[jay_shark]

Part Two :


a)6 pair from 13 hands : There are 13c6 ways to select which pairs . once we've selected the 6 pairs , there is now an odd card out . 13c6*(4c2)^6*28 = 2,241,727,488 . This number divided by 52c13 is 0.0035302... . Again multiply this number by 6 .

b)13c5*13c5*13c3*4c3*3 We multiply by 4c3 since there are 4 ways to select 3 suits from 4 .Also we may have something like 10 cards in spades and 3 cards in diamonds . So we have to add an additional 13c5*8c5*13c3*4c2 + 13c5*8c3*13c5*4c2 . In total we have 6 364 873 944 different ways this may occur . The probability is close to 1% if you take that number above divided by 52c13 . I'm omitting the 13 card straight flushes as well as the occasional time that one of the 5 card hands is a straight flush .

c) I have to think of an easy way to approximate c .

d)(4^13 -4)/52c13=0.0001056. Multiply this number by 13 .
e) 4/52c13 . If you want it in points , multiply it by 50 .

a,d and e are exact answers but b is pretty close.