[mykey1961]

[ QUOTE ]
I don't think I can explain this any clearer .

The probability 1 or 2 gets selected on the first draw is 2/10 , agree ?

Given that 1 or 2 has been selected , the other number gets removed from the vault so that there are 8 numbers left . The probability it's B should be 2/8 since he has two good choices out of 8 . Agree ?

However , B may hit 3 or 4 on the first draw rather than the second with probability 2/10 . Agree ?

Given that B hits a 3 or a 4 , we would remove the other number so that there are 8 numbers left . The probability A hits a 1 or a 2 from 8 possible choices is 2/8 . Agree ?

Now we add 2/10*2/8 + 2/10*2/8 = 0.1

Do the other examples the same way .

[/ QUOTE ]

I completely agree with your math there.

That's the exact same results I posted earlier:
Given that

Albert has 2 tickets therefore P(A|) = 2/10
Bill has 2 tickets therefore P(B|) = 2/10
Charley has 3 tickets therefore P(C|) = 3/10
Dennis has 3 tickets therefore P(D|) = 3/10

P(B|A) = 2/8
P(C|A) = 3/8
P(D|A) = 3/8
P(A|B) = 2/8
P(C|B) = 3/8
P(D|B) = 3/8
P(A|C) = 2/7
P(B|C) = 2/7
P(D|C) = 3/7
P(A|D) = 2/7
P(B|D) = 2/7
P(C|D) = 3/7

P(AB) = P(A|)*P(B|A) + P(B|)*P(A|B) = 2/10*2/8 + 2/10*2/8 = 1/10
P(AC) = P(A|)*P(C|A) + P(C|)*P(A|C) = 2/10*3/8 + 3/10*2/7 = 9/56
P(AD) = P(A|)*P(D|A) + P(D|)*P(A|D) = 2/10*3/8 + 3/10*2/7 = 9/56
P(BC) = P(B|)*P(C|B) + P(C|)*P(B|C) = 2/10*3/8 + 3/10*2/7 = 9/56
P(BD) = P(B|)*P(D|B) + P(D|)*P(B|D) = 2/10*3/8 + 3/10*2/7 = 9/56
P(CD) = P(C|)*P(D|C) + P(D|)*P(C|D) = 3/10*3/7 + 3/10*3/7 = 9/35

But what I'm having trouble with is why my alternate method of counting rather than calculating is giving different results.