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The Birthday Problem is very well detailled on Wikipedia:

the birthday problem or birthday paradox concerns the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday.

It’s counter-intuitive, but with only 23 randomly chosen people, the probability is 50%.

The math sound complex, but a Monte Carlo simulation can do wonders ; all we have to do is generate randomly birthdates, and count how often there are pairs who share a birthday.

Python was a bit slow, so I rewrote the simulation in Rust (`cargo add rand`

). It’s still not immediate. An easy trick would be to use `rayon`

in order to parallelize the computation, but we will find something better later on so this is not necessary.

```
extern crate rand;
use rand::prelude::*;
use rand::rngs::ThreadRng;
fn has_birthdays_on_same_day(nb_people: usize, rng: &mut ThreadRng) -> bool {
let mut counts = [0usize; 365];
for _ in 0..nb_people {
let day: usize = rng.gen_range(1..365);
counts[day] += 1;
if counts[day] == 2 {
return true;
}
}
false
}
fn main() {
let mut rng = rand::thread_rng();
// estimate the probability that 2 people share their birthdate,
// for 2 to 100 random people, with 10_000 iterations
let total_nb_iterations = 10_000;
for nb_people in 2..=100 {
let mut nb_birthdays_with_same_day = 0;
for _ in 0..total_nb_iterations {
if has_birthdays_on_same_day(nb_people, &mut rng) {
nb_birthdays_with_same_day += 1;
}
}
let p: f32 = nb_birthdays_with_same_day as f32 / total_nb_iterations as f32;
println!("{},{}", nb_people, p);
}
}
```

That’s short, but it turns out we can actually implement an analytic solution in a few lines (of Python). Since we are not performing multiple iterations, it’s in fact also much faster:

```
import matplotlib.pyplot as plt
nb_values = 60
p = 364/365
values = list(range(2, nb_values+1))
probas = []
for i in values:
nb_pairs = i*(i-1)/2
v = 1 - p ** nb_pairs
probas.append(v)
print("{} {:1.4f}".format(i, v))
plt.plot(values, probas)
plt.xlabel('Number of people')
plt.ylabel('Probability')
plt.title('Probability that 2 randomly selected people share a birthday')
plt.savefig("birthday-problem.png")
```

The results are similar:

Number of people | Probability that 2 people share their birthday |

2 | 0.0027 |

3 | 0.0082 |

4 | 0.0163 |

5 | 0.0271 |

6 | 0.0403 |

7 | 0.0560 |

8 | 0.0739 |

9 | 0.0940 |

10 | 0.1161 |

20 | 0.4062 |

23 |
0.5005 |

30 | 0.6968 |

40 | 0.8823 |

50 | 0.9653 |

60 | 0.9922 |