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123455

111555

the sample space is 0..9

row 1 can pick 6 numbers out of which 2 are repeated

row 2 can pick 6 numbers out of which 3 by 3 are repeated

I want to know what is the real probability that row 1 will match with 2 distinct numbers numbers row 2, and a repeated number (which is one of those first two distinct numbers if they match row 2). Warning, the 6th number is not obligated to match one of those two numbers.

The chance that row 1 matches row 2 with 2 numbers is:

C(5,2)C(10-5,0)/C(10,2)=10/45

C(2,2)C(10-2,3)/C(10,5)=56/252

The chance that row 1 repeated number matches row 2 numbers is:

C(5,1)C(10-5,1)/C(10,2)=25/45

C(2,1)C(10-2,4)/C(10,5)=140/252

how to combine these two probabilities to find out the real probability?

can I multiply them like this? 10/45 x 25/45 = 5/81? The result seems to low in comparison.

EDIT1:

But if I calculate like this I got a different answer:

1/10 x 1/9 x 1/8 x 1/7 x 1/6 x 1/10?

10!/5!(10-5)!10

126/5

final probabilty: 5/126

How to calculate?

EDIT2:

I've refined the first calculation of the repeated unit probability.

The logic is: the one repeated number can only match in 4 ways with the 2 distinct row2 numbers, or, the 4 distinct row1 numbers can match in 1 way (the repeated number), with the row2 2 distinct numbers.

The math is:

C(4,1)C(10-4,1)/C(10,2)=24/45

C(2,1)C(10-2,3)/C(10,4)=112/210

The question remains, the total row1 row2 match probability is (10/45) x (24/45) ?

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