[hanp]

In vuroth's "Play 3NT" thread the following question came up:

Suppose that LHO leads the diamond 5 against 3NT, RHO plays the jack. We can see the 2 and 3 but the ♦4 is missing. The opponents play 4th best leads, what is the chance that LHO has 5 diamonds?

This question is impossible to answer.

However, we can simplify the question: What is the number of 5-card diamond suits where the 5 is fourth best, compared to the number of 4-card suits where the 5 is fourth best?

The answer is easy: They are exactly equal. After all, for every 4-card suit you can add the ♦4 and you have a 5-card suit, and vice versa.

But what if LHO leads a higher spotcard? There is a simple but useful rule:

If we are missing n spotcards below LHO's lead, then the number of 5-card holdings is exactly n times as large as the number of 4-card holdings.

For example, if LHO led the 6 instead then there would be twice as many 5-card suits as 4-card suits. The computation is the same as above: for every 4-card suit you can add n different cards.