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Binomial Quiz Example

A multiple choice test has four possible answers to each of 16 questions. A student guesses the answer to each question, so the probability of getting a correct answer on any given question is one in four, a quarter, 1/4, 25% or fraction 0.25. The conditions of the binomial experiment are assumed to be met: n = 16 questions constitute the trials; each question results in one of two possible outcomes (correct or incorrect); the probability of being correct is 0.25 and is constant if no knowledge about the subject is assumed; the questions are answered independently if the student's answer to a question in no way influences his/her answer to another question.

First, we need to be able to use the binomial distribution constructor (and some std input/output, of course).

#include <boost/math/distributions/binomial.hpp>
  using boost::math::binomial;

#include <iostream>
  using std::cout; using std::endl;
  using std::ios; using std::flush; using std::left; using std::right; using std::fixed;
#include <iomanip>
  using std::setw; using std::setprecision;
#include <exception>
  using std::exception;

The number of correct answers, X, is distributed as a binomial random variable with binomial distribution parameters: questions n and success fraction probability p. So we construct a binomial distribution:

int questions = 16; // All the questions in the quiz.
int answers = 4; // Possible answers to each question.
double success_fraction = 1. / answers; // If a random guess, p = 1/4 = 0.25.
binomial quiz(questions, success_fraction);

and display the distribution parameters we used thus:

cout << "In a quiz with " << quiz.trials()
  << " questions and with a probability of guessing right of "
  << quiz.success_fraction() * 100 << " %"
  << " or 1 in " << static_cast<int>(1. / quiz.success_fraction()) << endl;

Show a few probabilities of just guessing:

cout << "Probability of getting none right is " << pdf(quiz, 0) << endl; // 0.010023
cout << "Probability of getting exactly one right is " << pdf(quiz, 1) << endl;
cout << "Probability of getting exactly two right is " << pdf(quiz, 2) << endl;
int pass_score = 11;
cout << "Probability of getting exactly " << pass_score << " answers right by chance is "
  << pdf(quiz, pass_score) << endl;
cout << "Probability of getting all " << questions << " answers right by chance is "
  << pdf(quiz, questions) << endl;

Probability of getting none right is 0.0100226
Probability of getting exactly one right is 0.0534538
Probability of getting exactly two right is 0.133635
Probability of getting exactly 11 right is 0.000247132
Probability of getting exactly all 16 answers right by chance is 2.32831e-010

These don't give any encouragement to guessers!

We can tabulate the 'getting exactly right' ( == ) probabilities thus:

cout << "\n" "Guessed Probability" << right << endl;
for (int successes = 0; successes <= questions; successes++)
{
  double probability = pdf(quiz, successes);
  cout << setw(2) << successes << "      " << probability << endl;
}
cout << endl;

Guessed Probability
 0      0.0100226
 1      0.0534538
 2      0.133635
 3      0.207876
 4      0.225199
 5      0.180159
 6      0.110097
 7      0.0524273
 8      0.0196602
 9      0.00582526
10      0.00135923
11      0.000247132
12      3.43239e-005
13      3.5204e-006
14      2.51457e-007
15      1.11759e-008
16      2.32831e-010

Then we can add the probabilities of some 'exactly right' like this:

cout << "Probability of getting none or one right is " << pdf(quiz, 0) + pdf(quiz, 1) << endl;

Probability of getting none or one right is 0.0634764

But if more than a couple of scores are involved, it is more convenient (and may be more accurate) to use the Cumulative Distribution Function (cdf) instead:

cout << "Probability of getting none or one right is " << cdf(quiz, 1) << endl;

Probability of getting none or one right is 0.0634764

Since the cdf is inclusive, we can get the probability of getting up to 10 right ( <= )

cout << "Probability of getting <= 10 right (to fail) is " << cdf(quiz, 10) << endl;

Probability of getting <= 10 right (to fail) is 0.999715

To get the probability of getting 11 or more right (to pass), it is tempting to use

1 - cdf(quiz, 10)

to get the probability of > 10

cout << "Probability of getting > 10 right (to pass) is " << 1 - cdf(quiz, 10) << endl;

Probability of getting > 10 right (to pass) is 0.000285239

But this should be resisted in favor of using the complement function. Why complements?

cout << "Probability of getting > 10 right (to pass) is " << cdf(complement(quiz, 10)) << endl;

Probability of getting > 10 right (to pass) is 0.000285239

And we can check that these two, <= 10 and > 10, add up to unity.

BOOST_ASSERT((cdf(quiz, 10) + cdf(complement(quiz, 10))) == 1.);

If we want a < rather than a <= test, because the CDF is inclusive, we must subtract one from the score.

cout << "Probability of getting less than " << pass_score
  << " (< " << pass_score << ") answers right by guessing is "
  << cdf(quiz, pass_score -1) << endl;

Probability of getting less than 11 (< 11) answers right by guessing is 0.999715

and similarly to get a >= rather than a > test we also need to subtract one from the score (and can again check the sum is unity). This is because if the cdf is inclusive, then its complement must be exclusive otherwise there would be one possible outcome counted twice!

cout << "Probability of getting at least " << pass_score
  << "(>= " << pass_score << ") answers right by guessing is "
  << cdf(complement(quiz, pass_score-1))
  << ", only 1 in " << 1/cdf(complement(quiz, pass_score-1)) << endl;

BOOST_ASSERT((cdf(quiz, pass_score -1) + cdf(complement(quiz, pass_score-1))) == 1);

Probability of getting at least 11 (>= 11) answers right by guessing is 0.000285239, only 1 in 3505.83

Finally we can tabulate some probabilities:

cout << "\n" "At most (<=)""\n""Guessed OK   Probability" << right << endl;
for (int score = 0; score <= questions; score++)
{
  cout << setw(2) << score << "           " << setprecision(10)
    << cdf(quiz, score) << endl;
}
cout << endl;

At most (<=)
Guessed OK   Probability
 0           0.01002259576
 1           0.0634764398
 2           0.1971110499
 3           0.4049871101
 4           0.6301861752
 5           0.8103454274
 6           0.9204427481
 7           0.9728700437
 8           0.9925302796
 9           0.9983555346
10           0.9997147608
11           0.9999618928
12           0.9999962167
13           0.9999997371
14           0.9999999886
15           0.9999999998
16           1

cout << "\n" "At least (>)""\n""Guessed OK   Probability" << right << endl;
for (int score = 0; score <= questions; score++)
{
  cout << setw(2) << score << "           "  << setprecision(10)
    << cdf(complement(quiz, score)) << endl;
}

At least (>)
Guessed OK   Probability
 0           0.9899774042
 1           0.9365235602
 2           0.8028889501
 3           0.5950128899
 4           0.3698138248
 5           0.1896545726
 6           0.07955725188
 7           0.02712995629
 8           0.00746972044
 9           0.001644465374
10           0.0002852391917
11           3.810715862e-005
12           3.783265129e-006
13           2.628657967e-007
14           1.140870154e-008
15           2.328306437e-010
16           0

We now consider the probabilities of ranges of correct guesses.

First, calculate the probability of getting a range of guesses right, by adding the exact probabilities of each from low ... high.

int low = 3; // Getting at least 3 right.
int high = 5; // Getting as most 5 right.
double sum = 0.;
for (int i = low; i <= high; i++)
{
  sum += pdf(quiz, i);
}
cout.precision(4);
cout << "Probability of getting between "
  << low << " and " << high << " answers right by guessing is "
  << sum  << endl; // 0.61323

Probability of getting between 3 and 5 answers right by guessing is 0.6132

Or, usually better, we can use the difference of cdfs instead:

cout << "Probability of getting between " << low << " and " << high << " answers right by guessing is "
  <<  cdf(quiz, high) - cdf(quiz, low - 1) << endl; // 0.61323

Probability of getting between 3 and 5 answers right by guessing is 0.6132

And we can also try a few more combinations of high and low choices:

low = 1; high = 6;
cout << "Probability of getting between " << low << " and " << high << " answers right by guessing is "
  <<  cdf(quiz, high) - cdf(quiz, low - 1) << endl; // 1 and 6 P= 0.91042
low = 1; high = 8;
cout << "Probability of getting between " << low << " and " << high << " answers right by guessing is "
  <<  cdf(quiz, high) - cdf(quiz, low - 1) << endl; // 1 <= x 8 P = 0.9825
low = 4; high = 4;
cout << "Probability of getting between " << low << " and " << high << " answers right by guessing is "
  <<  cdf(quiz, high) - cdf(quiz, low - 1) << endl; // 4 <= x 4 P = 0.22520

Probability of getting between 1 and 6 answers right by guessing is 0.9104
Probability of getting between 1 and 8 answers right by guessing is 0.9825
Probability of getting between 4 and 4 answers right by guessing is 0.2252
Using Binomial distribution moments

Using moments of the distribution, we can say more about the spread of results from guessing.

cout << "By guessing, on average, one can expect to get " << mean(quiz) << " correct answers." << endl;
cout << "Standard deviation is " << standard_deviation(quiz) << endl;
cout << "So about 2/3 will lie within 1 standard deviation and get between "
  <<  ceil(mean(quiz) - standard_deviation(quiz))  << " and "
  << floor(mean(quiz) + standard_deviation(quiz)) << " correct." << endl;
cout << "Mode (the most frequent) is " << mode(quiz) << endl;
cout << "Skewness is " << skewness(quiz) << endl;

By guessing, on average, one can expect to get 4 correct answers.
Standard deviation is 1.732
So about 2/3 will lie within 1 standard deviation and get between 3 and 5 correct.
Mode (the most frequent) is 4
Skewness is 0.2887
Quantiles

The quantiles (percentiles or percentage points) for a few probability levels:

cout << "Quartiles " << quantile(quiz, 0.25) << " to "
  << quantile(complement(quiz, 0.25)) << endl; // Quartiles
cout << "1 standard deviation " << quantile(quiz, 0.33) << " to "
  << quantile(quiz, 0.67) << endl; // 1 sd
cout << "Deciles " << quantile(quiz, 0.1)  << " to "
  << quantile(complement(quiz, 0.1))<< endl; // Deciles
cout << "5 to 95% " << quantile(quiz, 0.05)  << " to "
  << quantile(complement(quiz, 0.05))<< endl; // 5 to 95%
cout << "2.5 to 97.5% " << quantile(quiz, 0.025) << " to "
  <<  quantile(complement(quiz, 0.025)) << endl; // 2.5 to 97.5%
cout << "2 to 98% " << quantile(quiz, 0.02)  << " to "
  << quantile(complement(quiz, 0.02)) << endl; //  2 to 98%

cout << "If guessing then percentiles 1 to 99% will get " << quantile(quiz, 0.01)
  << " to " << quantile(complement(quiz, 0.01)) << " right." << endl;

Notice that these output integral values because the default policy is integer_round_outwards.

Quartiles 2 to 5
1 standard deviation 2 to 5
Deciles 1 to 6
5 to 95% 0 to 7
2.5 to 97.5% 0 to 8
2 to 98% 0 to 8

Quantiles values are controlled by the discrete quantile policy chosen. The default is integer_round_outwards, so the lower quantile is rounded down, and the upper quantile is rounded up.

But we might believe that the real values tell us a little more - see Understanding Discrete Quantile Policy.

We could control the policy for all distributions by

#define BOOST_MATH_DISCRETE_QUANTILE_POLICY real

at the head of the program would make this policy apply

to this one, and only, translation unit.

Or we can now create a (typedef for) policy that has discrete quantiles real (here avoiding any 'using namespaces ...' statements):

using boost::math::policies::policy;
using boost::math::policies::discrete_quantile;
using boost::math::policies::real;
using boost::math::policies::integer_round_outwards; // Default.
typedef boost::math::policies::policy<discrete_quantile<real> > real_quantile_policy;

Add a custom binomial distribution called

real_quantile_binomial

that uses

real_quantile_policy

using boost::math::binomial_distribution;
typedef binomial_distribution<double, real_quantile_policy> real_quantile_binomial;

Construct an object of this custom distribution:

real_quantile_binomial quiz_real(questions, success_fraction);

And use this to show some quantiles - that now have real rather than integer values.

cout << "Quartiles " << quantile(quiz, 0.25) << " to "
  << quantile(complement(quiz_real, 0.25)) << endl; // Quartiles 2 to 4.6212
cout << "1 standard deviation " << quantile(quiz_real, 0.33) << " to "
  << quantile(quiz_real, 0.67) << endl; // 1 sd 2.6654 4.194
cout << "Deciles " << quantile(quiz_real, 0.1)  << " to "
  << quantile(complement(quiz_real, 0.1))<< endl; // Deciles 1.3487 5.7583
cout << "5 to 95% " << quantile(quiz_real, 0.05)  << " to "
  << quantile(complement(quiz_real, 0.05))<< endl; // 5 to 95% 0.83739 6.4559
cout << "2.5 to 97.5% " << quantile(quiz_real, 0.025) << " to "
  <<  quantile(complement(quiz_real, 0.025)) << endl; // 2.5 to 97.5% 0.42806 7.0688
cout << "2 to 98% " << quantile(quiz_real, 0.02)  << " to "
  << quantile(complement(quiz_real, 0.02)) << endl; //  2 to 98% 0.31311 7.7880

cout << "If guessing, then percentiles 1 to 99% will get " << quantile(quiz_real, 0.01)
  << " to " << quantile(complement(quiz_real, 0.01)) << " right." << endl;

Real Quantiles
Quartiles 2 to 4.621
1 standard deviation 2.665 to 4.194
Deciles 1.349 to 5.758
5 to 95% 0.8374 to 6.456
2.5 to 97.5% 0.4281 to 7.069
2 to 98% 0.3131 to 7.252
If guessing then percentiles 1 to 99% will get 0 to 7.788 right.

See binomial_quiz_example.cpp for full source code and output.


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