Why Does My Probability Calculation Produce Unexpected Results?

Exploring Floating Point Error: A Simple Example

Conversations surrounding floating point variables often mention the potential for "error." This concept can become especially relevant when practicing problem-solving, such as in the context of the given puzzle.

Consider the following: to find the probability of obtaining exactly two successes in ten independent trials, you utilized the formula:

p_2x_success = pow(1-p, (double)8) * pow(p, (double)2) * (double)choose(8, 2);

The Problem of Floating Point Error

In this instance, floating point error becomes a potential issue. Floating point variables are utilized to represent real numbers on computers, but their representation is not exact due to limitations in storage space and processing capabilities. As a result, when performing floating point operations, errors can accumulate.

To illustrate this, let's visualize the function f(k) using a graph in logarithmic scale:

[Image of XY graph with X and Y in logarithmic scale]

If the computer could flawlessly represent 32-bit floats without rounding error, the result for every k would be zero. However, as k increases, we witness a growing error attributed to floating point error accumulation.

Replications in C

You can replicate this behavior in C by implementing the following code:

for (int k = 1; k < 1000; k++) {
  cout << "k = " << k << ", error = " << (pow(1-k, (double)8) * pow(k, (double)2) * (double)choose(8, 2)) << endl;
}

Conclusion

This simple example demonstrates the potential for floating point error when performing floating point operations. It is important to be aware of this phenomenon and consider it when implementing numerical computations.

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