# A ‘bridge’ math puzzle

** Published:**

I saw this puzzle in MIT’s Puzzle Corner and thought it was fun.

For background, bridge is a card game played by 4 players. Players who sit opposite one another are in a partnership, so there a 2 players in 2 partnerships.

**Problem**: 16 players, 8 men and 8 women, randomly take seats at 4 bridge tables. Is it more likely that all the partnerships will be same-sex, or that all partnerships will be opposite sex?

**Analysis**: There are 8 distinguishable red balls and 8 distinguishable blue balls, placed at random into 8 distinguishable slot-pairs, with each slot-pair having 2 distinguishable slots. Are there more ways to distribute the balls such that you end up with all different-color pairs or same-color pairs?

**Solution**: I worked this out with dynamic programming first, because I think that’s easier, and then looked at the combinatorics, to make it more beautiful.

For different-color pairings, I note that, when filling the first slot, you can pick any of the 8 red balls (8 choices), and any of the 8 blue balls (8 choices), and you can put them in red-blue or blue-red (2 choices). Then, when looking at the next slot, you have $7 \times 7 \times 2$ choices. In a dynamic programming sense, let $D(r, b)$ be the number of ways to place $r$ red and $b$ blue balls into $(r + b)/2$ slot-pairs with different colors in each slot-pair:

$$ \begin{equation*} D(r, b) = \begin{cases} 2rb \times D(r-1, b-1) &\text{ if $r>0$ and $r>0$} \\ 1 &\text{ if $r=b=0$} \end{cases} \end{equation*} $$

It’s pretty clear that $D(8,8) = (8!)^2 \times 2^8$.

From a combinatorics point of view, we can say that there are $8!$ ways to put the red balls in the “left” slots, and $8!$ ways to put the blue balls in the “right” slots, and then there are $2^8$ ways to permute the balls inside each slot-pair. This gives the same result: $(8!)^2 \times 2^8$.

For same-color pairings, I note that, when filling the first slot, you can pick either red or blue. You have 8 choices for the first ball, and then 7 choices for filling the other slot. If $S(r, b)$ is the number of ways to put $r$ red balls and $b$ blue balls into $(r + b)/2$ slot-pairs with all slot-pairs having the same colors, then we just said that:

$$ \begin{align*} S(8, 8) &= \#\{\text{ways if the first slot-pair is all red}\} + \#\{\text{ways if all blue}\} \\ &= 8 \times 7 \times S(6, 8) + 8 \times 7 \times S(8, 6) \end{align*} $$

And we can generalize that:

$$ \begin{equation*} S(r, b) = \begin{cases} r(r-1) S(r-2, b) + b(b-1) S(r, b-2) &\text{ if $r>0$ and $b>0$} \\ r(r-1) S(r-2, 0) &\text{ if $r>0$ and $b=0$} \\ b(b-1) S(0, b-2) &\text{ if $r=0$ and $b>0$} \\ 1 &\text{ if $r=0$ and $b=0$} \\ \end{cases} \end{equation*} $$

It wasn’t obvious to me what this comes out to!

From a combinatorics point of view, there are $\binom{8}{4}$ ways of choosing which slot-pairs are red, then $8!$ ways of arranging the red balls among those red slot-pairs, and then another $8!$ for arranging the blues, giving $\binom{8}{4} \times (8!)^2 = (8!)^3 / (4!)^2$. If you run the dynamic programming code, it comes to the same value.

Thus, the opposite-color pairings outnumber the same-color pairings by a factor of $2^8 \times (4!)^2 / 8! \approx 3.7$.

**Discussion**: Opposite-color arrangements only slightly outnumber same-color arrangements. This is a pretty subtle difference!

My friend Ralph Morrison suggested extending this problem for different numbers of tables. (There are $n=4$ tables in this problem.) If you do that, you can use Stirling’s approximation to show how the ratio of opposite-color to same-color arrangements changes with $n$.