244 lines
9.7 KiB
C++
244 lines
9.7 KiB
C++
// Copyright Paul A. 2007, 2010
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// Copyright John Maddock 2006
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// Use, modification and distribution are subject to the
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// Boost Software License, Version 1.0.
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// (See accompanying file LICENSE_1_0.txt
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// or copy at http://www.boost.org/LICENSE_1_0.txt)
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// Simple example of computing probabilities and quantiles for
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// a Bernoulli random variable representing the flipping of a coin.
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// http://mathworld.wolfram.com/CoinTossing.html
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// http://en.wikipedia.org/wiki/Bernoulli_trial
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// Weisstein, Eric W. "Dice." From MathWorld--A Wolfram Web Resource.
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// http://mathworld.wolfram.com/Dice.html
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// http://en.wikipedia.org/wiki/Bernoulli_distribution
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// http://mathworld.wolfram.com/BernoulliDistribution.html
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//
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// An idealized coin consists of a circular disk of zero thickness which,
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// when thrown in the air and allowed to fall, will rest with either side face up
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// ("heads" H or "tails" T) with equal probability. A coin is therefore a two-sided die.
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// Despite slight differences between the sides and nonzero thickness of actual coins,
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// the distribution of their tosses makes a good approximation to a p==1/2 Bernoulli distribution.
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//[binomial_coinflip_example1
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/*`An example of a [@http://en.wikipedia.org/wiki/Bernoulli_process Bernoulli process]
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is coin flipping.
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A variable in such a sequence may be called a Bernoulli variable.
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This example shows using the Binomial distribution to predict the probability
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of heads and tails when throwing a coin.
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The number of correct answers (say heads),
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X, is distributed as a binomial random variable
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with binomial distribution parameters number of trials (flips) n = 10 and probability (success_fraction) of getting a head p = 0.5 (a 'fair' coin).
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(Our coin is assumed fair, but we could easily change the success_fraction parameter p
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from 0.5 to some other value to simulate an unfair coin,
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say 0.6 for one with chewing gum on the tail,
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so it is more likely to fall tails down and heads up).
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First we need some includes and using statements to be able to use the binomial distribution, some std input and output, and get started:
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*/
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#include <boost/math/distributions/binomial.hpp>
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using boost::math::binomial;
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#include <iostream>
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using std::cout; using std::endl; using std::left;
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#include <iomanip>
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using std::setw;
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int main()
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{
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cout << "Using Binomial distribution to predict how many heads and tails." << endl;
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try
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{
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/*`
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See note [link coinflip_eg_catch with the catch block]
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about why a try and catch block is always a good idea.
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First, construct a binomial distribution with parameters success_fraction
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1/2, and how many flips.
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*/
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const double success_fraction = 0.5; // = 50% = 1/2 for a 'fair' coin.
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int flips = 10;
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binomial flip(flips, success_fraction);
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cout.precision(4);
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/*`
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Then some examples of using Binomial moments (and echoing the parameters).
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*/
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cout << "From " << flips << " one can expect to get on average "
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<< mean(flip) << " heads (or tails)." << endl;
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cout << "Mode is " << mode(flip) << endl;
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cout << "Standard deviation is " << standard_deviation(flip) << endl;
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cout << "So about 2/3 will lie within 1 standard deviation and get between "
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<< ceil(mean(flip) - standard_deviation(flip)) << " and "
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<< floor(mean(flip) + standard_deviation(flip)) << " correct." << endl;
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cout << "Skewness is " << skewness(flip) << endl;
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// Skewness of binomial distributions is only zero (symmetrical)
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// if success_fraction is exactly one half,
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// for example, when flipping 'fair' coins.
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cout << "Skewness if success_fraction is " << flip.success_fraction()
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<< " is " << skewness(flip) << endl << endl; // Expect zero for a 'fair' coin.
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/*`
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Now we show a variety of predictions on the probability of heads:
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*/
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cout << "For " << flip.trials() << " coin flips: " << endl;
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cout << "Probability of getting no heads is " << pdf(flip, 0) << endl;
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cout << "Probability of getting at least one head is " << 1. - pdf(flip, 0) << endl;
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/*`
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When we want to calculate the probability for a range or values we can sum the PDF's:
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*/
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cout << "Probability of getting 0 or 1 heads is "
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<< pdf(flip, 0) + pdf(flip, 1) << endl; // sum of exactly == probabilities
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/*`
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Or we can use the cdf.
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*/
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cout << "Probability of getting 0 or 1 (<= 1) heads is " << cdf(flip, 1) << endl;
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cout << "Probability of getting 9 or 10 heads is " << pdf(flip, 9) + pdf(flip, 10) << endl;
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/*`
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Note that using
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*/
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cout << "Probability of getting 9 or 10 heads is " << 1. - cdf(flip, 8) << endl;
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/*`
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is less accurate than using the complement
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*/
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cout << "Probability of getting 9 or 10 heads is " << cdf(complement(flip, 8)) << endl;
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/*`
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Since the subtraction may involve
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[@http://docs.sun.com/source/806-3568/ncg_goldberg.html cancellation error],
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where as `cdf(complement(flip, 8))`
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does not use such a subtraction internally, and so does not exhibit the problem.
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To get the probability for a range of heads, we can either add the pdfs for each number of heads
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*/
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cout << "Probability of between 4 and 6 heads (4 or 5 or 6) is "
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// P(X == 4) + P(X == 5) + P(X == 6)
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<< pdf(flip, 4) + pdf(flip, 5) + pdf(flip, 6) << endl;
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/*`
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But this is probably less efficient than using the cdf
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*/
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cout << "Probability of between 4 and 6 heads (4 or 5 or 6) is "
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// P(X <= 6) - P(X <= 3) == P(X < 4)
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<< cdf(flip, 6) - cdf(flip, 3) << endl;
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/*`
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Certainly for a bigger range like, 3 to 7
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*/
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cout << "Probability of between 3 and 7 heads (3, 4, 5, 6 or 7) is "
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// P(X <= 7) - P(X <= 2) == P(X < 3)
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<< cdf(flip, 7) - cdf(flip, 2) << endl;
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cout << endl;
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/*`
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Finally, print two tables of probability for the /exactly/ and /at least/ a number of heads.
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*/
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// Print a table of probability for the exactly a number of heads.
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cout << "Probability of getting exactly (==) heads" << endl;
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for (int successes = 0; successes <= flips; successes++)
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{ // Say success means getting a head (or equally success means getting a tail).
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double probability = pdf(flip, successes);
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cout << left << setw(2) << successes << " " << setw(10)
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<< probability << " or 1 in " << 1. / probability
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<< ", or " << probability * 100. << "%" << endl;
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} // for i
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cout << endl;
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// Tabulate the probability of getting between zero heads and 0 upto 10 heads.
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cout << "Probability of getting upto (<=) heads" << endl;
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for (int successes = 0; successes <= flips; successes++)
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{ // Say success means getting a head
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// (equally success could mean getting a tail).
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double probability = cdf(flip, successes); // P(X <= heads)
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cout << setw(2) << successes << " " << setw(10) << left
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<< probability << " or 1 in " << 1. / probability << ", or "
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<< probability * 100. << "%"<< endl;
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} // for i
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/*`
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The last (0 to 10 heads) must, of course, be 100% probability.
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*/
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double probability = 0.3;
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double q = quantile(flip, probability);
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std::cout << "Quantile (flip, " << probability << ") = " << q << std::endl; // Quantile (flip, 0.3) = 3
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probability = 0.6;
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q = quantile(flip, probability);
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std::cout << "Quantile (flip, " << probability << ") = " << q << std::endl; // Quantile (flip, 0.6) = 5
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}
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catch(const std::exception& e)
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{
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//
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/*`
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[#coinflip_eg_catch]
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It is always essential to include try & catch blocks because
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default policies are to throw exceptions on arguments that
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are out of domain or cause errors like numeric-overflow.
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Lacking try & catch blocks, the program will abort, whereas the
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message below from the thrown exception will give some helpful
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clues as to the cause of the problem.
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*/
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std::cout <<
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"\n""Message from thrown exception was:\n " << e.what() << std::endl;
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}
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//] [binomial_coinflip_example1]
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return 0;
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} // int main()
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// Output:
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//[binomial_coinflip_example_output
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/*`
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[pre
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Using Binomial distribution to predict how many heads and tails.
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From 10 one can expect to get on average 5 heads (or tails).
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Mode is 5
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Standard deviation is 1.581
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So about 2/3 will lie within 1 standard deviation and get between 4 and 6 correct.
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Skewness is 0
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Skewness if success_fraction is 0.5 is 0
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For 10 coin flips:
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Probability of getting no heads is 0.0009766
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Probability of getting at least one head is 0.999
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Probability of getting 0 or 1 heads is 0.01074
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Probability of getting 0 or 1 (<= 1) heads is 0.01074
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Probability of getting 9 or 10 heads is 0.01074
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Probability of getting 9 or 10 heads is 0.01074
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Probability of getting 9 or 10 heads is 0.01074
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Probability of between 4 and 6 heads (4 or 5 or 6) is 0.6562
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Probability of between 4 and 6 heads (4 or 5 or 6) is 0.6563
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Probability of between 3 and 7 heads (3, 4, 5, 6 or 7) is 0.8906
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Probability of getting exactly (==) heads
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0 0.0009766 or 1 in 1024, or 0.09766%
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1 0.009766 or 1 in 102.4, or 0.9766%
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2 0.04395 or 1 in 22.76, or 4.395%
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3 0.1172 or 1 in 8.533, or 11.72%
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4 0.2051 or 1 in 4.876, or 20.51%
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5 0.2461 or 1 in 4.063, or 24.61%
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6 0.2051 or 1 in 4.876, or 20.51%
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7 0.1172 or 1 in 8.533, or 11.72%
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8 0.04395 or 1 in 22.76, or 4.395%
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9 0.009766 or 1 in 102.4, or 0.9766%
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10 0.0009766 or 1 in 1024, or 0.09766%
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Probability of getting upto (<=) heads
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0 0.0009766 or 1 in 1024, or 0.09766%
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1 0.01074 or 1 in 93.09, or 1.074%
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2 0.05469 or 1 in 18.29, or 5.469%
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3 0.1719 or 1 in 5.818, or 17.19%
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4 0.377 or 1 in 2.653, or 37.7%
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5 0.623 or 1 in 1.605, or 62.3%
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6 0.8281 or 1 in 1.208, or 82.81%
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7 0.9453 or 1 in 1.058, or 94.53%
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8 0.9893 or 1 in 1.011, or 98.93%
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9 0.999 or 1 in 1.001, or 99.9%
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10 1 or 1 in 1, or 100%
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]
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*/
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//][/binomial_coinflip_example_output]
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