[Kaitlyn S]

North: H-JT9
West: H-AQ876 East plays H3 (standard count)
South: plays H2

West leads H7 against 1NT-3NT. Dummy plays the H9, East plays the H3 (standard count) and South plays the H2.

Simplistically, one could say that East either has H-543 or H-3 and they're almost equally likely.

One could also say that East has four possible singleton small holdings and four possible tripleton small holdings making them equally likely.

One could say that a 3-2 division of the remaining hearts is more likely than a 4-1 division so that East:543 is substantially more likely.

One could also say that while East's card is fixed (showing count) but South's card is restricted (he would play 2 from K2 but could play others from K542) so S-K2 is more likely.

One could say that if East had played the deuce, it could be from 542, 532, or 432 (and even though South's play eliminates two of those, that's irrelevant) so that if East played the deuce it would be more likely that East has three cards but since East played the H3 which has to be H-3 or H-543 they are essentially equally likely.

Any ideas?

[smerriman]

My non-expert thoughts which could well be wrong:

Kaitlyn S, on 2017-February-05, 14:22, said:

Simplistically, one could say that East either has H-543 or H-3 and they're almost equally likely.

Disagree for the reason that you know East has 13 cards. There are more hands where East has 543 (21C10 = 352716) vs 3 (21C12 = 293930). Of course, you know that South has a 1NT opener, so not all of those cases apply, but that probably puts things slightly more in favour of East being balanced too.

Kaitlyn S, on 2017-February-05, 14:22, said:

One could also say that East has four possible singleton small holdings and four possible tripleton small holdings making them equally likely.

Disagree; that relies on the same flawed assumption as above that singleton and tripleton are equally likely to begin with.

Kaitlyn S, on 2017-February-05, 14:22, said:

One could say that a 3-2 division of the remaining hearts is more likely than a 4-1 division so that East:543 is substantially more likely.

Agree for the similar 13-card reasons as described above.

Kaitlyn S, on 2017-February-05, 14:22, said:

One could also say that while East's card is fixed (showing count) but South's card is restricted (he would play 2 from K2 but could play others from K542) so S-K2 is more likely.

Disagree with the reasoning because while South *could* play others from K542, I believe he'd play the 2 100% of the time, thus restricted choice doesn't affect the probabilities for this case.

Kaitlyn S, on 2017-February-05, 14:22, said:

One could say that if East had played the deuce, it could be from 542, 532, or 432 (and even though South's play eliminates two of those, that's irrelevant) so that if East played the deuce it would be more likely that East has three cards but since East played the H3 which has to be H-3 or H-543 they are essentially equally likely.

Disagree that South's play is irrelevant; agree that if East had played the 2, 3 cards are more likely (but because 3 cards are more likely regardless); disagree with the implied idea of this case having anything to do with what happens when East plays the 3.

[kenberg]

I find this complex, and I say more in a later post.

My thoughts, although I cannot be certain without checking further.

E plays before seeing the deuce. We assume he plays the 3 from 345, from 32 and from 3 alone. We assume that S knows that EW play in this manner. We must take into account what S will do on the basis of this knowledge. We consider three possible player types SA,SB,SC.

We assume the play has gone 793


SA always plays the deuce if he has it.
SB never plays the deuce unless it is his only spot card.
SC, when he holds K542, sometimes plays the deuce, sometimes he randomly plays the 4 or 5.


We assume (for purposes of discussion) that EW know which style S is.

Now if we are playing against SB then we know SB started with K2.

So, at first glance, SA has a better strategy. But not really, because while he held the deuce on this hand, he will at other times be dealt the K65. But then SA will play the 6 or 5, and EW, since they know that they are playing against SA, will know that since he did not play the deuce he does not have the deuce.

Thus, although more detail is needed to make the complete argument, it is likely that SC has the best strategy. Whether he is dealt the deuce or not, and whether he plays the deuce or not, it will never be known with certainty what he holds.

Now where are we? Well, S will adopt some probabilistic strategy about which card to play when he is dealt 542. It is not obvious how frequently he should go low, how frequently he should go high. It would take some effort, I might give it a try later, to figure what he should do. How often he would go low might well influence EW.

Recap: Playing against SA, W will know a lot if S follows with the 4 or 5 but not much when he follows with the deuce, playing against SB, W will know a lot if S follows with the deuce but not much if he follows with the 4 or 5, so a shifty S might become an SC. In which case knowing his probability law could be useful. Maybe he plays the deuce half the time when he holds 542. Or maybe two thirds of the time. Or one third.


Added: Let's look at the more well known case where declarer has to play AT98 opposite K7654. He play the A getting 2,4,Q and then plays the 8 ans the 3 appears on his left. He would like to know whether W always plays the Q for QJ or never plays the Q from QJ. Most defenders know that from QJ tight is is best to play randomly select one or the other, with equal probabilities. The more W deviates from this, the more helpful it In the current situation, surely some sort of random play from 542 is no doubt right but I am less certain how often the 2 should be played.The more S knows about W's approach, the better it will be for him. I suppolse (very) strong games declarer knows the optimal percentages to play the deuce, and W assumes he uses that.

[SteveMoe]

So if we imagine West and north as our pair, then we have 8 cards and want to surmise a priori what the likelihood is each split occurs. Since the 3 signals and ODD number of cards, only 1,3, and 5 are in the relevant set. Wehn South plays the 2, only 1 and 3 remain. Now we know that 3-2 is 68% and 4-1 is 28% when the cards were dealt. True half of the symmetrical cases are eliminated because partner must have 3, not 2 or 1 not 4. However the relative probabilities still apply.

Se we expect (absent any other information (not inference)that partner is 68/96 or about 71% while partner holding a singleton is 29%.

However if there is solid information from the bidding (vacant spaces because partner preempted and has a 7-card side suit), or the play (a count on known cards) that affect the odds then we must adjust the estimate.

As for whether there's a restricted choice on the 2345 spots - definitely no. They are not significant cards and as a class can be represented by xxxx, meaning we attach no significance to the order of their play because they are not winning cards.

As for the relative probability of individual cases, we can estimate:

3-2 / 2-3 has 10 + 10 cases (count them) and therefore 68/20 or 3.4% each

1-4 / 4-1 has 5 + 5 cases (count them) and therefore 28/10 = 2.8% each.

Go to Richard Pavlicek's site to convince yourself. http://www.rpbridge....cgi-bin/xcc1.pl

[Kaitlyn S]

SteveMoe, on 2017-February-05, 22:00, said:

So if we imagine West and north as our pair, then we have 8 cards and want to surmise a priori what the likelihood is each split occurs. Since the 3 signals and ODD number of cards, only 1,3, and 5 are in the relevant set. Wehn South plays the 2, only 1 and 3 remain. Now we know that 3-2 is 68% and 4-1 is 28% when the cards were dealt. True half of the symmetrical cases are eliminated because partner must have 3, not 2 or 1 not 4. However the relative probabilities still apply.

Se we expect (absent any other information (not inference)that partner is 68/96 or about 71% while partner holding a singleton is 29%.

However, East is known not to have the ♥K. So couldn't you imagine the "small hearts" as a suit and ask whether they are 3-1 or 1-3 which are equal?

If that is my assumption, and I can't see anything wrong with that, then I am comparing East getting 12 other cards and South getting 9 other cards with East getting 10 other cards and South getting 11 other cards. If I ignore factors like the balanced hand for notrump and the high cards needed, I believe that 10-11 is 20% more likely than 12-9, so the odds of the three card holding in East should be 6/11.

(I get that by comparing 21!/912!x9!) to 21!/(11!x10!), I have to multiply the first figure by 12 to get 21!/(11!x9!) and then divide by 10 to get 21!/(11!x10!). I put this here to save anybody so inclined to figure it out the trouble of having to do so; they just need to make sure I didn't do something stupid.)

[SteveMoe]

Kaitlyn S, on 2017-February-05, 22:04, said:

However, East is known not to have the ♥K. So couldn't you imagine the "small hearts" as a suit and ask whether they are 3-1 or 1-3 which are equal?

Not quite.
If W holds the K, then we see that Kx is 13.57% and Kxxx is 11.3% a priori. Not so close to equal. 13.57/11.3 = 1.2, a 20% advantage for Kx over Kxxx.

[Kaitlyn S]

SteveMoe, on 2017-February-05, 22:13, said:

Not quite.
If W holds the K, then we see that Kx is 13.57% and Kxxx is 11.3% a priori. Not so close to equal. 13.57/11.3 = 1.2, a 20% advantage for Kx over Kxxx.

so you get a 20% edge and I get a 20% edge (see my prior post where I say 6 to 5.) By jove, I think we've got it! At least until somebody that knows something comes in and tells us something we hadn't thought of

[Kaitlyn S]

Actually I said 6/11, which is the same as 6 to 5.

[msjennifer]

[quote name='Kaitlyn S' timestamp='1486326166' post='912944']
North: H-JT9
West: H-AQ876 East plays H3 (standard count)
South: plays H2

West leads H7 against 1NT-3NT. Dummy plays the H9, East plays the H3 (standard count) and South plays the H2.

Simplistically, one could say that East either has H-543 or H-3 and they're almost equally likely.

One could also say that East has four possible singleton small holdings and four possible tripleton small holdings making them equally likely.

One could say that a 3-2 division of the remaining hearts is more likely than a 4-1 division so that East:543 is substantially more likely.

One could also say that while East's card is fixed (showing count) but South's card is restricted (he would play 2 from K2 but could play others from K542) so S-K2 is more likely.

One could say that if East had played the deuce, it could be from 542, 532, or 432 (and even though South's play eliminates two of those, that's irrelevant) so that if East played the deuce it would be more likely that East has three cards but since East played the H3 which has to be H-3 or H-543 they are essentially equally likely.

Any ideas?
Isn't it a headache ,all these combinations and percentages? And then,whether ,in actual play,the cards behave as per these figures ?

[kenberg]

The 4 and the 5 are interchangeable, the 2 is different.
Climb into the S hand holding 542 after the 3 is played.
If you play the 2, W will know with certainty was not dealt the 32 but be uncertain whether E was dealt 543 or stiff 3.
If you play the 5 or the 4, W wil know for certain that E was not dealt three small but will be uncertain whether E was dealt 32 or stiff 3.

So S can control which issue he wishes W to be uncertain about. This means the 2 is distinct from the 45.

It also means that W needs to climb into S's mind a bit. For example, suppose that S, after winning the trick, runs a finesse losing to W. What does S hope his T1 play has accomplished? If S played the 3 or 5 at T1 now W knows that E was dealt at most 2 hearts and so the A will not drop the K. If S played the 2 at T1 then W will be uncertain. Maybe the A will drop the K, maybe it won't.
Usually S will want W to try to drop his K when it won't drop, and he will want him to not drop his K when it does. So it appears that S should play the deuce at T1 when he was dealt K542.

Of course when S holds the K542 then, if W does not play the A now then he may never get it, since E has no more hearts. So maybe S wants W to play the Ace, establishing th K, maybe S wants W to not play the A, so it withers.

As posed, the question is simply about the odds of how cards are dealt. But I think it is more complex. In the QJ restricted choice the Q and J are indistinguishable. Here the play of the 2 carries different information than the play of the 4 or 5, and so we must think about which of these cards declarer might, if he has a choice, play based on the information he is trying to conceal.

[miamijd]

Kaitlyn S, on 2017-February-05, 14:22, said:

North: H-JT9
West: H-AQ876 East plays H3 (standard count)
South: plays H2

West leads H7 against 1NT-3NT. Dummy plays the H9, East plays the H3 (standard count) and South plays the H2.

Simplistically, one could say that East either has H-543 or H-3 and they're almost equally likely.

One could also say that East has four possible singleton small holdings and four possible tripleton small holdings making them equally likely.

One could say that a 3-2 division of the remaining hearts is more likely than a 4-1 division so that East:543 is substantially more likely.

One could also say that while East's card is fixed (showing count) but South's card is restricted (he would play 2 from K2 but could play others from K542) so S-K2 is more likely.

One could say that if East had played the deuce, it could be from 542, 532, or 432 (and even though South's play eliminates two of those, that's irrelevant) so that if East played the deuce it would be more likely that East has three cards but since East played the H3 which has to be H-3 or H-543 they are essentially equally likely.

Any ideas?

Yes, use a treatment such as Reverse Smith at NT, and partner can tell at trick 2 which holding he has.

Cheers,
Mike

[gszes]

SORRY for this but the FIRST consideration should be the opps bidding. Is there ANY clue that might help decipher if one declarer is more or less likely to hold four hearts?. I had to get that our of the way. Second we should use before deciding on trying restricted choice. Does your hand plus the, so far useless, bidding indicate we have the time and partner and we have the entries to delay making this choice. It is important to note just how far down the line restricted choice can become as a serious contender for decision making.
The overall odds of p having a singleton are significantly smaller than p starting with 3 hearts. The difference is so huge that restricted choice is a virtual non factor even with this highly limited set of conditions.

[sfi]

Restricted choice only applies when the player would play randomly from equals given a free choice.

Ken pretty much nailed it, but East isn't going to be choosing randomly. Therefore we really are only looking at the possibilities of 3 and 543, which are as you say fairly close in probability (with the 3-2 break holding a slight edge).

Where restricted choice comes in is South's play of the 2. With K2, they only have one option. With K542, the 2 is not equal to the 5 and the 4, but some of the time they would play a higher card to conceal the 2. You don't need to try and work out the odds, which is hard to do without knowing your opponent, but it is clear that the chance of the K542 holding should be discounted to cater for this. (A poor declarer would play the 2 most of the time, a reasonable one would always play a higher spot card, and an expert may or may not hide the 2 depending on the rest of the hand.)

The rest of the hand also needs to be taken into account. Is it likely declarer has a doubleton? A four-card suit? Can we beat the hand if declarer holds K542?

Absent any other useful information, partner holding 543 looks like a significant favourite.

[sfi]

gszes, on 2017-February-06, 15:46, said:

SORRY for this but the FIRST consideration should be the opps bidding. Is there ANY clue that might help decipher if one declarer is more or less likely to hold four hearts?.

It was given as a fairly unenlightening 1NT - 3NT.

Quote

The overall odds of p having a singleton are significantly smaller than p starting with 3 hearts. The difference is so huge that restricted choice is a virtual non factor even with this highly limited set of conditions.

But here we are only comparing the odds of one specific 4-1 break vs one specific 3-2 break. These odds are much closer, with SteveMoe correctly giving them as 2.8% to 3.4%. Easily close enough that it can be affected by other considerations.

[Phil]

Not restricted choice, but playing declarer for exactly K2 looks right to me.

With K42 I would think declarers optimal strategy is to hide the 2 and make it appear my partner has 32 dub.

[Kaitlyn S]

Phil, on 2017-February-06, 17:13, said:

Not restricted choice, but playing declarer for exactly K2 looks right to me.

With K42 I would think declarers optimal strategy is to hide the 2 and make it appear my partner has 32 dub.

Partner can't hold ♥K42, partner showed an odd number.

[Phil]

Yeah I meant K542.

[Kaitlyn S]

Perhaps somebody can tell me what is wrong with the following analysis:

You know that declarer has the ♥K. Either partner has three small hearts and declarer has one small (♥Kx) or declarer has three small hearts and partner has a singleton. Normally each combination would be almost equally likely except that because you have to split the non-hearts (12, 9) for East's singleton and (10,11) for East's tripleton, then each of East's tripleton small heart holdings is 20% more likely than each of East's singleton small holdings (calculations in an earlier post.) Since East can have four small singletons and four small tripletons, with each tripleton 20% more likely, the original deal odds make East's tripleton a 6:5 favorite.

Now, let's ignore South's choice of play for a minute. When East plays the 4 or 5, he can't have a tripleton (standard carding, showing count with the 2,3,4,5,K missing.) 75% of East's small tripletons follow with the deuce and 25% of East's small tripletons follow with the three. So, I think that before you see South's card, if East plays the 4 or 5, East can't have a tripleton, if East plays the 3, East has a tripleton 6/11 of the time, and if East plays the 2, East has a tripleton 18/23 of the time (comparing all of 542, 532, 432 against a singleton deuce with the singleton deuce being 5/6 as likely as any of the specific tripletons.)

Now let's look at South's play. South wants West to continue if he started with ♥Kxxx with ♥JT still in the dummy after trick 1. To make this happen, South has to create the illusion of a king that is dropping. Implying that East has a doubleton can never work because to make West think that declarer and East each have two cards when South has four and East followed suit would imply that West can't count to thirteen. Also, it's fruitless to make West think that East has a four-card holding because East would never play the ♥3 as a count card from four small. So IMO South's only chance is to make West think that East has three when East in fact only has one, and the deuce is the indicated play. (We are assuming expert South play with South knowing E-W's signalling methods.) Since South is going to signal correctly, we are looking at the true card deal odds - meaning that East's tripleton is still a 6-5 favorite when East plays the ♥3 and an 18-5 favorite when East plays the deuce. Increase those odds for the three by just a little bit when South is capable of making an ill-advised falsecard (or if South might not know East's signalling methods - after all, if South asks, he might give the show away, and asking with a doubleton king might be considered coffeehousing if the reason is to put doubt in West's mind.)

I've made a lot of assumptions in my calculations and any one of them might be wrong. If you have any ideas that should cause me to reconsider these odds, I'd be happy to hear about them.

[Phil]

Well, its hard to think about this in a vacuum. Is there another possible trick source? Are we concerned that conceding the heart trick is the 9th? You havent told us if we play Smith either, which can be a huge clue to partners holding too.

Note that with declarer originally holding 2 or 4 hearts theres no point to trying to get partner in. And for this reason Id call hiding the 2 by declarer a mandatory false card because if partner did have a doubleton we have to shift.

So, I dont agree with your premise.

[Kaitlyn S]

Phil, on 2017-February-06, 21:22, said:

Note that with declarer originally holding 2 or 4 hearts theres no point to trying to get partner in. And for this reason Id call hiding the 2 by declarer a mandatory false card because if partner did have a doubleton we have to shift.

So, I dont agree with your premise.

I couldn't see how it could help to imply a doubleton. I do now. Thank you.