Conditional Probability: Bridge Hand given north and south has 8 spades

Question

I apologise as I'm pretty certain I could find the actual answer myself on these boards (as in the solution to the problem) but if possible I'd like it if someone can confirm whether my thinking is correct.

In the card game bridge, the entire 52 cards are dealt out equally to 4 players, E/W/N/S if N and S have a total of 8 spades among them, what is the probability E has 3 of the remaining 5 spades?

This is example 2c on page 60 in "A First Course in probability 8th edition"

and gives the solution as $$\frac{{{5}\choose{3}}{{21}\choose{10}}}{{26}\choose{13}}\approx .339$$

Later in the book, it asks to recalculate this using conditional probability (question 3.3 page 102):

Compute the conditional probability that E has 3 spades given North and South have a combined total of 8 spades.

My answer is then as follows:

Let E be the event that E has 3 spades. and F the event that N and S have 8 spades.

Then $P(E)=\frac{{{5}\choose{3}}{{21}\choose{10}}}{{26}\choose{13}}$ as it is given that 26 cards have already been dealt to N and S, of which 8 of the 13 spades have also been dealt. This means for E to occur we have to choose 3 of the remaining 5 spades and any combination of 10 cards remaining from the 21 leftovers (that is 26 remaining cards minus the 5 spades). All are divided by the reduced sample space.

(This is one of the parts in which I want to be certain)

As we only care about E we can consider N and S to be one person which will be dealt 26 cards, 8 of which must be spades, giving $P(F)=\frac{{{13}\choose{8}}{{39}\choose{19}}}{{52}\choose{26}}$ this reasoning seems right to me as the two of them should receive any combination of 26 cards so long as 8 of them are spades. (We don't care if N has all 8 or if S has all 8.)

Then our conditional Probability is

$$P(E|F) = \frac{P(E\cap F)}{P(F)}.$$

My major issue is I'm not sure how I would go about calculating $P(E\cap F)$ in this instance. Logically I argue that the two events P and F are independent of each other, meaning $$P(E\cap F)=P(E)P(F)$$ and so making our final conditional probability $P(E|F)=P(E)$ which yes does give me the right answer, but unless the question specifically states that these are independent of each other, I would then have to prove that $P(E\cap F)=P(E)P(F)$ holds surely?

Any clarification would be greatly appreciated. thank you.

Answer 1

Calculating $Pr(E\cap F)$, let us define our sample space as all ways of distributing the cards where order within each hand doesn't matter and where north and south's hands are considered collectively as one. There are $\frac{52!}{26!13!13!}$ such distributions, each of which are equally likely to occur.

Let us count $|E\cap F|$ in this sample space. To do so, first choose which eight spades north/south got and then which 18 non-spades north/south got.

Then, choose which three spades from those remaining east got and which 10 non-spades from those remaining east got. All remaining cards will be given to west.

There are then $\binom{13}{8}\binom{39}{18}\binom{5}{3}\binom{21}{10}$ such arrangements implying the probability $Pr(E\cap F)$ is:

$$Pr(E\cap F)=\frac{\binom{13}{8}\binom{39}{18}\binom{5}{3}\binom{21}{10}}{52!/(26!13!13!)}=\frac{15152709}{276092852}\approx 0.05488$$

Continuing on to calculate $Pr(F)$, now we temporarily instead consider the sample space to be just the ways in which we divide the deck in half, i.e. giving 26 cards to north/south collectively where order within the hand is unimportant and giving the remaining cards to east/west. There are $\binom{52}{26}$ such ways to do so, each of which are equally likely to occur.

We count $|F|$ in this sample space as being $\binom{13}{8}\binom{39}{18}$, thus making $Pr(F)=\frac{\binom{13}{8}\binom{39}{18}}{\binom{52}{26}}=\frac{44681065}{276092852}\approx 0.161833$

Taking the ratio $\frac{Pr(E\cap F)}{Pr(F)}$ will arrive at the same answer as given before after simplifications.

$$Pr(E\mid F)=\frac{Pr(E\cap F)}{Pr(F)}=\frac{39}{115}\approx 0.3391304$$

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As an aside, let us look at $Pr(E)$ and inspect whether $E$ and $F$ truly are independent.

We let the sample space be all ways in which East is given a hand ignoring how the rest of the cards are distributed. There are $\binom{52}{13}$ ways in which this can happen. Choosing which three spades and which 10 nonspades gives us $\binom{13}{3}\binom{39}{10}$ possible hands and a probability of $Pr(E)=\frac{\binom{13}{3}\binom{39}{10}}{\binom{52}{13}}\approx 0.28633$

(notice that no mention of how spades are distributed among north/south appeared in the calculation of $Pr(E)$. You seem to have confused calculating $Pr(E)$ with $Pr(E\mid F)$)

We compare $Pr(E\cap F)\approx 0.05488$ to $Pr(E)Pr(F)\approx 0.161833\cdot 0.28633\approx 0.04634$. Indeed, even if we take the exact values we see that these are different numbers and therefore $E$ and $F$ are not independent as claimed.

Answer 2

For calculating $P(E\cap F)$ think of the number of ways that $13$ spades can be divided in this context such that exactly $3$ come in hands of $\bf E$ and exactly $2$ in hands of $\bf W$.

There are $\binom{13}3\binom{13}2\binom{26}8$ ways, and this on a total of $\binom{52}{13}$ ways.

This results in: $$P(E\cap F)=\frac{\binom{13}3\binom{13}2\binom{26}8}{\binom{52}{13}}$$

Sortlike you can find $P(F)$ as:$$P(F)=\frac{\binom{26}5\binom{26}8}{\binom{52}{13}}$$

Working this out with application of $P(E\mid F)P(F)=P(E\cap F)$ leads to:$$P(E\mid F)=\frac{13!13!5!21!}{3!10!2!11!26!}=\frac{\binom53\binom{21}{10}}{\binom{26}{13}}$$

Answer 3

The excellent answers provided by JMoraviz and drhab are already complete. Nevertheless, let me give a lengthier reasoning, which I believe could be helpful to novices.

Let the allocation of the players be fixed say N-E-S-W in the usual pattern. The problem tells us that the N-S pair has got 8 spades, which is equivalent to

the W-E pair has got 5 spades

Let "the W-E pair has got 5 spades" be the conditioning event $F$. Furthermore, let $E$ be the event

W has got 3 spades

Note first that we can distribute the 52 cards to the four players in $N={52\choose 26}{26\choose 13}{26\choose 13}$. This is because there are ${52\choose 26}$ ways to select 26 cards for the N-S pair and ${26\choose 13}$ to equally distribute the cards within this pair; then there are other ${26\choose 13}$ ways to distribute the other 26 among the W-E pair equally. By the Multiplication Principle, the sample space has thus $N$ elements, each having probability $1/N$.

Let us first compute $P(F)$. Here we are looking for the number of patterns 13-13-13-13 s.t. the pair W-E gets 5 spades overall (or, equivalently, the pair N-S gets 8 spades overall). The strategy is to select 26 cards for W-E s.t. 21 are non-spades and 5 are spades and then distribute these cards among W-E equally, i.e. 13-13. There are ${39\choose 21}$ ways to select 21 non-spades and ${13\choose 5}$ ways to select 5 spades. Thus, there are ${39\choose 21}{13\choose 5}$ ways to get 26 for the pair W-S and these can be distributed equally (i.e. 13-13) in ${26\choose 13}$ ways. Lastly, we are left with other 26 cards to be distributed between the members of N-S; this can be done in ${26\choose 13}$ ways. Thus the number of patterns 13-13-13-13 s.t. W-E gets 5 spades overall is ${39\choose 21}{13\choose 5}{26\choose 13}{26\choose 13}$ (again by the Multiplication Principle) and the probability of $F$ is

$$ P(F) = \frac{{39\choose 21}{13\choose 5}{26\choose 13}{26\choose 13}}{{52\choose 26}{26\choose 13}{26\choose 13}}. $$

(Wait don't simplify yet, they will cancel out later!)

Now $P(E\cap F)$. Here we are looking for the number of patterns 13-13-13-13 s.t. W-E gets 5 spades overall and s.t. W gets 3 spades. From the $39$ non-spades there are ${39\choose 21}$ ways to get 21 cards. For each of these, there are ${21\choose 10}$ ways to give W 10 non-spade cards. This completes the distribution of the non-spade cards between W-E. Now, there are ${13\choose 5}$ ways to select 5 spade cards and for each of these, there are ${5\choose 3}$ ways to select three spades for W (and thus 2 spades for E). This completes the distribution of 26 cards among W-E with the required condition. We are left with 26 cards and these can be distributed to N-S in ${26\choose 13}$ ways.

Thus the number of patterns 13-13-13-13 s.t. W and E get 5 spades overall and s.t. W gets 3 spades. is ${39\choose 21}{21\choose 10}{13\choose 5}{5\choose 3}{26\choose 13}$. The probability of $E\cap F$ is thus

$$ P(E\cap F) = \frac{{39\choose 21}{21\choose 10}{13\choose 5}{5\choose 3}{26\choose 13}}{{52\choose 26}{26\choose 13}{26\choose 13}}. $$

And finally we have what we are looking for

$$ P(E|F) = \frac{P(E\cap F)}{P(F)} = \frac{\frac{{39\choose 21}{21\choose 10}{13\choose 5}{5\choose 3}{26\choose 13}}{{52\choose 26}{26\choose 13}{26\choose 13}}}{\frac{{39\choose 21}{13\choose 5}{26\choose 13}{26\choose 13}}{{52\choose 26}{26\choose 13}{26\choose 13}}} = \frac{{21\choose 10}{5\choose 3}}{{26\choose 13}}. $$