han Posted August 13, 2007 Report Share Posted August 13, 2007 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? Quote Link to comment Share on other sites More sharing options...
bid_em_up Posted August 13, 2007 Report Share Posted August 13, 2007 Now I remember why I usually don't attempt to calculate percentages. :( Quote Link to comment Share on other sites More sharing options...
Trumpace Posted August 13, 2007 Report Share Posted August 13, 2007 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? Let me venture a guess: LHO has at 1 heart and 1 spade (11 vacant spaces). Rho has 4 hearts and 1 spade (8 vacant spaces). So, chances that LHO holds a specific card is 11/19 = (11/(11+8)) Quote Link to comment Share on other sites More sharing options...
ochinko Posted August 13, 2007 Report Share Posted August 13, 2007 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? Frances is quite right. It depends on how many cards you assume known in the defendants. With 4:6 which is minimum in order to know the distribution in hearts and spades it's 56.2%. With 5:7 it gets 57.1%, etc. I have my own calculator, but there are many. Quote Link to comment Share on other sites More sharing options...
han Posted August 13, 2007 Report Share Posted August 13, 2007 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? Let me venture a guess: LHO has at 1 heart and 1 spade (11 vacant spaces). Rho has 4 hearts and 1 spade (8 vacant spaces). So, chances that LHO holds a specific card is 11/19 = (11/(11+8)) First of all, there are 6 hearts out there, not 5. Secondly, Frances said 4+, not 4. Quote Link to comment Share on other sites More sharing options...
ochinko Posted August 13, 2007 Report Share Posted August 13, 2007 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? Let me venture a guess: LHO has at 1 heart and 1 spade (11 vacant spaces). Rho has 4 hearts and 1 spade (8 vacant spaces). So, chances that LHO holds a specific card is 11/19 = (11/(11+8)) Close. You have to draw one spade in order to see that they break 1:1, but you have to draw hearts three times in order to see that they are 4:2. So now you know that the one with 2 hearts has 9 slots to put ♣K, the other one - 7. Oops, my bad. The one with 4 hearts has 8 slots, not 7, because there's only one more card that is known, which makes it 52.9% to find the King in the guy with fewer hearts. Edit: I am not able to calculate how much the chance for finding both Kings in the hand with fewer hearts increases, but bid_em_up definitely has a point so maybe we should always go for the squeeze as being more attractive than a finesse. :( Quote Link to comment Share on other sites More sharing options...
Trumpace Posted August 13, 2007 Report Share Posted August 13, 2007 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? Let me venture a guess: LHO has at 1 heart and 1 spade (11 vacant spaces). Rho has 4 hearts and 1 spade (8 vacant spaces). So, chances that LHO holds a specific card is 11/19 = (11/(11+8)) First of all, there are 6 hearts out there, not 5. Secondly, Frances said 4+, not 4. I never said there were only 5 hearts. The information we know is: LHO has a heart. RHO has 4 or more (assumed). Spades split 1-1. So, LHO has 11 vacant spaces and RHO has 8. Chances that LHO holds the 6th heart is ~57% (assuming he is capable of leading singleton J etc), same as the chances that he has the CK, same as the chances that he has DK etc. This, is assuming LHO will lead a card at random from his 13 cards i.e we assume that the lead was a heart gives us no other information except the fact that LHO has at least one heart. Quote Link to comment Share on other sites More sharing options...
ralph23 Posted August 13, 2007 Author Report Share Posted August 13, 2007 ♣♦♥♠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? Quote Link to comment Share on other sites More sharing options...
han Posted August 13, 2007 Report Share Posted August 13, 2007 What's the inverse of 38.9%? Quote Link to comment Share on other sites More sharing options...
ralph23 Posted August 13, 2007 Author Report Share Posted August 13, 2007 What's the inverse of 38.9%? 61.1 ?? I.e. 61.1 + 38.9 = 100. They are inverses of each other wrt 100. Or maybe complement is the right term. Quote Link to comment Share on other sites More sharing options...
ochinko Posted August 14, 2007 Report Share Posted August 14, 2007 ♣♦♥♠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%. Quote Link to comment Share on other sites More sharing options...
ralph23 Posted August 14, 2007 Author Report Share Posted August 14, 2007 ♣♦♥♠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. Quote Link to comment Share on other sites More sharing options...
ochinko Posted August 14, 2007 Report Share Posted August 14, 2007 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. Quote Link to comment Share on other sites More sharing options...
Finch Posted August 14, 2007 Report Share Posted August 14, 2007 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) Quote Link to comment Share on other sites More sharing options...
Finch Posted August 14, 2007 Report Share Posted August 14, 2007 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%) Quote Link to comment Share on other sites More sharing options...
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