[bd71]

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Matchpoints. 2/1.

So we know partner is roughly 10-11 with 0/1 spades.

Questions:

1. At matchpoints, do you want to be in game? (At IMPs I think it's a no-brainer yes.)

2. If yes to #1, which game?

3. I have my own estimates for this, but would like validation. If partner has 1 spade, what probability is there the spades run w/ no losers? How about if he has no spades?

[Bbradley62]

3. If pard has one spade, chance of 3-3 split is 35.5% and chance of Jx is 16%, so the chance that your spades run is just over 50/50. If pard has no spades, it's 27% for Jxx plus about 1.5% for J9.

[aguahombre]

Probably doesn't matter to this discussion, but with what we will open 1S and rebid 2S with these days, 10 isn't enough.

What might matter however, is a thread some time ago where with 11/12 and two card spade support a lot of respected posters would have rebid 2N, rather than 3S. This possibility would alter our quick calculations at the table.

Anyway, I try 3NT, on the theory it won't make exactly 2NT.

[semeai]

1. I guess so. We're not even sure 2NT is making, so we might as well bid 3NT and have some upside when the spades run. It's also conceivable we make when the spades don't run if I still have my entry.

2. 3NT. I don't want to try for more tricks.

3. For 1 spade, 3-3 is approx 36% and 4-2 is approx 48% (remembered these). Jx-xxxx is 1/3 of all 4-2's (math or vacant spaces). So it's approx 36 + 48/3, so 52%. For 0 spades, 4-3 is approx 62% (looked this one up instead of remembering or calculating), and Jxx-xxxx is 3/7 of these (again, math or vacant spaces). So it's 3*62/7 which is approx 3*9, so 27%.

Added: I see I left out J9-xxxxx, but Bbradley got there.

[bd71]

semeai, on 2012-September-10, 09:54, said:

We're not even sure 2NT is making, so we might as well bid 3NT and have some upside when the spades run.

If we're playing IMPs, I think this logic is unassailable.

But even though (playing matchpoints) I bid 3N at the table using this exact logic, I'm now wondering now how well this applies in pairs? Should we worry about -200 at 3N losing to enough -100s (or -300s losing to -200s) that it would not offset the times when we bid the game bonus.

If the field matters to your thinking, this was a single-session open pairs at a sectional, with probably a better field than I've seen in a typical sectional. But would appreciate how you think different levels of play would affect "the right" decision.

For what it's worth, on this hand the spades did not run AND we took exactly 8 tricks anyway...oh well.

[semeai]

bd71, on 2012-September-10, 10:16, said:

If we're playing IMPs, I think this logic is unassailable.

But even though (playing matchpoints) I bid 3N at the table using this exact logic, I'm now wondering now how well this applies in pairs? Should we worry about -200 at 3N losing to enough -100s (or -300s losing to -200s) that it would not offset the times when we bid the game bonus.

If the field matters to your thinking, this was a single-session open pairs at a sectional, with probably a better field than I've seen in a typical sectional. But would appreciate how you think different levels of play would affect "the right" decision.

You're right that it's not as clear cut at matchpoints. The logic still applies to some degree unless the other tables will all be in 2NT or 3NT. I don't have a great sense of the extent to which that will be true, but we don't know if partner's decision was clear cut or whether (s)he had other options, and we also don't know if some will insist on spades with our hand.

Quote

For what it's worth, on this hand the spades did not run AND we took exactly 8 tricks anyway...oh well.

Ouch.

[jmcw]

This is a very strong hand in the context of the auction. I think it is a 3NT all day not close.

You have an "undisclosed" source of tricks, a likely entry in the ♦K and a working 9 and T in the other suits. It just doesn't get much better.

Didn't partner just invite!

[aguahombre]

Or, for bean counting, in our "box" of 11-14, we have nearly 16.

[semeai]

semeai, on 2012-September-10, 09:54, said:

For 1 spade, 3-3 is approx 36% and 4-2 is approx 48% (remembered these). Jx-xxxx is 1/3 of all 4-2's (math or vacant spaces). So it's approx 36 + 48/3, so 52%.

Actually, I guess it's 59% if all we know is partner has one spade: that one spade might be the ♠J (adding on 1/7 of the remaining 48%)!

[Bbradley62]

aguahombre, on 2012-September-10, 09:50, said:

Anyway, I try 3NT, on the theory it won't make exactly 2NT.

Of course, at matchpoints passing is better if it makes exactly 2NT or if even 2NT doesn't make.

[Bbradley62]

semeai, on 2012-September-10, 10:57, said:

Actually, I guess it's 59% if all we know is partner has one spade: that one spade might be the ♠J (adding on 1/7 of the remaining 48%)!

Good call!

[semeai]

I might also dispute this claim that partner has at most one spade. Can't partner have a notrumpy hand with two spades?

[aguahombre]

aguahombre, on 2012-September-10, 09:50, said:

What might matter however, is a thread some time ago where with 11/12 and two card spade support a lot of respected posters would have rebid 2N, rather than 3S. This possibility would alter our quick calculations at the table.

semeai, on 2012-September-10, 11:32, said:

I might also dispute this claim that partner has at most one spade. Can't partner have a notrumpy hand with two spades?

This is getting scary. We agree again; maybe we are both wrong.

[nofr]

Why not 4 spades ???
You havé 6 losing tricks
Y think your partner has 8 losing tricks
You havé together 14 losing tricks : you CAN try to Play a game, i think.

[Quantumcat]

Bbradley62, on 2012-September-10, 09:49, said:

3. If pard has one spade, chance of 3-3 split is 35.5% and chance of Jx is 16%, so the chance that your spades run is just over 50/50. If pard has no spades, it's 27% for Jxx plus about 1.5% for J9.

I think you might have your probability wrong, you don't add them together, but rather multiply their chances of not working. For example if there are four finesses to take and you need one to work to make your contract, you wouldn't add 50% + 50% + 50% + 50%, which would give 200%. You would multiply 0.5 X 0.5 X 0.5 X 0.5 = 0.0625 so the chances of making your contract would be 93.75%.

In the case of pard having one spade, using your quoted probabilities, 1-0.355 = 0.645 and 1-0.16 = 0.84, 0.645 * 0.84 = 0.5418, which gives a 45.82% chance of the spades running.

When pard has no spades, using your quoted probabilities we have 0.78 * 0.985 = 0.7683, so you have a 23.17% chance.

If you weight these two possibilities (pard with one spade and pard with no spades) equally, you'll get a total 45.82 * 0.5 + 23.17 * 0.5 = 34.495% chance of the spades running.

[han]

Quantumcat, Your comment could have been right if the topic was different, but here you just add the probabilities toghether.

Semai, I think the chance is a little better even because the opponents didn't bid.

[Quantumcat]

han, on 2012-September-13, 12:29, said:

Quantumcat, Your comment could have been right if the topic was different, but here you just add the probabilities toghether.

Thanks, you are right - a 3-3 split can't happen simultaneously with J doubleton so in the formula (A OR B) minus (A AND B) for finding the probability of either A or B happening, (A AND B) is zero. So you can't use the shortcut 1- (A OR B minus A AND B) = (1-A) AND (1-B).