[Dinarius]

I've been looking at this page...........

http://www.bridgehan...istribution.htm

I'm specifically interested in the first one - i.e. missing two cards.

What I want to know is this; why is 1 - 1 52% and not 50%? Why must one play for the drop?

E.g. Missing K,x....

The possible distributions are;

K,x......-

-.......K,x

K........x

x........K

Correct?

So, the finesse loses to 1 and 3, and wins in 2 and 4. i.e 50% Right?

And playing for the drop loses to 1 and 2, and wins in 3 and 4. i.e. 50% Right?

So, why in the above link is the drop a slight favourite at 52%?

Thanks.

D.

[1eyedjack]

Each of the four possible scenarios is not equally likely. This is because you start with a finite number of cards in each hand. There are more total number of hands where they break 1-1 than 2-0.

Another way of looking at it is that your decision to drop or finesse only arises after second hand has followed low. He then has one fewer cards remaining in his hand in which to hold the King than his partner who has yet to follow.

[Dinarius]

The second of your explanations is sheer elegance.

The theory of restricted choice (which I presume is what you're referring to) hadn't dawned on me.

Many thanks.

D.

[1eyedjack]

Hi - I was not referring to restricted choice.

There are some good books around addressed at bridge players that help the sums make sense.

Bridge odds for practical players, by Kelsey and Glauert. Actually that may be out of print now although there will be some second hand copies out there. But probably something written a bit more recently.

[manudude03]

Think of it as each opponent having 13 slots for a certain card to be placed. Give the first card (x) to say LHO, he will be down to 12 slots left while RHO has 13 slots left. Therefore the chances of the other card (K) being with RHO is 13/25. So the chances of it being x-K is 13/26 * 13/25 or 26%. This needs to be multiplied by 2 as it is equally likely to be K-x and so you get 52%.

You can extend this reasoning to get the probability of any break.

[fromageGB]

manudude03, on 2016-June-18, 05:57, said:

Think of it as each opponent having 13 slots for a certain card to be placed. Give the first card (x) to say LHO, he will be down to 12 slots left while RHO has 13 slots left. Therefore the chances of the other card (K) being with RHO is 13/25. So the chances of it being x-K is 13/26 * 13/25 or 26%. This needs to be multiplied by 2 as it is equally likely to be K-x and so you get 52%.

This seems overly complicated. You say the "chances of the other card (K) being with RHO is 13(slots)/25", so why are the odds on dropping the King not simply 13/25 = 52%? Do you need those last two sentences?

If you don't like the vacant spaces idea, and I don't really understand it (see next post), you can look at the odds of the finesse working, as long as LHO has followed low, by 1eyedjack's first post first paragraph :
100 deals, which for 2 missing cards, probability says are split (LHO-RHO) :
24 hands are 2-0, all these remain in the pot when LHO follows.
26 hands are K-x, but these are ruled out as LHO followed low.
26 hands are x-K and remain in the pot.
24 hands are 0-2, but are ruled out when LHO follows.
Once LHO follows, we have 24+26 = 50 deals remaining in the pot. Of these, finesse works 24 times, and drop 26 times.
There is a 26/50 = 52% chance of the drop working.
Is this calculation the right way to do it?

It's an area I am a bit hazy on.

[fromageGB]

To demonstrate my confusion, and hoping for clarification, consider the 3 missing cards scenario. You are missing the K and two small, and lead towards the AQ, with LHO following low. What are the odds now of the finesse working?

Probability says 100 deals are split (LHO-RHO) :
11 hands are 3-0, all remain in the pot when LHO follows.
11 hands are 0-3 and are ruled out as LHO follows.
39 hands are 2-1, and all remain the the pot.
39 hands are 1-2, but 13 of these are ruled out as LHO followed low, so 26 remain in the pot.
The pot considered now holds 11+39+26 = 76 hands.
Of these, the finesse works on all 11, plus 50% of 39, plus 0% of 26 = 30.5 hands.
So the odds of the finesse working is 30.5/76 = 40%.

Of the 76 hands, the drop works on 0% of 11, plus 50% of 39, plus 0% of 26 = 19.5 hands.
So the odds of the drop are 19.5/76 = 26%

By vacant spaces, LHO has 12 spaces for the K, and RHO 13, so the odds of the finesse working is 48%.

OK, the finesse works 40% of the time, or it works 48% of the time.
Or neither of these ?

Edit - the haze is developing into a deep fog.

[Stephen Tu]

Of the 2-1 splits, your counting is wrong above. The hook works 2/3, drop 1/3, not 1/2 of the time as you posted.

[wynsten]

Even splits are slightly more probable than you might expect, regardless of the number of cards involved. Suppose you are missing Kx of a suit. When the cards were originally dealt, if each of these were dealt in a different round, then the probability that one hand gets both is fifty-fifty. But in the rare case where both are dealt in the same round, the probability that one hand gets both is zero! This same reasoning works for any number of cards. Hence the rule that even splits are slightly favoured. On top of that, really lop-sided splits tend to be revealed in the bidding or early play. So when it is a close call, play for an even split.

[1eyedjack]

wynsten, on 2016-June-20, 15:19, said:

Even splits are slightly more probable than you might expect, regardless of the number of cards involved. Suppose you are missing Kx of a suit. When the cards were originally dealt, if each of these were dealt in a different round, then the probability that one hand gets both is fifty-fifty. But in the rare case where both are dealt in the same round, the probability that one hand gets both is zero! This same reasoning works for any number of cards. Hence the rule that even splits are slightly favoured. On top of that, really lop-sided splits tend to be revealed in the bidding or early play. So when it is a close call, play for an even split.

Did anyone else understand a word of this? I don't mean agree or disagree; just understand what is being said?

[runewell]

I'm not sure I buy any of the explanations presented here.

I think considerations of
Kx -
K x
x K
- Kx

are a good APPROXIMATION of what's going on.

To be completely accurate, I think there has to be consideration of how all the remaining suits are likely to break in the opponents' hands.
WARNING: MATH AHEAD.

The location of 11 hearts and 15 non-hearts is already known, so let's say you will draw for the hand on the left with the remaining 2 hearts and 24 non-hearts.
How many ways can you draw Kx of hearts? There are 2C2 = 1 way to draw Kx and 24C11 = 2,496,144 ways to draw the remaining non-hearts.
How many ways can you draw no hearts? There are 2C0 = 1 way to draw no hearts and 24C13 = 2,496,144 ways to draw the remaining non-hearts.
What about a stiff heart? There are 2C1 = 2 ways to draw a single heart (K or x) and 24C12 = 2,704,156 ways to draw the remaining non-hearts.

# Ways to get a 2-0 split = 2,496,144 + 2,496,144 = 4,992,288
# ways to get a 1-1 split = 2,704,156 x 2 = 5,408,312
Probability of 1-1 split = 5,408,312 / (4,992,288 + 5,408,312) = 5,408,312 / 10,400,600 = 52.0%.

[fromageGB]

1eyedjack, on 2016-June-20, 21:40, said:

Did anyone else understand a word of this? I don't mean agree or disagree; just understand what is being said?

Yes. For one hand you deal 4 cards in a "round", 13 times. If the K and the x were dealt in the same round, they cannot be in the same hand. If they were dealt in different rounds, they could be.

This argument implies 12 rounds of 50:50 and one which is split only, so the conclusion to be drawn - if you wish to draw one - is that the odds favour an even split of the 2 cards missing in your suit.

[fromageGB]

Stephen Tu, on 2016-June-20, 12:31, said:

Of the 2-1 splits, your counting is wrong above. The hook works 2/3, drop 1/3, not 1/2 of the time as you posted.

Thanks Stephen, but I am interested in how you derive this from probabilities. Care to make a derivation? My calculation says 40%, not 67%, (and not 48%). We are not talking of likelihood before a card has been played, but the likelihood after LHO has followed low.

Any statisticians/mathematicians to the rescue?

[manudude03]

fromageGB, on 2016-June-20, 09:35, said:

This seems overly complicated. You say the "chances of the other card (K) being with RHO is 13(slots)/25", so why are the odds on dropping the King not simply 13/25 = 52%? Do you need those last two sentences?

Maybe not in this precise example, but I was just making sure I didn't imply that it was a priori 52% that it was precisely x-K.

[fromageGB]

Stephen Tu, on 2016-June-20, 12:31, said:

Of the 2-1 splits, your counting is wrong above. The hook works 2/3, drop 1/3, not 1/2 of the time as you posted.

This must be wrong. I can envisage a break of LHO = x, and RHO = Kx. The finesse will not work. The drop will not work. 2/3 + 1/3 = 1, so it cannot be 2/3 and 1/3.

Before any card has been played, the chances of RHO=Kx are 26/100, and my calculation says after LHO follows low it is 26/76, = 34%. This ties in with my figure of finesse 40%, drop 26%, neither 34%, but the calculation method may be wrong. Help needed ...

[manudude03]

fromageGB, on 2016-June-21, 05:31, said:

This must be wrong. I can envisage a break of LHO = x, and RHO = Kx. The finesse will not work. The drop will not work. 2/3 + 1/3 = 1, so it cannot be 2/3 and 1/3.

Before any card has been played, the chances of RHO=Kx are 26/100, and my calculation says after LHO follows low it is 26/76, = 34%. This ties in with my figure of finesse 40%, drop 26%, neither 34%, but the calculation method may be wrong. Help needed ...

You also need to finesse when the suit broke 3-0. I think Stephen meant 2/3 of the time when LHO started with 2. I get around 65% (48/76) for the finesse of the cases when it matters (climbing to 74% if you consider restricted choice).

[Stephen Tu]

fromageGB, on 2016-June-21, 05:31, said:

This must be wrong. I can envisage a break of LHO = x, and RHO = Kx. The finesse will not work. The drop will not work. 2/3 + 1/3 = 1, so it cannot be 2/3 and 1/3.

Before any card has been played, the chances of RHO=Kx are 26/100, and my calculation says after LHO follows low it is 26/76, = 34%. This ties in with my figure of finesse 40%, drop 26%, neither 34%, but the calculation method may be wrong. Help needed ...

You kept on saying finesse and drop are 50-50 when lho has 2 and followed low. This is wrong.

If lho 2, say the missing cards are k23. There are 2 kx combos, k2 and k3, but only one xx combo, 23. So you win the hook in this case 2/3, not 50%. And drop only works 1/3 for this subset.


Overall, if looking only at the 76 cases that haven't been eliminated, (lho not void, lho not stiff k), the finesse works 37 times, the drop 13. The rest nothing works. Think about it. Out of the original 100, you expect finesse work half. 13 of those are the stiff k onside which you excluded. 50-13 is 37 winning cases left for the hook not 26.

[Stephen Tu]

manudude03, on 2016-June-21, 06:44, said:

You also need to finesse when the suit broke 3-0. I think Stephen meant 2/3 of the time when LHO started with 2. I get around 65% (48/76) for the finesse of the cases when it matters (climbing to 74% if you consider restricted choice).

How you get 65%?? It is 37/76. Restricted choice already is factored in.

Maybe you mean 37/50 for the hook, 74%, looking at only the cases it's possible to pick up the suit for no losers and the k didn't pop up stiff onside already.

[fromageGB]

Stephen Tu, on 2016-June-21, 09:46, said:

If lho 2, say the missing cards are k23. There are 2 kx combos, k2 and k3, but only one xx combo, 23. So you win the hook in this case 2/3, not 50%.

This is the bit I have trouble with. I will allow that the 2 or 3 could be played from the 32, and I think my calculation method has gone wrong is saying LHO played "low" rather than a specific card, when I am looking at specific cards in my calculation of hand numbers. Going back to my original post where I said LHO plays low, let me revise that to LHO plays the 2. Specifically the 2.

11 hands of 3-0 all remain in the pot. The finesse works on all these.
11 hands of 0-3 ruled out.
26 hands remain of the original 39 that split 2-1. The finesse works on 50% of them.
13 hands of 1-2 remain of the original 39. The finesse fails on all.
The pot of hands that are still possible holds 11+26+13 = 50 hands.
The finesse works on 11+13+0 = 24 out of 50, or 48%.

So this is not my original 40%, and funnily enough this 48% is the same as given by the simple vacant spaces method.
It is also the same as the 37/76 that you quote, and I am happy to accept it as identical, given rounded distribution percentages.

[Stephen Tu]

fromageGB, on 2016-June-22, 16:32, said:

This is the bit I have trouble with. I will allow that the 2 or 3 could be played from the 32, and I think my calculation method has gone wrong is saying LHO played "low" rather than a specific card, when I am looking at specific cards in my calculation of hand numbers. Going back to my original post where I said LHO plays low, let me revise that to LHO plays the 2. Specifically the 2.

Yeah, where your calculation goes wrong is that you don't seem to understand the principle of restricted choice. When LHO has both the 3 and the 2, he's going to play them at random, so the frequencies of these hands have to be halved in comparison to LHO holding K2 exactly, where LHO will not be playing them at random.

So you have to multiply all the 3-2 and K32 combos by half. Or treat the 2 and 3 as identical x's and notice that there are two Kx onsides and only one xx onside.

If you are up against a naive opponent who doesn't vary their low spot cards, and just always plays up the line, then you can then say that you are 50-50 when the 2 is played and the suit is 2-1 (rather than 1-2), because it's either 32 or K2. But now your success rate when the 3 is played is 100%, because it can't be 32 onside. So your overall success rate on these 2-1 splits is still 2/3 if you are considering both cases where opponent followed with the 2 and when they followed with the 3. Not 50-50. That's why restricted choice calculations just tend to tally up all possible holdings and treat the low spots as x's. If you want to calculate success percentage against opp who always plays their lowest spot, and specifically played the 2, then that's a different problem and you should specify you are trying to solve for that. But then your success rate when the 3 appears will be different, and combined will still end up in the same spot overall.

>11 hands of 3-0 all remain in the pot. The finesse works on all these.
But * half since restricted choice

>26 hands remain of the original 39 that split 2-1. The finesse works on 50% of them.
No. 13 hands of K2 onside, 13/2=6.5 hands of 32 onside because half the time he'll play the 3.

>13 hands of 1-2 remain of the original 39. The finesse fails on all.
>The pot of hands that are still possible holds 11+26+13 = 50 hands.

No, because of restricted choice, you are dealing with 38 hands. You need to exclude a portion where your opp chose to play the 3 holding both the 3 and 2.

>The finesse works on 11+13+0 = 24 out of 50, or 48%.
(5.5+13)/38 = 18.5/38 = 48.68% when you see a low card appear.

Quote

It is also the same as the 37/76 that you quote, and I am happy to accept it as identical, given rounded distribution percentages.

No, it's not identical. There's no rounding in this case, 78% and 22% are absolute exact percentages for 2-1 and 3-0 breaks. So out of 100, it's exactly 11 3-0 breaks and 13 singleton Ks excluded.

Your errors just happened to end up somewhat close to the right answer.