what is the probability of...

- 2 numbers matching 2 others from 0 to 9 = (1/10)^2?

This isn't necessarily true. Are you sampling with replacement? Does order matter?

Your question is incomplete; ref markr's post.

Assuming that you're drawing without replacement, that the 6 distinct numbers are also in the range 0-9, and that order doesn't matter, then I'd do it this way:

As you noted, the two equal numbers can be combined into a single distinct number.

So, we've got sets of 5 and 6 distinct numbers in the range 0-9, and we're looking for the probability that they overlap in exactly four places.

First, let's fix the set of 5 and compute the probability for selecting a set of 6.

There are C(10,6) ways to choose the set of 6.
There are C(5,4) ways to select 4 of the set of 5 to match.
There are C(10-5,2) ways to select the remaining two numbers that don't match.

Therefore, the probability is:
C(5,4) * C(10-5,2) / C(10,6) = 5 * 10 / 210 = 5/21

Let's do it the other way (fix the 6 and select the 5) to see that we get the same probability.

There are C(10,5) ways to choose the set of 5.
There are C(6,4) ways to select 4 of the set of 6 to match.
There are C(10-6,1) ways to select the remaining number that doesn't match.

Therefore, the probability is:
C(6,4) * C(10-6),1) / C(10,5) = 15 * 4 / 252 = 5/21

C(n,r) is the number of combinations of n things taken r at a time.

Thank you. The hypergeometric distribution

Multiply that by ~20% statistical probability to win 3rd category (win by 4 numbers) playing a repeated unit (2) and 4 distinct(1111) when a 6/49 draw draws all distinct units

as opposed to playing all distinct units when lottery draws all distinct units, which is 9/21 > 5/21 by x1.8 better. C(6,4)*C(10-6, 2) / C(10,6) = 9/21



Combined with the biggest statistic of ~50% of the time the lottery draws (21111), that is playing a repeated unit (21111) when the lottery draws the same and guessing 4 units, which chance is 0.099... C(5,4)*C(10-5, 1) / C(10,5) = 25/252

as opposed to the same statistic but playing all distinct units, which is C(6,4)*C(10-6, 1) / C(10,5) = 60/252=0.238... > 0.099... by x2.4

Playing all distinct units is better than playing one repeated unit and 4 distinct by:
1.8x20 + 2.4x50 = 156/70 = 2.2 better 70% (sample total) of the time?

Actually the answer is the hypergeometric distribution:
[kCx] [N-kCn-x] / [NCn]

5 units (6 out of which 2 are repeated which counts as one distinct), out of a space of 10 units, matching 4 units out of 6.

Which matches this problem:
http://easycalculation.com/statistics/learn-hypergeometric-distribution.php

N=10, n=6, k=5 and x=4
C(5,4)C(10-5,2)/C(10,6)=50/210

N=10, n=5, k=6 and x=4
C(6,4)C(10-6,1)/C(10,5)=60/252