What is the probability that the remaining two spades are distributed so that $B$ and $D$ have one spade apiece? In bridge, each of the four players $(A, B, C, D)$ receives $13$ cards. Suppose $A$ and $C$ have $11$ of the $13$ spades between them. What is the probability that the remaining two spades are distributed so that $B$ and $D$ have one spade apiece?
my Attempt
since there are $11$ of spades that were taken by $A$ & $C$, then the number of remaining cards is $52-11=41$.
The probablity that $B$ receives spades is then $1/41$.
The probablity that B receives spade is then $1/40$.
Then the probability that the remaining two spades are distributed so that $B$ and $D$ have one spade apiece is $(1/41)\cdot (1/40)$.
 A: Consider all situations where $A$ and $C$ hold exactly 11 spades inbetween them. We may safely assume then that they also holds 15 non-spades, which leaves 26 cards (of which 2 spades) for $B$ and $D$. 
We will split those cards into two piles to be distributed to $B$ and $D$. Put one of your two spades down. You still have 25 cards to distribute - 12 will be added to the pile with a spade in it already, and 13 to the other pile. There is a $\frac{12}{25}$ chance the other spade will be added to the pile with the spade. This translates to an $48$% chance for the spades to be together, ergo a $52$% chance for them to be split.
A: I'm a bridge player.  This problem almost certainly is intended to mean that $A$ and $C$ have exactly $11$ of the $13$ spades between them, no more, no less.
In that case, the chances that the spades are divided $1-1$ between the remaining two hands is exactly $52\%$.  Assign the first of the remaining two spades randomly to $B$ or $D$.  Then when we want to assign the second spade, there are $25$ vacant spaces remaining, and the second spade has an equal probability of being assigned to any of them.  Since $13$ of those $25$ spaces are in the hand that doesn't already contain a spade, the probability that the spades will divide $1-1$ between the remaining two hands is exactly $\frac{13}{25}=52\%$.
You can use a similar analysis if you really did mean that the last two spades also can go into hands $A$ and $C$.  The first spade "keeps you alive" with probability $\frac{26}{41}$ and if that happens, the second spade goes into the remaining "permitted" hand with probability $\frac{13}{40}$, so the answer is $\frac{169}{820} \approx 0.2061$.
