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 09-25-2017, 01:30 AM #1 PLOfish123 stranger   Join Date: May 2017 Posts: 7 Probability of a low in O8 I tried to calculate the odds of having at least 3 distinct low cards on the board using: (5 C 3) * (32/52) * (28/51) * (24/50) What am I doing wrong?
 09-25-2017, 02:31 AM #2 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,837 Re: Probability of a low in O8 We had a thread on this exact topic a short time ago. Here is the link. https://forumserver.twoplustwo.com/2...-post50823652/ If you still have questions after reading that thread, let us know.
09-25-2017, 03:06 AM   #3
PLOfish123
stranger

Join Date: May 2017
Posts: 7
Re: Probability of a low in O8

Quote:
 Originally Posted by heehaww C(8,3)*64 * C(20, 2) + C(8,4)*4^4 * 20 + C(8,5)*4^5 + 3*C(8,3)*(6*16*20 + 4*36 + 64) + 4*C(8,4)*6*64 / C(52,5)
Correct me if I'm wrong as I ask some questions:

Line 1 is 3 lows and 2 highs - where does the 64 come from?

Line 2 is 4 lows and 1 high, Line 3 is 5 lows and 0 highs? Again where did 4^4 and 4^5 come from?

Lines 4 and 5 I'm completely lost - I assume it includes the pairs/trips cases but no idea how the numbers were derived.

09-25-2017, 03:35 AM   #4
whosnext
Pooh-Bah

Join Date: Mar 2009
Location: California
Posts: 4,837
Re: Probability of a low in O8

Quote:
 Originally Posted by PLOfish123 Correct me if I'm wrong as I ask some questions: Line 1 is 3 lows and 2 highs - where does the 64 come from? Line 2 is 4 lows and 1 high, Line 3 is 5 lows and 0 highs? Again where did 4^4 and 4^5 come from? Lines 4 and 5 I'm completely lost - I assume it includes the pairs/trips cases but no idea how the numbers were derived. Thanks for your time.

Case 1: 3 distinct lows with counts [1,1,1]; other two cards are high cards
= C(8,3)*C(4,1)*C(4,1)*C(4,1)*C(20,2)
= 680,960

where the C(8,3) is the number of ways the three low ranks can be chosen from the eight total low ranks, C(4,1)^3 represents the possible suit combos of the three low cards, and C(20,2) is the number of ways the two high cards can be chosen from the 20 total high cards.

Case 2: 3 distinct low ranks with counts [2,1,1], other card is a high card
= C(8,3)*3*C(4,2)*C(4,1)*C(4,1)*C(20,1)
= 322,560

Case 3: 3 distinct low ranks with counts [2,2,1]
= C(8,3)*3*C(4,2)*C(4,2)*C(4,1)
= 24,192

Case 4: 3 distinct low ranks with counts [3,1,1]
= C(8,3)*3*C(4,3)*C(4,1)*C(4,1)
= 10,752

Case 5: 4 distinct low ranks with counts [1,1,1,1], other card is a high card
= C(8,4)*C(4,1)*C(4,1)*C(4,1)*C(4,1)*C(20,1)
= 358,400

Case 6: 4 distinct low ranks with counts [2,1,1,1]
= C(8,4)*4*C(4,2)*C(4,1)*C(4,1)*C(4,1)
= 107,520

Case 7: 5 distinct low ranks with counts [1,1,1,1,1]
= C(8,5)*C(4,1)*C(4,1)*C(4,1)*C(4,1)*C(4,1)
= 57,344

TOTAL = 1,561,728

Grand total of possible 5-card boards:
= C(52,5)
= 2,598,960

So the percentage of possible 5-card boards on which a low hand is available is:

= 1,561,728 / 2,598,960
= 60.09%

Let me know if you have any questions on the above. (The derivation contained in your quote is a condensed representation of these seven cases.)

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