[kereru67]

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

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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.

[steve2005]

16% for QJ♦ doubleton. or if catch singleton honour you can finesse with 67% odds for restricted choice. Not sure what that would work out to.
33% ♥ 3-3
50% spade finesse
stiff K♣ would help but very low odds, maybe ruff out clubs
So guess 3% of not using devil coup.
There are probably some other things can try but off QJ of trump and need more stuff can't expect great odds.
One thing going for devil coup is once one suit breaks odds of others breaking are higher than you have.

[gordontd]

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

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

Doesn't it still work with QJ tight on your right?

[sfi]

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

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

You need to factor in a few other things even if you find QJ of diamonds. There are still two spade losers to take care of, so you need to choose the spade hook or 3-3 hearts, as well as ruffing one of them. Given your odds, the Devil's Coup must be better.

[sfi]

steve2005, on 2018-February-18, 21:28, said:

16% for QJ♦ doubleton. or if catch singleton honour you can finesse with 67% odds for restricted choice.

A 4-1 break dooms you to a trump loser.

[rhm]

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

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

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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):

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Trick 1: ♠6, ♠2, ♠K ♠7
Trick 2: ♠4, ♠T, ♠3 ?

Plan the play

Rainer Herrmann

[nige1]

A similar coup

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Another

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[TheoKole]

I had a hand a few years ago that perhaps fits your description however, the bidding was unusual and I was almost forced to play it this way. I had opened 1NT and after a stayman 2 ♣ bid I bid 2 ♥ with 4432 hand shape. My partner then bid 3 NT. After this I corrected the contract to 4 ♠ since we played that stayman guaranteed a 4 card major. Then partner bid 5 ♥ and I was stumped for a bit trying to figure out any hand that would want to play in 3 NT but be willing to cue bid with a spade fit.

I eventually decided that my partner had made some kind of mistake and passed 5 ♥. Luckily partner actually did have a 4 card ♥ suit but had psyched having 4 ♠ wanting to play in 3 NT on his 2434 hand and trying to deter a ♠ lead. (it was matchpoints) 4 ♥ on the hand was almost laydown only requiring a 3-2 trump break so I had to make 5 ♥ or get a 0 for the board. Unfortunately, there were 2 top ♣ losers which did not go anywhere and I was missing QJxxx in the trump suit. I had 3 choices, playing for a devil's coup, playing for QJ tight in trumps or playing for one of the opponents to have a singleton honor and guessing the correct player.

Since the lead and the hands seemed to be balanced and from lack of bidding of the opponents I judged the trumps to be 3-2 which is mathematically probable. So I decided to play for a devil's coup, however I would never have done this play if I had been in the normal contract of 4 ♥ at matchpoints.

Cheers

[kereru67]

Intuitively it seems less of a stretch to me to try for QJ bare than to play for a very specific distribution, but it would be good to get actual figures. As pointed out the 6.8% chance of dropping QJ doesn't necessarily mean the contract is home, and clues from bidding and card play may sway the odds in favour of attempting a Devil's Coup. In Dorothy Truscott's example she didn't give any bidding (because "there might be children reading") but if East West are passing throughout, that makes it more likely that their hands are balanced. I also wouldn't make too many assumptions from the J♣ lead, it might be from JTxx or a doubleton in an unbid suit or even a singleton.

Is there an app that can calculate odds for things like this?