[Valardent]

Hi, you play in 6♠ in South after a transfer sequence.

Lead is ♥K, imps scoring.

IMPS

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Obviously, if ♦ are 3-3, slam will always win.


I think these are the best 2 lines (+ one variation) :

a) Unblock AQ♦, A♠, back to hand in ♣, and ♦K

b) Unblock AQ♦, back to hand in ♣, ♦K, and ♠finesse if ♦K didn't survive.

Variation :

Follow a) or b) depending whether ♦J falls on AQ or not.


Which one is best?

How calculate this 1) with enough accuracy at the table to choose the best one?
2) in the post mortem analyse to establish the percentage difference between the best 2 lines?




What about this way (focusing on the ♠ distributions)?

- If ♠ are 2-2, both lines are equivalent (50% chance)

- If ♠ are 3-1 or 1-3 :

Line a) wins when a defender holds x♠ and xx♦ or when ♠K is stiff behind ♠Ace

Line b) wins when a defender holds ♠Kxx(x) and xx♦in front of dummy

(If ♦J falls, line a) must be superior cos one wins also with Kxx and Jx♦in front of dummy)

- If ♠ are 4-0 or 0-4 :

Line b) wins when ♠ are in front of dummy with xx♦

Thoughts?

[aguahombre]

See post on using a calculator at the table to figure out the odds. Without it, and having to do something, I try the Ace of trump first.

[nige1]

Valardent asked for our thoughts on the play of 6♠ by South at IMPS.


IMO Cash ♥A, ♦AQ, ♣Q, advance ♦K.
- If both follow, discard a ♥ and run ♠9.
- If LHO ruffs, then over-ruff. Now...
- - If ♦J hasn't appeared, then ♣K, run ♠9.
- - If LHO had ♦J doubleton, then: cash ♠A, ♣K, advance ♦9.

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[JLOGIC]

Your analysis seems almost spot on Valardent. At the table I would think:

There are 3 cases of a spade and 2 small diamonds on my left.
There is 1 case of 3 spades and 2 small diamonds on my left.
There are 3 cases of Kxx spades and 2 small diamonds on my right.

There are 3 cases of 2 small spades on my left and 2 small diamonds.
There is 1 case of Kxxx of spades on your left with 2 small diamonds
There are 3 cases of Kx of spades on my right with 2 small dimonds

*** You seemed to forget 4-0 spades on your left with a doubleton diamond as a win for not cashing the SA

Calculating how many breaks is easy, for instance there are 3 possible combos of a stiff on your left, because there are 3 small stiffs LHO can have. There are 3 combos of Kx because K? the ? can only be one of the 3 small ones.


So, it is 7 cases to 7. Normally the edge would be to cashing the ace becuse it loses to a 4-0 break but picks up a 3-1 break, and a 3-1 combo is more likely than a 4-0. However, here it must be realized that LHO having short spades AND short diamonds is unlikely. Therefore, him having King third of spades and 2 small diamonds is far more likely than a stiff spade and 2 small diamonds.

So, keeping the ace in dummy is right imo. This is how I would consider it at the table, no fancy math is required and it wouldn't take you that long.

[Valardent]

JLOGIC, on 2011-August-31, 02:00, said:

Your analysis seems almost spot on Valardent. At the table I would think:

There are 3 cases of a spade and 2 small diamonds on my left.
There is 1 case of 3 spades and 2 small diamonds on my left.
There are 3 cases of Kxx spades and 2 small diamonds on my right.

There are 3 cases of 2 small spades on my left and 2 small diamonds.
There is 1 case of Kxxx of spades on your left with 2 small diamonds
There are 3 cases of Kx of spades on my right with 2 small dimonds

*** You seemed to forget 4-0 spades on your left with a doubleton diamond as a win for not cashing the SA

Calculating how many breaks is easy, for instance there are 3 possible combos of a stiff on your left, because there are 3 small stiffs LHO can have. There are 3 combos of Kx because K? the ? can only be one of the 3 small ones.


So, it is 7 cases to 7. Normally the edge would be to cashing the ace becuse it loses to a 4-0 break but picks up a 3-1 break, and a 3-1 combo is more likely than a 4-0. However, here it must be realized that LHO having short spades AND short diamonds is unlikely. Therefore, him having King third of spades and 2 small diamonds is far more likely than a stiff spade and 2 small diamonds.

So, keeping the ace in dummy is right imo. This is how I would consider it at the table, no fancy math is required and it wouldn't take you that long.

To sum it up (not bothering when ♠ are 2-2):

Line "♠A" is better in those 4 cases : _____ Line "finesse" is better in those 4 cases :

x♠ xx♦ in West (3 cases) (1)_____________ ♠Kxx xx♦ in West (3 cases) (2)

xxx♠ xx♦ in West (1 case) (3)____________ ♠Kxxx xx♦ in West (1 case) (4)


So (4) is a bit less frequent than (3), but (2) is much more frequent than (1)

When ♦J falls, playing "♠A" line wins also in (2), making "♠A" line best.

When ♦J does not fall, "finesse" line is best.

Anyone out there who can explain how to calculate the odds difference between (1) & (2) and (3) & (4) ?

[gnasher]

Valardent, on 2011-August-31, 11:50, said:

Anyone out there who can explain how to calculate the odds difference between (1) & (2) and (3) & (4) ?

This is how you do it:

With diamonds 2=4, the vacant spaces are initially 11:9. Each time we deal a spade to one of the players, the vacant spaces for that player are reduced by one. Hence:

Chance of a specific 3=1 break: 11/20 * 10/19 * 9/18 * 9/17 = 0.077
Chance of a specific 1=3 break: 11/20 * 9/19 * 8/18 * 7/17 = 0.048
Chance of a 4=0 break: 11/20 * 10/19 * 9/18 * 8/17 = 0.068
Chance of a 0=4 break: 9/20 * 8/19 * 7/18 * 6/17 = 0.026

The divisors are only relevant if you're comparing chances in two different suits. In this case you could ignore them, but to do exact calculations at the table you'd still have to be very good at mental arithmetic.

The other way to do this type of calculation is to use Richard Pavlicek's calculator at http://www.rpbridge.net/xsb2.htm

[gnasher]

Valardent, on 2011-August-31, 11:50, said:

When ♦J falls, playing "♠A" line wins also in (2), making "♠A" line best.

Unless ♦J unexpectedly falls when LHO has ♠Kx and ♦Jxxx. (OK, I know I'm dreaming.)

[inquiry]

gnasher, on 2011-August-31, 14:24, said:

This is how you do it:

With diamonds 2=4, the vacant spaces are initially 11:9. Each time we deal a spade to one of the players, the vacant spaces for that player are reduced by one. Hence:

Chance of a specific 3=1 break: 11/20 * 10/19 * 9/18 * 9/17 = 0.077
Chance of a specific 1=3 break: 11/20 * 9/19 * 8/18 * 7/17 = 0.048
Chance of a 4=0 break: 11/20 * 10/19 * 9/18 * 8/17 = 0.068
Chance of a 0=4 break: 9/20 * 8/19 * 7/18 * 6/17 = 0.026

The divisors are only relevant if you're comparing chances in two different suits. In this case you could ignore them, but to do exact calculations at the table you'd still have to be very good at mental arithmetic.

The other way to do this type of calculation is to use Richard Pavlicek's calculator at http://www.rpbridge.net/xsb2.htm

Actually, doesn't WEST's heart king (assumption he has ♥KQ) make vacant space after 2=4 diamond split is learned, 9:9?

[inquiry]

gnasher, on 2011-August-31, 14:24, said:

This is how you do it:

With diamonds 2=4, the vacant spaces are initially 11:9. Each time we deal a spade to one of the players, the vacant spaces for that player are reduced by one. Hence:

Chance of a specific 3=1 break: 11/20 * 10/19 * 9/18 * 9/17 = 0.077
Chance of a specific 1=3 break: 11/20 * 9/19 * 8/18 * 7/17 = 0.048
Chance of a 4=0 break: 11/20 * 10/19 * 9/18 * 8/17 = 0.068
Chance of a 0=4 break: 9/20 * 8/19 * 7/18 * 6/17 = 0.026

The divisors are only relevant if you're comparing chances in two different suits. In this case you could ignore them, but to do exact calculations at the table you'd still have to be very good at mental arithmetic.

The other way to do this type of calculation is to use Richard Pavlicek's calculator at http://www.rpbridge.net/xsb2.htm

Actually, doesn't WEST's heart king (assumption he has ♥KQ) make vacant space after 2=4 diamond split is learned, 9:9?

[gnasher]

inquiry, on 2011-August-31, 15:12, said:

Actually, doesn't WEST's heart king (assumption he has ♥KQ) make vacant space after 2=4 diamond split is learned, 9:9?

I don't know. We could probably expect a heart lead anyway. If he'd led the 10, would you make the same argument about the 109?

[semeai]

inquiry, on 2011-August-31, 15:11, said:

Actually, doesn't WEST's heart king (assumption he has ♥KQ) make vacant space after 2=4 diamond split is learned, 9:9?

gnasher, on 2011-August-31, 15:49, said:

I don't know. We could probably expect a heart lead anyway. If he'd led the 10, would you make the same argument about the 109?

I think inquiry is basically correct. If West leads ♥K whenever he has ♥KQ and never otherwise, then the lead gives us precisely the inference that he has ♥KQ (and no other info). This means we should take it into account for vacant spaces. I think it's not far from the truth that West leads ♥K whenever he has ♥KQ and never otherwise given our hands and the contract.

The ♥109 situation is much less clear. A lead from some sequence is not unlikely for this contract, so probably there's little to no adjustment to a vacant spaces calculation.