@w0lfshad3,

"No-no, I didn't restrict the player, remember I said the player can repeat all units: 999999 and alternate 898989 etc. This doesn't change the number of times the player and lottery match, but the ratio of the match number to the fail number, which means the probability is lower than 1/4."

Explain how the rules to the game that lead you to make this analysis:

a) because we're having sampling w/o replacement for the distinct units part, e.g.

lottery units: 1-2 4 6 8

player units: 1 4 6 8

we can apply the hypergeometric distribution:

Player) because it picks with 5 distinct units out of which 4 match 4 out of 5 distinct units of the lottery, we have C(5,4) way to pick success, C(10-5, 1) way to fail (that's the unit that always fails actually), and total sample space is C(10,5), thus the probability is:

C(5,4)*C(10-5,1)/C(10,5)=25/252

Lottery) because it picks with 5 distinct units out of which 4 match 4 out of 5 distinct units of the player, we have C(5,4) way to pick success, C(10-5, 1) way to fail (that's the unit that always fails actually), and total sample space is C(10,5), thus the probability is:

C(5,4)*C(10-5,1)/C(10,5)=25/252

differ from these rules that lead to a probability of 1/4:

- The sample space is the numbers 0 to 9.

- Five unique numbers are selected by the player and the lottery.

- The player's numbers contain no adjacent pairs (0 and 9 are adjacent).

- The lottery's numbers contain exactly one adjacent pair.

- What is the probability that the player matches exactly four of the lottery's numbers?

I don't want a defense of your formula or a critique of my result, but agreement or disagreement as to whether or not we're solving the same sub-problem. If we're not, explain how my rules need to be modified.