jakob_r Posted October 25, 2008 Report Share Posted October 25, 2008 [hv=d=n&v=e&n=sqt8543h5daqjcak5&s=sa96haj87dk5cj843]133|200|Scoring: IMP[/hv] we play in 6♠ by N and lead is ♣6 to 8, Q and A. how do you treat trumps when the ♠7 appears on your play of the ♠3 towards dummy? partner and me wondering if restricted choice factors in? thanks :lol: Quote Link to comment Share on other sites More sharing options...
nige1 Posted October 25, 2008 Report Share Posted October 25, 2008 [hv=d=n&v=e&n=sqt8543h5daqjcak5&s=sa96haj87dk5cj843]133|200|Scoring: IMPwe play in 6♠ by N and lead is ♣6 to 8, Q and A. how do you treat trumps when the ♠7 appears on your play of the ♠3 towards dummy? partner and me wondering if restricted choice factors in?[/hv]I guess to lead ♠T and run it unless East shows out or covers. Unless West is playing a deep game there is no danger of a ♣ ruff. Quote Link to comment Share on other sites More sharing options...
fred Posted October 26, 2008 Report Share Posted October 26, 2008 I am not really thinking about the whole hand, but as far as the spade suit is concerned you should win the Ace. Finessing twice gains when RHO has KJx or KJxx. Ace and another gains when LHO has KJx or KJ. The KJx holdings cancel out but KJ on your left is more likely that KJxx on your right. If you consider the 3 and 7 to be equals (which seems reasonable to me) then I suppose that "restricted choice" applies here, but you don't need to use conventional restricted choice thinking in order to solve this problem (or any problem for that matter). For most suit combinations, if you just consider against which layouts the various plans will succeed or fail (like I did here), you can arrive at the right answer without having to get involved in thinking about restricted choice. This is just as well since most players find restricted choice as it is usually taught to be unintuitive and confusing. Fred GitelmanBridge Base Inc.www.bridgebase.com Quote Link to comment Share on other sites More sharing options...
nige1 Posted October 26, 2008 Report Share Posted October 26, 2008 Sorry :lol: Fred is right. Quote Link to comment Share on other sites More sharing options...
MFA Posted October 27, 2008 Report Share Posted October 27, 2008 The swindle play is to start with a small from the ace, planning to insert the T and if necessary finesse against east later.Against some opponents it loses only to the stiff J or KJxx in east, but as west gets good enough to duck smoothly with Kx, this plan becomes less attractive. In the actual hand, one has to worry about a club ruff as well, of course. Quote Link to comment Share on other sites More sharing options...
effervesce Posted November 10, 2008 Report Share Posted November 10, 2008 <!-- NORTHSOUTH begin --><table border=1> <tr> <td> <table> <tr> <td>Dealer:</td> <td> North </td> </tr> <tr> <td>Vul:</td> <td> E/W </td> </tr> <tr> <td>Scoring:</td> <td> IMP </td> </tr> </table> </td> <td> <table border='1'> <tr> <th> <table> <tr> <th class='spades'>♠</th> <td> QT8543 </td> </tr> <tr> <th class='hearts'>♥</th> <td> 5 </td> </tr> <tr> <th class='diamonds'>♦</th> <td> AQJ </td> </tr> <tr> <th class='clubs'>♣</th> <td> AK5 </td> </tr> </table> </th> </tr> <tr> <th> <table> <tr> <th class='spades'>♠</th> <td> A96 </td> </tr> <tr> <th class='hearts'>♥</th> <td> AJ87 </td> </tr> <tr> <th class='diamonds'>♦</th> <td> K5 </td> </tr> <tr> <th class='clubs'>♣</th> <td> J843 </td> </tr> </table> </th> </tr> </table> </td> <td> </td> </tr> </table><!-- NORTHSOUTH end --> we play in 6♠ by N and lead is ♣6 to 8, Q and A. how do you treat trumps when the ♠7 appears on your play of the ♠3 towards dummy? partner and me wondering if restricted choice factors in? thanks :unsure: Looks like we want to play the spade suit for 1 loser. Outstanding cards are KJ72. The possibilities: - KJ72 ----- double hook2 KJ7 ----- double hookJ K72 ----- doublehook or A and anotherK J72 ----- doublehook or A and anotherJ2 K7 ----- anythingK2 J7 ----- anythingKJ 72 ----- A and anotherKJ2 7 ----- A and another The possible lines:DoublehookA and another Looks like double hook and A and another are equivalent for making the contract. Quote Link to comment Share on other sites More sharing options...
effervesce Posted November 10, 2008 Report Share Posted November 10, 2008 I am not really thinking about the whole hand, but as far as the spade suit is concerned you should win the Ace. Finessing twice gains when RHO has KJx or KJxx. Ace and another gains when LHO has KJx or KJ. The KJx holdings cancel out but KJ on your left is more likely that KJxx on your right. If you consider the 3 and 7 to be equals (which seems reasonable to me) then I suppose that "restricted choice" applies here, but you don't need to use conventional restricted choice thinking in order to solve this problem (or any problem for that matter). For most suit combinations, if you just consider against which layouts the various plans will succeed or fail (like I did here), you can arrive at the right answer without having to get involved in thinking about restricted choice. This is just as well since most players find restricted choice as it is usually taught to be unintuitive and confusing. Fred GitelmanBridge Base Inc.www.bridgebase.com Why is KJ with LHO more likely than KJxx with RHO? Both have exactly one layout. Quote Link to comment Share on other sites More sharing options...
655321 Posted November 10, 2008 Report Share Posted November 10, 2008 Why is KJ with LHO more likely than KJxx with RHO? Both have exactly one layout.Kelsey and Glauert in Bridge Odds for Practical players (a highly recommended book) explain about vacant places: KJ opposite 72:The King has 13/26 chances of being dealt to West, then the Jack has 12/25 chances, then the 7 has 13/24 chances of going to East, then the 2 has 12/23 chances of going to East. i.e. KJ doubleton with West (like all 6 of the 2-2 breaks) has a 6.78% chance. KJ72 with East is(13/26) * (12/25) *(11/24) * (10/23) = 4.78%. Incidentally, even though I agree with Fred that restricted choice is not the way to think about this combination, if you are going to list the cases without the holdings where East would have played the 2 (as you did in your earlier post), then you do need to halve the probabilities of the cases where East has the 2 and the 7, and could have played either card on the first round. Quote Link to comment Share on other sites More sharing options...
CSGibson Posted November 10, 2008 Report Share Posted November 10, 2008 I am not really thinking about the whole hand, but as far as the spade suit is concerned you should win the Ace. Finessing twice gains when RHO has KJx or KJxx. Ace and another gains when LHO has KJx or KJ. The KJx holdings cancel out but KJ on your left is more likely that KJxx on your right. If you consider the 3 and 7 to be equals (which seems reasonable to me) then I suppose that "restricted choice" applies here, but you don't need to use conventional restricted choice thinking in order to solve this problem (or any problem for that matter). For most suit combinations, if you just consider against which layouts the various plans will succeed or fail (like I did here), you can arrive at the right answer without having to get involved in thinking about restricted choice. This is just as well since most players find restricted choice as it is usually taught to be unintuitive and confusing. Fred GitelmanBridge Base Inc.www.bridgebase.com Why is KJ with LHO more likely than KJxx with RHO? Both have exactly one layout. Because presumably RHO has shown up with club length, making a few more vacant spaces available for LHO to have spade length. Just a guess, but it feels right. Quote Link to comment Share on other sites More sharing options...
pretzalz Posted November 10, 2008 Report Share Posted November 10, 2008 I am not really thinking about the whole hand, but as far as the spade suit is concerned you should win the Ace. Finessing twice gains when RHO has KJx or KJxx. Ace and another gains when LHO has KJx or KJ. The KJx holdings cancel out but KJ on your left is more likely that KJxx on your right. If you consider the 3 and 7 to be equals (which seems reasonable to me) then I suppose that "restricted choice" applies here, but you don't need to use conventional restricted choice thinking in order to solve this problem (or any problem for that matter). For most suit combinations, if you just consider against which layouts the various plans will succeed or fail (like I did here), you can arrive at the right answer without having to get involved in thinking about restricted choice. This is just as well since most players find restricted choice as it is usually taught to be unintuitive and confusing. Fred GitelmanBridge Base Inc.www.bridgebase.com Why is KJ with LHO more likely than KJxx with RHO? Both have exactly one layout. Because presumably RHO has shown up with club length, making a few more vacant spaces available for LHO to have spade length. Just a guess, but it feels right.KJ is more likely than KJxx because 22C11 is a bigger number than 22C9. Quote Link to comment Share on other sites More sharing options...
fred Posted November 10, 2008 Report Share Posted November 10, 2008 Why is KJ with LHO more likely than KJxx with RHO? Both have exactly one layout.There is an easy way to know this without having to do any post-kindergarten math: The more evenly a given case is divided, the more likely it is. You can use this truism only to compare cases in which the total number of cards are equal. So any given 2-2 case is more likely than any given 3-1 case which is more likely than any given 4-0 case, but you can't compare a 2-2 case and a 3-2 case using this rule (because 2+2=4 and 3+2=5). If you are thinking "how can this be true given that a 3-1 break (about 50%) is more likely than a 2-2 break (about 40%)?", the math required to show there is no contridiction is slightly more complicated: There are 8 3-1 cases but only 6 2-2 cases. So the odds of any given 3-1 case is 50% divided by 8 (about 6.2%) while the odds of any given 2-2 case is 40% divided by 6 (about 6.7%). Therefore, any given 2-2 case is more likely than any given 3-1 case. Back to the original question, any given 4-0 case happens about 5% of the time (about 10% divided by 2). So KJxx (5%) is less likely than KJ (6.7%). But if you don't care about the exact percentages (you don't need to in order to solve this suit combination problem), just use the rule above. Fred GitelmanBridge Base Inc.www.bridgebase.com Quote Link to comment Share on other sites More sharing options...
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