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This is the board : Board: [9d 8h 7c] My hand : [10d 7s] Opponents (unknown): [Xx Xx]

Cards remaining in the deck is 45. As there is still a turn and river to come, the total combination of hands is (n, k) = (45,2) = 990 combinations. Outs to improve hand: {J, 6, 10, 7} = (4+4+3+2) = 13 outs.

Is the following calculation of combining probabilities correct? It seems a rather large percentage.

Method 1: 
17.154 + 17.154 + 13.017 + 8.78 = 56.105 to improve hand on turn and river.     
Method 2:
170 + 170 + 129 + 87 = 556 / 990 = 56.16 to improve hand on turn and river.

Calculations:

[J] every [J] (=4) will complete a straight. Probability of catching a [J] on the turn and river is the objective. 52-7 = 45 cards left (4 [J]’s, 41 others)

Method 1
P([J] on turn OR river)
P([J] on turn) + P([J] on river) - P(both)
4/45 + 4/45 - ((choose(4,2)/choose(45,2))
8.88 + 8.88 - 0.606 = 17.154 %

Method 2
Determining number of hands
Any [J] will do, that is 4 cards.
Two [J]’s combinations: (n, k) = (4, 2) = 6 combinations 
One [J] combinations: 4*41 = 164
Total = 170
170/990*100 = 17.17

[6]: every [6] (=4) will complete a straight as well. Probability of catching a [6] on the turn and river is the objective. 52-7 = 45 cards left (4 [6]’s, 41 others)

Method 1
P([6] on turn OR river)
P([6] on turn) + P([6] on river) - P(both)
4/45 + 4/45 - ((choose(4,2)/choose(45,2))
8.88 + 8.88 - 0.606 = 17.154%

Method 2
Determining number of hands
Any [6] will do, that is 4 cards.
Two [6]’s combinations: (n, k) = (4, 2) = 6 combinations 
One [6] combinations: 4*41 = 164
Total = 170
170/990*100 = 17.17

[10]: as one [10] is taken, there are 3 [10]’s left in the deck. A [10] on the turn or river will give two pair and thus hand improvement: [10d] [10x] = [10d] [10x] [7s] [7c]

Method 1
P([10] on turn OR river)
P([10] on turn) + P([10] on river) - P(both)
3/45 + 3/45 - ((choose(3,2)/choose(45,2))
6.66 + 6.66 - 0.303 = 13.017%

Method 2
Determining number of hands
Only 3 [10] will do.
Two [10] combinations: (n, k) = (3, 2) = 3 combinations 
One [10] combinations: 3*42 = 126
Total = 129
129/990*100 = 13.03%

[7]: as two [7]’s are taken, there are 2 [7]’s left in the deck. A [7] on the turn or river will give 3 of a kind and thus hand improvement: [7x] = [7s] [7c] [7x]

Method 1
P([7] on turn OR river)
P([7] on turn) + P([7] on river) - P(both)
2/45 + 2/45 - ((choose(2,2)/choose(45,2))
4.44 + 4.44 - 0.303 = 8.78%

Method 2
Determining number of hands
Only 2 [7]’s will do.
Two [7]’s combinations: (n, k) = (2, 2) = 1 combination 
One [7] combinations: 2*43 = 86
Total = 87
87/990*100 = 8.78%
  • Cross posting is discouraged math.stackexchange.com/questions/2644495/… – paparazzo Feb 10 '18 at 17:01
  • What is the question? Probability of the turn or river being a J, T, 7 or 6? – Raymond Timmermans Feb 10 '18 at 17:39
  • There are not (45,2) possible turn and rivers, but (47,2) since the opponent's cards is not known. – Raymond Timmermans Feb 10 '18 at 17:39
  • @RaymondTimmermans Yes, the probability of the turn or river being a J, T, 7 or 6. So only the flop, turn and river will count as possibilities? – Bondeaux Feb 10 '18 at 17:49
  • I don't follow the last sentence. What do you mean? – Raymond Timmermans Feb 10 '18 at 18:32
1

I don't follow your math.

After the flop there are 47 unknown cards. You don't know what any opponent is holding.

Need to consider you can improve and still not win. T may not be good and 6 is dead to JT.

This is one that is easier for solve without combination in my opinion. Make the first or miss the first and make the second.

+13/47 + 13/46 * (1 - 13/47) = 48.1%

Believe it or not that is also the number for both cards hitting the flush. You are actually not supposed to add odds like and it is round about the answer is correct.

You can solve it using 1 - not = +1-(47-13)/47*(46-13)/46

Or you can solve it using combination.

If you have 13+ good outs you should call any bet on the flop. Mathematically you can jam the flop.

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