[andy_h]

A friend gave me this hand and told me that a few people chose different lines so he would be happy to hear other people's thoughts/guess.

Scoring: IMP

---

You're in 6♥ as South and received the lead of the ♦8 (Sorry, don't know the full auction). You play the King and RHO wins with the Ace. RHO then returns a small diamond. You ruff high and West pitches (I don't remember but if it affects your answer then say so). You have 12 tricks but now it comes down to how you handle the trump suit with two possibilities:

1) Play hearts from the top
2) Hook LHO for the heart ten on the first round

[655321]

Seems like a straightforward odds calculation.

I make it that finessing the heart (ten or eight?) is 54.9% (the 12/20 or 60% odds you get from vacant spaces for the ten to be onside, less the 5-0 break), and playing from the top is 69.8% (all the 3-2 and 2-3 breaks, and the 4-1 and 1-4 breaks where the ♥T is singleton).

[hanp]

I found the same as 653321, you can use the suit break calculator on www.rpbridge.net for this.

On a related note, playing in New Orleans I had ♥Kxxxx, dummy ♥AQ10. I don't remember anything else about the hand, except the auction:

1NT - (2C*) - 3H - (p)
4H - all pass

2C showed clubs and another suit, and for some reason I don't want to remember, we were not playing transfers at the 2- or 3-level.

LHO led a card and it became clear that (1) RHO had the black suits, and (2) playing 4H from rogerclee's side would likely have been better.

I ruffed the second trick in hand and because of communication problems I had to decide whether I was going to finesse immediately or pull trumps from the top. What should I do?

[Phil]

655321, on Aug 31 2010, 10:13 AM, said:

Seems like a straightforward odds calculation.

I make it that finessing the heart (ten or eight?) is 54.9% (the 12/20 or 60% odds you get from vacant spaces for the ten to be onside, less the 5-0 break), and playing from the top is 69.8% (all the 3-2 and 2-3 breaks, and the 4-1 and 1-4 breaks where the ♥T is singleton).

Looks like you used vacant spaces based on 12 left in West and 8 left in East to reach 69.8 / 54.9. However, East can't have a black suit void (no Lightner), and depending on bidding, LHO might have taken a spade call with Q-7th + a stiff, so at worst it seems that 9 to 5 would be the proper ratio, and it might be lower. On these assumptions the numbers start to creep closer to even. For instance:

12 / 8 = 69.7 / 54.8
11/7 = 69.2 / 55.7
10/6 = 68.2 / 56.7
9/5 = 66.7 / 58
8/4 = 63.8 / 59.6
7/3 = 58.3 / 61.7

[JLOGIC]

dammit

Oh I misread, the world is right, 655321 is still god.

I thought I got him one time

[hanp]

SOME OF THESE POSTERS ARE TRAINED VACANT SPACERS, DO NOT TRY THIS AT HOME!

Phil, I think that you are absolutely right that east probably has at least 3 spades and at least 1 club, and similarly for west. However, that does not mean that we can subtract 4 more vacant spaces from either side.

The difference with the diamond suit is that that suit is known to split exactly 5-1. We don't know how the black suits split, and even though both sides must have some cards in those suits, we not know the location of any exact black cards. The difference is subtle. You are correct that the black suited inferences do affect the odds, but the precise odds are much harder to compute. The easiest way to find the odds would probably be to use a simulation program.

[JLOGIC]

hanp, on Aug 31 2010, 11:54 AM, said:

SOME OF THESE POSTERS ARE TRAINED VACANT SPACERS, DO NOT TRY THIS AT HOME!

Phil, I think that you are absolutely right that east probably has at least 3 spades and at least 1 club, and similarly for west. However, that does not mean that we can subtract 4 more vacant spaces from either side.

The difference with the diamond suit is that that suit is known to split exactly 5-1. We don't know how the black suits split, and even though both sides must have some cards in those suits, we not know the location of any exact black cards. The difference is subtle. You are correct that the black suited inferences do affect the odds, but the precise odds are much harder to compute. The easiest way to find the odds would probably be to use a simulation program.

I made the same mistake as Phil once too, was promptly owned by either you or cherdano

[fred]

655321, on Aug 31 2010, 03:13 PM, said:

and playing from the top is 69.8% (all the 3-2 and 2-3 breaks, and the 4-1 and 1-4 breaks where the ♥T is singleton).

It is even better than that because you also make when RHO has 10xxx - you will take the marked finesse when LHO shows out on the second round (and after you unblock the King of spades).

That is worth another 4.33% according to the odds calculator in the BBO web-client (which also agrees with your other numbers).

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[655321]

fred, on Aug 31 2010, 12:20 PM, said:

It is even better than that because you also make when RHO has 10xxx - you will take the marked finesse when LHO shows out on the second round (and after you unblock the King of spades).

Nice, I didn't see that!

hanp, on Aug 31 2010, 10:24 AM, said:

and (2) playing 4H from rogerclee's side would likely have been better.

There are clever remarks to be made here, I am sure.

In a recent thread on r.g.b, there is a similar problem (with AKQT9x opposite a singleton, instead of your Kxxxx opposite AQT.) Richard Pavlevic says that playing from the top is best even when there are known to be more vacant spaces in front of the suit because you pick up against Jxxx only, and lose to J, Jx, Jxx offside. Your suit combination looks to be similar enough that (without doing the math) maybe the same logic applies.

(This r.g.b. post is also relevant to OP andy_h because Pavlevic then goes on to quote andy_h's BBO quote about pizza!)

[Phil]

hanp, on Aug 31 2010, 11:54 AM, said:

Phil, I think that you are absolutely right that east probably has at least 3 spades and at least 1 club, and similarly for west. However, that does not mean that we can subtract 4 more vacant spaces from either side.

Certainly 5-1 diamonds is 'known'. Isn't at least two spades and one club known as well?

It isn't as though our RHO flashed three black cards to us, but the inferernce seems strong enough to count them with what we know they have.

Would appreciate some more explanation, thanks.

[nigel_k]

From a simulation I got 54.8% for the finesse and 74.1% for A then Q.

If we assume East has at least one club and West's spades are worse than Qxxxxxx then it is 57.1% for the finesse and 70.5% for A then Q.

[gwnn]

I think this is like the question 'how many hearts does a 1H opening hold on average?'

If you think simply, you'd say that the opener has 5 hearts guaranteed and then one quarter of the rest, making for 7 hearts. This is clearly wrong.

If you think in a little more refined way you'd say there are 47 slots left out of which opener has only 8. Then it would be 5+8/47*8, or about 6.36. This is still clearly wrong!

The only correct way is to deal, say, 1 billion hands and discard all hands that do not have 5 hearts at least and then calculate an average from those hands that have not been discarded.

The two situations are analogous - Phil is trying to pre-deal 3 spades and 1 club to each hand and then deal randomly when he should deal a bunch of hands and discard the impossible ones.

The effects of the subtle error are, generally speaking, that the probabilities near the limit are unduly low (in this case, Phil's deals will have fewer cases with West having exactly 1club than in real life).

[jdonn]

gwnn, on Aug 31 2010, 09:56 PM, said:

The only correct way is to deal, say, 1 billion hands and discard all hands that do not have 5 hearts at least and then calculate an average from those hands that have not been discarded.

That way is not the only correct way because it's not correct. There are a finite amount of bridge hands, not an infinite amount The only correct way is to deal every possible bridge hand and discard the ones that can't apply.

[gwnn]

jdonn, on Sep 1 2010, 11:05 PM, said:

gwnn, on Aug 31 2010, 09:56 PM, said:

The only correct way is to deal, say, 1 billion hands and discard all hands that do not have 5 hearts at least and then calculate an average from those hands that have not been discarded.

That way is not the only correct way because it's not correct. There are a finite amount of bridge hands, not an infinite amount The only correct way is to deal every possible bridge hand and discard the ones that can't apply.

come on jdonn I didn't know you are a mathematician or, gulp, a philosopher.

[Dirk Kuijt]

FWIW, my simulation gives, of hands that open 1♥ in a 2/1 framework, the fraction of how many hearts are held:

9+ hearts: 0.0007
8 hearts: 0.0079
7 hearts: 0.0616
6 hearts: 0.2773
5 hearts: 0.6526

Setting hands with 9+ to exactly 9 hearts, gives an average number of hearts as 5.4269.

Your mileage may vary, especially depending on your view of opening 1NT with 5 hearts.

[hanp]

Phil, on Aug 31 2010, 07:55 PM, said:

hanp, on Aug 31 2010, 11:54 AM, said:

Phil, I think that you are absolutely right that east probably has at least 3 spades and at least 1 club, and similarly for west. However, that does not mean that we can subtract 4 more vacant spaces from either side.

Certainly 5-1 diamonds is 'known'. Isn't at least two spades and one club known as well?

It isn't as though our RHO flashed three black cards to us, but the inferernce seems strong enough to count them with what we know they have.

Would appreciate some more explanation, thanks.

Will come back to this later if somebody else doesn't.

edit: I see that gwnn has dealt with this rather well.