Distributions of Bridge Hands Probability

[tryingtoexcelatmath]

In Bridge we have 4 players, and each get 13 cards.

The most symmetrical hand is the (4, 3, 3, 3) hand which is 4 of 1 card type (spades) 3 hearts 3 diamonds and 3 clubs. (the card type order does not matter)

So the video says:
" We will now count these hands, and to do that we will first specify which suit in your hand has the 4 cards. Once we choose that suit, there are 13 choose 4 choices of the 4 cards in that suit because there are 13 cards of each suit in the deck and we take 4 of them. It doesnt matter which order the cards are dealt. So it is a collection not a sequence. In the remaining suits there are 13 choose 3 choices for the cards in each suit so by the multiplication principle. the total number of different hands with a 4, 3, 3, 3, distribution is 4 * 13 choose 4 * 13 choose 3 * 13 choose 3 * 13 choose 3"

TLDR:
4 * 13 choose 4 * 13 choose 3 * 13 choose 3 * 13 choose 3
we have 67B different hands w/ a total bridge hands being 52 choose 13 = 635B, so answer is 67/635 = 10.5%
10.5% of the time we will get a 4, 3, 3, 3 hand


In the next example (4, 4, 3, 2): the video states:

"Let's work out the number of hands with a 4, 4, 3, 2 distribution, To do this, we first choose which of the 2 suits will have 4 cards"

My first question here is why are we choosing the first 2 suits with 4 cards? In the example above for 4,3,3,3 we just multiplied by 4 in the beginning and that was it.

"There are 4 suits, we choose 2 of them, so there are 4 choose 2 = 6 possible choices of those suits, and in each of those suits, we have to choose 4 of the 13 cards, so there are 13 choose 4 choices for those cards. Now we have 2 suits remaining, one of them has 3 cards the other has 2 cards. First we choose the one that has 3 cards and there are 2 possible choices for that. Then once we've chosen that there are 13 choose 3 choices for the cards we hold in that suit. And Finally there are 13 choose 2 choices for the card in the remaining suit. So by the multiplication rule.."

TLDR:
(4 choose 2) * (13 choose 4) * (13 choose 4) * (2 choose 1) * (13 choose 3) * (13 choose 2)

How come in the second example of 4,4,3,2 we need a (4 choose 2) and then another (2 choose 1) for the choice of suits.

I understand all of the (13 choose x) because we have 13 cards in that suit and we are choosing X cards from each suit.

But in the first example given by the video they only have a 4 in the beginning (which I am assuming is 4 choose 1, we have 4 suits to choose from and we choose 1 of them)

How come in the first 4,3,3,3 example we don't have a (3 choose 1 suit) for the remaining 3 suits?

But in the 4,4,3,2 example they had a (4 choose 2) and then a (2 choose 1) to choose the different suits.

Why is that? What is the difference between the 4, 3, 3, 3, example and the 4, 4, 3, 2 example?



Video URL:

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[tryingtoexcelatmath]

Oh wait.. I just thought of it, I think I know why the video ignored it..

For the 4, 3, 3, 3 distribution we have

4 chose 1 suit
13 choose 4 cards
3 choose 3 suits = 1 <-- so it was ignored
13 choose 3 cards
13 choose 3 cards
13 choose 3 cards

I'm not sure if this is the correct reasoning or not. Would appreciate any help.

My question is why is it 3 choose 3 suits? Is it because the remaining 9 cards to choose are all in the same format (13 choose 3)?

Can someone help explain why we can lump these together instead of doing 3 choose 1 suit, 3 choose 1 suit, 3 choose 1 suit?


In the second example 4, 4, 3, 2, they also lumped the first 2 together.
4 choose 2 suits <-- they lumped together the 13 choose 4
13 choose 4 cards
13 choose 4 cards

What is the rationale behind this please?

 

[JeffM]

For 4-3-3-3 distribution, how many ways can we pick the long suit. Obviously 4. With respect to each suit, if n is the number of cards we have in the suit, we select n from 13 or

[MATH]\dbinom{13}{n} = \dfrac{13!}{n! * (13 - n)!}.[/MATH]
[MATH]4 \dbinom{13}{4} * \dbinom{13}{3} * \dbinom{13}{3} * \dbinom{13}{3} = \\ 4 * \dfrac{13!}{4! * 9!} * \left ( \dfrac{13!}{3! * 10!} \right )^3 \approx 66.906 * 10^8.[/MATH]As you realize, the possible number of hands is

[MATH]\dbinom{52}{13} = \dfrac{52!}{13! * (52 - 13)!} \approx 635.014 * 10^8[/MATH].

The probability of getting a 4-3-3-3 hand is approximately

[MATH]\dfrac{66.906 * \cancel {10^8}}{635.014 * \cancel {10^8}} \approx 10.54\%.[/MATH]
Any questions?

For the 4-4-3-2 distribution, things get a bit more complicated.

How many ways can we choose 2 suits out of 4 to be the long suits.

[MATH]\dbinom{4}{2} = \dfrac{4!}{2! * 2!} = 6.[/MATH]
What are they? SH, SD, SC, HD, HC, DC

Out of the two remaining suits, how many ways can we choose the suit with a doubleton?

Obviously 2.

The number of hands with 4-4-3-2 distribution is

[MATH]6 * 2 * \dbinom{13}{4} * \dbinom{13}{4} * \dbinom{13}{3} * \dbinom{13}{2} =\\ 12 * \left (\dbinom{13}{4} \right )^2 * \dbinom{13}{3} * \dbinom{13}{2} \approx 136.853 * 10^8.[/MATH]The probability of getting a 4-4-3-2 hand is approximately

[MATH]\dfrac{136.853 * \cancel {10^8}}{635.014 * \cancel {10^8}} \approx 21.55\%.[/MATH]

 

[tryingtoexcelatmath]

For 4-3-3-3 distribution, how many ways can we pick the long suit. Obviously 4. With respect to each suit, if n is the number of cards we have in the suit, we select n from 13 or

[MATH]\dbinom{13}{n} = \dfrac{13!}{n! * (13 - n)!}.[/MATH]
[MATH]4 \dbinom{13}{4} * \dbinom{13}{3} * \dbinom{13}{3} * \dbinom{13}{3} = \\ 4 * \dfrac{13!}{4! * 9!} * \left ( \dfrac{13!}{3! * 10!} \right )^3 \approx 66.906 * 10^8.[/MATH]As you realize, the possible number of hands is

[MATH]\dbinom{52}{13} = \dfrac{52!}{13! * (52 - 13)!} \approx 635.014 * 10^8[/MATH].

The probability of getting a 4-3-3-3 hand is approximately

[MATH]\dfrac{66.906}{635.014} \approx 10.54\%.[/MATH]
Any questions?

For the 4-4-3-2 distribution, things get a bit more complicated.

How many ways can we choose 2 suits out of 4 to be the long suits.

[MATH]\dbinom{4}{2} = \dfrac{4!}{2! * 2!} = 6.[/MATH]
What are they? SH, SD, SC, HD, HC, DC

Out of the two remaining suits, how many ways can we choose the suit with a doubleton?

Obviously 2.

The number of hands with 4-4-3-2 distribution is

[MATH]6 * 2 * \dbinom{13}{4} * \dbinom{13}{4} * \dbinom{13}{3} * \dbinom{13}{2} \approx 136.853 * 10^8.[/MATH]
The probability of getting a 4-4-3-2 hand is approximately

[MATH]\dfrac{136.853}{635.014} \approx 21.55\%.[/MATH]

Could you please help explain why they used 4 choose 2 for 4,4,3,2?

What is the math behind it? How come it is allowed to group (13 choose 4) and (13 choose 4) together.

We are making separate choices when we choose the first 4 and then when we choose the next 4. How come we can lump it together into 4 choose 2 instead of (4 choose 1) and another (3 choose 1).

Similarly for 4,3,3,3 how come we are allowed to use (3 choose 3 suites)? We have a grouping at the end of (13 choose 4 cards) (13 choose 4 cards) (13 choose 4 cards) and we lumped these all together with a (3 choose 3 suites) instead of a (3 choose 1) (2 choose 1) (1 choose 1).

Could someone explain why we are allowed to do that?

 

[JeffM]

Could you please help explain why they used 4 choose 2 for 4,4,3,2?

What is the math behind it? How come it is allowed to group (13 choose 4) and (13 choose 4) together.

We are making separate choices when we choose the first 4 and then when we choose the next 4. How come we can lump it together into 4 choose 2 instead of (4 choose 1) and another (3 choose 1).

Similarly for 4,3,3,3 how come we are allowed to use (3 choose 3 suites)? We have a grouping at the end of (13 choose 4 cards) (13 choose 4 cards) (13 choose 4 cards) and we lumped these all together with a (3 choose 3 suites) instead of a (3 choose 1) (2 choose 1) (1 choose 1).

Could someone explain why we are allowed to do that?

I did explain. We have two suits that are 4-4. We have a total of four suits. It is not so much choosing them because we do not choose our hand at bridge. It is a matter of counting. Here are the only possibilities, listing in descending rank order:
spades-hearts
spades-diamonds
spades-clubs
hearts-diamonds
hearts-clubs
diamonds-clubs.

That is six.

Now you can think about it in a different way, which is what I think you are doing.

The first suit to get four cards is spades. Then there are obviously three choices for the second four-card suit. That is 3 ways. Let's list them in rank order: spades-hearts, spades-diamonds, spades-clubs.

The first suit to get four cards is hearts. Then there are obviously three choices for the second four-card suit. That is 3 ways. Let's list them in rank order: spades-hearts, hearts-diamonds, hearts-clubs.

The first suit to get four cards is diamonds. Then there are obviously three choices for the second four-card suit. That is three ways.Let's list them in rank order: spades-diamonds, hearts-diamonds, diamonds-clubs.

The first suit to get four cards is clubs. That is three ways.Let's list them in rank order: spades-clubs, hearts-clubs, diamonds-clubs.

Add all those up and we get twelve. But we have no interest in which suit FIRST got four cards. We are just interested in which suits ULTIMATELY got four cards. Whether spades came first and clubs second or whether clubs came first and spades second is irrelevant. If you go through the combinations, you will see that each combination is listed twice. So we have to divide 12 by 2, which gives 6.

When order does not matter, it is deceptive to think about which came first and which came second. We don't care.

 

[tryingtoexcelatmath]

First, thank you Jeff, I am as dense as a rock.

I understand the explanation about sequence vs combination. I think what I don't understand is why it is allowed to group them together. In the 4,3,3,3 example, you get (4 choose 1 suit) and then a (3 choose 3 suit). Ignoring the (13 choose X cards).

How come you can group the 3 choose 3 suit together? Why not just group 4,3 and then 3,3, so we have 4 choose 2 and 2 choose 2?

Same with the second example of 4,4,3,2. Ok I get how you went through all the pairs, but why can we group 4 and 4 together and not the other two. Or why don't we group 4,4,3 --> into 4 choose 3?

I am sorry for my lack of understanding. I'm just having trouble connecting the dots.

 

[JeffM]

OK, I am not going to write a long answer because I want you to understand piece by piece and have a chance to ask questions.

We have a number of ways in which the suits may differ in length and a number of possible card combinations for each suit. We multiply all those numbers together to find the total number of hands with the specified distribution. So all the talk 4 choose 2 is about finding the right number to put into a series of multiplications.

So one way to think about this is that we compute five numbers and multiply them

Number of ways to specify the length of the suits

Number of ways to specify the cards in suit A

Number of ways to specify the cards in suit B

Number of ways to specify the cards in suit C

Number of ways to specify the cards in suit D

With me on that?

 

[tryingtoexcelatmath]

Yes with you on this.

 

[JeffM]

With respect to the first number, the number of ways that card lengths may be distributed, are you familiar with the notation that says 4 choose 2 can be (and usually is written)

[MATH]\dbinom{4}{2} = \dfrac{4!}{2! * (4 - 2)!}.[/MATH]
More generally, the number of ways to pick k things out of n distinct things without regard to order is

[MATH]\dbinom{n}{k} = \dfrac{n!}{k! * (n - k)!}.[/MATH]
Now let's think about this formula comes about.

If I have m distinct things, how many different orders can I put them in?

I have m choices for what goes into the first position, (m - 1) for what goes into the second position, (m - 2) for what goes into the third position and so on or

[MATH]m * (m - 1) \ ... \ 2 * 1 = m! \text { if } m > 0.[/MATH]
Make sense?

If I have n distinct things and select k of them, how many different orders can I put the k of them in.

Let's take some examples and see what we learn. Pick 3 out of 5. Five ways to pick the first, four ways to pick the second, three ways to pick the third. That is 60. How would we express that in the notation of factorials.

[MATH]60 = \dfrac{120}{2} = \dfrac{5 * 4 * 3 * 2 * 1}{2 * 1} = \dfrac{51}{(5 - 3)!}.[/MATH]
Let's try another. Pick four out of seven distinct things. How many ways can I order them? Seven for the first position, six for the second, five for the third position, and four for the last position. Multiplying 7 times 6 times times 5 times 4, we get 840. How do we express that in terms of factorials.

[MATH]840 = \dfrac{5040}{6} = \dfrac{7 * 6 * 5 * 4 * 3 * 2 * 1}{3 * 2 * 1} = \dfrac{7!}{(7 - 4)!}[/MATH]
Now you can try that with more numbers if you wish and persuade yourself that the general formula for the number of different orderings of k out of n distinct things is

[MATH]\dfrac{n!}{(n - k)!}.[/MATH]
But what if we do not care about the order of the k things. Well we know that we can order k distinct things k! ways. So if we do not care about their order, we have overcounted by k!. Thus, if we want to know how many ways we can select k things out of n distinct things WITHOUT REGARD TO ORDER, we have to go

[MATH]\dfrac{n!}{(n - k)!} \div k! = \dfrac{n!}{(n - k)!} * \dfrac{1}{k!} = \dfrac{n!}{k! * (n - k)!}.[/MATH]
The formulas are not arbitrary. Do you see now how they are derived?

I think we can complete this with one more reply if you are OK with all of this.

Now think about this. When we choose three suits of equal length, do we care about the order in which they are chosen or are we just interested in how many ways in which they can be chosen?

 

[tryingtoexcelatmath]

Yes I am ok with this.

"Now think about this. When we choose three suits of equal length, do we care about the order in which they are chosen or are we just interested in how many ways in which they can be chosen?"

I see. Can you finish the rest of your explanation but I do have a question for the quote above. Could you help explain later the equal length part?

I think that is where I am getting stuck. 4, 4, 3, 2 we combine the 4's because they are equal in length, is there a way to arrive at the same answer without combining the 4's? Similarly with the 4, 3, 3, 3, so we can combine them because they are equal length and we don't care about order. Is there a different way to arrive at the same conclusion without combining them?

I think my question is why combine equal lengths and is there a different way to do it separately but also arrive at the same conclusion.

 

[JeffM]

Yes I am ok with this.

"Now think about this. When we choose three suits of equal length, do we care about the order in which they are chosen or are we just interested in how many ways in which they can be chosen?"

I see. Can you finish the rest of your explanation but I do have a question for the quote above. Could you help explain later the equal length part?

I think that is where I am getting stuck. 4, 4, 3, 2 we combine the 4's because they are equal in length, is there a way to arrive at the same answer without combining the 4's? Similarly with the 4, 3, 3, 3, so we can combine them because they are equal length and we don't care about order. Is there a different way to arrive at the same conclusion without combining them?

I think my question is why combine equal lengths and is there a different way to do it separately but also arrive at the same conclusion.

Yes, we are now at the heart of your confusion. And I think it arises because of the word "combine," which is a vague word in this context.

We are talking about counting.

If we have 4-3-3-3 distribution, we have four possible suits that might be the one with four cards. We can show that as choose 1 out of 4.

[MATH]\dbinom{4}{1} = \dfrac{4!}{1! * (4 - 1)!} = \dfrac{4 * 3!}{3!} = 4.[/MATH]
Now we need to think about the three card suits. Well, we have already accounted for one suit so there are only three remaining suits. How many ways can we choose three 3-card suits from three available suits? Obviously just one way, and we can show that as

[MATH]\dbinom{4 - 1}{3} = \dfrac{(4 - 1)!}{3! * (3 - 3)!} = \dfrac{3!}{3! * 0!} = 1.[/MATH]
Do you see why I wrote 4 - 1? The point is that if spades is the 4-card suit and there are three suits with three cards, then those suits must be hearts, diamonds, and clubs. We do not ignore the 3-card suits; it is just that after we pick one of the four suits as the 4-card suit, there is only one way left to pick three suits out of the remaining ones. So what we are really doing is this computation

[MATH]\left ( \dbinom{4}{1} * \dbinom{(4 - 1)}{3} \right ) * \left \{ \left (\dbinom{13}{4} \right )^1 * \left ( \dbinom{13}{3} \right)^3 \right \}[/MATH]
Now let's think about the 4-4-3-2 hands.

There are now two 4-card suits. How many ways can we pick two from four.

[MATH]\dbinom{4}{2} = \dfrac{4!}{2! * (4 - 2)!} = \dfrac{4 * 3 * 2!}{2! * 2!} = \dfrac{4 * 3}{2} = 6[/MATH]
But that leaves just two suits remaining. How many ways can we pick the one suit with three cards from the remaining two.

[MATH]\dbinom{4 - 2}{1} = \dbinom{2}{1} = \dfrac{2!}{1! * (2 - 1)!} = 2.[/MATH]
That is, if we pick spades and clubs as the 4-card suits, that leaves just two choices for the 3-card suit, hearts or diamonds.

Once we make that pick for the 3-card suit, how many options do we have for the 2-card suit. Just one, which works because because we have only one 2-card suit.

[MATH]\left ( \dbinom{4}{2} * \dbinom{(4 - 2)}{1} * \dbinom{(4 - 2) - 1}{1}\right ) * \\ \left \{ \left (\dbinom{13}{4} \right )^2 * \left ( \dbinom{13}{3} \right)^1 * \left ( \dbinom{13}{2} \right)^1\right \}[/MATH]We are always accounting for four suits. Look at the "denominators" in the left bracket: they add up to 4. Look at the exponents in the right bracket: they add up to 4. And the numerators in the left bracket are the number of suits remaining to pick from.

How about the 7-3-2-1 distribution

[MATH]\left ( \dbinom{4}{1} * \dbinom{(4 - 1)}{1} * \dbinom{(4 - 1) - 1}{1} * \dbinom{\{(4 - 1) - 1\} - 1}{1} \right )\\ \left \{ \left (\dbinom{13}{7} \right )^1 * \left ( \dbinom{13}{3} \right)^1 * \left ( \dbinom{13}{2} \right)^1 * \left ( \dbinom{13}{1} \right)^1 \right \}[/MATH]
Does this help?

 

[tryingtoexcelatmath]

Yes it does, sorry but I can't really find the 'why' in your explanation. I understand everything you wrote and it is very very clear thank you very much.
But I still don't get the why.

For example: "There are now two 4-card suits. How many ways can we pick two from four."
Why do we start with picking 2 suits? Is it because out of the 4 group of card distributions, the first two both have 4 cards?

If so.. why is that allowed to "pick two from four" if the first 2 distribution of cards have the same number of cards.

Similarly if we have 4,5,2,4 <-- in this distribution even if the 1st 4 is first and the 2nd 4 is last, I am assuming we can still use "pick two from four"
Why is that? That is what I don't understand.

Right now I just memorized "if the card numbers are the same, we group them together" but I still dont get why it is allowed.

Ok wait wait, I just read your explanation for the 10th time..

Is it because mathematically we are grouping together the 'cards' part?

So if we have 4, 4, 3, 2, we have:
4 choose 2 suits, 2 choose 1 suits, 1 choose 1 suits
And we also have
13 choose 4 cards, 13 choose 4 cards, 13 choose 3 cards, 13 choose 2 cards
Or it is written as:
(13 choose 4 cards)^2 , 13 choose 3 cards, 13 choose 2 cards

Is that why? because we have (13 choose 4 cards)^2 ? So we also have (4 choose 2 suits) instead of (4 choose 1 suits) (4 choose 1 suits)?

But then.. it would be a different outcome if we did (4 choose 1) (4 choose 1) vs (4 choose 2) <-- not equal

Usually in math if we can group an equation, we can also ungroup it. For example: (x+1)(x−1) = x^2 - 1 <-- left side is ok, right side is ok

So.. I think I am still not understanding 'why'?
For example: Why is that allowed to "pick two from four" if the first 2 distribution of cards have the same number of cards.

 

[JeffM]

There is a lot in your most recent post. I may decide to answer it in pieces. So first you are correct. The first part of the analysis has to do with the number of ways the lengths can be distributed among the four suits. The second part has to do with the number of ways the thirteen cards in a suit can be allocated to x slots in a suit. Let's focus for now on the first part of the analysis.

With a 4-3-3-3 distribution, we have exactly four distinct ways that one suit can have 4 cards. That suit could be spades or hearts or diamonds or clubs. And for each of those 4 ways, there is exactly one way that three suits can each have 3 cards. If spades is the suit with 4 cards, then the other three suits must be hearts, diamonds, and clubs. If hearts is the suit with 4 cards, then the other three suits must be spades, diamonds, and clubs. If diamonds is the suit with 4 cards, then the other three suits must be spades, hearts, and clubs. If clubs is the suit with 4 cards, then the other three suits must be spades, hearts, and diamonds. There are just 4 * 1 = 4 ways to have 4-3-3-3 distribution.

We arrived at that answer by counting. What does that have to do with binomial coefficients? We say to ourselves, we must pick one suit from four to have 4 card length, and then, for each of those, we must pick three suits from the three that are left to have 3 card length. The picking is all in our minds to let us do systematic counting.

[MATH]\dbinom{4}{1} * \dbinom{3}{3} = \dfrac{4!}{1! * 3!} * \dfrac{3!}{3! * 0!} = \dfrac{4 * 3!}{3!} * \dfrac{1}{1}[ =4.[/MATH].

The first coefficient says how many different ways can we pick one suit from four. The second coefficient says how many different ways can we pick three suits from the three that remain.

Now I suspect that one thing that is bothering you is why we started with the 4 card suit. That seems arbitrary. (The probable reason is that bridge players discuss distributions in descending order of length.) It is arbitrary. Let's start with the 3 card suits.

The number of different ways that we can pick three suits out of four is [MATH]\dbinom{4}{3} = 4.[/MATH]
Let's check. Spades-hearts-diamonds, or spades-hearts-clubs, or spades-diamonds-clubs, or hearts-diamonds-clubs.

So, for any one of those, how many ways can we pick one suit out of the one suit remaining.

That number is [MATH]\dbinom{1}{1} = 1.[/MATH]
And 4 * 1 = 4. We get the same answer whether we start with the longer suit or the shorter suit.

How about 4-4-3-2 distribution.

Let's try going in descending order.

How many ways can we pick the two 4-card suits from four? [MATH]\dbinom{4}{2} = 6.[/MATH]
How many ways can we pick the one 3-card suit from the remaining two? [MATH]\dbinom{2}{1} = 2.[/MATH]
How many ways can we pick the one 2-card suit from the remaining one. [MATH]\dbinom{1}{1} = 1.[/MATH]
6 * 2 * 1 = 12.

Let's check.

S=4, H=4, D=3, C=2
S=4, H=4, D=2, C=3
S=4, H=3, D=4, C=2
S=4, H=3, D=2, C=4
S=4, H=2, D=4, C=3
S=4, H=2, D=3, C=4
S=3, H=4, D=4, C=2
S=3, H=4, D=2, C=4
S=3, H=2, D=4, C=4
S=2, H=4, D=4, C=3
S=2, H=4, D=3, C=4
S=2, H=3, D=4, C=4

You can begin to see why we do not do detailed enumeration for this kind of counting. But now let's try in ascending order.

How many ways can we pick the one 2-card suit out of the four suits? [MATH]\dbinom{4}{1} = 4.[/MATH]
How many ways can we pick the one 3-card suit out of the three remaining suits? [MATH]\dbinom{3}{1} = 3.[/MATH]
How many ways can we pick the two 4-card suits out of the two remaining suits? [MATH]\dbinom{2}{2} = 1.[/MATH]
4 * 3 * 1 = 12. Same answer.

Switching gears. It is true that some mathematical operations are not affected by reversing order.

[MATH]3 * 2 = 2 * 3.[/MATH]
But that is not universally true of mathematical operations. Some, not all, operations have that special property called "commutativity."

[MATH]4 \div 8 \ne 8 \div 4.[/MATH]
[MATH]3^2 = 9 \ne 8 = 2^3.[/MATH]
So it is no surprise that

[MATH]\dbinom{4}{1} * \dbinom{4}{1} = \left ( \dbinom{4}{1} \right )^2 \ne \dbinom{4}{2}.[/MATH]
Before we go on to talk about allocating cards in slots within each suit, please ask more questions wherever anything is unclear.

 

[tryingtoexcelatmath]

Oh I think I get it. (sorry for any typos I’m using my iPhone)

From your example here:

For 4,4,3,2

I think I see why it is 4 choose 2.

There are 6 distinct ways to allocate the 2 4’s in the 12 choices below:

S=4, H=4, D=3, C=2 here
S=4, H=4, D=2, C=3
S=4, H=3, D=4, C=2 here
S=4, H=3, D=2, C=4 here
S=4, H=2, D=4, C=3
S=4, H=2, D=3, C=4
S=3, H=4, D=4, C=2 here
S=3, H=4, D=2, C=4 here
S=3, H=2, D=4, C=4 here
S=2, H=4, D=4, C=3
S=2, H=4, D=3, C=4
S=2, H=3, D=4, C=4

how do you count the 3 and 2’s?

It looks like the 3’s and 2’s have 4 different places each?

For 4,3,3,3
S=4 H=3 D=3 C=3
S=3 H=4 D=3 C=3
S=3 H=3 D=4 C=3
S=3 H=3 D=3 C=4

so there’s 4 different ways to have the number 4

Again it looks like there’s more than 1 distinct sequence for the 3’s so I’m not sure how to count it.

 

[JeffM]

Oh I think I get it. (sorry for any typos I’m using my iPhone)

From your example here:

For 4,4,3,2

I think I see why it is 4 choose 2.

There are 6 distinct ways to allocate the 2 4’s in the 12 choices below:

S=4, H=4, D=3, C=2 here
S=4, H=4, D=2, C=3
S=4, H=3, D=4, C=2 here
S=4, H=3, D=2, C=4 here
S=4, H=2, D=4, C=3
S=4, H=2, D=3, C=4
S=3, H=4, D=4, C=2 here
S=3, H=4, D=2, C=4 here
S=3, H=2, D=4, C=4 here
S=2, H=4, D=4, C=3
S=2, H=4, D=3, C=4
S=2, H=3, D=4, C=4

how do you count the 3 and 2’s?

There are three 3's and three 2's. 6 + 3 + 2 = 12. That is interesting, but I do not see how it generalizes to handle problems of this kind.

It looks like the 3’s and 2’s have 4 different places each?

Exactly if we start with the 3-card suit in a 4-4-3-2 distribution, there are 4 ways in which just one suit can have 3 cards. But if we start the analysis that way, how many options are there for the one 2-card suit? Just 3. We no longer have four suits to choose from after we have chosen one to be the 3-card suit.

if we start with the 2-card suit in a 4-4-3-2 distribution, there are 4 ways in which just one suit can have 2 cards. But if we start the analysis that way, how many options are there for the one 3-card suit? Just 3. We no longer have four suits to choose from after we have chosen one to be the 3-card suit.

So we can start with any length of suit, but, once we have done so, that constrains later choices.

For 4,3,3,3
S=4 H=3 D=3 C=3
S=3 H=4 D=3 C=3
S=3 H=3 D=4 C=3
S=3 H=3 D=3 C=4

so there’s 4 different ways to have the number 4

Again it looks like there’s more than 1 distinct sequence for the 3’s so I’m not sure how to count it.

You are absolutely correct, there are four distinct combinations of three suits. But if we look at any one of them, how many choices are there for the fourth suit. Just 1. We get 4 * 1 = 4.

There are four distinct choices of of one 4-card suit. And if we look at any one of them, how many options are left for the remaining three card suits. Just 1. We get 4 * 1 = 4.

There is nothing sacred in which length to consider first, which to consider next, and so on. However we proceed, the initial choice affects the second choice, which affects the third choice. The mathematical result is the same.

[MATH]\dbinom{4}{1} * \dbinom{4 - 1}{3} = \dfrac{4! * 3!}{1! * 3! * 3! * 0!} = \dfrac{4 * 3! * 3!}{3! * 3!} = 4.[/MATH]
[MATH]\dbinom{4}{3} * \dbinom{4 - 3}{1} = \dfrac{4! * 1!}{3! * 1! * 1! * 0!} = \dfrac{4 * 3!}{3!} = 4.[/MATH]

 

[tryingtoexcelatmath]

When you wrote them all out I think I can see why we group them together because the outcome is the same.

4,4,3,2,

S=4, H=4, D=3, C=2 four
S=4, H=4, D=2, C=3
S=4, H=3, D=4, C=2 four
S=4, H=3, D=2, C=4 four
S=4, H=2, D=4, C=3
S=4, H=2, D=3, C=4
S=3, H=4, D=4, C=2 four
S=3, H=4, D=2, C=4 four
S=3, H=2, D=4, C=4 four
S=2, H=4, D=4, C=3
S=2, H=4, D=3, C=4
S=2, H=3, D=4, C=4

If we start with the 4, I see which ones are distinctly 4

But which one's are for 3's?

S=4, H=4, D=3, C=2 four
S=4, H=4, D=2, C=3 <--- here C is 3
S=4, H=3, D=4, C=2 four
S=4, H=3, D=2, C=4 four
S=4, H=2, D=4, C=3
S=4, H=2, D=3, C=4 <--- here D is 3
S=3, H=4, D=4, C=2 four
S=3, H=4, D=2, C=4 four
S=3, H=2, D=4, C=4 four
S=2, H=4, D=4, C=3
S=2, H=4, D=3, C=4
S=2, H=3, D=4, C=4 <--- here H is 3

But isn't 3's suppose to be only 2 choices?

Additionally for 2's:

S=4, H=4, D=3, C=2 four
S=4, H=4, D=2, C=3 <--- here C is 3
S=4, H=3, D=4, C=2 four
S=4, H=3, D=2, C=4 four
S=4, H=2, D=4, C=3 <--- here H is 2
S=4, H=2, D=3, C=4 <--- here D is 3
S=3, H=4, D=4, C=2 four
S=3, H=4, D=2, C=4 four
S=3, H=2, D=4, C=4 four
S=2, H=4, D=4, C=3 <-- here S is 2
S=2, H=4, D=3, C=4
S=2, H=3, D=4, C=4 <--- here H is 3

Isn't 2's suppose to only be 1 choice?

Or am I counting the wrong rows?

I don't really understand how to count the 3's and 2's after counting the 4's

 

[JeffM]

Let's look first at spades. The first six lines show 4 spades. The next three lines show 3. The last three lines show 2.

Now let's look at hearts. Lines 1, 2, 7, 8, 10 and 11 show four hearts. That adds up to six. Lines 3, 4, and 12 show three hearts. Lines 5, 6, and 9 show two hearts.

How about diamonds? Lines 3, 5, 7, 9, 10, and 12 show four diamonds. That adds up to six. Lines 1, 6, and 11 show three diamonds. And lines 2, 4, and 8 show 2 diamonds.

Do you suspect what we shall find if we look at clubs? Lines 4, 6, 8, 9, 11, and 12 show four clubs. That adds up to six. Lines 2, 5, and 10 show three clubs. And lines 1, 3, and 7 show 2 clubs.

We are shuffling these lengths around among the four suits.

I am going to start over. If you were asked how many possible hands have four spades, two hearts, three diamonds, and four clubs, the answer would be

[MATH]\dbinom{13}{4} * \dbinom{13}{2} * \dbinom{13}{3} * \dbinom{13}{4} = \left ( \dbinom{13}{4} \right )^2 * \dbinom{13}{3} * \dbinom{13}{2} = x.[/MATH]
Do you see why?

And do you see that you would get the exact same answer of x if you were asked how many possible hands have 2 spades, four hearts, four diamonds, and three clubs?

So to answer the question of how many possible hands have 4-4-3-2 distribution, we multiply x by the number of ways we can shuffle those lengths among the four suits. Still with me?

OK. Let's assume that we do this. We have four balls labeled S, H, D, and C. They represent the four suits. And we have three buckets labeled 4, 3, and 2. Your job is to pick two balls to go in the 4-bucket, one ball to go in the 3-bucket, and one ball to go in the 2-bucket. How many different ways can you do this?

Let's say we start with the 4-bucket. There are 6 ways to choose 4 from 6. Now there are two balls left and two buckets left. Let's say we work on the 3-bucket. how many ways can we pick one ball out of two to go in the 3-bucket? Two obviously. Now how many ways can we pick a ball to go in the 2-bucket? There is only 1 ball left so the answer is 1.
6 * 2 * 1 = 12.

There are six different sequences in which you can fill buckets. Try some of the others. You will always end up with twelve different ways to fill the three different buckets with four different balls.

The number of balls is always four. The number of buckets is the number of different suit lengths. The number of balls to go into each bucket is the number of suits of that length. Thinking about the process in physical terms may help. Try 7-3-2-1. In fact, you can do the physical process if it helps.

 

[tryingtoexcelatmath]

Let's look first at spades. The first six lines show 4 spades. The next three lines show 3. The last three lines show 2.

Now let's look at hearts. Lines 1, 2, 7, 8, 10 and 11 show four hearts. That adds up to six. Lines 3, 4, and 12 show three hearts. Lines 5, 6, and 9 show two hearts.

How about diamonds? Lines 3, 5, 7, 9, 10, and 12 show four diamonds. That adds up to six. Lines 1, 6, and 11 show three diamonds. And lines 2, 4, and 8 show 2 diamonds.

Do you suspect what we shall find if we look at clubs? Lines 4, 6, 8, 9, 11, and 12 show four clubs. That adds up to six. Lines 2, 5, and 10 show three clubs. And lines 1, 3, and 7 show 2 clubs.

We are shuffling these lengths around among the four suits.

This part definitely makes sense. Is it just a coincidence that the 6 lines show one of the 4 suits?

Because (4 choose 2) is = 6, is this just a coincidence?

Because there are 3 lines that show 3, and 3 lines that show 2.

After (4 choose 2) for 3,2 it is (2 choose 1), then (1 choose 1), which = 2 and 1 respectively.

So does that mean it is just a coincidence that we have 6 lines showing 4 suits?

Because the other (2 choose 1) and (1 choose 1) does not match the number of remaining lines.

 

[JeffM]

I cannot answer your question about "coincidences." Every one of the twelve lines shows two 4-card suits, one 3-card suit, and one 2-card suit. You can persuade yourself by looking and experimenting that no possibility was left out and no possibilty was counted more than once. There are twelve.

I should perhaps have ignored your question about "counting fours" and "counting threes" and "counting twos" because every single one of the twelve has TWO 4-card suits, ONE 3-card suit, and ONE 2-card suit. So the question does not make sense in the abstract. It does, however, make sense with respect to a given suit. With respect to clubs, say, there are six distributions where clubs has 4-cards, three where clubs has 3-cards, and three where clubs has 2-cards. As it must, that adds up to twelve.

So when you ask whether

[MATH]\dbinom{4}{2} * \dfrac{2}{1} * \dfrac{1}{1} = 6 * 2 * 1 = 12 = 6 + 3 + 3[/MATH]
is a "coincidence," I have not a clue what is meant. It is not necessary that one number making up a sum is also a factor of a product that equals that sum.

[MATH]7 + 5 = 12 = 3 * 4.[/MATH]
I doubt that it is possible to tie the individual summands of a sum to individual factors of that sum in any generalizable way. If it is possible, that would lead deep into number theory, where I would be a hopeless because uninformed guide.

You seem to want to find a way to count that avoids combinatorics. I believe the only alternative is exhaustive, non-duplicative listing, which is far less efficient than combinatorics. Please think about (or even experiment with) the balls and the buckets to understand the logic behind combinatorics rather than worrying about whether certain things are coincidences.

 

[JeffM]

To help you with your physical experiment, instead of balls and buckets, get four plates, the four aces from a deck of playing cards, and some pieces of paper.

Now I am going off in a seperate direction to talk about suit lengths in a bridge hand, which contains 13 cards.

I cannot have all four suits be of equal length because 4 does not divide 13 equally. But I can have three suits of equal length: 0-0-0-13, 1-1-1-10, 2-2-2-7, 3-3-3-4. Three suits of equal length, one of longer length. With me? Take two of the plates and put a piece of paper labeled E below one plate and a piece of paper labeled L below the other. We first pick three aces to go on the plate labeled E to represent the three suits of equal length. How many different ways can you do that? With respect to any single one of the set of three aces, how many different ways can you pick an ace to put on the plate labeled L to represent the longer suit? So how many different ways can we have a distribution with three suits of equal length? Now repeat the experiment by first picking one ace to to go on the plate labeled L. How many ways can you do that? With respect to any given ace on plate L, how many ways do you have to pick the three aces to go on plate E? Do you see that what you have done mathematically in the first case is

[MATH]\dbinom{4}{3} * \dbinom{4 - 3}{1} = 4 * 1 = 4.[/MATH]
What you did mathematically in the second case is

[MATH]\dbinom{4}{1} * \dbinom{4 - 1}{3} = 4 * 1.[/MATH]
Can you have two suits of equal length? Sure. So their sum will be an even number. When we subtract an even number from 13 to determine the combined length of the remaining two suits, we get an odd number. But if the sum of two numbers is odd, then one number is even and one is odd. In short, we can have two suits of equal length, one suit of an odd length, and one suit of even length. For example, we can have 5-5-3-0 or 5-5-2-1. So now we have three different lengths. Take three plates. Label one E for equal length, one O for odd length, and one E for even length. Again do the experiment with aces and plates. You will soon persuade yourself that the number of different ways to arrange the aces on the plates is

[MATH]\dbinom{4}{1} * \dbinom{4-1}{1} * \dbinom{4-1-1}{2} = 4 * 3 * 1 = 12 =\\ \dbinom{4}{1} * \dbinom{4 -1}{2} * \dbinom{4-1-2}{1} = 4 * 3 * 1 = 12 =\\ \dbinom{4}{2} * \dbinom{4-2}{1} * \dbinom{4-2-1}{1} = 12.[/MATH]Of course, you can have all four suits of different lengths. Now you need four plates. Label them L for longest, ML for middle but longer, MS for middle but shorter, and S for shortest.

Start experimenting. How many ways can you pick an ace to go on plate L? Given one of those choices, how many ways can you pick an ace to go on plate ML? Given that pair of choices, how many ways can you pick an ace to go on plate MS? Given that triple sequence of choices, how many ways can you pick an ace to go on plate S?

The answer is [MATH]\dbinom{4}{1} * \dbinom{4-1}{1} * \dbinom{4-1-1}{1} * \dbinom{4-1-1-1}{1} = 4 * 3 * 2 * 1 = 24.[/MATH]
But experiment. Try a different sequence of plates.

Mathematicians dislike intensely the notion that mathematics is a physical science because that notion has been incorrect for millennia. Epistemologists dislike intensely the idea that physical science is inductive because that is not the logic of modern science, merely the practice. But we can treat counting as an inductive physical science using plates and cards if we have to to convince ourselves that the abstract logic is correct.