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.
Let me describe the seven cases, which should answer your questions.
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.)
