"My computer tells me that $\\pi$ is *approximatively*"
]
]
},
},
{
{
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"## 1.2 Buffon’s needle\n",
"## Buffon’s needle\n",
"\n",
"\n",
"Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
"Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
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"## 1.3 Using a surface fraction argument\n",
"## Using a surface fraction argument\n",
"\n",
"\n",
"A method that is easier to understand and does not make use of the sin function is based on the\n",
"A method that is easier to understand and does not make use of the sin function is based on the fact that if *X ∼ U(0, 1)* and *Y ∼ U(0, 1)*, then *P\\[X<sup>2</sup> + Y<sup>2</sup> ≤ 1\\] = π/4* (see [\"Monte Carlo method\" on Wikipedia)](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
"fact that if *X ∼ U(0, 1)* and *Y ∼ U(0, 1)*, then *P\\[X<sup>2</sup> + Y<sup>2</sup> ≤ 1\\] = π/4* (see [\"Monte Carlo method\" on Wikipedia)](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
]
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"It is then straightforward to obtain a (not really good) approximation to *π* by counting how\n",
"It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how many times, on average, X<sup>2</sup> + Y<sup>2</sup>\n",
"many times, on average, X<sup>2</sup> + Y<sup>2</sup>\n",
