Approach

Hi all

Here’s a deal from the Wednesday teams night at Piedmont. The discussion invoked almost physical violence. In an uncontested auction South gets to declare 6. West leads a small . What’s the best approach?

North
A93
AT82
A
QJT84
South
KQJ5
K654
J3
AK6

Two paths: take the safety play in ( King and then small towards the Ten), or, if the lead smells like a singleton play trumps from top.

Do the statistics perhaps provide an answer? Any 4-1 trump split occurs 28.26% of the time, but you can only tackle four trumps in West, that leaves 14.13%. Compare that to the splitting exactly 1-5, which is close to 7.27%. So it looks like we’ve got a winner, safety play beats singleton 2-1.

But…

Bayes Theorem (credits to Geoff for explaining me):

P( singleton | lead) = ( P( singleton) * P( lead | singleton) ) / P( lead)

Read this as:

The odds the being a singleton given a lead are equal to the odds for a singleton , multiplied by the odds for a lead given a singleton, divided by the odds for a lead.

We’ll assume West always leads a singleton if he has one, so:

P( lead | singleton) = 1
P( singleton) = 0.0727
P( lead) = 0.0727 + 1/3 * 0.9273

That last part means that in 1/3 of the remaining space will be the lead chosen. We’ll ignore a broken trump holding as lead option.

This all calculates to more than 19%. So compared to the safety play, this is the superior path.

Well, this was pretty much an eye opener for me, very counter intuitive.

West Dealer
NS Vul
North
A93
AT82
A
QJT84
West
4
Q93
9842
97532
East
T8762
J7
KQT765
South
KQJ5
K654
J3
AK6

At our table Geoff (East) decided his hand looked like a weak two in . After South’s take out double I took maximum advantage of the vulnerability and put up a massive wall by jumping to 6. Let them figure out if and what slam they have at the 6-level! North took her plus and doubled. When the defence slipped up by not leading trumps twice, we got away for -500. Not a bad deal if you get your odds right.

One thought on “Approach

  1. Just wondering (and being picky)
    I think the chances on the spade lead should be closer to 1/3. Wouldn’t it be correct to take into account the chance of a club singleton that isn’t lead? This would lead to:
    0.0727+1/3*(0.9273-0.1414)=0.335 and a subsequent chance of a spade singleton of 21.7%

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