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Can you calculate the odds at the table?

#1 User is offline   kgr 

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Posted 2006-September-16, 04:59

(This is coming from Justin's blog)
Your holding in H's:

In a 4S contract you have stripped other suits and from the bidding you know LHO has HQ and some H length (3+). LHO is on lead and plays the HQ.

Comment from the blog ('I' is LHO)


Here there is not restricted choice on the Q, assuming I am a good defender, I know that declarer has already placed me with the Q for my bid. Accordingly, with QJ I MUST play the queen, otherwise Aaron will know that I still have it in my hand. This means it's not clear what the percentage play is in hearts. To figure it out we need to calculate the number of possible QJx(x) combinations and weigh it against the number of Qxx(x) combinations. If I have 3 hearts, there are 4 QJx's possible and 6 Qxx's possible. If I have 4 hearts there are 6 QJxx's possible and 4 Qxxx's possible!

My question:
How do you calculate these odds at the table? Is there any easy logical way to do this?

#2 User is offline   cherdano 

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Posted 2006-September-16, 12:06

I don't know how to calculate THESE odds, as they are wrong. ;)
The easiest way is just:


If LHO has 4 hearts, the hearts I don't know about split 3=3, so it is 50-50. If LHO has 3 hearts, they split 2=4, so RHO is a 2:1 favourite to hold the J.

If you want to count Justin's way, there are 10 QJxx holdings versus 10 Qxxx holdings, and 5 QJx holdings versus 10 Qxx holding. Only thing you need to know that there are 10 ways to pick 2 cards from 5 (the 5 xxxxx available).

The easiest way to count losers is to line up the people who talk about loser count, and count them. -Kieran Dyke

#3 User is offline   PhantomSac 

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Posted 2006-September-16, 17:11

delete this please.
The artist formerly known as jlall

#4 Guest_Jlall_*

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Posted 2006-September-16, 17:13

Good catch, thanks. I have added an update to the post. For some reason I thought there were 6 missing hearts, not 7, in which case my math was correct.

As for the original question, Arend is right that it's very easy to figure out these odds at the table. I am not a math person, and have taken no math past the high school level, so this is very low level.

I think of things in terms of possible combinations, but using arends method if there were 6 missing hearts you could say if LHO has 4 hearts, there are 3 unknown to RHOs 2 making him 60-40 to have the jack, but if LHO has 3 then he has 2 unknown hearts to RHOs 3, making it 40-60. It's all the same, just use what youre comfortable with.

#5 User is offline   kgr 

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Posted 2006-September-17, 02:20

Seems like Arend's way is much easier for this one (at least for me).

Now for another question:

Qxxx Jxx
You are LHO and know that declarer has the A. You have to break this suit. It is best to play the Q? Why is that?

PS: I think that this was Justin's first post since a long time. If this makes him post more often again then my question was VERY interesting :)

#6 User is offline   cnszsun 

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Posted 2006-September-17, 03:01

With any one combination of Qxxx, Jxxx, QJxx, you have to exit with an honor, otherwise if you exit with a small card declarer has to play honor split, then he will always make. So, your only chance is to play an honor Q or J. A good declarer will play your hand as Qxxx according to restricted choice (you may exit with Q or J if your hand is QJxx).
Michael Sun

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