[Tatung42]

I would like to discuss this hand that I saw on a bridge puzzle webpage. South is playing at 7♠.

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The only way to make the contract is to do a "dummy reversal" where you ruff north's three diamond losers before pulling trump. The webpage suggested that the dummy reversal is the correct way to play the hand.


However, if I were playing the hand without being able to see east and west, I would play it as follows:
- Draw two rounds of trump.
- Play clubs and hope that they break 3-3
- If clubs do break 3-3, I would pull the last trump, then win the 4th club trick for my 13th trick.
- Even clubs don't break 3-3, there is still a chance that I can trump the 4th club for the 13th trick, if the player with the short clubs also has the short spades.


I have a about a 36% change of clubs breaking 3-3. Then I have about a 32% chance that the player with the short clubs also has the short spades. This gives me about a 68% chance to make the contract.


The "dummy reversal" method also has about a 68% chance to make the contact because it fails if trump are not split 3-2. But when it fails, there is a chance to go down two (or more) tricks because the person with long spades could potentially also have the long diamonds and win extra tricks.


Does anyone agree or disagree with my assessment? I just get annoyed by these bridge puzzles, where the “correct” way to solve the puzzle is not the correct way to actually play the hand.

[antonylee]

Tatung42, on 2012-September-12, 00:13, said:

Then I have about a 32% chance that the player with the short clubs also has the short spades.

I assume you got this 32% by saying that the player with short clubs has short spades half of the times (half of the remaining 100-36=64%, i.e. 32%). But this is incorrect, as (by a vacant places argument) he will have short spades less than half of the times.

To be precise, a simulation tells me that your line wins only 59% of the times.

[Quantumcat]

You are also assuming in your line that trumps are 3-2, so whatever percentage you come up with, the dummy reversal line is 100% (both cases assuming are 3-2 spade break, but dummy reversal needs no more than this, whereas you do).

[Tatung42]

Thanks for the replies.

Quantumcat, on 2012-September-12, 21:02, said:

You are also assuming in your line that trumps are 3-2, so whatever percentage you come up with, the dummy reversal line is 100% (both cases assuming are 3-2 spade break, but dummy reversal needs no more than this, whereas you do).

I am pretty sure that my line of play would work against even a 5-0 trump split? I can flush trump with the 5 trump in south's hand, either after setting up the 4th club (In the case of a 3-3 split) or after trumping the 4th club with north's trump (in the case of an unbalanced split). My line of play wouldn't work against a bad trump split if the hand with the long trump is also the hand with the short clubs, but it would also lose in this case even with a 3-2 trump split.

antonylee, on 2012-September-12, 00:34, said:

I assume you got this 32% by saying that the player with short clubs has short spades half of the times (half of the remaining 100-36=64%, i.e. 32%). But this is incorrect, as (by a vacant places argument) he will have short spades less than half of the times.

To be precise, a simulation tells me that your line wins only 59% of the times.

Ok thanks, that is why i said "about a 32% chance", I knew it was a little bit less, but I didn't think it would be 9% less. I didn't have an easy way to calculate it. What simulator do you use? I've been looking for something to do calculations like this.

I am curious what % of the hands does the dummy reversal strategy go down more than 1 trick due to a bad trump split? If it is more than 9%, my strategy would still be superior in MP right? In IMPs the "dummy reversal" would always be the superior strategy.

[antonylee]

https://github.com/anntzer/redeal

from redeal import *
predeal = {"N": H('QJT 432 A432 Q32'), "S": H('AK987 AKQ 5 AK54')}

win1 = win2 = 0

def do(deal):
    global win1, win2
    if len(deal.east.spades) in (2, 3):
        win1 += 1
    if (len(deal.east.clubs) == 3 or
        len(deal.east.clubs) < 3 and len(deal.east.spades) < 3 or
        len(deal.west.clubs) < 3 and len(deal.west.spades) < 3):
        win2 += 1

def final(n_tries):
    print(win1, win2)

or just from the command line (with the version of the code I've just uploaded):

$ redeal -n100 -N'QJT 432 A432 Q32' -S'AK987 AKQ 5 AK54' --initial 'self.win1 = self.win2 = 0' --do 'self.win1 += len(deal.east.spades) in (2, 3); self.win2 += len(deal.east.clubs) == 3 or len(deal.east.clubs) < 3 and len(deal.east.spades) < 3 or len(deal.west.clubs) < 3 and len(deal.west.spades) < 3' --final 'print(self.win1, self.win2)'

Too lazy to work out the down 2 case.

[Stephen Tu]

Tatung42, on 2012-September-13, 00:17, said:

Ok thanks, that is why i said "about a 32% chance", I knew it was a little bit less, but I didn't think it would be 9% less. I didn't have an easy way to calculate it. What simulator do you use? I've been looking for something to do calculations like this.

Use Pavlicek's dual suit break calculator on the following page:
http://www.rpbridge.net/rpca.htm
The card combo analyzer on that page is also very useful. He also has an older DOS mode card combo analyzer on his utilities page:
http://www.rpbridge.net/rput.htm
Great thing on all is you can bias by vacant spaces if necessary.

This one is simple enough to calculate directly and exactly if you want:
let's give one hand the long clubs, then *2 later since either east or west can have these:
4 clubs, 3 spades: 15 club combos * 10 spade combos * 15c6 (rest of cards) = 750750
4 clubs, 4 spades: 15 * 5 * 15c5 = 225225
4 clubs, 5 spades 15 * 1 * 15c4 = 20475
5 clubs, 3 spades 6 * 10 * 15c5 = 180180
5 clubs, 4 spades 6 * 5 * 15c4 = 40950
5 clubs, 5 spades 6 * 1 * 15c3 = 2730
6 clubs, 3 spades 1 * 10 * 15c4 = 13650
6 clubs, 4 spades 1 * 5 * 15c3 = 2275
6 clubs, 5 spades 1 * 1 * 15c2 = 105
Total: 1236340 hands.
*2 = 2472680 since either east or west can have both black suits. Total possible hands is 26c13 = 10400600
2472680/10400600 = 23.774%.
Which matches the result of Pavlicek's calculator.
You subtract a tiny percentage if you think East will lightner with void clubs, but that's fairly insignificant.