how do i do this 2-7 tripple draw math

[NYplayer]

lets say i am dealt 235 K A and draw 2. what is the chance i make an 8 or better on the next draw? what's the math behind it so i can do other calculations?

thanks

[lippy]

to get to an 8, you need a 4,6,7,8 with both cards... so 16 outs.

16 outs with 47 unknowns and 15 outs with 46 unknowns.

16/47 * 15/46

[*TT*]

[ QUOTE ]
to get to an 8, you need a 4,6,7,8 with both cards... so 16 outs.

16 outs with 47 unknowns and 15 outs with 46 unknowns.

16/47 * 15/46

[/ QUOTE ]


Consider this - Hypothetically you catch a 7 on the first cards draw, this give us 12 remaining outs (not 15). But what if you don't catch any cards on the first card's draw? And of course the second card's draw could be a perfect card, or it could be a blank. And then we need to factor in what happens on successive streets - oi vey! I think you can begin to see that its a very hard question to answer precisely.

TT [img]/images/graemlins/club.gif[/img]

[2461Badugi]

He asked what the chance was on the next draw.

It's not hard to answer precisely even including all the draws, it's just somewhat complex.

[Notorious G.O.B.]

Yep, and then you can factor in the likelihood of catching exactly a 4 and a six, and subtract that from the first number. I got approximately .111- .0074= .1036

[NYplayer]

I just did the math out in a spread sheet. i have to get a 4,6,7, or 8 on the first card. there's a 8.5% chance of each for a total of 34%. if you get a 4 or 6 you have 8 outs (a 7 or an 8) and if you get a 7 or 8 you have 12 outs (4, 6 and 8 or 7). so half that 34% you have 17% to improve and the other half you have 26% to improve. do out the multiplication and it turns out you have 7.4% chance to make an 8 with a 235. break even is about 12:1

[*TT*]

[ QUOTE ]
break even is about 12:1

[/ QUOTE ]

Thats what I found as well for a 2 cards on the 3rd draw, its 12:1 with 16 outs.

PS: Sorry I misunderstood in my previous responce, I thought you were talking about 2 draws for 2 cards.. my bad.

TT [img]/images/graemlins/club.gif[/img]

[Notorious G.O.B.]

good call, I screwed up a bit. My bad.

[F8thless]

I would approach it as how many ways are there to make each hand that you would consider favorable (16 ways to make 47 etc, will vary based on any dead cards). Then divide each of those possibilities by the total possible 2 card combos ((52-5)*(52-5-1)/2) for the example you mentioned. Then finaly add up all individual two card combinations. You can also look at how many ways to draw 1 good card, and then one you consider bad, or how many ways to catch two bricks (1 - all good 2 card and good 1 draws should equal all bad two card draws)

I think it's similar to looking at hold'em, (52*51)/2 = 1326 possible hands, 6 ways to make AA etc.

Anyone else confirm my way or know of an easier way?

[MarkGritter]

Here are the two-card draws that make you a pat hand:
47
48
67
68
78

We don't know that any cards are dead, so they are all equally likely--- they each come in 4*4 combinations (four cards of each suit, order doesn't matter) for a total of 5*16 = 80 distinct two-card combinations.

There are 52 - 5 = 47 unknown cards, so there are 47C2 = 47*46/2 = 1081 possible two-card combinations you could draw.

This gives odds of 1001:80 or about 12.5 to 1 (7.4 percent) Which happily agrees with the numbers you got on your spreadsheet. (But some of the other calculations in this thread are wrong.)

I find that it is much easier to work with combinations rather than trying to do it card by card. It is straightforward to extend this method to calculate the total chances of improvement or to take dead cards into account.

(On the other hand, some problems are more easily addressed card-by-card. For example, if you have 72JKQ, what are the chances of completely bricking your next draw? The combinations are harder to count for this one but solving it card by card is trivial.)