There were four corners to consider: two in the top row and two in the bottom row. So first, what was the probability that neither corner in the top row contained a 1? Of the five spaces in that row, two were corners. Since the 1 was equally likely to be in any of them, the probability it wasn’t in a corner was 3/5.

Next, what was the probability that the 1 wasn’t in a corner in the bottom row? Since this had to be a Latin square, you couldn’t have a 1 in the same column as the 1 in the top row. That left four remaining places, two of which were corners. So the probability there wasn’t a 1 in a corner was 2/4, or 1/2.


Putting this all together, the probability that no corner contained a 1 was (3/5)·(1/2), or 3/10

Many solvers generalized this to the case of an N-by-N Latin square. The probability of not having a corner 1 in the top row was (N−2)/N. From there, the probability of not having a corner 1 in the bottom row was (N−3)/(N−1). Multiplying these together gave a total probability of [(N−2)(N−3)]/[(N)/(N−1)].

For extra credit, you had to find the probability that two of the four corners in the randomly chosen five-by-five Latin square had the same number. The probability of either pair of corners matching was 1/4. Then, to find the probability that at least one pair matched, you could add the probabilities of either pair matching and subtract the probability they both matched (since this had been double counted). Naively, you might have expected this to be 1/4 + 1/4 ﹣1/16, or 7/16. But this being the extra credit, and therefore trickier, this answer was wrong.

Why? As it turned out the probabilities of each pair matching weren’t independent. In other words, the probability that both corners had matching numbers wasn’t (1/4)·(1/4), or 1/16, but rather 1/28. Having one pair of matching corners meant the other pair of corners was less likely to match, since there were fewer possible overall arrangements.

  • Optiver3
  • MIcompany
Forgot password