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### Randomly Setting the Table

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Date: 1/25/99 at 15:11:44
From: Whitney Rooney
Subject: Probability

John has 4 place settings. Each consists of a cup, a saucer, and a
plate; and each setting is a different color (red, green, blue, and
yellow). If he places the settings randomly around a table, what is
the probability that at least one place setting -- cup, saucer, and
plate -- is of the same color?

Most of us got 1/16 for an answer. We made a tree diagram and counted
the possibilities that the colors would match. We assumed that the
first layer is plates and it doesn't matter how you lay them out.
Then there are 16 ways to lay out the saucers, (4 ways at each four
plates), then there are 4 ways to lay out the cups at each of the 16
plate and saucer combinations. So there are 64 combinations, and 4
have the same color, so 4/64 = 1/16.

I took squares of paper and colored the sides. For the first square
(plates), the order of the colors doesn't matter. The next square is
bigger and is the saucers. The colors can be in 6 different orders,
and placed around the plates in 24 ways, since 6x4 = 24. The
possibility that any match is 15/24. The next step was to realize that
there is a different probability for third match when all 4 colors of
the plates and saucers match, 2 of the colors match or just 1 color
matches.

It got real complicated, but out of 576 combinations, 123 should match
at least one color.

My friend and I spent hours Saturday trying to figure this out with
our parents.

I think the problem is like the marbles in a jar that you answered in
the archives, because what you draw changes the probability of the
next draw.

Whitney
```

```
Date: 01/25/99 at 16:27:23
From: Steve Rooney
Subject: Re: probability

I would like to expand a bit on our thinking above. I would appreciate
any comments you may have time to give.

I divided the problem into levels 1, 2, and 3. In the first level,
plates, the order of placement doesn't affect the probability. Level 2
is the placement of the saucers, and level 3 is the placement of the
cups.

I divided level 2 into the probability that one and only one color
match, that 2 colors match, and that all 4 colors match. The total
probability that there is a match at level 2 is 15/24. This consists
of the probability of one and only one match (8/24), of two matches
only (6/24), and of all four matches (1/24).

If one and only one color matches of the plates and saucers, the
probability that the cup will match any one of these is 6/24. So the
chance of all three matching is 8/24 * 6/24 = 48/576.

If at level 2, there are 2 matches, (e.g. both the red and green
plates and saucers match), we have 6 out of 24 ways to match two
items, and 10 unique ways for each of those to match with a third
item, so the probability of a complete match is 60/576.

The possibility of three matches at level 2 is 0, for if 3 plates
matched the saucers, then the fourth will also match.

The probability that all four match at level 2 is 1/24. If all four
match, then the probability that any one of the remaining cups
completes a match is 15/24. So the probability of a complete match is
15/24 * 1/24 = 15/576.

Thus the total probability is 48/576 + 60/576 + 15/576 = 123/576,
which is approximately .21.
```

```
Date: 01/26/99 at 20:19:05
From: Doctor Anthony
Subject: Re: probability

I will go through the calculation to find the probability that all
three items match in at least one position. (This would also include
2, 3, or 4 positions being matched.)

We can approach this in a different way, by considering what happens
as we increase the number of positions at the table from 1 to 2 to 3
to 4.

I will show the table of the calculations first and then explain how
the entries are made. I am assuming that the colour of the cup is
fixed and we then allocate a saucer and a plate at random. The number
of colours is taken to equal the number of places. I will describe the
process in detail below. Although the explanation is necessarily long-
winded, making up the table was done very quickly.

No. of   Total   0 Correct  1 Correct  2 Correct  3 Correct  4 Correct
Places   Ways    ---------  ---------  ---------  ---------   --------
-------  ----
1        1        0           1
2        4        3           0          1
3       36       26           9          0          1
4      576      453         104         18          0          1
----------------------------------------------------------------------

With 1 place, we initially have 1 colour and so the settings MUST be
correct.

With 2 places, we start with 2 colours. The plates could be chosen in
2 different orders and also the saucers, giving 4 possibilities with
the settings. With just 2 colours, there is 1 way only for 2 to be
correct. There cannot be just 1 correct with only 2 colours, so the
column under 1 correct shows 0. By subtraction, the number of ways
with 0 correct is 3.

With 3 places, we start with 3 colours. The plates could be arranged
in 3! = 6 ways and also the saucers in 6 ways, giving 36 ways with 3
colours. Using the fact that 3 arrangements existed, giving 0 correct
places when only 2 colours were used, we can think of adding a
complete correct place to these 3 to get 1 correct place. And the
correct colour could be chosen in 3 ways giving 3 x 3 = 9 with 1
correct. We cannot have 2 correct, and then 3 correct can clearly be
chosen in 1 way only. By subtraction, the number under 0 correct is
26.

Finally, we start with 4 colours. Now the 26 under the 0 correct
column for 3 places can be added to by a correct placing and then
entered under the 1 correct column. But the correct place can be
chosen in 4 ways, giving 4 * 26 = 104 under the 1 correct column.
There is only 1 possibility under the 4 correct column and zero under
the 3 correct column. For 2 correct, we can see that 18 = 3 * 6 is the
number. The following should illustrate why.

G and Y are to be the incorrect placings, R and B are correct.

RB|GY   RB|GY   RB|GY
RB|GY   RB|YG   RB|YG
RB|YG   RB|YG   RB|GY

This shows 3 ways that the incorrect placings can be arranged. The two
correct colours, R and B in example above, can be chosen in C(4,2) = 6
ways. So the total for 2 correct colours is 3 * 6 = 18.

Therefore, we have the total with 4 placings giving at least one
correct setting:

104 + 18 + 1 = 123

And the required probability is 123/576 = 41/192 = 0.21354.

This answer agrees with the calculation (though not using the same
method) as that which you developed.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 01/30/99 at 10:08:16
From: Doctor Anthony
Subject: Re: probability

Returning to the problem of the cups, saucers and plates, I think that
a general method can be developed such that the probabilities in the
cases of 6, 8 or 12 place settings can easily be calculated,
especially with the aid of a spreadsheet like Excel. I will go through
a table for a total of six place settings, but clearly the method can
be extended indefinitely without invoking advanced theorems of
combinatorial theory.

The basic idea is very simple. If you have the results for n place
settings, then the results for n+1 settings is obtained from the
previous line for, say, r correct settings by imagining that you add a
correct setting to the existing arrangements (thus increasing the
number of correct settings to r+1), but you multiply the previous

number of ways of choosing r+1 from n+1     C(n+1,r+1)
----------------------------------------  =  ----------
number of ways of choosing r from n           C(n,r)

I reproduce the table I gave you before: the numbers fit the
calculation just described. The number for 0-correct is always found
by subtraction from the total for that line after the other values
have been filled in. The number for n correct is always 1 and the
number for n-1 correct is always 0.

No. of   Total   0 Correct  1 Correct  2 Correct  3 Correct  4 Correct
Places   Ways    ---------  ---------  ---------  ---------   --------
-------  ----
1        1        0           1
2        4        3           0          1
3       36       26           9          0          1
4      576      453         104         18          0          1
----------------------------------------------------------------------

The 9 in line for 3 places is obtained from 3 in line 2 by multiplying
3 by the factor C(3,1)/C(2,0) = 3.

The 104 in line line 4 is obtained from 26 in previous line by
multiplying 26 by the factor C(4,1)/C(3,0) = 4.

The 18 in line 4 is obtained from 9 in line 3 by multiplying 9 by the
factor C(4,2)/C(3,1) = 2.

I shall now extend the table to six places:

No. of  Total 0-Corr. 1-Corr 2-Corr 3-Corr 4-Corr 5-Corr 6-Corr
Places  Ways  -----  ------- ------ ------- ----- ------ ------
------- ----
1      1      0       1
2      4      3       0       1
3     36     26       9       0      1
4    576    453     104      18      0       1
5  14400  11844    2265     260     30       0      1
6 518400 439975   71064    6795    520      45      0     1
-----------------------------------------------------------------

For example,

520 = 260 x C(6,3)/C(5,2)
6795 = 2265 x C(6,2)/C(5,1)

Now the probability of at least one correct setting is

78425/518400  =  0.15128

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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