C(6,2)*36 + C(6,3) = 4340.

Flop 2 from {6,8} with 1 from {A,J}: C(6,2)*6 = 90.

Flop straight 8,9, or T high: 3*4^3 = 192.

-------------------------------------------------------
4340 + 90 + 192 = 4622

4622/C(48,3) =~ 26.7%.

[Moose747]

You may want to add in a few more flops. If 68o picks up a flush draw or an open-ended straight draw and a backdoor flush draw, it's slightly ahead on the flop. I imagine these wouldn't have a huge effect on the result.

[Antoine]

Wow I am impressed, can you explain me your calcul please ?

6*C(36,2) + C(6,2)*36 + C(6,3) = 4340.
What is 36.2 ? 6.2 ? C ??

I am sorry but I would love know to calcul like you.
Thanks to help me.

[BruceZ]

[ QUOTE ]
You may want to add in a few more flops. If 68o picks up a flush draw or an open-ended straight draw and a backdoor flush draw, it's slightly ahead on the flop. I imagine these wouldn't have a huge effect on the result.

[/ QUOTE ]

The OP wanted to ignore the effect of the turn and river, so draws are no good.

[BruceZ]

[ QUOTE ]
Wow I am impressed, can you explain me your calcul please ?

6*C(36,2) + C(6,2)*36 + C(6,3) = 4340.
What is 36.2 ? 6.2 ? C ??

[/ QUOTE ]

C(36,2) = 36*35/2 is the number of ways to select 2 cards out of the 36 that are not 6,8,J,A. 6*C(36,2) is the number of flops with one 6 or 8, and 2 cards that are not 6,8,A,J. C(6,2)*36 = 6*5/2 * 36 is the number of flops with two 6's, or two 8's, or a 6 and an 8, and 1 card that is not 6,8,J,A. C(6,3) = 6*5*4/3! = 6*5*4/(3*2*1) is the number of flops with all 3 cards that are 6 or 8. In general, C(n,k) = n!/(n-k)!/k! = n*(n-1)*(n-2)*.../ [k*(k-1)*(k-2)*...*1] where the numerator has k terms.

[bachfan]

[ QUOTE ]
How often will 68o outflop AJo (ignoring the effect on the hand's outcome of the turn and river)? And how precisely does one figure this? I want to assume that the four hole cards are known.

Also, is there a utility available that will calculate hand matchups assuming only three community cards (rather than five) are dealt?

[/ QUOTE ]

I wrote an online utility for exactly this purpose. To see the graph of the odds of each hand on the flop, click here . If you'd like to play with it, at the bottom of the page is a link to edit the simulation. To get the graph, you click "Graph Equity Distribution".

Let me know if this fits the bill.

Cheers,
bachfan

P.S This is general purpose, so you can do multiple hands, and you can see flop, turn, or river distributions for hold'em, omaha, or omaha-8.

[deepdowntruth]

I missed some of the responses to this thread as I was away from the internet for a while. Thanks to those who took the time to answer. I have a follow-up question.

Let's create two groups of flops: Flops for which AJo is ahead, and flops for which AJo is behind.

For each group, what is the average win percentage for AJo? That is, for flops on which AJo is ahead, how often will it remain ahead through the turn and river; and for flops on which AJo is behind, how often will it end up winning by the river?

Also, if there is any reasonable way to show your work, that would be appreciated.

Thanks!