kereru67, on 2018-February-18, 20:45, said:

This is the example given in Dorothy Truscott's "Winning Declarer Play".

North-South have overbid to 7

♦ (6 is easy). The line given is:

West leads

♣J. Declarer wins

♣A, then cashes 3

♥, ending in dummy.

♣ ruff,

♠ to K,

♣ ruff,

♠A,

♠ ruff,

♣ ruff,

♠7. Now if west ruffs low, declarer overruffs with

♦9 and AK become good. If West ruffs with

♦Q, declarer overruffs with

♦K then finesses against East's

♦J.

For this play to work, the following conditions have to be present:

(1)

♥ must split 3-3 (odds = 36%)

(2)

♣ must split 4-4 (odds = 33%)

(3)

♠ must split 4-3 (odds = 70%)

(4)

♦must split 3-2 (obviously they will be if conditions 1-3 are met), with Q and J in different hands (odds = 50%)

So my estimate is that the odds of this line succeeding are about 0.36 * 0.33 * 0.7 * 0.5 = about 4%.

The more prosaic play is to lead out the AK of trumps. This will work if trumps split 3-2 (odds = 68%) and the QJ is doubleton (odds = 10%), i.e. odds about 6.8%.

So it seems to me the Devil's Coup is never a mathematically sound play, unless I've miscalculated.

Your calculations assume that (1) (2) and (3) are mathematically independent events.

They are not.

They are correlated, because everyone is dealt exactly 13 cards.

The correct way of looking at this question is to ask , given the missing 26 outstanding cards, what is the likelihood that West leading the

♣J was dealt a 4333 distribution with exactly 4 clubs and one diamond honor.

Doing this you would get that there are 10,400,600 ways how you can select 13 cards from 26.

A priory given the missing cards in the different suit there are 490,000 ways West could have a 4333 distribution with 4 clubs.

But since we need him to hold exactly one diamond honor this reduces West to 294,000 ways

So a priory the odds are 294,000 / 10,400,600 or 2.8%

(slightly better because with the more extreme distributions West might have been tempted to bid)

However, we have not yet taken into account the opening lead against the grand slam.

This marks West with the the ten and 9 in clubs and East with the king of clubs.

Does this affect the odds?

Now there are only 22 cards left of which West will get ten.

This reduces the number of hands West can now hold dramatically.

And we start the exercise all over again:

There are now 646646 ways we can deal 10 cards out of 22 to West.

However there are now only 16800 ways West can be dealt 4333 with 4 clubs and precisely one diamond honor.

The odds have dropped to 2.6%

The chance for someone holding QJ tight in diamonds is about 6.8%.

So the overall chances to win, after 2 rounds of trumps, then test the hearts and if these do not break take the spade finesse is a bit better.

But who wants to make a grand slam on Queen Jack tight in trumps when you can play for the Devil's Coup?

Anyway there are other positions, where there is no alternative (taken from another book):

Trick 1:

♠6,

♠2,

♠K

♠7

Trick 2:

♠4,

♠T,

♠3 ?

Plan the play

Rainer Herrmann

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