The sock drawer

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This Probability problem is taken from Mosteller, 1966/87, p1.


The Problem

A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is 1/2

(a) How small can the number of socks in the drawer be?
(b) How small if the number of black socks is even?

Known Solutions

Here are some different ways that people have used to approach this problem.

Part A: The smallest number of socks in the drawer

Teams in the VMT Spring Fest '07 used different ways to organize their thinking about this problem. One approach is to count and organize your cases. Since the problem asks for the smallest number of total socks in the drawer so that the probability of drawing two red socks at random fits a particular value (1/2), one could start trying different arrangement of numbers checking which ones "would spark a .5 result" (as Team 9 put it). This would be like having an imaginary drawer and an unlimited number of red and black socks to try, like this picture from Team 2 indicated:

File:Team2 Session1.jpg

After trying different cases, several teams found that three red socks and 1 black sock worked (for a total of 4 socks). Here is how Team 3 showed that this combination of socks works:

File:TeamC partA.jpg

  Name the socks R1, R2, R3, and B1. The combinations (of possible drawings) are
   R1 & R2, 
   R1 & R3, 
   R2 & R3, 
   R1 & B1, 
   R2 & B1, and 
   R3 & B1. 

As one can see, there are 3 cases (R1&R2, R1&R3, R2&R3) that work (drawing two red socks in a row) out of 6 possible ones for a probability of 1/2. Team 6 also showed that this case works in a very similar way.

Another way of approaching this problem is to note how the drawing of the socks works in each step (drawing) using a Tree_diagram. Team 1 tried that and their tree looks pretty much like this one:

            /        \
           /           [black]
    start \
           \            [red]
            \         /

Each node in this tree, consider as an independent event, has a probability that depends on how many red socks and black socks are in the drawer at the moment a sock is taken out. For example, when we start with 2 red socks and 1 black sock, here are the probabilities of each event:

                                    [red, p=1/2]
                     [red, p=2/3] *
                   /               \
                  /                 [black, p=1/2]
 [2 red, 1 black] 
                  \                 [red, p=1]
                   \               /
                    [black, p=1/3]*       
                                    [black, p=0]  

Based on this values and following the formula for conditional probability we can calculate the probability of drawing 2 red socks at random for this case: (2/3)*(1/2) = (1/3). Since the probability is not 1/2 this case does not satisfy the condition of the problem, but we can use this example to think about the general case. As Team 5 showed, the probability of drawing two red socks without replacement can be represented in this way:

  number of red socks         number of red socks-1          1
 --------------------   *  ----------------------------  =  ---
 total number of socks      total number of socks -1         2

So the problem is transformed into finding the small combination of number of red socks and total number of socks that satisfy this equation. Several teams found 4 to be the smallest total number of socks, with 3 red ones and 1 black: (3/4)*(2/3) = 6/12 or 50% just like the other method had suggested. Now... is this the smallest number? Not many teams reported a reason for this combination of socks to be the smallest but they probably tried all the smaller combinations of socks and found them to produce a probability different than 1/2. Here is what that table might look like

   #socks    #red  #black  Prob 2 reds
  -------   -----  ------  -----------
     3        2      1         ?
     4        2      2         ?
     4        3      1        1/2 (!)

Some teams wondered about whether the problem implied replacement or not. Team 3 wrote the following note about drawing the socks with replacement:

 if your pulling two socks out at a time then the number of red socks has to equal 
 the number of black socks. the smallest number of red socks you can have is 2 
 because your pulling out 2 socks. So 2 red socks and 2 black socks is 4 socks total

What do you think?

Part B: How small if the black socks are even?

The same two approaches that were used for part A can be pursued to find what the smallest number of socks would be if the number of black socks is an even number. For example the following table shows all the cases with 2, 4 and 6 black socks and their probabilities, until a "successful case" was found with 21 socks (15 red socks and 6 black socks):


Using the formula that Team 5 presented for part A, a similar table can be constructed. In this case we are tracking the variables in the formula until we find a successful case:


When applying this values to the formula, Team 5 showed how a probability of 1/2 was produced:

 1*2=2     2/2=1
 2*3=6     6/2=3
 3*4=12   12/2=6
 .          .
 .          .
 20*21=420     210
 15    14        210
---- *  ---- = --------
 21    20        420 
 If the number of black socks is even, the smallest number of socks you can have is 15 red ones and 6 black ones.  
 When you multiply this by 14/20, you get a fraction of 1/2 (210/420)


A number of extensions to the two original questions can be explored. One could ask variations of the two original questions or imagine totally new but related situations. Below are some examples.

Is there a pattern?

Question b asks about the smallest number of socks in the drawer if the number of black socks is even and the probability of drawing two red socks at random is 1/2. From the solutions presented above we know that smallest number is 21, with 15 red socks and 6 balck ones, but what would be the next even number of black socks that would satisfy the problem? How about the next one after that? And better yet... is there a pattern of these numbers? Is there a systematic way to find those even values for b?

One way to approach this is to start with the formula of the conditional probability that some teams used and develop it algebraically. To simplify things, lets say we have r red socks and b black socks in the drawer for a total of (r+b) socks so we can rewrite the formula this way:

   r        r-1        1
 ----- *  -------  =  ---
 (r+b)      r+b-1      2
 which simplifies to:
 (r-b)(r-b-1) = 2b^2

So, this gave us a simpler equation to work with. There are two additional things we know that can be used to keep simplifying this equation. First, b needs to be an even number which means that it has to be twice the value of some integer, say k. i.e. b = 2*k. Also, (r-b) and (r-b-1) are consecutive numbers like 23 and 24. Using these two insights we come to the realizatin that:

 b = sqrt((r-b)(r-b-1)/2)  is an integer

After making some additional considerations and replacing r-b and (r-b-1)/2 with m^2 and n^2 we get the equation

 2n^2 + 1 = m^2

In other words, we reduced the problem to finding pairs of number (n,m) where n is even and m is odd, and they satisfy the equation 2n^2 + 1 = m^2

An investigation with a spreadsheet program reveals the following pairs

 3^2    =  2*2^2 + 1
 17^2   =  2*12^2 +1
 99^2   =  2*70^2 + 1
 577^2  =  2*408^2 + 1 

In other words

    n     m     b=(m*n)   r
   2      3      6        15
   12     17     204      493
   70     99     6930     16731
   408    577    235416   568345

Can you generate other pairs in a systematic way?

Why is this a problem?

One thing that makes this problem interesting is that you have to create the "situation" that fits a particular value of probability. Usually probability problems are presented the other way around, where you are given a situation and asked for the probability of an event. As you can see int he "extensions" section, this type of problem can lead to very rich explorations.

Who has worked on this problem?

Here are all the teams' Summary pages about the Socks Drawer problem as part of the VMT Spring Fest 2007:

Team 1,   Team 2,   Team 3,   Team 4
Team 5,   Team 6,   Team 7,   Team 9
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