Enigmatic Code
Programming Enigma Puzzles
Tantalizer 270: Cupid’s arrow
Posted by on 8 June 2022
From New Scientist #821, 23rd November 1972 [link]
With eight handsome bachelors and eight ravishing spinsters in the office, prospects look bright for matrimony. The snag is, however, all are fanatical Catholics or Protestants and the chances against a random bride and groom being of the same religion are nine to seven. In a random mixed marriage, the groom would probably not be a Catholic, even though there are more Catholic men than Protestant women.
How many of the 16 are Protestants?
[tantalizer270]
Enigmatic Code 
Odds of “nine to seven against” are a probability of 9/(9 + 7) = 9/16.
So of the 64 possible male/female pairs 9/16 of them (= 36) are mismatched (so 28 of them are matched).
If there are x Catholic men and y Protestant women (so x > y), then the number of mismatched pairs is:
And we are looking for M = 36. (This is enough to find the answer to the puzzle).
Additionally we are told that in a random mixed marriage (i.e. one of the 36 mismatched pairs) the groom would probably not be Catholic.
There are xy mismatched pairs with a Catholic groom, so we need xy < 18.
from enigma import (irange, subsets, printf) # choose x and y; x > y for (y, x) in subsets(irange(0, 8), size=2): xy = x * y if xy < 18 and xy + (8 - x) * (8 - y) == 36: # calculate C and P (C, P) = (x + 8 - y, 8 - x + y) # output solution printf("C={C} P={P} [x={x} y={y}]")Solution: There are 7 Protestants.
We have 3 Catholic (and 5 Protestant) men, and 6 Catholic (and 2 Protestant) women.
There are 3×2 (= 6) mismatched pairs with a Catholic groom, and 5×6 (= 30) mismatched pairs with a Protestant groom.
So there are 36 mismatched pairs in total (as required) and fewer than half with a Catholic groom.