I have 120 blocks. Each block is one of two different materials, 3 different colors, 4 different sizes and 5 different shapes. No two blocks are exactly the same of all four properties. I take two blocks at random. What is the probability the two blocks have exactly two of the four properties the same?

I try with R to replicate this:

replicateExperiment<-function(i,stateSpace) length(which(colDiffs(stateSpace[sample(1:120,2),])==0))==2

I get something more like $P(\text{2 out of 4 characteristics are the same)}\approx .3$.

My question is, how can we compute the number of blocks that have exactly 2 characteristics in common out of all 120 boxes.

What I have done so far:

Let's start with having three in common out of 4 characteristics (4 out of 4 is not possible). There are 4 possible ways to share 3 out of 4 characteristics.

I will denote the characteristics as $\{M,C,Si,Sh\}$ so $P(B_2=\{1,1,1,0\}|B_1=\{1,1,1,0\})$ is the probability that box 2 shares its first three characteristics with box 1 given that box one has characteristics $\{1,1,1,*.*\}$.

Then we have

$$P(B_2 \text{ has 3 elements in common with } B_1)=P(B_2=\{1,1,1,0\}|B_1=\{1,1,1,0\})*P(B_1=\{1,1,1,0\})+P(B_2=\{1,1,0,1\}|B_1=\{1,1,0,1\})*P(B_1=\{1,1,0,1\})+P(B_2=\{1,0,1,1\}|B_1=\{1,0,1,1\})*P(B_1=\{1,0,1,1\})+P(B_2=\{0,1,1,1\}|B_1=\{0,1,1,1\})*P(B_1=\{0,1,1,1\})$$

$$P(B_2 \text{ has 3 elements in common with } B_1)=(4*5*24+3*4*30+2*3*40+1*2*60)\frac{1}{120*119}\approx0.084$$ which is close enough to the value from the simulation code (modified to count the number of 3-ties).

The problem:

Now, I use the same reasoning to compute:

$$P(B_2 \text{ has 2 or more elements in common with } B_1)=(19*20*6+11*12*10+9*10*12+14*15*8+7*8*15+5*6*20)\frac{1}{120*119}\approx0.54$$

This is not close at all to the value from the simulation code (modified to count all $2^+$ ties).


Here's an ultra kludgy explanation: you need to generate vectors of the form (x4, x3, x2, x1) (where xi = characteristic) and each is drawn from its own domain.

Consider that you want exactly 2 characteristics to match, so given one box, the other one should have 2 characteristics that match (so theres only one way to set those values) and 2 characteristics that don't. (so xi's domain's size minus one ways). And, happily, these are all mutually exclusive.

Vary the characteristics in pairs (and forsaking all elegance): (2-1)((3-1)+(4-1)+(5-1)) + (2)((3)+(4)) + (12) = 35 (second boxes for each first box);

35/119 is about .29

  • $\begingroup$ This is so elegant, I need to take a breath! $\endgroup$ – user603 Feb 2 '16 at 19:57

Suppose I draw two boxes with the same color, size, and shape (but different material). That pair of boxes is counted in the term for the same {color, size}, but also in the term for the same {color, shape}, and in the term for the same {size, shape}....

  • $\begingroup$ Ok, I understand the problem, but how to compute the total number of doubly counted boxes? Or what formula to use to avoid double counting them in the first place? $\endgroup$ – user603 Feb 1 '16 at 22:16
  • $\begingroup$ Same basic way that you count the singly counted boxes, just subtract them instead of adding. :) $\endgroup$ – Danica Feb 1 '16 at 22:18

I have answered it here:

This is my answer.


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