# Probability - reduced sample space

In a bridge game, 52 cards are dealt equally to players E, W, N, S.

If N and S have a total of 8 spades among them, what is the probability that E has 3 of the remaining 5 spades?

$$\frac{(^5C_3)(^{21}C_{10})}{^{26}C_{13}}=0.339$$

Reasoning goes:

1. Choose 3 spades for E (first term in numerator)
2. Choose the other 10 cards out of the 21 cards (second term in numerator)
3. Denominator: choose 13 out of 26 cards for E.

And so my question is: what is the reduced sample space here actually? Why is it 26 cards as the sample space for both numerator and the denominator? Why not 52 cards (i.e. all four players)?

In a bridge game, $52$ cards are dealt equally to players $\color\red{E}$, $\color\green{W}$, $\color\orange{N}$, $\color\purple{S}$.

The number of combinations in which $\color\orange{N}$ and $\color\purple{S}$ have a total of $8$ spades among them:

$\sum\limits_{n=0}^{8}\color\orange{\binom{13}{n}\cdot\binom{52-13}{13-n}}\cdot\color\purple{\binom{13-n}{8-n}\cdot\binom{52-(13-n)}{13-(8-n)}}\cdot\color\red{\binom{52-13-13}{13}}\cdot\color\green{\binom{52-13-13-13}{13}}$

Which evaluates to $371637481296340124590207883400$.

The number of combinations in which $\color\red{E}$ has $3$ of the remaining spades:

$\sum\limits_{n=0}^{8}\color\orange{\binom{13}{n}\cdot\binom{52-13}{13-n}}\cdot\color\purple{\binom{13-n}{8-n}\cdot\binom{52-(13-n)}{13-(8-n)}}\cdot\color\red{\binom{13-8}{3}\cdot\binom{52-13-13-(13-8)}{13-3}}\cdot\color\green{\binom{52-13-13-13}{13}}$

Which evaluates to $126033580613541433556679195240$.

The probability is therefore $\frac{126033580613541433556679195240}{371637481296340124590207883400}\approx33.91\%$.

And the size of the reduced sample space is the denominator in the fraction above.

You need to find the probability that E has 3 of the remaining 5 spades. Now between them N and S have a total of 8 spades, out of their combined total of 26 cards. This is why you have a reduced sample space, since you are now looking for the probability of the spades being distributed in a "smaller deck" of 26 cards between E and W, and can ignore the "deck of 26 cards" held by N and S.

Then we have that $^5C_3$ is the number of different combinations of $3$ spades out of the remaining $5$ (which is the number of spades we need for E). Then out of the $26-5=21$ remaining non-spade cards you have $^{21}C_{10}$ combinations to choose from to make up Es other $10$ cards. So now multiply these together to give the number of all possible hands that give E 3 spades, and then divide by the total number of combinations of $13$ cards from $26$, which is $^{26}C_{13}$, to give the probability, which gives the answer you have.

• thank you for your answer. But this begs the question - why is E and W automatically chosen to be the reduced sample space? why can't I just reduce the sample space to, for example, just E alone? Sep 14, 2016 at 15:23
• For N and W you can disregard their $26$ cards since their $8$ spades have been allotted to them. Now you still have to divide up the spades between E and W, so you must look at the different number of hands out of $26$ that arise from E having 3 spades and W having 2 spades. You can't do this by just thinking of E alone, that would leave you with a sample space of $13$ but with the remaining spades distributed among $26$ cards, which wouldn't make sense. Sep 14, 2016 at 15:40
• thank you for that comment, really informative! Sep 14, 2016 at 15:49
• No problem. Thanks :) Sep 28, 2016 at 18:01

Another way to solve the problem using a reduced sample space focuses only on the $5$ spades

Each of E and W have $13$ slots available, and we want $3$ spades for E, $2$ for $W$

$\boxed{.}\boxed{.}\boxed{s}\boxed{.}\boxed{s}\boxed{s}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\quad\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{s}\boxed{.}\boxed{s}\boxed{.}\boxed{.}\quad$

Thus $Pr = \dfrac{\binom{13}{3}\binom{13}{2}}{\binom{26}{5}}$