[jakob_r]

Scoring: IMP

---

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

[nige1]

jakob_r, on Oct 25 2008, 05:48 PM, said:

Scoring: IMP

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?

---

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.

[fred]

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 Gitelman
Bridge Base Inc.
www.bridgebase.com

[nige1]

Sorry Fred is right.

[MFA]

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.

[effervesce]

jakob_r, on Oct 25 2008, 05:48 PM, said:

<!-- 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

Looks like we want to play the spade suit for 1 loser.

Outstanding cards are KJ72. The possibilities:

- KJ72 ----- double hook
2 KJ7 ----- double hook
J K72 ----- doublehook or A and another
K J72 ----- doublehook or A and another
J2 K7 ----- anything
K2 J7 ----- anything
KJ 72 ----- A and another
KJ2 7 ----- A and another

The possible lines:
Doublehook
A and another

Looks like double hook and A and another are equivalent for making the contract.

[effervesce]

fred, on Oct 25 2008, 07:41 PM, said:

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 Gitelman
Bridge Base Inc.
www.bridgebase.com

Why is KJ with LHO more likely than KJxx with RHO? Both have exactly one layout.

[655321]

effervesce, on Nov 9 2008, 09:58 PM, said:

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.

[CSGibson]

effervesce, on Nov 9 2008, 07:58 PM, said:

fred, on Oct 25 2008, 07:41 PM, said:

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 Gitelman
Bridge 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.

[pretzalz]

CSGibson, on Nov 10 2008, 01:40 AM, said:

effervesce, on Nov 9 2008, 07:58 PM, said:

fred, on Oct 25 2008, 07:41 PM, said:

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 Gitelman
Bridge 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.

[fred]

effervesce, on Nov 10 2008, 02:58 AM, said:

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 Gitelman
Bridge Base Inc.
www.bridgebase.com