Strange passage in Reese on Play

[quiddity]

In "Reese on Play", in Chapter 1 ('The Simple Probabilities'), Reese writes:

 

"There is another factor to be considered. The further the play has advanced, the more likely are the even divisions. For example, if in the middle of the play a declarer with a combined seven cards in a suit plays two rounds, to which all follow, it is better than evens that the two outstanding cards will break 1-1; in other words a 3-3 break has become more likely than 4-2."

 

How can this be right?

[mikeh]

In "Reese on Play", in Chapter 1 ('The Simple Probabilities'), Reese writes:

 

"There is another factor to be considered. The further the play has advanced, the more likely are the even divisions. For example, if in the middle of the play a declarer with a combined seven cards in a suit plays two rounds, to which all follow, it is better than evens that the two outstanding cards will break 1-1; in other words a 3-3 break has become more likely than 4-2."

 

How can this be right?

I think it depends on whether the opps give reliable count signals :)

 

Sorry for the facetious reply

[ArtK78]

Far be it from me to take issue with Terrence Reese, but I believe the statement is incorrect.

 

I don't have a table of suit distribution probabilities handy, but I found this online:

 

The chances of 6 outstanding cards breaking in a particular manner are as follows:

 

4-2 or 2-4 - 48.5%; 3-3 - 35.5%; 5-1 or 1-5 - 14.5%; 6-0 or 0-6 - 1.5%.

 

Assuming that these are approximately correct, once both opponents have followed to two rounds you can eliminate the 5-1 and 6-0 breaks from consideration. But, all other things being equal, the relative probabilities of the remaining cards breaking 2-0 or 0-2 as opposed to 1-1 are the same as the relative probabilities of the original distribution of the suit being 4-2 or 2-4 as opposed to 3-3 - 57.7% for an original 4-2 or 2-4 distribution, and 42.3% for an original 3-3 distribution.

 

On the other hand, if you find out the distribution of the other suits is fairly well balanced, the chance of a 3-3 break of this suit increases. But that is not what Reese is discussing.

[jdonn]

On the other hand, if you find out the distribution of the other suits is fairly well balanced, the chance of a 3-3 break of this suit increases.  But that is not what Reese is discussing.

I believe that's exactly what he is discussing, and on that basis his claim is quite correct.

 

Think of it like this. With 5 tricks remaining, if you haven't seen one of the suit it can't be 6-0. After another trick it can't be 5-1. So there is at least a sense in which he is definitely right.

[ArtK78]

On the other hand, if you find out the distribution of the other suits is fairly well balanced, the chance of a 3-3 break of this suit increases.  But that is not what Reese is discussing.

I believe that's exactly what he is discussing, and on that basis his claim is quite correct.

 

Think of it like this. With 5 tricks remaining, if you haven't seen one of the suit it can't be 6-0. After another trick it can't be 5-1. So there is at least a sense in which he is definitely right.

In the paragraph quoted in the OP, Reese makes no comment about how the other suits are breaking - only the suit in question. For all we know, the remaining three suits may have broken quite badly. The only thing Reese said is that all followed to the first two rounds of the suit in question. While this does eliminate the possibility of a 6-0 or 5-1 break, it doesn't change the relative probability of a 3-3 or 4-2 break. So, while the absolute probability of a 3-3 break has increased from 35.5% to 42.3%, the relative probability of a 3-3 break versus a 4-2 break has not changed. The odds are still roughly 4-3 in favor of the 4-2 break.

 

If the point is that the absolute probability of a 3-3 break has increased, yes, that is true. But it is still considerably more probable that the suit will break 4-2 as opposed to 3-3. So his point that it is better than even that the remaining cards in the suit will break 1-1 is just not true.

[kfay]

Far be it from me to take issue with Terrence Reese, but I believe the statement is incorrect.

 

I don't have a table of suit distribution probabilities handy, but I found this online:

 

The chances of 6 outstanding cards breaking in a particular manner are as follows:

 

4-2 or 2-4 - 48.5%; 3-3 - 35.5%; 5-1 or 1-5 - 14.5%; 6-0 or 0-6 - 1.5%.

 

Assuming that these are approximately correct, once both opponents have followed to two rounds you can eliminate the 5-1 and 6-0 breaks from consideration. But, all other things being equal, the relative probabilities of the remaining cards breaking 2-0 or 0-2 as opposed to 1-1 are the same as the relative probabilities of the original distribution of the suit being 4-2 or 2-4 as opposed to 3-3 - 57.7% for an original 4-2 or 2-4 distribution, and 42.3% for an original 3-3 distribution.

 

On the other hand, if you find out the distribution of the other suits is fairly well balanced, the chance of a 3-3 break of this suit increases. But that is not what Reese is discussing.

Well it depends on what cards we are looking for. If the holdings are, say,

 

AK103

 

Q42

 

Then all Jx combinations have also been eliminated and in fact the chance of a 3-3 split is slightly more than 52%.

[EricK]

At the start of the hand the 4-2 break is more likely than the 3-3 break. But if you come down to the last three tricks and none of this suit have been discarded then the 3-3 break is now certain.

 

So, do the odds of a 3-3 break go relatively smoothly from "less probable" to "certain", or is there a large jump right at the end with 3-3 being less probable nearly all the way? If the former is the case then that would make Reese right, wouldn't it?

[Echognome]

Well it depends on what cards we are looking for. If the holdings are, say,

 

AK103

 

Q42

 

Then all Jx combinations have also been eliminated and in fact the chance of a 3-3 split is slightly more than 52%.

I don't think this is quite right. What if were "looking for" the 7 and it hadn't appeared. Then all of the 7x combinations have also been eliminated. The same can be said for any particular card you are missing.

 

Although it might change our chance of taking 4 tricks, I can't see how it would make a 3-3 split any more likely.

[kfay]

Well it depends on what cards we are looking for.  If the holdings are, say,

 

AK103

 

Q42

 

Then all Jx combinations have also been eliminated and in fact the chance of a 3-3 split is slightly more than 52%.

I don't think this is quite right. What if were "looking for" the 7 and it hadn't appeared. Then all of the 7x combinations have also been eliminated. The same can be said for any particular card you are missing.

 

Although it might change our chance of taking 4 tricks, I can't see how it would make a 3-3 split any more likely.

What?

 

The point is that the 7 isn't a significant card. This is true.

[Echognome]

What?

 

The point is that the 7 isn't a significant card.  This is true.

My question to you, is how does a "significant card" versus any other card going to affect the split of all the remaining cards?

 

Edit: Let me put it another way. The only reason the jack is "significant" is that it affects the number of tricks we can take. It's not significant when it comes to the split of the cards.

 

Edit2: Thanks Adam. I see your point (and thus Kevin's). It's the underlying assumption of how the opponents will play the cards and how they won't play the J from Jxx randomly. It's a fair enough assumption, although not a certainty.

[awm]

So, do the odds of a 3-3 break go relatively smoothly from "less probable" to "certain", or is there a large jump right at the end with 3-3 being less probable nearly all the way? If the former is the case then that would make Reese right, wouldn't it?

 

It depends on the opponents' discarding tactics. If it's known that neither opponent will ever discard this suit unless he absolutely has to (i.e. has only this suit) then there is a large jump right at the end. If opponents discard in a somewhat random way, or are potentially guarding other suits, then it will be more smooth.

 

My question to you, is how does a "significant card" versus any other card going to affect the split of all the remaining cards?

 

The difference is that we assume opponents will not randomly play the jack when they don't have to. So if the jack falls in two rounds, it is definite that the suit is 4-2. Thus if the jack doesn't fall, the chance of the suit breaking 4-2 diminishes (i.e. any 3-3 break is still possible, but not every 4-2 break is). Watching the seven is not the same, because there are many holdings where the 7 is played in the first two rounds of the suit without a 4-2 break (even 765, they might play the seven in the first two rounds). So it doesn't really give any information about the suit break in the way that seeing that jack might.

[Trumpace]

Maybe Reese was referring to the fact that:

 

if you play two rounds, both follow, and on the third round, you see one of the opps follow (them playing second to the trick), then a 3-3 break is more likely.

 

 

For instance in:

 

AKQT opposite xxx

 

You cash AK, then play x towards QT. If LHO follows, drop is better than finesse.

[jdonn]

In the paragraph quoted in the OP, Reese makes no comment about how the other suits are breaking - only the suit in question. For all we know, the remaining three suits may have broken quite badly. The only thing Reese said is that all followed to the first two rounds of the suit in question. While this does eliminate the possibility of a 6-0 or 5-1 break, it doesn't change the relative probability of a 3-3 or 4-2 break. So, while the absolute probability of a 3-3 break has increased from 35.5% to 42.3%, the relative probability of a 3-3 break versus a 4-2 break has not changed. The odds are still roughly 4-3 in favor of the 4-2 break.

First point, I am willing to give Reese the benefit of the doubt that he knew if you hold AK opposite xx and play them off, and the suit breaks 8-1, that does not make it more likely some other suit will break evenly simply because you have played some tricks.

 

Secondly, I admit I'm not sure about this. However I suspect the relative odds of a 3-3 to 4-2 break DO change once you have played off enough tricks to exclude the 5-1 and 6-0 breaks. But I admit I'm not sure. Are you? If you know how, by all means prove me wrong or right.

[dburn]

What?

 

The point is that the 7 isn't a significant card.  This is true.

My question to you, is how does a "significant card" versus any other card going to affect the split of all the remaining cards?

 

Edit: Let me put it another way. The only reason the jack is "significant" is that it affects the number of tricks we can take. It's not significant when it comes to the split of the cards.

In perfectly straightforward fashion. If with

 

AK104

 

Q32

 

you play off the ace and king and the jack does not appear (as it will not unless it is singleton or doubleton), then the possible remaining divisions of cards are simply:

 

West has Jx (and had Jxxx originally);

East has Jx (and had Jxxx originally);

West has the bare jack (and had Jxx originally);

East has the bare jack (and had Jxx originally).

 

Now, a priori the chance that West or East was originally dealt Jxx is around 36% (the chance of a 3-3 break). The chance that West or East was originally dealt Jxxx is around 32% (two thirds of the chance of a 4-2 break). Thus, the chance that the suit was originally 3-3 (and is now 1-1) is 36/68, or around 53%.

 

It is true that if you began with

 

AKJ4

 

Q32

 

and played off the ace and queen whilst specifically looking for, say, the seven, then if the seven has not appeared, the chance that the suit was originally 3-3 has also risen from around 36% to around 52%. This may come in useful someday, although it is difficult to imagine in precisely what circumstances.

[awm]

It is true that if you began with

 

AKJ4

 

Q32

 

and played off the ace and queen whilst specifically looking for, say, the seven, then if the seven has not appeared, the chance that the suit was originally 3-3 has also risen from around 36% to around 52%. This may come in useful someday, although it is difficult to imagine in precisely what circumstances.

This is not true.

 

You know that no one started with doubleton seven of course. So the possibilities are:

 

E or W started with 7xx, never played the 7, and now has bare 7

E or W started with 7xxx, never played the 7, and now has 7x

 

But the probabilities you gave only apply if you assume that the seven would never be played unless "forced." In fact, supposing that when I have three or four small spot cards which can never possibly win a trick I just play randomly, the odds that I would've played the seven in two rounds from 7xx are 2/3, whereas playing the 7 from 7xxx has only probability 2/4.

 

The upshot is that the odds of a 3-3 break have only improved in that 5-1 and 6-0 have been ruled out (or in other words, not at all relative to the odds of a 4-2 break) if you assume that the opponents simply follow suit with random cards. The probabilities only shift substantially if there are certain cards that they would always/never play if they could.

[dburn]

It is true that if you began with

 

AKJ4

 

Q32

 

and played off the ace and queen whilst specifically looking for, say, the seven, then if the seven has not appeared, the chance that the suit was originally 3-3 has also risen from around 36% to around 52%. This may come in useful someday, although it is difficult to imagine in precisely what circumstances.

This is not true.

Indeed it isn't - my apologies; I should have strengthened the suit to

 

AKJ8

 

Q109.

 

Even then, if an opponent would play randomly (or in suit-preference or count fashion) instead of bottom-up from small cards, then what I said is still not true for more or less the reasons that awm gives. However, the vast majority of opponents do not play like that - at least, not at the beginner or even the intermediate level.

 

A related question that may assist Echognome in considering what are significant cards is this one: you begin with

 

AK43

 

Q5

 

and play off three rounds, all following. Which defender (assuming that both opponents will play insignificant cards at random) is more likely to have the remaining card in the suit?

[whereagles]

I think the point is, the further you play on, if bad breaks were about, they would have shown already. So stuff rates to break ok.

[ArtK78]

In the paragraph quoted in the OP, Reese makes no comment about how the other suits are breaking - only the suit in question.  For all we know, the remaining three suits may have broken quite badly.  The only thing Reese said is that all followed to the first two rounds of the suit in question.  While this does eliminate the possibility of a 6-0 or 5-1 break, it doesn't change the relative probability of a 3-3 or 4-2 break.  So, while the absolute probability of a 3-3 break has increased from 35.5% to 42.3%, the relative probability of a 3-3 break versus a 4-2 break has not changed.  The odds are still roughly 4-3 in favor of the 4-2 break.

First point, I am willing to give Reese the benefit of the doubt that he knew if you hold AK opposite xx and play them off, and the suit breaks 8-1, that does not make it more likely some other suit will break evenly simply because you have played some tricks.

 

Secondly, I admit I'm not sure about this. However I suspect the relative odds of a 3-3 to 4-2 break DO change once you have played off enough tricks to exclude the 5-1 and 6-0 breaks. But I admit I'm not sure. Are you? If you know how, by all means prove me wrong or right.

Yes, I am sure. But my assumption is that nothing has occurred which effects the odds of the division of the suit. Some posters posited that one of the missing 6 cards was a significant card, as in a situation where the only high card missing is the J. Another poster mentioned that if you have played 10 tricks without playing the suit and neither opponent has discarded a card in the suit, then the remaining cards in the suit must split 3-3. Obviously, when you play other cards or play cards in the suit and things happen or do not happen (the J falls or doesn't fall, or no one ever discards a card in the suit) the odds can change.

 

I am assuming that nothing has occurred which effects the odds of the division of the suit. In that case, the fact that you play 2 rounds of the suit with both opponents following eliminates the possibilities of a 6-0 break and a 5-1 break, but it does not effect the a priori relative probability of a 4-2 break versus a 3-3 break.

 

HOWEVER, when you play the third round of the suit and the next player follows, things have changed. Now you can eliminate half of the possible 4-2 breaks. At this point, the odds of a 3-3 break are higher than the odds of the only remaining possible 4-2 break. The odds in favor of the 3-3 break are now 35.5 to 21.2, or in excess of 3-2 in favor of the 3-3 break.

[jdonn]

Um, you posted that you are sure, then spent three paragraphs not proving why. You went off on a tangent in the first paragraph, simply repeated your claim in the second, and made a completely incorrect claim in the third.

 

Your third paragraph is the same argument by which beginners try to drop the queen doubleton offside with AKJx and xxxx. They cash the ace then lead toward the king, lho follows, they say "well if the suit is 3-2 onside, it's as likely RHO's last card is the missing x as it is the missing Q, so my play is a 50-50 guess."

[gwnn]

This thread reminded me of

 

http://forums.bridgebase.com/index.php?showtopic=18964

[barmar]

I think the point is, the further you play on, if bad breaks were about, they would have shown already. So stuff rates to break ok.

That's how I interpreted it as well. In suit contracts, opponents are likely to lead their short suits in an attempt to get a ruff. At NT, they'll lead their long suits to try to set them up. So towards the end, you mainly have the even suits left.

[Echognome]

As to the original question and to a point made earlier, if there are 3 tricks left and neither defender has played the suit then it is a certainty that they split 3:3. If there are 4 tricks left then the odds are 4:3 in favor of a 3:3 split. The odds start out at against a 3:3 split. I leave it as an exercise to show this relationship is monotonic and to calculate the number of remaining tricks left when the odds become greater than "evens".

[H_KARLUK]

In "Reese on Play", in Chapter 1 ('The Simple Probabilities'), Reese writes:

 

"There is another factor to be considered.  The further the play has advanced, the more likely are the even divisions.  For example, if in the middle of the play a declarer with a combined seven cards in a suit plays two rounds, to which all follow, it is better than evens that the two outstanding cards will break 1-1; in other words a 3-3 break has become more likely than 4-2." 

 

How can this be right?

When opponents    Division   Probability

hold

      6 cards                  3-3            36%

      6 cards                  4-2            48%

      6 cards                  5-1            15%

      6 cards                  6-0              1%

 

A point worth remembering is that when th opponents have an odd number of cards in a suit th most probable division is th most even one. When opponents have an even number of cards, however, an uneven break is th most probable.

There is no need to commit this table to memory, but a knowledge of th main points can be useful in helping to determine th best line of play. It is immediately seen, for instance, that a simple finesse (50 %) is a better shot than relying on a 3-3 break (36%), but a worse bet than playing for a 3-2 break (68 %).

 

Finesse or drop

When opponents    Probability of an honor card being

have                        Singleton    Doubleton    Trebleton

    6 cards                  2.5%              16%              36%

 

When opponents have

2 cards               Play for th drop

3 or 4 cards        Finesse against th king but not against th queen or knave

5 or 6 cards        Finesse against th king or queen but not against th knave

7 cards               Finesse against th king, queen or knave

[Trumpace]

 

 

HOWEVER, when you play the third round of the suit and the next player follows, things have changed. Now you can eliminate half of the possible 4-2 breaks. At this point, the odds of a 3-3 break are higher than the odds of the only remaining possible 4-2 break. The odds in favor of the 3-3 break are now 35.5 to 21.2, or in excess of 3-2 in favor of the 3-3 break.

 

 

Your third paragraph is the same argument by which beginners try to drop the queen doubleton offside with AKJx and xxxx. They cash the ace then lead toward the king, lho follows, they say "well if the suit is 3-2 onside, it's as likely RHO's last card is the missing x as it is the missing Q, so my play is a 50-50 guess."

 

Art's reasoning seems right to me... What am I missing?

[awm]

As to the original question and to a point made earlier, if there are 3 tricks left and neither defender has played the suit then it is a certainty that they split 3:3. If there are 4 tricks left then the odds are 4:3 in favor of a 3:3 split. The odds start out at against a 3:3 split. I leave it as an exercise to show this relationship is monotonic and to calculate the number of remaining tricks left when the odds become greater than "evens".

Not true.

 

The problem is that there is a difference between the following processes:

 

(1) Deal out eight cards to two people, of which six are clubs. Check the club break.

 

(2) Deal out twenty-six cards to two people, of which six are clubs. Ask each person to discard nine non-clubs if possible. Check the club break.

 

In the first case, the odds of a 3-3 club break are 4/7 and a 4-2 club break is 3/7, exactly the odds given by Echognome and 4:3 in favor of a 3-3 split.

 

But in the second case, the odds of a 3-3 club break before the discards were 35.5% and a 4-2 break was 48.5%. The discards of non-clubs don't change anything here (except eliminating the 5-1 and 6-0 break possibilities) so the odds remain the same, roughly 4:3 in favor of a 4-2 split!