Double Finesses Probabilities?

[sdebois]

Hi There,

 

Here are two suit combinations from Kelsey's "Matchpoint Bridge" (p. 48-49):

[tt]

 1)  A Q T 4  2)  A J T 4

 

   7 6   7 6

[/tt]

Assume that in both cases, we begin the suit by finessing the ten, losing to the J/Q respectively. Kelsey now makes 3 claims:

• At the outset, the odds of making more than one trick is 76 percent for both combinations,
  
• After loosing the first finesse in combination 1), the second has a 52 percent chance,
  
• After lossing the first finesse in combination 2), the second has a 68 percent chance.
  

This is the explanation provided:

 

"Since the overall chances of making more than one trick by taking two finesses are the same (76 per cent) for each combination, it seems strange that the second finesse should have so much better a chance on [combination 2] rather than [combination 1]. The basic reason is that with [combination 1] you used up a larger slice of your chances on the first round. The losing finesse of the ten used up 50 of the overall 76 per cent, leaving 26 out of the remaining 50, i.e., 52 per cent, for the second Finesse. With [combination 2] only 24 of the overall 76 per cent were used on the first round (the chance of West having both honours). That leaves 52 out of the remaining 76, i.e., 68.4 per cent, for the second finesse."

 

This is impenetrable to me. It seems to me that when I count the placement of outstanding honors, in both cases I find that from the outset, I win 2+ tricks whenever east do not have both honors (3 in 4 cases), and that after loosing to easts honor, only one of the 3 winning cases are left (exactly one of the two "splitted honors" cases).

 

If someone would care to clarify Kelseys argument and demonstrate the error of mine, I'd be most grateful. Also, why are Kelseys numbers 52, 76 etc. rather than 50, 75 etc.?

 

I believe the context for the problem is immaterial, but here it is anyway:

[tt]

E-W/S

 

 J 7 6 4 3

 K 6 5

 A Q T 4 (A J T 4)

 J

 

 A K T 8 5 2

 A 7 2

 7 6

 8 5

 

 1s - p - 4s - a.p.

 

 Lead: CK

 

 CK J 7 5

 HJ K 9 2

  x S3 x A

  x T J D7

  4 A 5 H3

 ?

[/tt]

[Wolferl]

Case 1:

 

East may hold K, J, both honnors or no honnor, each holding

with roughly equal chance. After loosing the first round finesse

to the J, only two of the four initial distributions are still possible,

so finding West with the King is now about 50%.

 

Case 2:

 

Now we have only 3 possible distributions for east: no honnor (25%),

Q or J (50%), QJ (25%). Say you loose a finesse to the Q or J.

One or both honnors with east are still possible, but the former

is 2 to 1 favorite (50% to 25%).

 

 

This is another application on restricted choise since Q and J are

considered equal.

 

Why 76%? This may be explained with the principle of remaining

spaces. If you deal one player the Q (a specific card), there are

now only 12 remaining spaces with this player with the other

one still having 13. So the odds are 13 to 12 for him to hold the

other honnor.

[inquiry]

Hi Debois

 

Let's see if I can explain the odds to you simply.

 

Do you know the odds of a 2-0 versus a 1-1 split is a suit? Assume you are missing Kx in a suit, do you hook or play for the drop?

 

The odds are

West East

Kx -- 24%

K x 26%

x K 26%

-- Kx 24%

 

So playing for the "drop" is 52% (single king either hand). The distribution of the missing QJ follows the same odds.

 

QJ -- 24%

Q J 26%

J Q 26%

-- QJ 24%

 

Odds of both hook losing is only 24%, so the odds of one of them being successful is 76%. Really, you should do the more complicated math to look at all the combination of cards… but this works. This is where the 76% comes from. This also works for the missing KJ. But the QJ are equals, while the KJ are not. So the od

 

After the first hook loses, you eliminate the possibility where both Honors were on side is no longer possible. So the odds change, the old odds and new odds are (were QQ is used to show equal honors, ….

 

QQ -- 24% 0%

Q Q 52% 68.4%

-- QQ 24% 31.6%

 

But with unequal honors, the original holdings were

 

KJ -- 24% 0%

J K 26% 0%

K J 26% 52%

-- KJ 24% 48%

 

In this case, the first two possibilities are no longer possible, since East won the Jack on first hook. When you correct the remaining odds, there you can see where the 52 comes from. with two possibilities removed, the remaining odds (orginally 50%) now has to be recalculated. 26/50 = 52%, 24/50 = 48%.

 

Hope that helps.

 

Ben