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			238 lines
		
	
	
		
			9.3 KiB
		
	
	
	
		
			C++
		
	
	
	
	
	
			
		
		
	
	
			238 lines
		
	
	
		
			9.3 KiB
		
	
	
	
		
			C++
		
	
	
	
	
	
| // Copyright Paul A. 2007, 2010
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| // Copyright John Maddock 2006
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| 
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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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| 
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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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| 
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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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| 
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| //[binomial_coinflip_example1
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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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| #include <boost/math/distributions/binomial.hpp>
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|   using boost::math::binomial;
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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| /*`
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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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| 
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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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|   }
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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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| 
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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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| 
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| // Output:
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| 
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| //[binomial_coinflip_example_output
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| /*`
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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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| 
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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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| 
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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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| 
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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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