Probability Check
Need a check. Two Players.
What is the probability that if neither has a pair, there will be one rank matched , e.g., (A5 and A8) or (A5 and K5)?
(I ain't gonna tell you what I got unless I'm right 😀)
2 Replies
I will give this a try using my brute-force combinatorial method:
Case 1: The four cards have 1 rank
Total 2-person 2-card deals
= C(13,1)*C(4,4)*C(3,1)
=39
Clearly in this case the 2 players cannot both not have pairs.
=0
Clearly in this case the 2 players cannot both not have pairs and they share exactly one rank.
=0
Case 2A: The four cards have 2 ranks, 2 of each rank
Total 2-person 2-card deals
= C(13,2)*C(4,2)*C(4,2)*C(3,1)
= 8424
Total 2-person 2-card deals in which both players do not have a pair
= C(13,2)*C(4,2)*C(4,2)*C(2,1)
= 5616
Total 2-person 2-card deals in which both players do not have a pair and they share exactly one rank
=0
Case 2B: The four cards have 2 ranks, 3 of one rank and 1 of the other rank
Total 2-person 2-card deals
= C(13,2)*C(2,1)*C(4,3)*C(4,1)*C(3,1)
= 7488
Total 2-person 2-card deals in which both players do not have a pair
= 0
Total 2-person 2-card deals in which both players do not have a pair and they share exactly one rank
=0
Case 3: The four cards have 3 ranks
Total 2-person 2-card deals
= C(13,3)*C(3,1)*C(4,2)*C(4,1)*C(4,1)*C(3,1)
= 247,104
Total 2-person 2-card deals in which both players do not have a pair
= C(13,3)*C(3,1)*C(4,2)*C(4,1)*C(4,1)*C(2,1)
= 164,736
Total 2-person 2-card deals in which both players do not have a pair and they share exactly one rank
= C(13,3)*C(3,1)*C(4,2)*C(4,1)*C(4,1)*C(2,1)
= 164,736
Case 4: The four cards have 4 ranks
Total 2-person 2-card deals
= C(13,4)*C(4,1)*C(4,1)*C(4,1)*C(4,1)*C(3,1)
= 549,120
Total 2-person 2-card deals in which both players do not have a pair
= C(13,4)*C(4,1)*C(4,1)*C(4,1)*C(4,1)*C(3,1)
= 549,120
Total 2-person 2-card deals in which both players do not have a pair and they share exactly one rank
=0
-----
Subtotals:
Total number of 2-person 2-card deals
= 812,175
Total 2-person 2-card deals in which both players do not have a pair
=719,472
Total 2-person 2-card deals in which both players do not have a pair and they share exactly one rank
=164,736
-----
So the desired probability is
= 164,736 / 719,472
= 0.22896791
Looks good. I did it a little simpler.
Given no pairs: The number of ways villain can have a match is 2 * 3 * 44 = 264
The total number hands villain can have given there are no pairs is 1225 - (11*6+2*3) = 1153
Prob. = 264/1153 = 0.22896791
I also got this result (to 4 decimals) by simulation.
