[gnasher]

whereagles, on 2011-July-28, 06:04, said:

Come on, you're making this WAY too complicated. The shift in odds due to LHO having this or that particular card is minimal and even considering he must have 5 diamonds, they don't shift that much (see e.g. Borel's book "mathematical theory of bridge").

Inquiry wasn't asking how to play this particular hand. He was asking whether the knowledge that a player must have a given honour affects the odds of another suit's breaking. That's an interesting question, with an instructive answer.

Quote

In the case at hand I would just lead a heart towards dummy and watch RHO's count card. That's probably way more telling than 1000 sims.

How come all of your opponents always give count? If I were East, it wouldn't even occur to signal my heart count here.

[wclass___]

gnasher, on 2011-July-28, 02:27, said:

Here's another thought experiment.

Take the original deal, but assume that we know West has 3 spades and 7 diamonds, leaving him with space for only 3 cards in hearts and clubs.

If we don't know where ♥K is, West's possible heart holdings are:
1 void
7 singletons
21 doubletons
35 trebletons
The chance that East has 2 clubs is proportional to 7/64 ~= 0.11

If we know West has ♥K, West's possible heart holdings are:
0 voids
1 singleton
6 doubletons
15 trebletons
The chance that East has 2 clubs is proportional to 1/22 ~= 0.04

So, giving West ♥K reduces the chance that East has two clubs.

As i staded with my 1st replay in this thread i think that odds calculation do change, but not in the way as to adding one vacant space for LHO. I misread your post thinking that your post disagrees with my statement, after re-reading your post i realized it is about something else.

Anyway after you play ♥ to Q, K with LHO is not a restriced card. And there aren't 21 doubletons possible anymore.

Or are you trying to say that odds at point one when you just see a hand and know LHO has ♥K is different from point where we play ♥ to dummy and learn that LHO has ♥K?

[wclass___]

gnasher, on 2011-July-28, 03:50, said:

That's completely different, because LHO chooses which card to name.

This is the correct comparison: you ask LHO to shuffle his cards, then you look at a randomly selected card. If that isn't ♠K, the probability that he was dealt ♠K is reduced.

Suppose that he lets you see 12 randomly selected cards, and none of them is ♠K. Do you really think that the finesse is still 50%?

IMO there is quite a big difference between my example and asking opponent to show his entire hand. Anyway, lets say West opened with 1♣ showing ♣T and nothing else, contract is 7NT by North, and you receive non-spade lead from East. Do you think that chances for successful finesse are now 48%? Do you think your chances after you play your non-spade top trick and learn that west really did have ♣T are still 48%? (Opponents keep ♥K and 3 other hearts.)

[gnasher]

Now I'm not sure what you're arguing. Maybe it's better if I tell you what I think, then you can tell me if (and how) you disagree. I think that this is correct:

- If we know that diamonds are 5-3, and we don't care where ♥K is, the vacant spaces in the two hands are West: 8, East: 10. That is, the chance that West will hold a specific club is 8/18.
- If we know that diamonds are 5-3, and we are require West to hold ♥K, the vacant spaces in the two hands are West: 7, East: 10. That is, the chance West will hold a specific club as well as ♥K is 7/17.

[wclass___]

gnasher, on 2011-July-28, 07:41, said:

- If we know that diamonds are 5-3, and we are require West to hold ♥K, the vacant spaces in the two hands are West: 7, East: 10. That is, the chance West will hold a specific club as well as ♥K is 7/17.

I was asking if you think that vacant spaces after you play ♥ to dummy (LHO coming up with king) in the two hands are West: 7, East: 10.

Similarly in 7NT whether odds that finesse is on after 11 tricks are played are still 48%, assuming other things don't matter, like possible spade lead or not discarding club T.

Also i was asking your opinion about situation where we don't know if LHO has ♥K and play small ♥ to dummy learning that LHO has ♥K, what are vacant spaces in that case?

[semeai]

wclass___, on 2011-July-28, 08:05, said:

Also i was asking your opinion about situation where we don't know if LHO has ♥K and play small ♥ to dummy learning that LHO has ♥K, what are vacant spaces in that case?

This is not what inquiry was asking. The answer to this question is likely* very close to, but not exactly the same as, the answer to that question.

This question is more complicated because LHO would not always play ♥K if he has it.

Probability of LHO holding a specific club right now given that diamonds are 5-3 and all we've done is play a heart up and see ♥K pop:

prob(hearts K-6)*p_1*(7/11) + prob(hearts K1-5)*p_2*(6/11) + prob(hearts K2-4)*p_3*(5/11) + prob(hearts K3-3)*p_4*(4/11) + prob(hearts K4-2)*p_5*(3/11) + prob(hearts K5-1)*p_6*(2/11) + prob(hearts K6-0)*p_7*(1/11)

divided by

prob(hearts K-6)*p_1 + ... + prob(hearts K6-0)*p_7

Here by prob(hearts K2-4) I mean the probability that hearts are Kxx - xxxx given diamonds 5-3 already. Here p_n is the probability that LHO plays the K when he has n of them. For example p_1 is 1 and p_2 is pretty close to 1. Let me say that the rest are all equal for simplicity.** Note I'm ignoring reasons other than length that may distinguish playing ♥K or not for LHO.

I won't do the calculation out, but the first two terms (hearts K-6 or K&1-5) are not so likely and p_n is not too small, so the rest of the terms dominate. There the p_n's are assumed to be all the same, and they drop out of the numerator and denominator, giving us:

prob(hearts K-6)*(7/11) + ... + prob(hearts K6-0)*(1/11)

divided by

prob(hearts K-6) + ... + prob(hearts K6-0)

Which is the answer to the sort of question inquiry asked. It's also exactly equal to 7/17, the vacant spaces answer, which I invite you to check or prove.

*This question has no definitive numerical answer because it depends on the probability that LHO will play ♥K from various holdings.

**This is where the ♥K is different from say the ♥7. LHO's motivations for playing ♥K or not are not likely to change much as the number of spot cards increases beyond 2, but the likelihood he'd play a random spot like the 7 would decrease if he had more random spots to go with it.

[wclass___]

I have found answers to my questions.

wclass___, on 2011-July-28, 08:05, said:

I was asking if you think that vacant spaces after you play ♥ to dummy (LHO coming up with king) in the two hands are West: 7, East: 10.

Yes, 7-10, plus information that RHO isn't void in ♥.

wclass___, on 2011-July-28, 08:05, said:

Similarly in 7NT whether odds that finesse is on after 11 tricks are played are still 48%, assuming other things don't matter, like possible spade lead or not discarding club T.

Yes, 48%.

wclass___, on 2011-July-28, 08:05, said:

Also i was asking your opinion about situation where we don't know if LHO has ♥K and play small ♥ to dummy learning that LHO has ♥K, what are vacant spaces in that case?

8-10 plus information that opponents are not void in ♥

[semeai]

wclass___, on 2011-July-28, 08:05, said:

Also i was asking your opinion about situation where we don't know if LHO has ♥K and play small ♥ to dummy learning that LHO has ♥K, what are vacant spaces in that case?

wclass___, on 2011-July-28, 10:01, said:

8-10 plus information that opponents are not void in ♥

This answer is correct only if LHO plays his heart randomly. If you read the previous post of mine, this is equivalent to assuming p_n is 1/n.

As I said in the second footnote (egad!), LHO's motivations for playing ♥K are not likely to change much as the number of spot cards increases beyond 2, and even for 1 spot card, i.e. Kx, it's likely pretty close to 1, not 1/2.

[inquiry]

wclass___, on 2011-July-27, 19:50, said:

---

7NT with a spade lead, you know that LHO has say ♣T. Does your chances for finesse really worsens? What if the lead is non-spade and you cash your non-spade top tricks and learn that LHO really did have ♣T, does your chances are still worse than 50%?!

I think same applies to given hand, where if you want, you can play ♥A, ♥ to Queen learning what you already assumed.

The ♣T is a meaningless card (other than proving West is not void in clubs). West can volunteer to play that club at anytime without a problem and conveying no useful information. So no, the fact that WEST has the ♣T and/or shows it to us, is meaningless. I think the analogy you are trying to draw is this....

In this hand if WE FORCE west to have the club TEN (give it to him and then deal the remaining 25 defensive cards randomly), West is less likely than East to have the king of spades 12/25th's or 48%), just as the arguement on the posted hand is if WEST has five (or six) diamonds adding the king of hearts further decreases the chances of WEST having the club jack *7/17 or 6/16th). But that analogy falls apart, Your position is that "you know LHO has say ♣T." Here you know that BECAUSE he has choose to show it to you (drop it on the table, play it, tell you he has it, what-ever). This does not affect the odds of the spade king at all (unless you searched through the deck, gave it to him before dealing the other cards, and then dealt the others randomly which of course would be cheating).

The heart King is not some random card. The bidding and the need to make this contract requires you place that card in West hand. Thus you can treat the long diamonds and the heart king as "know cards" in the problem hand. But not in your example, where the club ten, or any card the defense can choose to show without direct force is not a known card despite you seeing it. Thus the situations between the problem in the OP and your "example" are are not the same. The heart king situation simply does not apply in your example hand. There is nothing necessary or implied about any specific non-spade, including the club ten. During the play, you may learn that East has at least six spades to West maximum of three or vise versa, which of course will affect the odds of the spade hook winning. But, the location of the club TEN has no bearing on it at all. The difference is the heart KING is not some random heart. We expect East will have some hearts (after all West did not show a red two suiter), we expect west to have some hearts (after all he did not show a diamond-black two suiter).

So as far as your example with 48% chance after West shows up with the club TEN. Nonsense. The club ten is an unimportant card, it is any random card. It's being in WEST hand has no bearing whatsoever on the placement of the spade king. Only if you can get a count on spades by their following suit to the other three suits, does any card they play in clubs, diamonds or hearts have any influence whatsoever on the actual chances the spade finesse will win.

[wclass___]

inquiry, on 2011-July-28, 11:20, said:

So as far as your example with 48% chance after West shows up with the club TEN. Nonsense. The club ten is an unimportant card, it is any random card. It's being in WEST hand has no bearing whatsoever on the placement of the spade king. Only if you can get a count on spades by their following suit to the other three suits, does any card they play in clubs, diamonds or hearts have any influence whatsoever on the actual chances the spade finesse will win.

Did you read this?

Quote

Anyway, lets say West opened with 1♣ showing ♣T and nothing else, contract is 7NT by North, and you receive non-spade lead from East. Do you think that chances for successful finesse are now 48%? Do you think your chances after you play your non-spade top trick and learn that west really did have ♣T are still 48%? (Opponents keep ♥K and 3 other hearts.)

[Phil]

Ben, I don't think your analysis is correct.

The club 10 isn't random; its known. Its as if West's pass is conventional and "promises 1+ ♣". When West "shows us" the ♣T (versus simply shuffing his cards and pulling out the ♣T at random), this is a deterministic choice. Accordingly, hands with East holding five clubs gets thrown out of the (very large) pile of possible layouts of EW.

Furthermore, you are greatly mixing things up when you talk about 'dropping the card on the table' against 'telling us he has it'. The former is random; the latter isn't.

(edit - I see w_class said the same thing above).

By the way, I will refer to the math genii on this one, but I think this is the same as Month Hall / Bayesian where the 'actual' % of the ♠K being on is really 48/98, not 48/100, so the 'real %' is 48.97%.

Please correct me if this isn't right.

[semeai]

Phil, on 2011-July-28, 11:45, said:

By the way, I will refer to the math genii on this one, but I think this is the same as Month Hall / Bayesian where the 'actual' % of the ♠K being on is really 48/98, not 48/100, so the 'real %' is 48.97%.

Please correct me if this isn't right.

It used to be 13/26. Now it's 12/25 = 0.48. The denominator did change, as it does in Bayesian situations, just not as you're doing it here.

[inquiry]

I did miss the following statement

Quote

Anyway, lets say West opened with 1♣ showing ♣T and nothing else,

An odd agreement, no doubt. But let's accept that opening bid. In normal bridge, people make opening all the time that promse something, like 1NT promising 15-17. If you are missing 16 total hcp, and they hold only one ACE between them, who has the ACE? That points the way precisely. Your stipulation that an opening bid shows the club TEN will have a less dramatic, but definate effect I think.

Alright, so here we "know" that WEST hold the ♣Ten from BIDDING agreement they have. This is then similar to the situation on the problem hand with one HUGE exception. We can cash all our clubs and find out EXACTLY how many clubs east and west held. If West held only the club ten, the chance of the ♠K with WEST goes up (12/21). If West held all five clubs, the chances go way down (8/21). You can discover this stuff before trying the spade hook, but take the hook you will, so it matters little in the long run. During the bidding, even if I knew both NS hands 100% and I knew that WEST held the ♣Ten I would agree that the chances that WEST also held the spade king would slightly decreased. But that is immaterial, and as a calculation a waste of time.

The hand in the OP here is different. If EW did not bid, you would have no reason to play west for extra long diamonds. Nor consider anything other than try to drop the club jack in three rounds (3-3 clubs or Jx or J), as there is precious little time to try and discover what line to take. The bidding, however, marks WEST with a.) long diamonds, and b.) hopefully the heart king (or you don't make), and c.) Probably a spade honor or two. Here discovery is not available (well, you can "discover" if you have a chance to make by playing a heart towards the queen. The issue is how good is hooking East for the club jack versus the play you might have taken if WEST has passed.

I think the a prior chance WEST holds the spade king in your example (before any cards are played) is in fact 48% (not sure how phil get 49%) given your stipulation. I think playing a few rounds of clubs will give a different a posteriori calculation.

[Phil]

inquiry, on 2011-July-28, 23:07, said:

I think the a prior chance WEST holds the spade king in your example (before any cards are played) is in fact 48% (not sure how phil get 49%) given your stipulation. I think playing a few rounds of clubs will give a different a posteriori calculation.

I suddenly feel as though I have swam far away from the shore.

In the 7N problem, before we are given any information, the hook is 50/50. When West pulls the ♣10 from his hand, he is in effect saying, "the spade hook just became 48%, since I (West) have 12 of the remaining 25 and East has 13 of the 25.

Let's revisit the Monty Hall problem. Our a priori odds are 1/3. When we are given the choice of switching, our a posteriori chances of getting the door right are 2/3 (the remainder from our original choice), and that the 'switched' door inherits the % from the disclosed 'losing' door.

Here, the a priori of the spade hook is 50%. Nothing has changed for East. However, now west is 48% (12/25). What I thought I was doing correctly was that the relevant ratios are 48 and 50 (or 48/98) or 48.9%

But I guess it does. A posteriori trumps a priori and the whole equals the sum of its parts. East must inherit the %'s gained from West's accounted for vacant space.

Right?

[semeai]

Phil, on 2011-July-29, 10:35, said:

I suddenly feel as though I have swam far away from the shore.

In the 7N problem, before we are given any information, the hook is 50/50. When West pulls the ♣10 from his hand, he is in effect saying, "the spade hook just became 48%, since I (West) have 12 of the remaining 25 and East has 13 of the 25.

Let's revisit the Monty Hall problem. Our a priori odds are 1/3. When we are given the choice of switching, our a posteriori chances of getting the door right are 2/3 (the remainder from our original choice), and that the 'switched' door inherits the % from the disclosed 'losing' door.

Here, the a priori of the spade hook is 50%. Nothing has changed for East. However, now west is 48% (12/25). What I thought I was doing correctly was that the relevant ratios are 48 and 50 (or 48/98) or 48.9%

But I guess it does. A posteriori trumps a priori and the whole equals the sum of its parts. East must inherit the %'s gained from West's accounted for vacant space.

Right?

East's odds did go up, from 13/26 = .50 to 13/25 = .52 when West's odds went down from 13/26 = .50 to 12/25 = .48.

This is the same as the Monty Hall situation, or a variant rather. There's a goat, a sheep, and a car. You pick door #1, Monty has control of doors #2 and #3. By the vacant doors principle, your odds are 1/3 and his are 2/3 of having the car.

You ask him point blank "do you have the goat behind one of your doors" (just like him having to bid 1♣ if he has ♣10 and pass otherwise) and he says, fortunately, yes. Now your odds are 1/2 and his are 1/2 since the vacant doors are 1 to 1 now. His odds decreased from 2/3 to 1/2 and yours increased from 1/3 to 1/2.