[1175]

In the duplicate IMPs, I had some hope for a good score when (at my table) the Robot, playing 3NT, can take all 13 tricks, but only takes 9. I hadn't counted on play like this at other tables:

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One would think that the Robot would see the risk of what subsequently happened, and avoid going down.

[smerriman]

"Can take all 13 tricks" is true, but a bit misleading, isn't it? Correct me if I'm wrong, but diamond to the ace and club finesse for exactly 9 tricks seems like the best line.*

The reason it played a heart at trick 9 will just be down to the awful free robots again, who simulate a very small number of hands. Out of the 252 ways to arrange the opponents' last 5 cards, playing a heart only loses on 6 I believe (when one opponent holds all four spades and not a heart honor), so there's a decent chance a free robot won't deal any of those, leading it to think there's no difference between a heart and spade.

* I'll correct myself, perhaps diamond finesse is better; I was thinking high handles a stiff King with South as well as the 50-50 club finesse, which is better than a finesse alone, but it probably costs more as you can still make with a losing diamond finesse if hearts are 4-4.

That's another issue with the free robots - their declarer logic is double dummy only. They play the diamond ace, assuming that they'll immediately see the opponents' hands after that - if hearts are 4-4, they'll continue with another diamond, otherwise take the club finesse. Advanced robots use a single-dummy algorithm, so don't switch plans midstream like this. Unfortunately, that single dummy algorithm only kicks in after 2 tricks, since it's too slow before that, so it's possible they'll also play the ace of diamonds. If I force it to cash the king of spades first, then an advanced robot takes the diamond finesse, while a free robot still plays the ace.

[johnu]

1175, on 2024-June-15, 09:06, said:

In the duplicate IMPs, I had some hope for a good score when (at my table) the Robot, playing 3NT, can take all 13 tricks, but only takes 9. I hadn't counted on play like this at other tables:

---

One would think that the Robot would see the risk of what subsequently happened, and avoid going down.

Good play by West. A lessor declarer might have gone down 2.

[1175]

smerriman, on 2024-June-15, 15:02, said:

* I'll correct myself, perhaps diamond finesse is better; I was thinking high handles a stiff King with South as well as the 50-50 club finesse, which is better than a finesse alone, but it probably costs more as you can still make with a losing diamond finesse if hearts are 4-4.

I can try to answer that. Obviously, two fully unknown hands have 10,400,600 arrangements (26! / (13! * 13!)). I have a Python script that builds every one of those hands, from which I can use various text processing tools to filter and count them.

Suppose that we give North KQJ of ♥ (for the lead), and assume that North would have bid with: 6, 7, or 8 ♠, 6, 7, or 8 ♥, and various combinations of 5-5-3-0 and 5-5-2-1 distributions. Those assumptions reduce the number of possible North-South hands to 604,097.

First, we should figure out what to do when North shows out on the first diamond (17,150 hands). I have 70 hands where North has 4-3-0-6 distribution (this assumes that North would lead a ♥ with KQJ tight rather than a ♣ with Q86543 - something the robot might actually do). Either play (♦A or finesse) will work here. If declarer goes up with the ♦A, when South shows out on the first club, declarer should not finesse (North would win and unblock the hearts), but win and knock out the ♦K, and hope for either 4-4 hearts or heart blockage. If North has four hearts (6580 hands),taking the marked losing diamond finesse still makes the hand. Going up with the ♦A only wins when South has the ♣Q (only 1750 hands - North must have four clubs if North has no diamonds, which makes the club finesse much less than even money) or when South shows out on the first club, and declarer wins and reverts to diamonds (280 hands). If South has five hearts (10,500 hands), playing the ♦A gains only when the club finesse works (only 3500 hands - exactly one-third of the time).

To sum that up, if North shows out on the first diamond, playing the ♦A wins if North has three hearts (70), four hearts with either the club finesse on or declarer able to revert (1750 + 280), or five hearts with the club finesse on (3500). Taking the marked losing finesse works when North has three hearts (70) or four hearts (6580). So if North shows out of the first diamond, taking the marked losing finesse wins 6650-5600.

Finessing the ♦K works slightly more than half the time (306,116 / 604,097, or 50.67 percent). North has a slightly greater chance of holding the ♦K, because we have removed many distributional North hands from consideration, while South could still hold 6, 7, or even 8 spades, for example. The line of play also succeeds when South has the ♦K, and North has three or four hearts, an additional 123,181 cases. So taking the diamond finesse has a total success chance of 429,297 / 604,697, or 71.06 percent.

Going up with the ♦A (the wrong play even if North shows out) works when either North (36,386 hands) or South (37,531 hands) has a singleton ♦K. Let's assume that declarer sticks with the ♦A even if North shows out, but still reverts to knocking out the ♦K if South shows out on the first club. The club finesse works (assuming declarer did not find a singleton ♦K) on 269,223 hands, and hearts divide 4-4 when South shows out on the first club 955 times. Going up with the ♦A will succeed 56.96 percent of the time ((36,386 + 37,531 + 269,223 + 955) / 604,097).

Obviously, I have made a lot of assumptions (about when North bids) in this analysis, but it seems fairly clear that taking the diamond finesse offers a significantly higher chance of making the contract (not to mention overtricks, which still add up) than playing a diamond to the A and taking the club finesse.

[1175]

smerriman, on 2024-June-15, 15:02, said:

The reason it played a heart at trick 9 will just be down to the awful free robots again, who simulate a very small number of hands. Out of the 252 ways to arrange the opponents' last 5 cards, playing a heart only loses on 6 I believe (when one opponent holds all four spades and not a heart honor), so there's a decent chance a free robot won't deal any of those, leading it to think there's no difference between a heart and spade.

I think that even the free robots should have some logic that cashes winners unless playing another card offers the chance of a better result. Instead of the Robot saying (metaphorically) "the sample I looked at didn't find any hands where playing a heart loses," it should instead have logic to discover "playing a heart never gains a trick, so I will just cash the spade."

[cherdano]

Isn't the best line obviously cashing the top clubs, then taking the diamond finesse if the club queen did not appear?

[smerriman]

cherdano, on 2024-June-18, 15:11, said:

Isn't the best line obviously cashing the top clubs, then taking the diamond finesse if the club queen did not appear?

The clubs are blocked - that will only get you 8 tricks if the queen drops.

1175, on 2024-June-17, 00:24, said:

I think that even the free robots should have some logic that cashes winners unless playing another card offers the chance of a better result. Instead of the Robot saying (metaphorically) "the sample I looked at didn't find any hands where playing a heart loses," it should instead have logic to discover "playing a heart never gains a trick, so I will just cash the spade."

The robots could do a lot of things. But "if all else equal, cash top tricks" is a human concept; the robot's algorithm has no concept of what cashing top tricks mean, and there are certainly examples when you'd want the robot to do the exact opposite.

Indeed, give it a two way finesse situation and an ace as the only stopper in a side suit; your rule would tell it to cash the ace first, which of course would be a disaster. Trying to introduce human-like rules usually makes things worse; it has to be fixed by resolving the fundamental issues.

[1175]

smerriman, on 2024-June-18, 15:35, said:

Indeed, give it a two way finesse situation and an ace as the only stopper in a side suit; your rule would tell it to cash the ace first, which of course would be a disaster. Trying to introduce human-like rules usually makes things worse; it has to be fixed by resolving the fundamental issues.

In that case, wouldn't taking the finesse potentially produce more tricks (presumably, even a small number of simulations would show that)? If so, then the Robot should not cash the ace. I want the Robot to eliminate plays that can never take additional tricks (as here).

[smerriman]

1175, on 2024-June-18, 16:01, said:

In that case, wouldn't taking the finesse potentially produce more tricks (presumably, even a small number of simulations would show that)? If so, then the Robot should not cash the ace. I want the Robot to eliminate plays that can never take additional tricks (as here).

No, I was talking about a two-way finesse - e.g. AJ9x vs KT8x. Double dummy will tell it that playing a low card will take four tricks in this suit, as will playing a side ace, but of course you can't risk cashing the side ace if that leaves you wide open should the finesse fail.

Of course, this is something also resolved by the non-free robot's single dummy algorithm.

[smerriman]

In fact, even simple finesses would break entirely with that logic, wouldn't they? It'd virtually never take a finesse anymore, since if the card is onside, double dummy will confirm that it was just as good to cash outside top tricks first before taking the finesse.

[1175]

smerriman, on 2024-June-18, 16:24, said:

In fact, even simple finesses would break entirely with that logic, wouldn't they? It'd virtually never take a finesse anymore, since if the card is onside, double dummy will confirm that it was just as good to cash outside top tricks first before taking the finesse.

I obviously don't understand the entirety of how the Robot works. Doesn't the Robot see that a losing finesse will cost multiple tricks in those cases?

In any case, at least at IMP scoring, some consideration towards making the contract (particularly for games and slams) should influence the play of the hand.

[smerriman]

Free robots only see what the double dummy score is for each card in the hand they're leading from. If they're on lead with A-xx in hand opposite x-AQ in dummy, then no matter how many ways you simulate what the opponents hold, all three cards will give identical scores double dummy.

They do take into account IMPs and making contracts, but still based on how many tricks double dummy says they are making on each sample. If the samples are flawed, so are the results.

[1175]

smerriman, on 2024-June-19, 01:47, said:

Free robots only see what the double dummy score is for each card in the hand they're leading from. If they're on lead with A-xx in hand opposite x-AQ in dummy, then no matter how many ways you simulate what the opponents hold, all three cards will give identical scores double dummy.

That sounds terrible. But what about the play from dummy? Let me expand your example:

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What does the Robot use to determine what card to play from South, and what does the Robot use (assuming it chose to lead a heart towards dummy) to determine what card to play from North?

[johnu]

cherdano, on 2024-June-18, 15:11, said:

Isn't the best line obviously cashing the top clubs, then taking the diamond finesse if the club queen did not appear?

You need 4 club tricks (added to 3 spade tricks 1 heart trick, 1 diamond trick) to avoid taking the diamond finesse. You need specifically ♣Q8 to take 4 club tricks without a finesse.

Also, cashing 2 top clubs and not dropping the ♣Q sets up a club trick for the defense (good falsecard available with the ♣8), so if hearts are 4-4, you've found a way to go down if the diamond finesse is off.

[smerriman]

1175, on 2024-June-19, 06:02, said:

That sounds terrible. But what about the play from dummy? Let me expand your example:

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What does the Robot use to determine what card to play from South, and what does the Robot use (assuming it chose to lead a heart towards dummy) to determine what card to play from North?

From South, it deals out various ways of arranging the unknown cards to the opponents, and then calculates the double dummy score of each card in its hand. When it has dealt Kx or Kxx to East, this score is 2 tricks for all cards. In other cases, the score is 3 tricks for all cards. After adding the results, it confirms that all three cards score identically no matter what, so it doesn't matter what it plays. *

If it leads a heart, it deals out various ways of arranging the unknown cards to the opponents. When it has dealt Kx or Kxx to West, it sees that Q takes 3 tricks, and A takes 2 tricks. When it has dealt xxx to West, it sees that Q takes 0 tricks, and A takes 3 tricks, and so on. Each case occurs a different number of times in the sample. What it does then depends on scoring; for example if it needs exactly 2 tricks to make game at IMPs, it will see that playing the Q will average a negative number of IMPs compared with the Ace, but if it's just for an overtrick / MPs, the finesse gains on average.

* I thought it would play randomly here, but it seems I was wrong, and it always plays a heart. This is definitely not because it foresees the finesse; that's part of the single dummy logic only. Will have to experiment more to guess what tiebreak criteria it's using..