please check my math

[bosulli]

I have calculated the P of a hitting a flush draw, inside straight draw, 2 pair or trips on the flop holding cards of (6s7s).

For an inside straight draw, combinations of (4,5), (5,8) or (8,9) are needed. There are 36 pair combinations of the needed cards. There are 1,728 needed three card combinations. 19,600 possible 3 card combinations. P = 11.7%.

For 4 card flush draw [11/50*10/49*39/48*3]=10.9%. Got this from Yao King’s book “Weighing the Odds…”

For 2 pair, there are 9 combinations of (6,7), and 432 possible 3 card combinations. P=2.2%.

For trips, there are 3 combinations of (6,6) & 3 of (7,7), and 288 possible 3 card combinations. P= 1.5%.

Therefore the P of the first statement is 26.3%.

Right????

[AaronBrown]

Close.

First of all, you are looking at an open-ended straight draw, not an inside straight draw. Counting the number of ways to get them is a bit complicated. Start with just one, like 5,8. There are four 5's, four 8's and 34 cards that are not 4, 5, 8 or 9. 4x4x34 = 544. You can multiply this number by 3 to get 1,632. You can also get an open ended straight draw with a flop like 5,5,8 and you can get a straight with a flop like 4,5,8. Do you want to count these or not?

Flush is correct but a little over 1 out of 16 of your straight draws will also be flush draws. So you can't add the probabilities.

For two pair, there are indeed 9 combinations of 6,7, but you multiply this by 44, not 46. There are 44 cards that are not 6's or 7's. The other 2 cards give you a full house, and you have to figure this separately, you can't just multiply.

The same comment applies to trips.

[bosulli]

Sorry, for the confusion, I meant open ended straight in my post, not inside straight.

Thanks for your help, and please critique this and tell me error of my ways.

I need a (4,5),(5,8) or (8,9) to give me the desired draw. There are 16 combinations of each. Therefore, (3x16) or 48 two card combinations. I don’t care what the third card is. There are 48 possible cards to make up the third card. So, there are (48x48) or 2,304 possible flop combinations that will give me the open ended straight. P = 11.8%.

[BruceZ]

[ QUOTE ]
Sorry, for the confusion, I meant open ended straight in my post, not inside straight.

Thanks for your help, and please critique this and tell me error of my ways.

I need a (4,5),(5,8) or (8,9) to give me the desired draw. There are 16 combinations of each. Therefore, (3x16) or 48 two card combinations. I don’t care what the third card is. There are 48 possible cards to make up the third card. So, there are (48x48) or 2,304 possible flop combinations that will give me the open ended straight. P = 11.8%.

[/ QUOTE ]

This post and its associated links has everything you want to know. The numbers in that post are all very solid, as they have been verified by different people using different methods, as well as by computer simulation.

Some of these are tricky, and there are some subtleties. For example, you are double counting cases where the flop pairs, and you are not excluding made straights or made flushes, so yes you do need to worry about what the 3rd card is. You cannot just multiply by 48, or you will be double counting many combinations.

The following is the calculation for 8-out straight draws which I extracted from the above link for you, since it was buried pretty deep. This includes double belly buster straight draws, but it excludes made flushes.

[ 3*(15*34 + 1*27 + 2*6*4) + 4*4*4*2 - 2 ] / C(50,3) = 9.60%.

That is, there are 3 2-card combinations, 15 of these are unsuited, and there are 48-8-6 = 34 ways to pick the last card so that it doesn’t complete the straight and doesn’t pair the board. There is 1 2-card combination that is suited in our suit, and there are 34-7=27 ways to pick the last card so it doesn’t complete the straight, doesn’t pair the board, and doesn’t make a flush. We subtract 7 instead of 9 since the 2 flush cards that make a straight were already subtracted. The third term is for the paired boards. There are 2 ranks that can pair, 6 ways to make the pair, and 4 ways to choose the other card. Then we add the 4*4*4*2 double belly busters, and subtract 2 for the double belly buster draws that are flushes.

[bosulli]

[ QUOTE ]


[ 3*(15*34 + 1*27 + 2*6*4) + 4*4*4*2 - 2 ] / C(50,3) = 9.60%.

That is, there are 3 2-card combinations, 15 of these are unsuited, and there are 48-8-6 = 34 ways to pick the last card so that it doesn’t complete the straight and doesn’t pair the board. There is 1 2-card combination that is suited in our suit, and there are 34-7=27 ways to pick the last card so it doesn’t complete the straight, doesn’t pair the board, and doesn’t make a flush. We subtract 7 instead of 9 since the 2 flush cards that make a straight were already subtracted. The third term is for the paired boards. There are 2 ranks that can pair, 6 ways to make the pair, and 4 ways to choose the other card. Then we add the 4*4*4*2 double belly busters, and subtract 2 for the double belly buster draws that are flushes.

[/ QUOTE ]

Thanks, I will have to absorb this.

BTW, what is a "double belly buster"?

[BruceZ]

[ QUOTE ]
BTW, what is a "double belly buster"?

[/ QUOTE ]

Say you hold JT, and the flop comes AQ8. Then either a K or a 9 will complete a straight. If the flop comes K9<font color="red">7</font>, then either a Q or an <font color="red">8</font> will complete a straight. In either case, you have 8 outs from the 2 separate inside straight draws. An inside straight draw is called a "gut-shot" draw, so 2 gut-shot draws are called a "double gut-shot" draw or a "double belly-buster" draw. There will always be 2 possible double belly-buster flops for every connector from 54 to JT. For 43 and QJ, there is only 1 possible double belly-buster flop, and the remaining connectors have none. It is also possible with any starting hand having <font color="red">1-3 gaps, or 5 gaps.</font>

[BruceZ]

This has already been corrected in the original post. Thanks BogusPomp.

[ QUOTE ]
If the flop comes K98, then either a Q or a 7 will complete a straight.

[/ QUOTE ]

That's true, but for a double belly-buster it should be:

<font color="blue">If the flop comes K97, then either a Q or an 8 will complete a straight.</font>

Also, a double belly-buster is possible with 1-3 gappers or 5 gappers, but 4 gappers can only flop open-ended draws or 4 out draws.