Texas Hold'em Poker Probability - Pocket Pair vs. 2 Overcards(not suited, not connected)

Say if you have pocket pair 77, the other player X has AJ (not suited).  What's your probability of winning before the flop?

Possible remaining 5 card combinations: 1,712,304=48*47*46*45*44/5!=C(48,5)

Let's try to calculate the upper bound of the probability by listing out the possible cases.

The remaining 5 cards:

a) Without 7, A, J: 658,008=C(40,5)  the number of remaining cards without 7AJ is 40.

b) With a 7, but no AJ: 182,780=2*C(40,4)  two possible 7

c) With a 7, and only one A or J: 118,560=2*6*C(40,3)  two possible 7, six possible AJ

d) With a 7, and only (AA, JJ, AJ): 23,400=2*(3+3+9)*C(40,2)  3 possible AA, 3 JJ, 9 AJ

e) With 77, and but no AAA(JJJ): 15,178=C(46,3)-2  combination of the remaining 46 cards minus 2 for AAA/JJJ

f) Without 7, but with any AJ combinations: assuming your winning rate is q.

If we don't consider the other flush or straight possibilities for player X to win, the sum from a to f would be the upper bound for you to win.

The sum is 997,926+, the probability is about 58.28%+q, this is the upper bound of the probability for you to win.

Player X could get a TJQKA straight:

In case a, without 7,A,J, but 

  a.1) TQKyz: 4*4*4*C(28,2)  yz not in (7,A,J,T,Q,K)

  a.2) TTQKy,TQQKy TQKKy:  6*4*4*C(28,1)*3

  a.3) TTQQK, TTQKK,TQQKK: 6*6*4*3

Sum of (a.1,a.2,a.3) is 32,688, the probability is about 1.91%. 

In case b,

  b.1) 7TQKy: 2*4*4*4*C(28,1)  y not in (7,A,J,T,Q,K)

  b.2) 7TTQK,7TQQK,7TQKK:  2*6*4*4*3

Sum of (b.1,b.2) is 3,584, the probability is about 0.21%. 

  

In case c,

  c.1) 7TQKA, 7TQKJ: 2*4*4*4*3*2=768, the probability is about 0.04%. 

So, these cases will lower your winning probability by 2.16% to 56.12%+q without considering you win with a flush.