Calculating the odds... A question

[inquiry]

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The lead is Q♦.

This is one of the hands deleted from a recent thread. The point of the hand when it was posted was that the declarer correctly guessed how to play clubs. In that thread, I pointed out (correctly I hope) the correct line of play based on percentages. A few assumptions about the hand...

 

West has five diamonds at a minimum, and likely six.

West has to have the heart king

West probably has a spade honor or two for his bid as well

[/hv]

 

The question is, in calculating vacant space, can you include in the subset of cards WEST holds the heart king (which has to be there) along with the diamonds? That doesn't seem too much of a stretch. How about spade honors? Would you factor the heart king with west along with the diamonds in calculating the best line (ignoring heart king does not affect the correct line in this example.

[semeai]

The question is, in calculating vacant space, can you include in the subset of cards WEST holds the heart king (which has to be there) along with the diamonds? That doesn't seem too much of a stretch.

 

Yes, include ♥K. Think of it this way: you're enumerating hands which are consistent with the bidding and allow you to make the contract. All such enumerations will have (at least) 5 diamonds and the heart King with West.

 

In fact, the heart King is easier to deal with than the diamonds for the vacant space calculation. It's a unique card. In the diamond suit, the fact that West has QJ and at least 3 lower cards is slightly harder, as it could instead be QJ and 4 lower cards, etc. The proper way to do things would be to find the relative probabilities of 5 diamonds to the QJ with West and three with East, 6 diamonds to the QJ with West and two with East, etc, and then do your vacant space calculation for each (including ♥K with West) and weight by the relative probabilities.

 

How about spade honors?

 

If you think for the hand to be consistent with the bidding West must have at least one spade honor, there should be a correction to the vacant space calculation. As with diamonds above, however, you can't just put one spade with West. Instead, you compute the relative probability of a 1-1 split (of which there are two) and a 2-0 split (given the diamonds and heart King!) and do your vacant space calculations for each case, and then weight the answers by the relative probabilities.

 

To be completely proper, you'd also want to exclude hands with a 5 card major from West, as well as anything else that would change the bidding or play up to this point. This shouldn't have much of an effect on this hand, but such considerations are the key point when vacant space computations are flawed (e.g. when someone has led a suit and you use that suit for vacant space purposes, you're forgetting to exclude some percentage of the cases in which they have a longer or equal length side suit, and it does end up mattering).

[wclass___]

The question is, in calculating vacant space, can you include in the subset of cards WEST holds the heart king (which has to be there) along with the diamonds?

No. Correct procedure would be to eliminate the impossible holdings (void with LHO) from calculated odds using given vacant spaces, but not by adding one more vacant space.

[semeai]

No. Correct procedure would be to eliminate the impossible holdings from calculated odds using given vacant spaces, but not by adding [subtracting] one more vacant space.

 

This is equivalent in this case with the ♥K

 

No. Correct procedure would be to eliminate the impossible holdings (void with LHO) from calculated odds using given vacant spaces, but not by adding one more vacant space.

 

No, eliminating void with LHO doesn't go far enough. You also need to eliminate ♥1072 with West and ♥KJ54 with East, and many others. Eliminating all such holdings is equivalent to just placing ♥K with West and moving on. It's a well-defined single card.

[wclass___]

http://www.rpbridge.net/7z75.htm

[semeai]

http://www.rpbridge.net/7z75.htm

 

Not the same. A low card is one of many. The relevant passage you seem to be referring to:

 

As further evidence that an opponent following suit is volunteered information, suppose there are only two cards left when you play this simple holding.

 

♥AQ

♥32

 

Also assume that you previously learned that each opponent began with four hearts, and only hearts remain. Obviously, your chance of the finesse winning is 50 percent. Now suppose you lead the ♥2 and West plays a low heart. Does this affect your chances? One might conjecture that, after West plays a low heart, he has only one unknown card left; while East has two unknown cards. Therefore, the missing king is more likely to be with East by 2-to-1 odds. Surely, this is nonsense because West's play was volunteered. The low heart was his choice, not yours. Since West would always play his low heart from ♥Kx the information is meaningless, and your chances are still 50 percent. In other words, percentages do not change when the only new information is an opponent following suit.

[wclass___]

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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.

[semeai]

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7NT with a non spade lead, you know that LHO has say ♣T. Does your chances for finesse really worsens?

 

How do I know this? Did he fumble his cards and drop them on the ground and one flipped face up and it was ♣10? In this case, yes my odds for the finesse are now 12/25.

 

Did he lead ♣10? In this case no. The information is (almost entirely)* irrelevant, cf the Monty Hall Problem. He led one of many irrelevant cards which he is happy to show me (the ♠K he is not going to show me).

 

The actual situation is different. We can must (to make the contract) place the ♥K in West's hand and it was not irrelevantly volunteered. The probabilities for different heart distributions are different now because the heart suit is now effectively 12 cards.

 

A simple thought experiment:

In a three card ending, you need clubs to be 2-2. There are 4 clubs out, and the ♠AK. Left hand opponent's bid places him with both ♠AK. Is this irrelevant?

 

Added: This example is not so related. I'll give a better one below.

 

A slightly less simple thought experiment which is a bit more related:

In a five card ending, we're missing ♣Q1098 and ♠KJ10987. Dummy has ♠AQ ♣432 and declarer has ♠32 ♣AKJ. We need all the tricks, and pitched ♣5-7 and ♠6 from dummy on the run of our diamonds and hearts. We play club A, spade to Q, spade A (drops K, mandatory falsecard possible). Now we play a club. All have followed throughout. What is the probability we'll succeed on the drop in clubs? On the hook in clubs?

 

* Leading the ♣10 suggests certain club holdings, and also suggests certain holdings that are attractive to lead from were not as likely to exist in other suits, so this is not exactly correct.

[semeai]

Better thought experiment:

 

♠32 ♣AQ

 

♠AQ ♣32

 

We're missing ♠KJ109 and ♣KJ and ♦32, which were the opponents' only original spades and clubs were the only cards the opponents held that were not unintentionally faced and turned into penalty cards earlier (i.e. they don't matter for any calculations), and we know from the bidding that the diamond distribution is split (i.e. 1-1 now). We need the last 4 tricks and are in hand. We play a club toward the Q and LHO follows with the J. What do we do? (Playing BAM in 7NXX)

 

What about missing ♠KJ10 ♣KJ ♦432, knowing diamonds are 1-2?

[wclass___]

How do I know this? Did he fumble his cards and drop them on the ground and one flipped face up and it was ♣10? In this case, yes my odds for the finesse are now 12/25.

No. After you receive lead, you ask LHO to name one card from his hand and he names ♣T, so you can tell partner afterwards that he shouldn't bid 48% slams. :D

[wclass___]

Better thought experiment:

 

♠32 ♣AQ

 

♠AQ ♣32

 

We're missing ♠KJ109 and ♣KJ and ♦32, which were the opponents' only original spades and clubs, and we know from the bidding that the diamond distribution is split (i.e. 1-1 now). We need the last 4 tricks and are in hand. We play a club toward the Q and LHO follows with the J. What do we do? (Playing BAM in 7NXX)

 

What about missing ♠KJ10 ♣KJ ♦432, knowing diamonds are 1-2?

 

 

♣J gives us useful information that suit doesn't split 0-2, but seriously what does this have to do with this topic where we discuss how knowing one exact card in advance changes odds? If you meant that we now that LHO has ♣J in advance than by ''eliminating impossible holdings'' we arrive to the same solution.

[semeai]

♣J gives us useful information that suit doesn't split 0-2, but seriously what does this have to do with this topic where we discuss how knowing one exact card in advance changes odds? If you meant that we now that LHO has ♣J in advance than by ''eliminating impossible holdings'' we arrive to the same solution.

 

No, the point is you have to place ♠K with RHO to make, so that does alter the vacant space calculations. This is exactly the same as placing the ♥K in the original hand.

[benlessard]

Another way to look at it is that the more Hearts West has the less clubs he will have and vice versa. The more Hearts hes got the better the chances of him having the K of H. So when hes got the K of H he is more likely to be short in clubs.

 

This is not useful for getting exact % but its a quick shortcut for those who are unsure about how vacant spaces worked.

[gnasher]

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.

[hotShot]

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The lead is Q♦.

This is one of the hands deleted from a recent thread. The point of the hand when it was posted was that the declarer correctly guessed how to play clubs. In that thread, I pointed out (correctly I hope) the correct line of play based on percentages. A few assumptions about the hand...

 

West has five diamonds at a minimum, and likely six.

West has to have the heart king

West probably has a spade honor or two for his bid as well

[/hv]

 

Shouldn't you include into your assumptions that West is unlikely to have single kings in ♥ or ♠?

Leading ♦Q suggest the hope to have 2 entries to use the developed ♦ suit.

So West would need 2 out of:

♣Jxxx, ♥Kx, ♥Kxx, ♠Kx, ♠Kxx or ♠QJx

Assuming at least 2 cards in the majors 2-2-5-4 is the only shape that allows 4thJ♣, if it's likely 6♦ that shape is impossible.

[gnasher]

No. After you receive lead, you ask LHO to name one card from his hand and he names ♣T, so you can tell partner afterwards that he shouldn't bid 48% slams. :D

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%?

[wclass___]

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.

Suppose you win ♦ lead and play ♥ to Q, LHO winning and RHO playing some small card. Additional information that you will have learnt is that RHO isn't void, but some doubletons and tribletons are now impossible, thus changeing odds. hmm?

[whereagles]

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").

 

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.

[gnasher]

Suppose you win ♦ lead and play ♥ to Q, LHO winning and RHO playing some small card. Additional information that you will have learnt is that RHO isn't void, but some doubletons and tribletons are now impossible, thus changeing odds. hmm?

 

Which doubletons and tripletons are now impossible? All the small hearts are equivalent, so all RHO's combinations of small cards are still possible. (Or, if you prefer, you can do a restricted-choice calculation to reach the same result.)

[vianu2]

The point of the hand when it was posted was that the declarer correctly guessed how to play clubs.

My point there was not only that the declarer correctly guessed clubs but the tempo he used. I still don't know if a good declarer, guessing 4♣ to E , will take the ace ♦ and finesse ♣ at the second tric.

I suggested another line, winning the ♦ in your hand, cashing ace♣, returning a ♥ to the dummy and then finesse the ♣ next time when u get to the dummy.

Does it make any difference? More, i would play ace♥ then ♥ for the simple reason of a stiff k♥in the west's hand.

In addition,if W ducks the heart i could get new info's.

[gnasher]

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.

 

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___]

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___]

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___]

- 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?