The Tuesday Child Puzzle via Monte Carlo (repost)

I get more than a few hits for this old post, so here is a rerun.

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There has been a lot of chatter about the Tuesday Child puzzle:

I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?

Everyone's instinct is to say: Tuesday has nothing to do with it, so the answer is simply 1/2.

That answer is wrong. Here is the trouble with everyone's intuition: they read the problem as: I have one son born on a Tuesday, what is the probability that my next child will be a son? There the answer is 1/2. But that is not the problem at hand--here both children are already born. It's a different problem

The answer is actually 13/27 = 0.481, not 1/2.

Let's accept that for now. If you Google you'll find a lot of proofs, some with tables and some using complicated Bayesian analysis. I'll get the result later, via simulation, but for now we'll assume it is correct. But the way to think about it is this: there are lots of ways that two children can be born on any of seven days, say Boy on Tuesday and Girl on Saturday, all with equal probability, and exactly 1/4 of them have two boys. But as I place constraints, such as Boy on Tuesday, many of the possibilities are eliminated and the probabilities change. For example, I can place a very tight constraint: I have two children, one is a son born Tuesday and the other is a son born Saturday. What is the probability I have two sons? Why it is one of course, because all arrangements except Boy on Tuesday and Boy on Saturday have been eliminated by the constraints.

Of course Tuesday really has nothing to do with it. What is relevant is that a boy is constrained to be born on one specific day—any day would do--this could just as easily be the Friday Child Puzzle. It is the limitation that one of the children is a boy born on one specific day out of seven that is relevant.

To see this, assume you asked:

I have two children. One is a boy born on a Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday. What is the probability I have two boys?

Clearly this is the same as simply asking: I have two children, one is a boy, what is the probability that the other is a boy? Here the answer is clear—the possible arrangements given that we have at least one boy are: BB, BG, GB. They occur with equal probability, so the probability of BB (two boys) is 1/3.

We could generalize the puzzle this way:

I have two children. One is a boy born no later than day N where N is 1..7. What is the probability I have two boys?

Let's call that probability P(N).

So the original form of the puzzle is: what is P(1)? The answer we are accepting (for now) is not 1/2 but 13/27.

The second form of the puzzle, where the constraint is a boy born on any of the seven days, could be stated this way: what is P(7)? That we just demonstrated is 1/3.

What is the meaning of P(0)? This would mean that the first boy wasn't born on any day. This pathological case becomes, simply, what is the probability that the other child is a boy, which is our beloved 1/2.

Let's make a prediction. We have:


P(0) = 1/2 = 14/28
P(1) = 13/27
P(2) = 12/26
P(3) = 11/25
P(4) = 10/24
P(5) = 9/23
P(6) = 8/22
P(7) = 7/21 = 1/3

The red values P(2) through P(6) come from a prediction based on an obvious pattern. Let us remember what this means. P(1) is the probability of two boys given that one son is born on one specific day, say Tuesday. P(2) is the probability given that one son is born on one of two days, say Tuesday or Wednesday. Etc., etc., etc.

I wrote a Monte Carlo simulation for this problem and the results are:


Days   Prob of 2 boys
-------------------
 P(1)       0.4811
 P(2)       0.4615
 P(3)       0.4399
 P(4)       0.4167
 P(5)       0.3911
 P(6)       0.3631
 P(7)       0.3336

These are good approximations to the predicted fractions above. The zero row is not in the output because it is a pathological case and the code won't handle it.

Note that P(1), as advertised, is 13/27.

The probability (that the second child is a boy) drops smoothly from P(0) = 1/2 (when in effect there is no first child) to 1/3 when the first child is a boy, born any day.

The JAVA program is given below. ¹

public class TuesdayChild {

//constant defining a boy baby
   private static final int BOY = 0;

//method that randomly selects a sex
   private static int randomSex() {
     return (int)(Integer.MAX_VALUE*Math.random()) % 2;
   }

//method that randomly selects a day, 0--6
  private static int randomDay() {
    return (int)(Integer.MAX_VALUE*Math.random()) % 7;

  }

//main method
   public static void main(String arg[]) {
     TuesdayChild tchild = new TuesdayChild();

//how many trials per iteration
     int numTrial = 10000000;

//hold the results for each case. We will vary the
//number of days one boy is constrained to be born on
//from 1 (corresponding to the problem as stated) to 7
     double results[] = new double[7];

//loop over the constraint days 

    for (int numDays = 1; numDays <= 7; numDays++) { 

      int passCount = 0; //number of trials passing constraint
      int twoBoyCount = 0; //subset that have two boys

//now loop over the trials
     for (int i = 0; i < numTrial; i++) { 

      Trial trial = tchild.new Trial(numDays); 

      if (trial.keepTrial()) { 

        passCount++; 

        if (trial.twoBoys()) {

          twoBoyCount++; 

        } 

      } 

    } 

    results[numDays-1] = ((double)twoBoyCount)/passCount; 

  } 

//now print the results

   System.out.println("Days Prob 2 boys");
   System.out.println("-------------------");
   for (int numDays = 1; numDays <= 7; numDays++) {

      System.out.println(String.format("%d %8.4f", numDays, results[numDays-1])); 

    } 

  } 

//class for a single trial
   class Trial {
//sex of child 1 and child 2
   int sex1 = randomSex();
   int sex2 = randomSex();

//day of birth child 1 and child 2
   int day1 = randomDay();
   int day2 = randomDay();

//this will determine how many days we constrain the birth
//of one boy. It can be 1--7. At one, one boy will be constrained
//to be born on one day, such as Tuesday. This is analogous to the
//puzzle as stated. If it is two then one boy is constrained to be
//born on two days, say Tuesday or Wednesday. If it is seven, one boy
//is constrained to be born on any day. This is the same as simply saying
//you have one boy. The answer for that case should be 1/3.
   int _max;

   public Trial(int max) {
     _max = max;
  }

//see if we keep this trial because at least one of the two children
//was a boy born within the constrained number of days
   public boolean keepTrial() {
     return ((sex1==BOY)&&(day1 < _max)) || (sex2==BOY)&&((day2 < _max)); } 

//see if this trial has two boys
   public boolean twoBoys() {
     return (sex1 == BOY) && (sex2 == BOY);
     }
   }
}

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UPDATE: 

Here is a tabular solution. We'll use the first letters of each day, except with R for Thursday and Q for Sunday. And we also use first child - second child ordering, for example B_TG_M means the first child was a boy born on Tuesday and the second was a girl born on Monday. We can easily enumerate all the possibilities with a boy on Tuesday:

B_TB_T

B_TB_Q  B_TB_M  B_TB_W  B_TB_R  B_TB_F   B_TB_S          

B_QB_T  B_MB_T  B_WB_T  B_RB_T  B_FB_T  B_SB_T 

B_TG_Q  B_TG_M  B_TG_T  B_TG_W  B_TG_R  B_TG_F  B_TG_S

G_QB_T  G_MB_T  G_TB_T  G_WB_T  G_RB_T  G_FB_T  G_SB_T

Each of these occurs with equal probability. ³ There are 27 total. Those with two boys are shown in red. There are 13 of those. So the answer, again, is 13/27.

This problem would be quite different is it were stated:

I have two children. The first (oldest) was a boy born on a Tuesday. What is the probability I have two boys?

Now the enumeration is

B_TB_T

B_TB_Q  B_TB_M  B_TB_W  B_TB_R  B_TB_F   B_TB_S          

B_TG_Q  B_TG_M  B_TG_T  B_TG_W  B_TG_R  B_TG_F  B_TG_S

And we get the intuitive result of 7/14, or 1/2.

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¹ God programs in JAVA, and uses the blessed ²  K&R brace style, just as surely as Jesus speaks in early 17th century English. However, His use of comments and exception handling is  purely anthropomorphic.
² This is one of those special times when "blessed" is a two syllable word.
³ In physics terms the macrostate is "we have two children and at least one is a son born on Tuesday" and there are 27 corresponding microstates. So the entropy of our macrostate is S = k ln(27), where k is the Boltzmann  constant.