Question 1195816: https://docs.google.com/document/d/e/2PACX-1vQytc1_zv8tSsDtgSW_muwH2juU0fELm_6xPfkm1MXabMkOX5zJi-CW5H6jSjseGG3Ay4mvM7LIEq9I/pub
Here's my question(any better way to ask questions with images? please share).
I want to find probability for various cases. I've calculated successfully for some, I know the answer of all as these are classic problems, but I don't know the logic behind the calculation. So, I am asking with you.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The example shown on the URL you provided is explained clearly; I'm not sure what your question is exactly.
The calculation for all cases (different poker hands) has the same denominator, C(52,5), because that is the number of combinations of 5 cards you can be dealt from a 52-card deck.
For the calculation for the full house shown in the URL, the numerator is the product of the numbers of ways you can fulfill the requirements of the problem:
(1) Choose 1 of the 13 ranks to be the one of which you are going to get 3: C(13,1)
(2) Choose 3 of the 4 cards of that rank: C(4,3)
(3) Choose 1 of the OTHER 12 ranks to be the one of which you are going to get 2: C(12,1)
(4) Choose 2 of the 4 cards of that rank: C(4,2)
The numerator is C(13,1)*C(4,3)*C(12,1)*C(4,2).
The calculations will be similar for the other cases.
And most if not all of them can be set up in different ways that of course get the same result. For example, the calculation for the full house could go like this:
(1) Choose 2 of the 13 ranks to be the ones you are going to get: C(13,2)
(2) Choose 1 of those 2 ranks to be the one of which you are going to get 3: C(2,1)
(3) Choose 3 of the 4 cards of the first rank: C(4,3)
(4) Choose 2 of the 4 cards of the second rank: C(4,2)
The numerator is then C(13,2)*C(2,1)*C(4,3)*C(4,2).
That calculation is slightly different than before, because it uses slightly different logic; but the result of the calculation is the same.
If you need more help with this topic, make new post(s) showing specific cases (i.e., for different poker hands), SHOWING how you tried to set up the calculations.
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Added to response after message from student....
*** Note you can skip this section and go to my second revision further below
*** for an easier and more complete description of how to compute the
*** probabilities for the different cases for the 5-digit number problem
The "choose" method is not applicable to the problem with the 5-digit numbers -- at least, not directly.
For this problem, you need to look at the probability that each of the 5 digits in the number satisfies the conditions of the problem.
For one pair, the correct answer is 0.5040, and your calculation was this:
10C1*9C1*8C1*7C1/10,000=0.504
It is only by chance that your calculation gives the right answer.
To get the answer by the correct method, you need to consider all the ways you can get one pair of identical digits in a 5-digit number. Note for each digit there are 10 possibilities, so the denominator in every case is 10^5 = 100000.
(1) 2nd digit matches first: 10*1*9*8*7
(2) 3rd digit matches one of the first two: 10*9*2*8*7
(3) 4th digit matches one of the first three: 10*9*8*3*7
(4) 5th digit matches one of the first four: 10*9*8*7*4
Using the common factor 10*9*8*7, the numerator of the probability fraction is
10*9*8*7*(1+2+3+4) = 50400
and the answer is 50400/100000 = 0.5040
The other cases can be solved by similar methods -- although I have not been able to find the given probabilities for some of the cases. Apparently some of the orders in which the required digits can be found are escaping me.
But I can see how a couple of the others are done....
All different: The numerator is 10*9*8*7*6 = 30240; the probability is 30240/10000 = 0.3024
4 of a kind:
(1) 5th digit is different: 10*1*1*1*9
(2) 4th digit is different: 10*1*1*9*1
(3) 3rd digit is different: 10*1*9*1*1
(4) first two digits are different; third can match either of the first two; last two match third: 10*9*2*1*1
Again using the common factor 10*9*1*1, the numerator of the probability fraction is
10*9*1*1(1+1+1+2) = 450
and the probability is 450/100000 = .0045
5 of a kind:
This one is easy; the last four digits must match the first. The numerator is 10*1*1*1*1; and the probability is 10/100000 = .0001
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Second revision....
For each case, determine the number of different positions in the 5-digit number that the identical digits can be.
(1) All different
There is only one possibility for the positions to be occupied by the 5 different digits -- they must be in positions 12345. The numerator for this single possibility is 10*9*8*7*6 = 30240; the probability is 30240/100000 = 0.3024.
(2) 1 pair
You need to choose 2 of the 5 positions to be the ones containing the identical digits; the number of ways you can do that is C(5,2)=10. The numerator in each case is 10*9*8*7*1 = 5040; the probability is (10*5040)/100000 = 50400/100000 = 0.5040.
(3) 2 pair
This one is the trickiest of the bunch. You need to choose 2 of the 5 positions for one of the pairs, then 2 of the remaining 3 positions for the other pair; but the two pairs could switch places. So the number of different combinations of positions for the two pairs is ((C(5,2)*C(3,2))/2=15. The numerator for each of those cases is 10*9*8*1*1 = 720; the probability is (15*720)/100000 = 10800/100000 = 0.1080.
(4) 3 of a kind
You need to choose 3 of the 5 positions to be the ones containing the three identical digits; the number of ways you can do that is C(5,3)=10. The numerator for each of those cases is 10*1*1*9*8 = 720; the probability is (10*720)/100000 = 7200/100000 = 0.072.
(5) Full House
Very similar to the case for 3 of a kind. You again need to choose 3 of the 5 positions to be the ones containing the three identical digits; the number of ways of doing that is C(5,3)=10. But here the last two digits need to match, so the numerator for each of these cases is 10*1*1*9*1 = 90; the probability is (10*90)/100000 = 900/100000 = 0.009.
(6) 4 of a kind
This one is easy; you need to choose 1 of the 5 positions for the digit that is different from the rest; there are obviously 5 different ways to do that. The numerator in each case is 10*9*1*1*1 = 90; the probability is (5*90)/100000 = 450/100000 = 0.0045
(7) 5 of a kind
And this one is the easiest of them all. There is only one way to choose all 5 of the positions to be the ones containing the single digit; the numerator is 10*1*1*1*1=10; the probability is (1*10)/100000 = 10/100000 = 0.0001
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