Deuce to Seven Triple Draw Questions

[paulish]

I've been reading up on Negreanu's chapter in SS2. And I wondered if any of you guys could help me out with some numbers here:

<font color="red"> Q1 </font> How many unique 5 card combination are possible

<font color="red"> Q2 </font> How many pat 7 &amp; 8 hands are there?

<font color="red"> Q3 </font> How many "one card 7 &amp; 8 draw"-hands (excluding 8 draws with a straight draw) are there?

<font color="red"> Q4 </font> How many hands are there in total in the following Two-Card Draw groups:
a) 234, 235, 237, 245, 247 &amp; 257?
b) 236, 246, 256 &amp; 267?
c) 247, 248, 357, 358, 367 &amp; 457?

<font color="red"> Q5 </font> How many Three card draw hands are there (23, 24, 25 &amp; 27)?

[fnord_too]

This question is better suited to the probability forum. I will answer the first one for you.

The way to find the total number of 5 card hands is to use the choose function. There are 52 cards from which you may choose 5. If order counted, that would be:
52*51*50*49*48 combinations
But the order you get the cards does not matter, so divide by the number of ways you can be dealt the same 5 cards:
5*4*3*2*1

So, 52 choose 5 is (52*51*50*49*48)/(5*4*3*2*1) = 2,598,960

All of your questions fall under the field of combinatorics. You should pick up a book on (or just look for instruction on the web for) combinatorics and probability. The sorts of questions you are asking are ideal for learning those things, and if you are not sure of your work just post it in the probability forum and someone will comment on it.

[MarkGritter]

I provided some rough estimates here--- they do not distinguish between good and bad draws, though.

http://forumserver.twoplustwo.com/sh...Number=3167336

[MarkGritter]

Working these out is not to hard, at least for the good draws...

For pat hands, just total up the number of variations (SS2 provides a helpful chart) and multiply by (4^5-4) to account for all the different suit possibilities, minus the four flushes.

For 1-card draws, we need to account for all possible 5th cards, but exclude those that make a pat hand already counted. So for 2347x, for example, the x can be 23479TJQKA but not 568, or 36 possible cards. (52-4-12) But to avoid double-counting you need to split into two cases.

2347x where x is not a pair (9TJQKA): 24 x's, 4^4 suits for 2347 = 6144 possibilities.

2347x where x pairs (2347): 4 different pairs possible, 4^3 suits for non-paired cards, 4C2 suits for paired cards = 4 * 64 * 6 = 1536 possibilities.

So, each 1-card draw to a 7 counts for 7680 possibilities, total up the number of 1-card draws. 1-card draws to an 8 can be counted similarly, just the number of remaining cards needs to be adjusted (since now there are 4 cards that could create a pat hand.)

2-card and 3-card draws are just elaborations of this same method, although it is slightly more complicated to count the possible discards and avoid double-counting pairs.

I.e., for 27xxx you need to consider:

27xxx where x &gt;= 9 (4^2 * 24C3 = 32384 possibilities.)
27pxx where x &gt;= 9, p = 2 or 7
27ppx "
27ppp "

[paulish]

thx! [img]/images/graemlins/smile.gif[/img]