Wednesday, May 13, 2015

Don't upgrade to a splinter

Playing with an occassional partner in a strong club game, I was West on this hand:

N Deals
None Vul
 Q87
 983
 A6
 QT743
17
 KJ52
 AQ64
 J872
 9
N
WE
S
 AT643
 KJ
 KQ
 J862
8
1114
7
 9
 T752
 T9543
 AK5



















North passed and partner (East) opened the hand 1S.  As West, I had 4 choices:

  1. Jacoby 2NT showing 4 trumps and forcing to game
  2. 3S showing 4 spades and an invitational hand
  3. 3H (fit-jump) showing good hearts, 3+ spades and an invitational hand
  4. 4C showing 4 trumps, club shortness and 12-14 points.
Your call?

At the table, I forgot about option #3.

I decided to upgrade my hand and splinter with 4C.  Partner, with no wastage in clubs, got excited and we ended up one level too high, in 5S.  When spades turned out to be 3-1, we were down 1.  Everyone else in the room was bidding and making 4S.

Splinters are very well-defined bids and are there to help you find slams holding fewer than 30 high card points.  Because of this, though, they work only when you stay within the parameters.  Just add one point to my hand to make it 12 points (by changing the J of spades to the Q of spades) and note that 5S is totally safe.  Add 3 points to my hand (by changing the Jack of diamonds to the Ace of diamonds) and note that 6S is on whenever spades are 2-2 (a 52% slam).


We would not have had this disaster if I had upgraded my hand and bid Jacoby 2NT.  Partner with a semi-balanced minimum would have bid 4S.


Don't upgrade to a splinter.

Wednesday, May 6, 2015

Ian McEwan explains Restricted Choice

The principle of Restricted Choice is probably the best-known bridge maxim that many players don't quite understand.  Most bridge players can explain the logic behind finesses, split-honors, and "eight ever, nine never".  But restricted choice will throw your average club player for a loop.

Imagine my surprise then at seeing it laid out extremely nicely as a vignette in Ian McEwan's novel "The Sweet Tooth".

The way the principle is usually explained is in terms of an adaptation of the Monty Hall show.  You get onto Monty's show and you are shown three doors. Behind one of the doors is a car.  Behind the other two doors are goats. The way the show works is this. You get to pick one of the doors.  Now, Monty opens one of the other doors and shows you that it has a goat and gives you a choice. You can either go with your original choice or you can switch to the remaining door. Should you switch or not? This question was famously posed to Marilyn Von Savant who got the answer right, but got pilloried for it by many pompous toffs who couldn't get their heads around the logic.

So, what's the logic? I find it easiest to explain this assuming that there is a long corridor full of doors. A thousand doors, say. You go in and pick one of the doors. What's the chance that there's a car behind that door?  That's right. 1 in a 1000. Now, Monty who knows which door the car is behind comes along and opens 998 of the remaining doors and shows you 998 goats. Do you switch to the door he didn't open or do you stick with your original 1 in a 1000 chance? Of course you switch! Monty's essentially telegraphed to you which door the car is behind because he carefully avoided that door.  Monty's choice was restricted -- you now have a 999 in 1000 choice of getting the car! You are paying off to the remote possibility that you happened to pick the right door on the first try.  Reduce the 1000 to 3, and the logic is the same. You had a 1/3 choice of picking the right door, but after Monty opens the door with a goat, your odds go up to 2/3 if you switch.

In bridge terms, you apply this principle when the QJ of a suit are missing and you hold:

A1098

K763

in your two hands.  When you plop down the Ace, your left-hand opponent (LHO) drops the Queen.  Should you finesse the 10 on the way back, or should you hope that LHO has the Queen-Jack tight? The principle of restricted choice says that LHO's choice was restricted, and so your percentage play is to finesse.

With that primer, onto Ian McEwan's inspired vignette.  Here's the setup.  A jealous husband follows his wife and her lover to a hotel where he sees them vanish around the corridor.  He wants to catch them in flagrante, so he waits a little bit and then prepares to break down the door.  But there are three rooms: 401, 402 and 403.  Behind one of them is his cheating wife, but the odds of picking the right one are only 1 in 3.  He waits to see if he can hear any sounds, but he can hear nothing from any of the rooms.  As he is debating which door to choose, he sees two housekeepers approaching.  "Let's work on one of the two empty rooms," one maid tells another.  Thinking quickly, the husband positions himself in front of 401.  Now, the maid's choice is restricted -- seeing the guest blocking her way into 401, she will choose to work on either 402 or 403, whichever is empty.  She opens the door to 403, and our hero knowing that his odds have increased to 2 in 3 now, breaks down the door to 402.  His mathematical savvy is rewarded by the dubious prize of catching his wife in bed with another man.

A remarkably savvy mathematical vignette in a book of fiction aimed at the masses!  Atonement, here I come!