Bridge probability problem From volume 1 of Feller:
North and South have 10 trumps between them ( trumps being cards of a specific suit)
(a) find the probability that all three remaining trumps are in the same hand (that is, either East or West has no trumps) 
(b) if it is known that the king of trumps is included among the three, what is the probability that he is "unguarded" ( that is, one player has to king, the other the remaining two trumps )
I am stuck with (a).  It seems there is probability $2 * \frac{\binom{52-26-3}{10}}{\binom{52-26}{13}}$ for this scenario but the solution decreases this by $\frac{11}{50}$ which I do not see the origin of.
 A: Assuming a 52 cards deck and 13 cards per player, there are 26 cards in East and West's hands.
There are 3 trumps among those 26 cards. The question a) can be interpreted as: drawing 13 of those 26 cards, what are the chances that no trumps will be drawn by either player?
The probability of one player drawing none is $\frac{23}{26}\times \cdots \times \frac{11}{14} = \frac{13\times 12\times 11}{26\times 25 \times 24}=\frac{11}{100}$.
The probability of either player drawing none is twice that, or $\frac{11}{50}$.
A: $26$ remaining cards. You can distribute them 
between east and west in $\binom{26}{13}$ ways.
Then $\binom{2}{1}$ for which of them who gets the 
three remaining trumps and then $\binom{23}{10}$
for dealing the $10$ remaining cards to the one who got the trumps.
$$\frac{\binom{2}{1}\binom{23}{10}}{\binom{26}{13}}$$
So your solution looks right to me.
A: Simply focus on the $3$ trumps, and the $26$ available slots with $E$ and $W$ .
The first can go anywhere, but the next two are restricted to the same person, $$Pr = \frac{12}{25}\cdot\frac{11}{24} = \frac{11}{50}$$
