YAY

[Trumpace]

[hv=n=skxx&w=s&e=s&s=sjxxxx]399|300|Trump suit.[/hv]

 

You have to play this trump suit for maximum two losers.

[bid_em_up]

Does either hand have either the 10 or the 9?

[Trumpace]

Does either hand have either the 10 or the 9?

Unfortunately, no.

 

8 and 7 were there though. Something like

 

K8x

J7xxx

 

if that changes things.

[ralph23]

I am going to assume that the trumps are split 3-2. If they aren't, I think I'm in a world of hurt. So I'm ignoring that possibility.

 

So, on this assumption: Another way of stating this problem is, that I must not allow the defense to capture both my King and my Jack (with their Ace and Queen, respectively). Or, alternatively, I must win a trick with either the King or the Jack.

 

Assuming the 3-2 split, I can always do this if either (1) the Ace is with West or (2) the Queen is with East. A priori, that combination will occur 75% of the time, and the only time I lose is finding East with the Ace, and West with the Queen (1 chance out of 4 for this combination).

 

Can I do better than 75% by making some silly looking play like, low from each hand on the first round of trumps? Or, looked at another way, do I have to lose, just because West unfortunately has the Queen and East unfortunately has the Ace?

 

I think I can do better.

 

Working on the assumption that they are 3-2, let's play low from each hand on the first round and someone will win with e.g. the 9.

 

Now they fool around in another suit and eventually South gets the lead again.

 

Now low toward the King. If West has Queen doubleton initially, and East the Ace, then I win on this combo as I cover the Queen and East must take his Ace. Now my Jack is good. (I would have lost on this layout if I'd started with lo-to-the-King).

 

What do I give up to try this gambit? Well, if East started with AQx, then my simplistic plan of low-to-the-King, then later low-to-the-Jack would have worked. But now it won't.

 

So I think I have a trade-off: Which is more likely? West with Qx, or East with AQx? I think Qx is more likely.

 

An interesting problem, indeed, Trumpace!!

[Stephen Tu]

Having the 8 & 7 changes things. Think about what the right play is if LHO plays T or 9 first round.

[ralph23]

Having the 8 & 7 changes things. Think about what the right play is if LHO plays T or 9 first round.

I think it just increases the odds that lefty started with either AT/9 dub, or QT/9 dub. Or maybe he's just falsecarding with Tx or 9x.

 

If he's not falsecarding, then {AT, A9, QT or Q9} is more likely that T9 tight. So I'm still playing the same way, low from both NS on trick one.

 

NB - If he has T9x tripleton and got cute, then I'm always going to succeed, so that case is ignored.

[Guest Jlall]

Having the 8 & 7 changes things.  Think about what the right play is if LHO plays T or 9 first round.

I think it just increases the odds that lefty started with either AT/9 dub, or QT/9 dub. Or maybe he's just falsecarding with Tx or 9x.

 

If he's not falsecarding, then {AT, A9, QT or Q9} is more likely that T9 tight. So I'm still playing the same way, low from both NS on trick one.

 

NB - If he has T9x tripleton and got cute, then I'm always going to succeed, so that case is ignored.

what about...stiff ten or stiff 9...

[ralph23]

Having the 8 & 7 changes things.  Think about what the right play is if LHO plays T or 9 first round.

I think it just increases the odds that lefty started with either AT/9 dub, or QT/9 dub. Or maybe he's just falsecarding with Tx or 9x.

 

If he's not falsecarding, then {AT, A9, QT or Q9} is more likely that T9 tight. So I'm still playing the same way, low from both NS on trick one.

 

NB - If he has T9x tripleton and got cute, then I'm always going to succeed, so that case is ignored.

what about...stiff ten or stiff 9...

If it's a singleton: Then they've split 4-1 and my hypothesis was, they split 3-2 or I'm sunk.

 

I've not considered chances of success with 4-1 splits.

[han]

Which is more likely? West with Qx, or East with AQx? I think Qx is more likely.

I think that they are equally likely.

[Finch]

Which is more likely? West with Qx, or East with AQx? I think Qx is more likely.

I think that they are equally likely.

Yes.

The way to see this is to count the 'x's. Each of those holdings - Qx and AQx has exactly one 'x', so there will be the same number of ways of holding each of them (3, in this case). And as both of them assume a 3-2 break, each individual holding is equally likely.

[vuroth]

Which is more likely? West with Qx, or East with AQx? I think Qx is more likely.

I think that they are equally likely.

Yes.

The way to see this is to count the 'x's. Each of those holdings - Qx and AQx has exactly one 'x', so there will be the same number of ways of holding each of them (3, in this case). And as both of them assume a 3-2 break, each individual holding is equally likely.

Brilliant!

[ralph23]

OK, I see it now... it's interesting. I have a slightly different way of coming to the conclusion:

 

1. The big honors can be divided 4 ways.

2. Q with West (and A with East) is just as likely as AQ with East.

3. We're only considering 3-2 splits, so the candidates are:

a Qx Axx

b Qxx Ax

c xxx AQ

d xx AQx

 

b and c aren't being considered; so a and d are equally likely.

 

SO, if they split 3-2, there's no advantage to ducking a low one in both hands first versus just playing the "normal" way.

 

If you also take 4-1 into account, then what if West plays the 9 or T (we'll just call either of them the "N", for ease of presentation), and disregard falsecarding, then we compare the likelihood of

 

N ... AQNx

 

versus

 

NN ... AQx.

 

A. Odds of 4-1 split: 28.26%

Odds of lefty holding the stiff: 14.13%

Odds of it being an N (40% of above): 5.66%

 

B. Odds of a 3-2 split: 67.83%

Odds of lefty having the doubleton: 33.91%

Odds of his having specifically NN (10 percent of above): 3.39%

 

Restricted Choice wins again. If N appears from Lefty, rise with the King, then play to finesse Righty out of his N.

[Stephen Tu]

a Qx Axx

d xx AQx

 

so a and d are equally likely.

 

You are still missing something.

QT A9x

Q9 ATx

 

are different from

Qx AT9

 

T9 AQx

is different from

Tx AQ9 & 9x AQT.

 

Restricted choice is applicable in some of these cases also, not just the 4-1 case.

[ralph23]

a Qx Axx

d xx AQx

 

so a and d are equally likely.

 

You are still missing something.

QT A9x

Q9 ATx

 

are different from

Qx AT9

 

T9 AQx

is different from

Tx AQ9 & 9x AQT.

 

Restricted choice is applicable in some of these cases also, not just the 4-1 case.

The question posed was, how do you play the suit?

 

Answer given was low to the King...

 

If that loses, then next time, low to the Jack.

 

Exception: If lefty plays 9 (or ten) on the first round, play the King and then plan to finesse righty out of his ten (or 9).

 

A. Would you play it differently??

 

B. Or if not --- are you asking, what if the exception doesn't apply (i.e. lefty played a low trump), but next round, righty plays the ten (or 9)?

[Stephen Tu]

Your final conclusion is right, but your analysis of the winning/losing cases seemed incomplete. Before you were saying "lose with Qx" on left, win with "AQx" on right, and not enumerating cases where LHO might follow with T or 9 first round.

 

In reality, you are supposed to win with QT, Q9, Tx, 9x, (as well as cases where LHO has A), but lose to Qx, T9 on your left. And you should also win with stiff T/9 on the left.

 

Of these 6 3-2 cases, the low to K line picks up 4, assuming a hook on the way back if the T/9 popped. Your original suggestion of duck completely 1st trick no matter whether T, 9, x appears only picks up 3 of them, as well as also blowing the 1-4 split cases.

[ralph23]

>>>Your final conclusion is right, but your analysis of the winning/losing cases seemed incomplete. Before you were saying "lose with Qx" on left, win with "AQx" on right, and not enumerating cases where LHO might follow with T or 9 first round.

 

Well, and then if you read the latter posting (August 7, about 1.00 PM), and not the earlier posting (August 6, about 2.00 PM), you will see that I did in fact examine specifically -- in the latter posting -- the case of LHO following with "N".

 

I even made up the symbol "N" for this purpose.

 

>>>In reality, you are supposed to win with QT, Q9, Tx, 9x, (as well as cases where LHO has A), but lose to Qx, T9 on your left. And you should also win with stiff T/9 on the left.

 

Yes, true. My stated strategy in the latter post follows this outcome matrix exactly.

 

1. QT on left. Play the King, later finesse for the nine. That will drop the Queen without wasting the Jack.

2. Q9 on left. The same as #1 above.

3. Tx on left. Whether or not lefty falsecards, your Jack will always score.

4. 9x on left. Same as #3 above.

5. Qx on left. Your King gets snarfed up, so does your Jack on the way back.

6. T9 on your left. Your King gets snarfed up, so does your eight on the way back.

7. Stiff N on the left. Clearly, the stated strategy of finessing against the missing N (presumed to be on the right) is going to work here.

 

>>>Of these 6 3-2 cases, the low to K line picks up 4, assuming a hook on the way back if the T/9 popped. Your original suggestion of duck completely 1st trick no matter whether T, 9, x appears only picks up 3 of them, as well as also blowing the 1-4 split cases.

 

And so this was therefore my reason for making the second post, modifying the original suggestion, doncha think???

 

..........

SO -- here's a follow-up. :D . Suppose Lefty plays small when you lead low to the King, and Righty plays the N (the ten or the nine) underneath the King! What are now the odds that you will succeed in your goal of only losing two trump tricks?

Edited August 7, 2007 by ralph23