William Cordwell wrote:
There are (4 C 2) = 6 ways to pick which two suits you have; this also fixes the suits of your partner. You have, from these two suits, (26 C 13) ways of getting your 13 cards; your partner also has (26 C 13) ways of getting his 13 cards from the other two suits. You then divide by (52 C 13) ways of getting your hand randomly, and divide also by (39 C 13) ways of your partner getting his hand randomly from the remaining 39 cards.
Thus, P = 6 * (26 C 13) * (26 C 13) / [ (52 C 13) * (39 C 13) ] = 21850/173640942909 = (acc'ding to Mathematica)1.25834377733429657622523386340107748418239154034425 x 10^(-7)(forty places)
I'm not convinced that this is right, but I'd like to know why, if not. What are the odds of getting the same answer to 14 digits, but differing otherwise? :-)
In a problem like this? Very high. :-) You've over-counted a few deals: the ones in which you and your partner have only *one* suit apiece. You say "pick which two suits you have", but in that situation there's more than one way to do that picking while still getting the same deal. You have therefore counted those deals multiple times (twice, as it turns out): once for each way of picking "which two suits you have". -- g