### by Danny Roth

#### Published Sunday April 25 2021 (link)

George and Martha are playing bridge with an invited married couple. Before play starts, the players have to cut for partners. Each player draws a card from a standard pack and those drawing the two highest-ranking cards play together against the other two. For this purpose, the rank order is ♠ A, ♥ A, ♦ A, ♣ A, ♠ K, ♥ K etc. down to ♦ 3, ♣ 3, ♠ 2, ♥ 2, ♦ 2, ♣ 2 (the lowest).

George drew the first card, then Martha drew a lower-ranking card. “That is interesting!” commented the male visitor to his wife. “The probability that we shall play together is now P. Had Martha drawn the ♥ 7 instead of her actual card, that chance would have been reduced to P/2, and had she drawn the ♥ 6, the chance would have been reduced to P/3.

Which cards did George and Martha draw?

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Here is a manual solution:

The married couple will play together if their cards are both above George’s or both below Martha’s.  Hence the probability of this happening is the sum of the probabilities of these two outcomes.

If the cards are labelled from 0 to 51 in low to high rank order, there will be $$x$$ cards below card $$x$$ and $$51 – x$$ cards above it.  Denoting George’s card by $$g$$ and Martha’s card by $$m$$, the probability ($$p$$) of the couple playing together is hence $p=\frac{(51-g)(50-g)}{50\times49} + \frac{m(m-1)}{50\times49}$Applying this formula for the 6 of Hearts ($$p=P/3$$ for $$m = 18$$) and the 7 of Hearts ($$p=P/2$$ for $$m=22$$) gives:$2450P=3(51-g)(50-g)+918$$2450P=2(51-g)(50-g)+924$Eliminating $$P$$ now allows us to produce an equation for $$g$$:$(51-g)(50-g) = 6$Multiplying this equation by 4 and rearranging it now gives$4g^2-404g+10201=(2g-101)^2=5^2$from which $$g$$ can be determined as 48 (since the alternative root, 53, is too large). Back substitution shows that $$m=31$$.

The solution is hence the ‘Ace of Clubs’ and the ‘9 of Spades’.

Hi Brian, in the 2nd equation you must mean 924 iso 462.

I can’t get used to the dot for multiplication. For me 50.49 is closer to 50 than 51 and for you it’s closer to 51.

Yes, it can be a problem when the context isn’t clear (I avoided it in the
manual solution for this reason).

Not very exiting employ Python just to loop up a solution for a set of equations. I found some thrill in how to report the answer though. I’m sure it can be done in a better way than mine.

Hi Erling,

I like your effort on the output.

Thanks Brian,

The reason I added that I thought it could be done in a better way, was that I tried to get away with a list rather than a dictionary, like John has done below. But I found that I could not rely on the list’s ordering; it did not come out the same way it was built..

With the dictionary in place, however, it would have been more consistent to use it also in the first conversion, from card-faces to rank-order:

..and replace ‘22’ and ‘18’ in line 17 and 18 with M1 and M2. (I am not asking for that change to be done. It’s just for the record.)