[Stephen Tu]

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matchpoints, blue ribbon pairs

3♥ was singleton or void.

West leads a spade, you put up the Q, east covers K and you win. Cash another spade, you find they are 2-2. On the second club, East plays an honor. Play restricted choice finesse or not, and why?

I did the math on this and think I know the answer, but I want to see if people come up with something different.

[Zelandakh]

Are you sure playing on clubs first is the best approach?

[billw55]

Zelandakh, on 2012-November-28, 05:47, said:

Are you sure playing on clubs first is the best approach?

Well, it does offer a chance to dispose the heart loser before opponents can cash it.

[rhm]

If my calculations are correct:

If you finesse you should make roughly:

50% of the time 12 tricks
16.6% " 11 tricks
33.3% " 10 tricks

Average number of expected tricks: 11.166 tricks

If you play for the drop:

25% of the time 12 tricks
58.3% " 11 tricks
16.6% " 10 tricks

Average number of expected tricks: 11.083 tricks

This is an approximation. I worked with 4-2 break twice as likely as a 3-3 club break and assuming any diamond honor to be as likely with East as with West.
So if my calculations are correct (a big if) the finesse gains approximately 1/12 of a trick.
Whether this means you should take it is a different question.

Rainer Herrmann

[rhm]

deleted

[Phil]

I'm having a hard time seeing how rhm came up with these calculations.

Can you elaborate?

[mikeh]

I also don't understand the math.

Btw, if we hook, how does this hand play on a diamond back?

On a weird day we might go down if we hook, tho many LHO's would have led a diamond from KQx.

So is rainer assuming that RHO won't return a diamond? If he does, maybe we should be popping A, to get rid of our heart, and thus ensure 10 tricks. But he can't be saying that, else his percentages for taking 10 or 11 tricks make no sense.

I get the 50% for the 12 tricks on the hook: we win 2-1, and if we win we are 75% for 12 tricks. I'd estimate maybe a trifle more, since some LHOs lead the diamond K from KQ. But a good defender, and this is the Blue Ribbon, will usually not be aggressive on these auctions.

As for what I'd do: I am not even going to try the math. I am in an event where the quality of play is usually very good at every table, especially after the first cut. I think I need to rack up good scores whenever I can and here I have a chance....so I hook the club.

[PhilKing]

mikeh, on 2012-November-28, 10:33, said:

I also don't understand the math.

Btw, if we hook, how does this hand play on a diamond back?

We rise with the ace and claim ten tricks, obviously. We always make exactly ten trick when the club loses, hence 33.33%.

I think his figure for 11 tricks is basically 1/4 x 2/3 (16.66%) - ie the club finesse wins but West has both diamond honours, but higher against an overly aggressive leader.

There is a vacant spaces issue hence the word "roughly" followed by a decimal.

[rhm]

PhilKing, on 2012-November-28, 11:03, said:

We rise with the ace and claim ten tricks, obviously. We always make exactly ten trick when the club loses, hence 33.33%.

I think his figure for 11 tricks is basically 1/4 x 2/3 (16.66%) - ie the club finesse wins but West has both diamond honours, but higher against an overly aggressive leader.

There is a vacant spaces issue hence the word "roughly" followed by a decimal.

Thanks for elaborating correctly.

You just have to look at the correct line of play for each case once when you play for the drop and once for the finesse in clubs.
Usually this means taking the double finesse in diamonds, unless you took a failing club finesse and a diamond return, in which case you have to rise with the ace to discard your heart on the fourth club.

Approximate and roughly is also correctly explained. The true probabilities for restricted choice are not exactly two to one, lead considerations and vacant spaces have not been taken into account, but I believe the additional precision to be gained here looks rather small to me.

Rainer Herrmann