Restricted Reasoning

[lamford]

[hv=pc=n&s=sak8654ha7dk32c86&n=sqj32hk54dj84caq3&d=s&v=b&b=7&a=1sp2n(Jacoby)p3s(non-min)p4cp4hp4sppp]266|200[/hv]

Teams.

 

West leads the jack of clubs. You try the queen but East wins and returns a club to the nine and your ace. You ruff a club, and they appear to be 4-4, draw trumps (West having two and East discarding a heart). You play three rounds of hearts, all following. East playing nine, jack, queen, while you ruff. Now you cross to a top spade and play a diamond. East plays the nine. What now? And what do you intend to do on the next round of diamonds whoever leads it?

[Lovera]

Hi, seems to me that needs to count shape and cards in suits. Probably the diamond suit is 5-2 on opps.

[smerriman]

Hi, seems to me that needs to count shape and cards in suits. Probably the diamond suit is 5-2 on opps.

Based on the tricks to date, you already know diamonds split 4-3.

 

I would play the king, then the jack.

 

If we duck the first round, we have a straight 50-50 guess on the next round as to whether East holds the Ace or Queen, as everything is symmetric between those two options.

 

Playing the king the first round therefore has to be better.

 

East can play the 9 from T9x to give us a losing option, but there are actually no winning options to us inserting the 8; East having Q9x would be a Grosvenor.

 

Edit - actually this isn't correct at all, since East playing x from Q9x loses if declarer can cover it with a smaller x.. hmm, that makes it trickier since the x matters.

[Cyberyeti]

Based on the tricks to date, you already know diamonds split 4-3.

 

err - why ?

 

W has 4 clubs and 2 spades, if E is playing true cards, E has ♥QJ9 so west has 5 of those and thus only 2 diamonds. If this is what's going on, ducking the diamond cannot lose, if not you have to guess right on the second one.

 

How good is E ?

[smerriman]

E has ♥QJ9 so west has 5 of those

East discarded a heart on the trump.

[smerriman]

Second attempt.

 

The relevant cases are T9x and Q9x, where x<7 (would play the 7 otherwise). East will always play high from T9x.

 

Suppose the answer to this question is that declarer should always play the J.

 

East, knowing this, can safely play the 9 from Q9x. Restricted choice means that declarer only wins 1/3 of the time (while playing the 8 would have won 2/3 of the time).

 

Suppose the answer to this question is that declarer should always play the 8.

 

East, knowing this, must now always play low from Q9x - hoping that declarer doesn't have a higher x. So when East does play the 9, declarer will always lose (while playing the J would have always won).

 

So if East knows what you're going to do, it's better to do the opposite. Next up is calculating the equilibrium for a randomised strategy.. or I may read the Bridgewinners thread first to see if I'm totally off-track :)

[Cyberyeti]

East discarded a heart on the trump.

 

Ah, OK, missed that, but why on earth pitch a heart when you can disguise the count by pitching a safe club

[smerriman]

OK, lots of math ahead, either this is right or I'm going to look very silly, but hey:

 

Suppose that declarer plays the J with probability p.

 

Suppose East holds precisely Q95 of diamonds. If East plays low, declarer can cover cheaply 7/10 of the time and always win; the other 3/10 he can't and will always fail.

 

So East should play the 9 if p > 3/10.

 

Going over each holding we get:

 

Q93: 0

Q92: 0

Q942: 0

Q932: 0

Q94: 1/10

Q952: 1/6

Q95: 3/10

Q962: 1/2

Q96: 3/5

 

Each of these 9 holdings is equally likely, and contributes 2/3 towards the queen being held, and 1/3 towards the queen not being held by restricted choice. Or basically, if we multiply by 27, we can count 2 for each queen, and 1 for each non-queen.

 

So we just need to sort out what the optimal p is in each range.

 

If p>3/5, East will play the 9 in all cases. East has the queen 18/27 times, which is a majority, so declarer wants to minimise p. The best p in this range is thus 3/5, and declarer wins 18/27 * 2/5 + 9/27 * 3/5 = 7/15.

 

If p is between 1/2 and 3/5, we rule out the Q96 case (but still include T96), so the denominator becomes 25. East has the queen 16/25 times, so again we want to minimise p to be 1/2. Declarer wins 16/25 * 1/2 + 9/25 * 1/2 = 1/2.

 

If p is between 3/10 and 1/2, East has the queen 14/23 times, so p = 3/10 and declarer wins 14/23 * 7/10 + 9/23 * 3/10 = 25/46.

 

If p is between 1/6 and 3/10, East has the queen 12/21 times, so p = 1/6 and declarer wins 12/21 * 5/6 + 9/21 * 1/6 = 23/42.

 

If p is between 1/10 and 1/6, East has the queen 10/19 times, so p = 1/10 and declarer wins 10/19 * 9/10 + 9/19 * 1/10 = 99/190.

 

If p is less than 1/10, East has the queen 8/17 times, which is now less than half, so declarer wants to maximise p, so p = 1/10 again and declarer wins 8/17 * 9/10 + 9/17 * 1/10 = 81/170.

 

Note those last two endpoints don't match up for p = 1/10, because it gives East a choice of two options.. it's possible I've made a mistake but I think it's just because we're solely restricting to the case where declarer has 32 in the final outcome, whereas East is optimising all cases.

 

So overall declarer should play the Jack with a probability somewhere between 1/6 and 3/10 of the time, and will succeed in 23/42 of the relevant cases.

 

[late edit] Found a mistake; that's assuming declarer doesn't vary p based on the spots he himself holds. So need to factor that in too somehow..

[Lovera]

It needs to play for A10 in W and Q in E covering with K in ♦ suit.

[lamford]

Much analysed on Bridgewinners. My final thoughts:

 

Where is the ten of hearts?

 

I think it is overwhelmingly likely to be with West. East played six, nine, jack, queen and if he had the ten, there would have been four ways to play the nine, ten, jack and queen, but without it only one way. He would also not have thrown the six of hearts if he had QJT96.

 

We can therefore assume that East is at least 80% to have four diamonds, and probably closer to 100%. The possible 35 layouts of the seven diamonds are as follows.

 

West East

a) AQT 9765

b) AQ9 T765

c) AQ7 T965

d) AQ6 T975

e) AQ5 T976

f) AT9 Q765

g) AT7 Q965

h) AT6 Q975

i) AT5 Q976

j) A97 QT65

k) A96 QT75

l) A95 QT76

m) A76 QT95

n) A75 QT96

o) A65 QT97

p) QT9 A765

q) QT7 A965

r) QT6 A975

s) QT5 A976

t) Q97 AT65

u) Q96 AT75

v) Q95 AT76

w) Q76 AT95

x) Q75 AT96

y) Q65 AT97

z) T97 AQ65

aa) T96 AQ75

bb) T95 AQ76

cc) T76 AQ95

dd) T75 AQ96

ee) T65 AQ97

ff) 976 AQT5

gg) 975 AQT6

hh) 965 AQT7

ii) 765 AQT9

 

Kit’s strategy is to always play the king, and then always play the jack on the way back. He succeeds in 20 of the 35 cases immediately, and a further 5 on the second round. Total success rate 71.4%.

 

My strategy would be to duck if East plays the six, and then to play the king if East wins and returns the suit. If East plays the seven, nine or ten on the first round, I play the king and if it loses, I play the jack on the return of the suit. I win in the 20 cases that the ace is onside. The other 15 cases are interesting. To do better than Kit I need six of them. Let us say that East never plays the five, plays the ten or nine when he has both, and plays the seven when he has it but not both the nine and ten. These are the layouts which can be made (other than the 20 with the ace onside which are gin).

 

West East

a) AQT 9765

b) AQ9 T765

c) AQ7 T965

d) AQ6 T975

e) AQ5 T976

f) AT7 Q965

g) A97 QT65

 

So, my strategy is no worse off, and if East finds the nine and ten from the last two holdings all the time it is the same. But a learned contributor thought this would be a Grosvenor. Theoretically Kit is right, that his strategy is equally optimal, but only if the opponents know your strategy.