[ochinko]

ralph23, on Aug 14 2007, 01:11 AM, said:

♣♦♥♠
Let's go s..l..o..w....e.....r......

1. Assumption (or Hypothesis) 1: West has 6♥; well, in this case we are always going down, so let's ignore this one.

2. Assumption 2: West has 5♥, and East has 1♥.

In this assumed or hypothesized case #2, West has 6 cards known to be in the majors, and East has 2 known to be in the majors.

Therefore, of the available 18 minor card slots, West has 7 of those and East has 11 of those.

Therefore, if Assumption 2 is true, then West has 7 chances out of 18 to "draw" any minor suit card, and East has 11 chances out of 18.

So, if Assumption 2 is true, then the odds of East having any particular minor card -- whether the King of ♣ or the 6 of ♦ -- is 11/18 (or 61.1%). West has the inverse of 38.9%.

Anyone disagree with this so far?

I do. What lawful method would allow you to know 4 cards more in West's hand if spades are 1:1, and hearts are 5:1?

In the absence of opponents' bids you can only arrive at that conclusion if you draw spades once, and hearts 2 times. By that time you'll know 3 cards from East's hand, and 3 from West's. Because you know that the remaining 3 hearts are in West there are 3 slots less in his hand for putting any unknown card, not 4 as you say. So the odds are not 11/18 to 7/18 but 10/17 to 7/17 or 58.8% to 41.2%.

[ralph23]

ochinko, on Aug 14 2007, 01:53 AM, said:

ralph23, on Aug 14 2007, 01:11 AM, said:

♣♦♥♠
Let's go s..l..o..w....e.....r......

1.  Assumption (or Hypothesis) 1: West has 6♥; well, in this case we are always going down, so let's ignore this one.

2.  Assumption 2: West has 5♥, and East has 1♥.

In this assumed or hypothesized case #2, West has 6 cards known to be in the majors, and East has 2 known to be in the majors.

Therefore, of the available 18 minor card slots, West has 7 of those and East has 11 of those.

Therefore, if Assumption 2 is true, then West has 7 chances out of 18 to "draw" any minor suit card, and East has 11 chances out of 18.

So, if Assumption 2 is true, then the odds of East having any particular minor card -- whether the King of ♣ or the 6 of ♦ -- is 11/18 (or 61.1%). West has the inverse of 38.9%.

Anyone disagree with this so far?

I do. What lawful method would allow you to know 4 cards more in West's hand if spades are 1:1, and hearts are 5:1?

In the absence of opponents' bids you can only arrive at that conclusion if you draw spades once, and hearts 2 times. By that time you'll know 3 cards from East's hand, and 3 from West's. Because you know that the remaining 3 hearts are in West there are 3 slots less in his hand for putting any unknown card, not 4 as you say. So the odds are not 11/18 to 7/18 but 10/17 to 7/17 or 58.8% to 41.2%.

Well, do you agree at least that EW share 18 minor suit cards ???

Proof: NS have 8 minor suit cards. There are 26 minor suit cards in all. 26-8 = 18.

[ochinko]

ralph23, on Aug 14 2007, 04:12 PM, said:

Well, do you agree at least that EW share 18 minor suit cards ???

Proof: NS have 8 minor suit cards. There are 26 minor suit cards in all. 26-8 = 18.

That was true when we saw the dummy but not anymore, because if East started with only one heart, on the second heart he had to discard a minor suit card, as we drew his only spade.

So if we know that West had 5 hearts, and East had 1, this is only because we saw East discarding a minor suit card after which defenders have 17 between them.

I am not nitpicking, I just focus on the a posteriori probabilities as I find them to have more practical value.

[FrancesHinden]

Hannie, on Aug 13 2007, 09:18 PM, said:

FrancesHinden, on Aug 13 2007, 02:10 PM, said:

If East has 4+ hearts, then the chance of West having the CK is 57%, once trumps are 1-1.

How did you get this Frances?

When I check it I also got 57%, but I thought it was a lot of work, using Pavlicek's dual calculator and then adding the percentages and even dividing them. Did you use a clever trick?

It took me a bit of work as well. Probably the same about of work.

57% was the sum of

% chance when hearts are 2-4 * chance that hearts are 2-4 given trumps are 1-1
+
% chance when hearts are 1-5 * chance that hearts are 1-5 given trumps are 1-1
+
% chance when when hearts are 0-6 * chance that hearts are 0-6 given trumps are 1-1

divided by

(hearts 2-4 given trumps 1-1 + hearts 1-5 given trumps 1-1 + hearts 0-6 given trumps 1-1)

(I have a little spreadsheet that does much the same as Pavlicek's calculator)

[FrancesHinden]

ralph23, on Aug 13 2007, 08:19 PM, said:

FrancesHinden, on Aug 13 2007, 02:10 PM, said:

ochinko, on Aug 13 2007, 01:19 PM, said:

Second case: East has longer hearts. Now if one of the defenders has both Kings he is squeezed, and when we ruff our last heart all we need to do is to cash both Aces. (50% total)

If East has 4+ hearts, then the chance of West having the CK is 57%, once trumps are 1-1.

If East was dealt precisely 4♥, and the probability of West having the King of ♣ is X% in that case (where X > 50), then is the probability of West having the King of ♣ even higher when East was dealt precisely 5♥?

Same reasoning of course for a 6-0 split, except those get special treatment because West led a ♥.....

Yes. 57% is the answer for 4+ hearts, not for 4 hearts (for exactly 4 hearts it is closer to 55%)