[FrancesHinden]

I understood it that if all leads concede 13 tricks (say) then all leads get one point i.e. no tie-breaking algorithm.

Of course the biggest problem with the analysis is that 'most number of tricks' is a long long way from 'best lead'. Can you not differentiate between 'beats the contract' and 'doesn't beat the contract' ?

(I've carefully missed out my rant on the benefits of DD vs single D analysis for once)

[TimG]

hrothgar, on Jul 15 2008, 12:18 PM, said:

I would expect the sum of the number of points to be less than or equal 116 (One point each time there was a "best" lead with some sort of tie breaking algorithm)

If a heart and a spade both result in 5 tricks for the defense (and diamonds and clubs produce fewer tricks for the defense) then both the heart and the spade get a point.

I think it is just a coincidence that the sum is 3 x 116. The table could read:

spade jack: 51
spade nine: 60
spade eight: 60
spade six: 60
spade four: 60

etc.

and have the same meaning, if I understand correctly.

[han]

hrothgar, on Jul 15 2008, 12:18 PM, said:

Silly question:

I would expect the sum of the number of points to be less than or equal 116 (One point each time there was a "best" lead with some sort of tie breaking algorithm)

This is not a question, this is a statement.

Quote

The total number of points actually sums to 348 (or precisely 3 x 116)

OK, not very interesting to my mind but I trust it does.

[han]

Yes what Tim said is correct.

[Fluffy]

So all in all it is close.

I would lead a spade and don't think any more, I hate looking silly when partner has 4+ spades.

[lonesome31]

Prefer ♥ lead,