Declarer play. Distribution of outstanding cards
What's the split?
You've worked out that the opposition have a certain number of cards in a suit. But how they will split? For example, if they've got four cards altogether, what are the chances that they have exactly two cards each?
The general principle:
- when the opponents have an odd number of cards, they will most likely divide as evenly as possible; but
- when the opponents have an even number of cards, they will most likely not divide evenly.
- . . .except that two cards are a slight favourite to divide 1-1
The full table
| 1002. Distribution of defence's cards in a suit | |||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Total cards | Most EVEN split | UNEVEN splits | |||||||||||||||||||||
| split | probability | split | probability | ||||||||||||||||||||
| 2 | 1-1 | 52% | 2-0 | 48% | |||||||||||||||||||
| 3 | 2-1 | 78% | 3-0 | 22% | |||||||||||||||||||
| 4 | 2-2 | 40% | 3-1 4-0 |
50% 10% |
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| 5 | 3-2 | 68% | 4-1 5-0 |
28% 4% |
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| 6 | 3-3 | 36% | 4-2 5-1 6-0 |
48% 15% 1% |
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| 7 | 4-3 | 62% | 5-2 6-1 7-0 |
31% 6% 1% |
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| 8 | 4-4 | 33% ? | 5-3 6-2 7-1 8-0 |
46% ? 15% ? 4% ? 1% ? |
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| No need to remember this chart, just note that with an even number of cards, an even split is the less likely, but with an odd number of cards the most even split is the most likely. | |||||||||||||||||||||||
| Add to your customised cribsheet | |||||||||||||||||||||||
