## Problem no. 34, 2008

### The problem

Any bridge hand has either three suits with an even number of cards and one with odd, e.g. 4-4-4-1, or three suits with an odd number and one with even, e.g. 5-3-3-2. We call the first type even, the second type odd.

Recently, I read a study, made by T.C. Pant of India, regarding odd and even hands. He had looked at 2000 randomly dealt deals to see what the result would be if he looked at a full deal: How many hands were odd, how many were even? His result was that 63% of the time, three hands were of one type, the fourth of the other, e.g. three odd and one even. 28% of the time there were two of each. In the remaining 9% all four hands were of the same type, e.g. odd.

He also wrote: "If you, as declarer, have the same parity as dummy, and later learn that one of the opponents also has that type, the chance is roughly 90% that the remaining hand is of the opposite type. That knowledge may come in handy when you plan the play."

This surely sounded interesting, so I took a deck, shuffled and dealt. Then I looked at North's and South's hand. Both were odd. But then it struck me that on this very deal, East and West must have different parity. If West's hand is even, East must be odd, and vice versa. It was impossible that both were odd or both were even.

### Solution

A bridge hand must be either even or odd, as we know, but that is also true for how any suit is distributed between the four hands. Both consist of thirteen cards. In the deal I happened to get, both the declarer and the dummy had the same odd suit, let's say spades. The sum of the cards in all suits of the declaring side is therefore even. If we present it in a table, it looks like this:

 Spades Hearts Diamonds Clubs South Odd Even Even Even North Odd Even Even Even East ?? ?? ?? ?? West ?? ?? ?? ??

Thirteen is an odd number. In order for the sum to be odd in all suit columns, East and West must have different parity in all suits. We must enter one 'Odd' and one 'Even' in all four columns. Besides, East and West must have three suits of one type, on of the other. It is thererfore impossible that both have the same parity. One must be odd, and the other even.

If North and South have their odd suit in different suits, there are three possibilities. If the odd suits are spades and hearts, respectively, the table will be like this:

 Spades Hearts Diamonds Clubs South Odd Even Even Even North Even Odd Even Even East ?? ?? ?? ?? West ?? ?? ?? ??

If both East and West have odd hands, we will get two odd hands and two even; if both East and West are even, we will get four even hands; and if one is odd ant the other even, we will get three odd and one even. All those possibilities can be fitted into the table, since two of the suit columns are odd and two are even. When all four are even, there is only one possibility.

We learn one more thing from the table, namely that if there are three odd hands, there must also be three odd suits. Why? Because we then shall enter 'odd' in ten of the cells, 'even' in the remaining six. That is only possible if we get three odd lines and one even, both for the suits and the players.

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