down 2 down 1 or making

[sceptic]

playing imps

 

I had this AJ109 in dummy and 74 in hand

 

I took one finese playing 4 to the J it lost

 

I am in 3NT and have 7 tricks no other source of tricks

 

I am not 100% sure other clues are there to make the decision

 

so should I juat take the ace and go down 1 with, what I considered no chance of dropping the remaining honour

 

or should I take the finese and if it makes I get 9 tricks or go down 2

[pclayton]

At IMPs, unless you have a really strong reason not to try for your game, you have to take the chance to make. In this circumstance, you can work out the IMP odds and the expected payoff.

[goobers]

I mean, if there are no other clues, then it looks like a second finesse will work 75% of the time, giving you 9 tricks for your 3NT.

 

So you have a 75% chance of making this.

[kenrexford]

At first, I was prepared to make a joke. The post title suggests a smart-ass response of MAKING. That's always best.

 

After reading the question, I'll stick with my joke.

[BebopKid]

At first, I was prepared to make a joke. The post title suggests a smart-ass response of MAKING. That's always best.

 

After reading the question, I'll stick with my joke.

LOL. I like it.

[goobers]

Lol.

[jtfanclub]

I mean, if there are no other clues, then it looks like a second finesse will work 75% of the time, giving you 9 tricks for your 3NT.

 

So you have a 75% chance of making this.

Not true. This is the old box game.

 

The odds of either finesse working BEFORE you tried it was 75%.

 

Assuming no stiff honors:

 

A. KQ........None

B. K..........Q

C. Q..........K

D. None.....KQ

 

Before you finesse, A, B and C would work, and D would fail. 75%.

 

But when you finesse the first time, and it fails to the King, A and B are no longer possible, so you're down to C or D...a 50-50 shot.

 

There are other factors, of course...Restricted Choice, 13 cards per hand, etc. etc. But it does come down to only slightly more than 50%.

 

In IMPs, you go for it. In MPs, it mainly depends upon where everybody else is going to be.

 

Suppose you ended up in 3NT but you expect much of the field to be in 2NT, which will end up making of course. Now Down 1 and Down 2 are about the same thing (close to bottom), so you may as well finesse. If, on the other hand, you expect lots of people to be in 4♠ down 2, so Down 1 and Making are about the same (close to top), just settle for the safety play.

[kgr]

I mean, if there are no other clues, then it looks like a second finesse will work 75% of the time, giving you 9 tricks for your 3NT.

 

So you have a 75% chance of making this.

After the 1st finesse lost I think this is 50%

[goobers]

That probability isn't correct. I know it looks right, but it isn't. You're 75% to make this second finesse.

 

It's conditional probability. I can try to work it out more demonstratively when I get home, but not right now at work.

[kenrexford]

What?!?!

 

A priori, you have four possibilities:

 

KQ offside

K offside

Q offside

Both onside.

 

If the first finesse fails, we can eliminate both onside. We can do this a posteriori because we are allowed to check on this.

 

So, when the checking for KQ onside is completed, a test we are allowed to do, we now know that we have a 66% chance of making.

 

Not 75% Not 50% 66%.

 

Of course, this leaves out some factors, like the fact that some people will stick in an honor from Hx(x)(x) on the first round as a nice play, this suggesting that HH(x)(x)(x) is more likely to be behind dummy. But, barring a nice play by LHO as a possibility not taken because of unavailability, 66%.

[han]

You are both incorrect. I think that the chance that the second finesse makes is roughly 2/3 = ~67%. If you play against a beginner who always wins with the lowest card then the chance is roughly 50%.

 

 

[i notice that Ken wrote the same, his response wasn't up yet]

[cherdano]

Wow. I agree with every bit of every post by Ken in this thread.

 

We should have a beer, Ken!

 

Arend

[kenrexford]

Wow. I agree with every bit of every post by Ken in this thread.

 

We should have a beer, Ken!

 

Arend

I'm afraid we won't get to drink that much if we only drink when you agree with me 100%. ;)

[goobers]

Oops. I meant 66%. Way to do division.

 

Anyway, I knew it had to be better than 50%. Thanks for the clarification.

[kgr]

A priori, you have four possibilities:

 

KQ offside

K offside

Q offside

Both onside.

 

If the first finesse fails, we can eliminate both onside.  We can do this a posteriori because we are allowed to check on this.

 

So, when the checking for KQ onside is completed, a test we are allowed to do, we now know that we have a 66% chance of making.

 

Not 75%  Not 50%  66%.

Sorry, but I don't understand this.

You say:

A priori, you have four possibilities:

KQ offside

K offside

Q offside

Both onside.

 

This is the same as saying:

 

KQ offside

Q onside and K offside

K onside and Q offside

Both onside.

 

If the first finesse fails to f.i. the K then both the 3rd and 4th possibility are eliminated giving us a 50% chance for the 2nd finesse?

[kenrexford]

Try the finesse four times.

 

On one of those four times, it will win the first time. On three of those four times, it will lose the first time.

 

Now, you prepare to take a second finesse every time the first lost.

 

Three times the first lost.

 

Of those three, only one will fail the second time. Two will succeed.

 

Thus, the second finesse is 66-33 to succeed.

[Double !]

I knew I has seen this card combination in print and found the exact reference in literally 10 seconds. The bidding dinosaur had also, at times, been a bookworm when much younger and actually had time to read. I try to provide references to support assertions (when I can).

 

In this case, I refer the reader to page 49 of H.W. Kelsey's book, "Match-Point Bridge", paperback edition, 1979, where he asserts that the second finesse will work 68.4% of the time. Hopefully, the many mathematicians among you all can explain how Kelsey derived that exact percentage.

 

I know: 66%, 67%, 68.4% not a significant difference. The point is one more source supports the position that the second finesse is the percentage play.

 

DHL

[brianshark]

I suspect the exact percentage (which is around the 2/3 mark) depends on how many cards the opps hold in the suit in question. Just a guess though.

[Finch]

It's another example of restricted choice (which is also just another example of conditional probabilities). When RHO wins the first trick with the king (say) he is twice as likely to have the king only, than both the K or the Q as half the time he might have won the first trick with the Queen. That gives you 2/3 probabilities.

 

In fact, the probability may be slightly less than that, depending on the dummy. If dummy (the AJ109 suit) has no outside entry, then the second finesse is much much worse against good opponents, as LHO would have played an honour on the first round (other than with six of them). How much worse depends on whether LHO knows what your length is: holding Kxxx inserting the king looks very silly if partner has singleton Q...

[kenrexford]

Kelsey is probably figuring in the reality that, in many layouts, LHO cannot play small (Kx/Qx).

[inquiry]

Kelsey is probably figuring in the reality that, in many layouts, LHO cannot play small (Kx/Qx).

I will have to think on this claim a while. When LHO follows to the second round in this suit, you care correct Hx as well as small singleton is removed. That leaves nine possible holdings for LHO (where x is any small card and H is the honor not lost at the first trick).

 

xx

xxx

xxxx

xxxxx

Hxx

Hxxx

Hxxxx

Hxxxxx

 

Since there is only five "x's", there is only one possible way LHO can hold xxxxx (all the smalls and none of the two bigs), so that 1 combination. Likewise, there is only two way LHO can hold Hxxxxx (his partner has to hold the singleton K or Q == the other honor. Using such there is ten ways he could hold Hxxxx (his parter holds the either of the H and any one of the five little "x's").

 

If you think of it this way, there is a grand total (after EAST wins the first trick and west follows LOW to the second) of 78 possible combinations (numbers shown below -- with those where the "winning" play of finessee again is in bold).

 

10 xx

10 xxx

5   xxxx

1   xxxxx

20 Hxx

20 Hxxx

10 Hxxxx

2   Hxxxxx

 

If you total the number of winning cases (where the finessee) works, you find 52 of them. 52/78 = 0.66667, the expected value for this restricted choice play -- even taking into account WEST can not have Kx or Qx nor x.

 

So to answer the question in this thread, assume you are vul, lets see how you do...

 

1. Play the ACE on second round. You drop the second honor offside one time out of 78, and you go down 1 the other 77 times, for an average of -91.03 per play.

 

2. Hook the second time. You will make 52 times out of 78 and go down two 26 times out of 78 for an average result of +333.33 points per board.

 

Which looks better to you? I know what you are thinking, imps scale is not linear, so what would be the expectet result for both plays if our opponents were in the same contract. Fair enough. Assume your opponents are in 3NT and take the second finessee how do you fair?

 

On those hands where you play ACE on second round. If RHO has KQ doubleton, you win 12 imps, and on the other 25 boards where the second honor is offside, you win 3 imps. However, on the 52 boards where the honor is onside, you go down one (-100) while they make so you lose 12 imps on each. The net average loss per board is -6.88 imps for this line of play.

 

If you choose the same play as your opponent (finesse), you of course average 0 imps. Now imagine your opponent rested in 1NT, making either +90 or +150 how do the various plays do? On the 52 hands where hook wins, they are +150, on the others they are +90. If you play the ACE on second round you are +600 once, and win 10 imps on that one, but you are -100 on all the others, losing 5 imps on all the other 77 boards, for essentially a certain 5 imp/board loss.

 

On the other hand, if you finesse on the second round, you win 10 imps on 52 boards (520 imps) and lose 7 imps on 26 boards (-182 imps). 520-182 = net +338 imps, divided by 78 = average gain of 4.33 imps.

 

So, in each case, risking down two is much, much better than playing for the fairly sure down two.. don't you think?

[kgr]

This was interesting. Hopes this makes me understand restricted choice better.