Singletons in bridge hands

[Douglas]

Yes, I am 8 years late in answering this original posting.

The probability of having a singleton in a hand of a deal
at bridge is 0.32

The probability to have at least one singleton is 0.308...

True. 0.295584 + 0.012909 + 0.000004 = 0.308497...

These are the probabilities of a single singleton in the other
three hands + two singletons + three singletons.

0.32 is the expected number, which differs since you might
have 2 singletons (or 3 since this is a math group).

Meaningless nonsense.

...and thus 1.28 that there is a singleton amongst the four
hands.

That (meaningless number) was apparently calculated on the
assumption that different hands to having singletons are
exclusive events, which is far from true.

I fully agree.

Easy follow on question: How much does the known occurrence
of a singleton in South's hand change the probability for
East to have a singleton?

Easy?

It seems it depends completely on the shape of South's hand.

The two most extreme probabilities:

South has a 4441 hand. There is a 0.110357 probability of a
singleton in the other 39 cards in this deal. Therefor, any
one of the other hands has 1/3rd of that probability.


South has a 10210 hand. There is a 0.191909 probability of a
singleton in the other 39 cards in this deal. Therefor, any
one of the other hands has 1/3rd of that probability.

Douglas

[Douglas]

Post by Douglas
Yes, I am 8 years late in answering this original posting.

The probability of having a singleton in a hand of a deal
at bridge is 0.32

The probability to have at least one singleton is 0.308...

True. 0.295584 + 0.012909 + 0.000004 = 0.308497...
These are the probabilities of a single singleton in the other
three hands + two singletons + three singletons.

0.32 is the expected number, which differs since you might
have 2 singletons (or 3 since this is a math group).

Meaningless nonsense.

...and thus 1.28 that there is a singleton amongst the four
hands.

That (meaningless number) was apparently calculated on the
assumption that different hands to having singletons are
exclusive events, which is far from true.

I fully agree.

Easy follow on question: How much does the known occurrence
of a singleton in South's hand change the probability for
East to have a singleton?

Easy?
It seems it depends completely on the shape of South's hand.
South has a 4441 hand. There is a 0.110357 probability of a
singleton in the other 39 cards in this deal. Therefor, any
one of the other hands has 1/3rd of that probability.
South has a 10210 hand. There is a 0.191909 probability of a
singleton in the other 39 cards in this deal. Therefor, any
one of the other hands has 1/3rd of that probability.
Douglas
These are the probabilities of a single singleton in the other
three hands + two singletons + three singletons.

I concentrated on the numbers so much, I messed up this paragraph.
Below is the corrected paragraph.

These are the probabilities of a single singleton in one hand +
two singletons + three singletons.

Douglas

[Charles Brenner]

Post by Douglas
Yes, I am 8 years late in answering this original posting.

The probability of having a singleton in a hand of a deal
at bridge is 0.32

The probability to have at least one singleton is 0.308...

True. 0.295584 + 0.012909 + 0.000004 = 0.308497...
These are the probabilities of a single singleton in the other
three hands + two singletons + three singletons.

0.32 is the expected number, which differs since you might
have 2 singletons (or 3 since this is a math group).

Meaningless nonsense.

Certainly meaningless as snipped. Possibly the complete context would tells us what "the" and "differs" refer to. Maybe not - I've become aware that sometimes my writing is overly terse. I hope I write more clearly now than I did in 2011.

Post by Douglas

...and thus 1.28 that there is a singleton amongst the four
hands.

That (meaningless number) was apparently calculated on the
assumption that different hands to having singletons are
exclusive events, which is far from true.

I fully agree.

Easy follow on question: How much does the known occurrence
of a singleton in South's hand change the probability for
East to have a singleton?

Easy?

Hard to guess whether I meant that as tongue in cheek since a lot of arithmetic would be needed, or expressing the fact that it's only arithmetic and we have machines to do arithmetic, or hinting that there's a clever shortcut.

Post by Douglas
It seems it depends completely on the shape of South's hand.

Dividing the calculation into cases according to South's distribution is a reasonable approach.

Post by Douglas
South has a 4441 hand. There is a 0.110357 probability of a
singleton in the other 39 cards in this deal. Therefor, any
one of the other hands has 1/3rd of that probability.
South has a 10210 hand. There is a 0.191909 probability of a
singleton in the other 39 cards in this deal. Therefor, any
one of the other hands has 1/3rd of that probability.

South is 10-2-1-0? Then 0.19 seems wrong. Unless the 10-card suit is 10-3-0-0 around the table someone has a singleton in it, so that seems like about 96% (1-(1/3)^3). Since the suit might be 10-1-1-1, the chance for any of the other players to have a singleton is more than 1/3 of that; it's 0.39.

[Charles Brenner]

Post by Charles Brenner

Post by Douglas
Yes, I am 8 years late in answering this original posting.

The probability of having a singleton in a hand of a deal
at bridge is 0.32

The probability to have at least one singleton is 0.308...

True. 0.295584 + 0.012909 + 0.000004 = 0.308497...
These are the probabilities of a single singleton in the other
three hands + two singletons + three singletons.

0.32 is the expected number, which differs since you might
have 2 singletons (or 3 since this is a math group).

Meaningless nonsense.

Certainly meaningless as snipped. Possibly the complete context would tells us what "the" and "differs" refer to. Maybe not - I've become aware that sometimes my writing is overly terse. I hope I write more clearly now than I did in 2011.

I see (and stand by) my meaning now.

The context of the original conversation years ago must have been that someone posted 0.32 as the chance to have a singleton somewhere around the table. That number is a bit higher than the correct 0.308 and I was pointing out that the reason for the error is that 0.32 arises by making the mistake of double-counting deals that include 2 (or 3) singletons.

Perhaps if we start with Doug's calculation

0.295584 + 0.012909 + 0.000004 = 0.308497...

and modify it to

0.295584 + 2x0.012909 + 3x0.000004,

we'll get to 0.32.

[Douglas]

Post by Douglas
It seems it depends completely on the shape of South's hand.
South has a 4441 hand. There is a 0.110357 probability of a
singleton in the other 39 cards in this deal. Therefor, any
one of the other hands has 1/3rd of that probability.
South has a 10210 hand. There is a 0.191909 probability of a
singleton in the other 39 cards in this deal. Therefor, any
one of the other hands has 1/3rd of that probability.
Douglas

I revised my spreadsheet to analyze this question as a matrix
rather than as a vector sequence. I obviously had at least
one logic error in my original posting.

First, holding a 10-2-1-0 hand: I have never found estimates
based on proportions to work for me with multiple combinations.
Combin(3,1)*combin(36,12) = 0.4623. Add the other three suits,
and the final probability is 0.5446/3 = 0.1815 for at least
one singleton in one of the other hands.

Next, holding a 4-4-4-1 hand: (3*Combin(9,1)*combin(30,12))
*Combin(12,1)*Combin(27,12) = 0.3132/3 = 0.1044 for at least
one singleton in one of the other hands..

Lastly, I note the special case I did not notice the first
time around. A 12-1-0-0 hand has a probability of 1 for the
12-card suit. I very rarely run into a 1 for an individual
probability. The same total probability for this hand is
1.0566/3 = 0.3522.

Douglas

[Charles Brenner]

Restoring necessary context:

1. The main question we're dealing with which I posed is, given that South is dealt a singleton what's the chance of East being dealt a singleton?

2. A secondary question which you posed is, if South is 10-2-1-0 what's the chance of East being dealt a singleton?

Re 2, I wrote that East's chance of holding a singleton spade (South's 10-card suit) is 0.39.
---------

Post by Douglas

Post by Douglas
It seems it depends completely on the shape of South's hand.

I didn't mention before but the way you phrased the above made me fear that you might think there is no answer to my question but rather many answers, one for each distribution of South. I hope I misunderstood and that we agree there is a single answer (and one way to compute it is to make a separate calculation for each of the 296 possible distributions South may hold and then take a suitable weighted average).

Post by Douglas

Post by Douglas
South has a 4441 hand. There is a 0.110357 probability of a
singleton in the other 39 cards in this deal. Therefor, any
one of the other hands has 1/3rd of that probability.
South has a 10210 hand. There is a 0.191909 probability of a
singleton in the other 39 cards in this deal. Therefor, any
one of the other hands has 1/3rd of that probability.
Douglas

I'll snip your private thoughts that are not comprehensible as written.

Post by Douglas
Combin(3,1)*combin(36,12) = 0.4623.

(You omitted to write division by combin(39,13) in the formula, but I agree with 0.46.)

But from what ensues you apparently lost track of what you are calculating. 0.46 (not 0.39 as I wrote before when I left out 13 cards) is in fact the chance that East has a singleton spade given that South is 10-2-1-0. Therefore the answer to question 2, the chance that East has any singleton, is slightly more ...

Post by Douglas
Add the other three suits,
and the final probability is 0.5446/3 = 0.1815 for at least
one singleton in one of the other hands.

so dividing by 3 above is nonsense. I tried to explain in my previous post why it's obvious even without calculating that, when South has 10 spades, East has a huge chance to have a singleton. But if you prefer to calculate, here are some numbers. Assuming South has 10 spades the probabilities are

0.284 that E is void in spades,
0.462 that E has a singleton spade,
0.222 that E has a doubleton spade,
0.031 that E has the other 3 spades.
Note that the above probabilities add to 1, as they must. Your proposal that the biggest one is somewhat under 0.18 doesn't pass the smell test.

Incidentally apparently you obtained 0.5446 simply by adding the chances of a singleton in each of the four suits which neglects double counting. I.e. it's 0.46 and 0.04 that East has a singleton spade or heart respectively. But that doesn't mean it's 0.5 to have a singleton major. East's chance to have a singleton heart when *not holding a singleton spade* must be about (1-0.46)x0.04 (& similar correction for the other suits) hence it would be more accurate to round 0.5446 to 0.5.

[Douglas]

So, your final conclusion is East, specifically East, has an approximate 50/50 chance to have singleton in his/her hand if south has a 10-2-1-0 hand. Do I understand that correctly?

If so, does not North and West each have exactly the same 50/50 chance? I cannot get that to add up to 1.

I await your answer with interest.

Douglas

[Charles Brenner]

Post by Douglas
So, your final conclusion is East, specifically East, has an approximate 50/50 chance to have singleton in his/her hand if south has a 10-2-1-0 hand. Do I understand that correctly?

You understand correctly.

Post by Douglas
If so, does not North and West each have exactly the same 50/50 chance?

Yes they do.

Post by Douglas
I cannot get that to add up to 1.
I await your answer with interest.

I'll do my best. You're having a blind spot. Our recent thread began with you styling my 2011 comment about "expected number" as "nonsense". I'm glad I now have your attention and interest in learning the difference between "expected number" and "probability".

You have just implied that the probabilities of East, North, and West to have a singleton must add to 1 (though I suspect you really meant that they cannot sum to more than 1 as per your own number the total is less than 1). Why would you think they cannot? If I told you that each hand is 50% to have more red cards than black, would you similarly argue that I must be wrong because the three probabilities total more than 1?

In effect you seem to think that adding two probabilities should always result in a probability. That's necessarily so when the two probabilities relate to mutually exclusive events, but is not so in general.

As a reductio ad absurdum suppose we deal from a simplified deck with 1-card suits. Each player gets a single card. The probability of East to have a singleton is 1. Same for North, South, and West. The total of the probabilities is 4. Do you agree? If not, where do you claim I went wrong?

Of course 4 is not a probability. It does have a name though. It's called the "expected" number of singletons in a deal. ("Expected" means about the same as "average", except that "expected" generally applies to a mathematically defined situation where it makes sense to average over an infinite number of trials.)

Returning to 52 card deals, again it is possible for more than one hand in a deal to have a singleton. If each of East, West, and North is 50% to have a singleton then over 1000 deals we expect 1500 hands with a singleton. 1500/1000 or 1.5 is then the expected number of hands with a singleton per deal. Nothing paradoxical or peculiar.

[Douglas]

Post by Charles Brenner
In effect you seem to think that adding two probabilities
should always result in a probability.

Change "should" to "can" in this sentence, and I fully agree
with it. But I currently see this as merely another rabbit
hole you want me to chase you down in this particular thread.

I will make this my final effort with you: I choose to
consider the 12-1-0-0 bridge deal for illustration. It has
the unique characteristic of ensuring there is at least one
singleton in one of the other three bridge hands in a
single deal. That is probability 1.00.

Now we need to assign this certainty to those three hands.
Does any one hand have any better chance at having the
minimum possible singleton than another? No. Therefore,
each has the same chance. Hence, 1/3 + 1/3 + 1/3 = 1.

Douglas

[ais523]

Post by Douglas

Post by Charles Brenner
In effect you seem to think that adding two probabilities
should always result in a probability.

Change "should" to "can" in this sentence, and I fully agree
with it. But I currently see this as merely another rabbit
hole you want me to chase you down in this particular thread.
I will make this my final effort with you: I choose to
consider the 12-1-0-0 bridge deal for illustration. It has
the unique characteristic of ensuring there is at least one
singleton in one of the other three bridge hands in a
single deal. That is probability 1.00.
Now we need to assign this certainty to those three hands.
Does any one hand have any better chance at having the
minimum possible singleton than another? No. Therefore,
each has the same chance. Hence, 1/3 + 1/3 + 1/3 = 1.

Err, in this case, each of the other three hands has a more than 1/3
chance to contain a singleton, and the total probability is more than 1.

The reason is that there might be more than one singleton among the
other hands. For example, say we have a 12=1=0=0 hand; someone will have
a singleton spade, up to two people might have a singleton club or
diamond, we have a singleton heart but up to two other players might
also have one. So the average number of singletons among the other
three players is more than 1.

The average number of singletons held by a particular, specific other
player is 1/3 the average number of singletons among the other three
players (by symmetry). The probability that a particular, specific other
player holds at least one singleton is slightly less than this (because
the average counts hands with two singletons twice and with three
singletons three times, whereas the probability counts these cases only
once).

In cases where multiple results can't happen, your reasoning is correct.
For example, each of the other players has a 1/3 chance of holding a
spade singleton, 1/3 of the 100% chance that a spade singleton exists;
the reason your reasoning works here is that multiple spade singletons
are impossible when someone has a 12=1=0=0 hand, neither within the same
hand or between two different hands.

[Charles Brenner]

Post by Douglas

Post by Charles Brenner
In effect you seem to think that adding two probabilities
should always result in a probability.

Change "should" to "can" in this sentence, and I fully agree
with it. But I currently see this as merely another rabbit
hole you want me to chase you down in this particular thread.

That's disappointing as it testifies that you didn't manage to understand even the simple examples I provided you. What was your reaction to my reduction ad absurdum - the 4 card deck? Plug your ears and bellow "I'm not listening"?

Post by Douglas
I will make this my final effort with you: I choose to
consider the 12-1-0-0 bridge deal for illustration. It has
the unique characteristic of ensuring there is at least one
singleton in one of the other three bridge hands in a
single deal. That is probability 1.00.
Now we need to assign this certainty to those three hands.
Does any one hand have any better chance at having the
minimum possible singleton than another? No. Therefore,
each has the same chance. Hence, 1/3 + 1/3 + 1/3 = 1.

ais523 handled this well -- fortunately as it's not rewarding to me to play with you the one-sided game where you write but don't read.

[Douglas]

Post by Douglas
Lastly, I note the special case I did not notice the first
time around. A 12-1-0-0 hand has a probability of 1 for the
12-card suit. I very rarely run into a 1 for an individual
probability. The same total probability for this hand is
1.0566/3 = 0.3522.

This is the final paragraph of a recent posting of mine in
this thread. Do you suppose this is the specific probability
results you refer to?

What I referred to with merely 1/3 of 1 was to refute
Brenner's assertion that the probabilities did not need to
be allocated among the three non-south hands. No more, no
less.

Douglas

[Charles Brenner]

Post by Douglas
What I referred to with merely 1/3 of 1 was to refute
Brenner's assertion that the probabilities did not need to
be allocated among the three non-south hands. No more, no
less.

From those words no one could understand what you think Brenner asserted. (What "the"? Why plural?) From the discussion history one could guess that you imagine if Brenner denies that "all men are rich" he must be claiming that "no man is rich."