<< What is the probability there is some pair of suits your bridge hand does not include, and your partner's hand does not include the other pair?

...

My answer to this agrees with Gareth's and then Joshua's post -- as a fraction in lowest terms, p/q. My strategy was this: (how many ways you & your partner can each have hands in which exactly two suits appear) + (how many ways you can have one complete suit, & partner's hand shows exactly 2 suits) + (vice versa) + (the 12 ways each of you can have one full suit) Here's the Mathematica session: ------------------------------------------------------------------------------------------------------------ In[1]:= c = 26!/13!^2 Out[1]= 10400600 In[2]:= h22 = 4!((c-2)/2)^2 Out[2]= 649034632545624 In[3]:= h12 = 4*3!*(c-2)/2 Out[3]= 124807176 In[4]:= h21 = h12 Out[4]= 124807176 In[5]:= total = h22 + h12+h21 + 12 Out[5]= 649034882159988 In[6]:= npairhands = 52!/(13!^2 26!) Out[6]= 5157850293780050462400 In[7]:= prob = total/npairhands 54086240179999 Out[7]= --------------------- 429820857815004205200 Out[8]= 1.2583437773342733107186438969809670960643871973497 * 10^-7 [to 50 decimal places]. -------------------------------------------------------------------------------------------- Is this solution now in doubt? --Dan