Monthly Archives: February 2023

The Tea-Cup Problem

Here’s a little problem to test your skills at combinatorial probability.

You have a set of six cups and saucers. Two are NC State red R, two are UNC light blue b, and two are Duke dark blue B. You place the saucers in a line on the table RRbbBB. Then a blind man comes in and puts the saucers on the cups in random order. Let M be the number of cups that match the color of the saucer they are on. Your job is to compute the distribution of M.

To get you started I will specify a probability space which is the first step in solving any problem of this type. I once thought it was good to number the cups but a student in my class this year taught me it was better to treat the two cups of a given color as indistinguishable so we have 6!/(2!2!2!) = 90 outcomes instead of 720. To help check the solution note that not only should the probabilities sum to 1, but we must have EM = 6(1/3)=2. In the next paragraph I will start to reveal the solution starting at 6 and working down, so if you want to discover it on your on you should stop scrolling.

P(M=6)=1/90. In our probability space there is only one outcome where the cups all match, which is better than the situation when the cups are numbered and there are 2 x 2 x 2.

P(M=5)=0. If say the 2R match and 2b match then the 2B must match so 5 is impossible

P(M=4)=12/90. Matching 2-2-0 is impossible by the reasoning for 5, so we must have 2-1-1. There are 3 ways to pick the color with two matches, and for each color with only one match 2 choices of where the matching cup is. The rest of the outcome is now forced, e.g., RRbBBb.

P(M=3)=16/90. Matching 2-1-0 is impossible so we must have 1-1-1. We can pick the locations of the matching cups in 2 x 2 x 2 ways. The other three nonmatching cups must be either BRb or bBR

P(M=2)=27/90. We can have 2-0-0. Once we pick the double matching color in 3 ways the rest is forced, e.g. RRBBbb. We can have 1-1-0. We can pick the color with no match in 3 ways and 2 x 2 ways for the location of the matching cups. Suppose dark blue has no match. Then the two B cups must be on R and b, but there are 2 ways to put R and b on the B sauces for a total of 24 + 3 = 27.

P(M=1) =24/90. We can pick the location of the matching cup in 6 ways. Suppose it is the first R saucer. The second Red saucer can be B or b (2 ways). If it is B then we have bb on the Blue saucers, and we can have RB or BR on the b sauces (x2). If it is B then we have BB on the blue saucers and we have two possibilities on the B saucers, but autocorrect in Word will not let me type them.

P(M=0) =10/90. We can have BB on the Red saucers and then must have RR on b and bb on B. The situation is similar for bb on Red. This gives 2 outcomes. If we have {Bb} on the red saucers then we must have {Rb} on b and {Rb} on B where the set braces indicate we have not specified the order, so there are 2 x 2 x 2 = 8 outcomes.

1+12+16+27+24+10=90, 6 x 1+12 x 4+3 x16 +2 x 27+1 x 24 = 180 (so the mean is 2).