Alisha M. answered • 11/29/16
Tutor
5
(8)
Specializing in Dietetics, Nutrition, Anatomy & Physiology, Algebra
The number of possible hands of five cards is correct: 2,598,960.
To count the ways of getting two pairs, let's look at the number of ways to get just one pair (meaning what are the possible combinations of suits for a single pair): you can get spade-club, spade-diamond, spade-heart, club-diamond, club-heart, and finally diamond-heart. That's six ways to get a single pair of cards.
Now all we need do is count the total number of ways to get two pair, disregarding suits. Let's say a pair of aces and 2's can be written "Ax2".
With aces as one of the pairs, then, the possible combinations of pairs with one pair being aces are
Ax2, Ax3, Ax4, Ax5, Ax6, Ax7, Ax8, Ax9, Ax10, AxJ, AxQ, and AxK. That's 12 ways. (We eliminated two pairs of aces because that's four of a kind.)
With deuces being one of the pairs, we first eliminate 2xA, since we already counted Ax2 in the "one pair is aces" list. We also eliminate 2x2. We are left with
2x3, 2x4, 2x5, 2x6, 2x7, 2x8, 2x9, 2x10, 2xJ, 2xQ, and 2xK. That's 11 more ways.
There's a pattern. For the 3's we eliminate 3xA, 3x2, and 3x3. The other 10 ways are 3x4, ..., 3xK.
Thus for the 4s we have 9 combinations, for the 5's we have 8 combinations, etc. on up to the Queens we have 1 combination. (We have zero combinations for the Kings because two pairs of kings are 4 of a kind, and all the other combinations were already counted, if you look at the last combination of each list.)
The total number of ways of getting 2 pairs, then (not including suits) is 12 + 11 + 10 + ... + 1 = 78.
But remember that, for each of these number-combinations, we have 6 ways of getting each pair, if you take suits into account. Thus the total ways of getting two pairs is 78*6*6 = 2808.
We must include our fifth card possibilities now. After getting our two pair, there are 48 cards left in the deck...but we must exclude the two cards of each number that would make our hand a full house. That is, if we had a pair of Aces and 9s, we'd need to exclude the remaining two Aces and the remaining two 9s so as not to get a full house.
This leaves 44 cards in the deck that will leave us with only two pair.
Now our grand total possible number of ways to draw two pair in 5 cards is 2802 * 44 = 123.552. The probability, then, of getting two pair is 123,552/2,598,960 = 0.04754, which is about 1 in 21.
To count the ways of getting two pairs, let's look at the number of ways to get just one pair (meaning what are the possible combinations of suits for a single pair): you can get spade-club, spade-diamond, spade-heart, club-diamond, club-heart, and finally diamond-heart. That's six ways to get a single pair of cards.
Now all we need do is count the total number of ways to get two pair, disregarding suits. Let's say a pair of aces and 2's can be written "Ax2".
With aces as one of the pairs, then, the possible combinations of pairs with one pair being aces are
Ax2, Ax3, Ax4, Ax5, Ax6, Ax7, Ax8, Ax9, Ax10, AxJ, AxQ, and AxK. That's 12 ways. (We eliminated two pairs of aces because that's four of a kind.)
With deuces being one of the pairs, we first eliminate 2xA, since we already counted Ax2 in the "one pair is aces" list. We also eliminate 2x2. We are left with
2x3, 2x4, 2x5, 2x6, 2x7, 2x8, 2x9, 2x10, 2xJ, 2xQ, and 2xK. That's 11 more ways.
There's a pattern. For the 3's we eliminate 3xA, 3x2, and 3x3. The other 10 ways are 3x4, ..., 3xK.
Thus for the 4s we have 9 combinations, for the 5's we have 8 combinations, etc. on up to the Queens we have 1 combination. (We have zero combinations for the Kings because two pairs of kings are 4 of a kind, and all the other combinations were already counted, if you look at the last combination of each list.)
The total number of ways of getting 2 pairs, then (not including suits) is 12 + 11 + 10 + ... + 1 = 78.
But remember that, for each of these number-combinations, we have 6 ways of getting each pair, if you take suits into account. Thus the total ways of getting two pairs is 78*6*6 = 2808.
We must include our fifth card possibilities now. After getting our two pair, there are 48 cards left in the deck...but we must exclude the two cards of each number that would make our hand a full house. That is, if we had a pair of Aces and 9s, we'd need to exclude the remaining two Aces and the remaining two 9s so as not to get a full house.
This leaves 44 cards in the deck that will leave us with only two pair.
Now our grand total possible number of ways to draw two pair in 5 cards is 2802 * 44 = 123.552. The probability, then, of getting two pair is 123,552/2,598,960 = 0.04754, which is about 1 in 21.
