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Randomly Setting the Table

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 

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.

Thanks for your help!

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 

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 

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 

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

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 
answer by the ratio:

    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
Associated Topics:
High School Probability

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