88 lines
1.9 KiB
Markdown
88 lines
1.9 KiB
Markdown
---
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id: 5900f3d01000cf542c50fee3
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title: 'Problem 100: Arranged probability'
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challengeType: 5
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forumTopicId: 301724
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dashedName: problem-100-arranged-probability
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---
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# --description--
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If a box contains twenty-one colored discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs.
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$${P(BB)} = \frac{15}{21}×\frac{14}{20} = \frac{1}{2}$$
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The next such arrangement, for which there is exactly a 50% chance of taking two blue discs at random, is a box containing eighty-five blue discs and thirty-five red discs.
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By finding the first arrangement to contain over `limit` discs in total, determine the number of blue discs that the box would contain.
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# --hints--
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`arrangedProbability(20)` should return a number.
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```js
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assert(typeof arrangedProbability(10) === 'number');
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```
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`arrangedProbability(20)` should return `15`.
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```js
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assert.strictEqual(arrangedProbability(20), 15);
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```
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`arrangedProbability(100)` should return `85`.
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```js
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assert.strictEqual(arrangedProbability(100), 85);
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```
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`arrangedProbability(100000)` should return `97513`.
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```js
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assert.strictEqual(arrangedProbability(100000), 97513);
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```
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`arrangedProbability(1000000000)` should return `3822685023`.
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```js
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assert.strictEqual(arrangedProbability(1000000000), 3822685023);
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```
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`arrangedProbability(1000000000000)` should return `756872327473`.
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```js
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assert.strictEqual(arrangedProbability(1000000000000), 756872327473);
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```
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# --seed--
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## --seed-contents--
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```js
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function arrangedProbability(limit) {
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return true;
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}
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arrangedProbability(20);
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```
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# --solutions--
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```js
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function arrangedProbability(limit) {
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// Based on https://www.mathblog.dk/project-euler-100-blue-discs-two-blue/
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let blue = 15;
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let discs = 21;
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while (discs < limit) {
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const nextBlue = 3 * blue + 2 * discs - 2;
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const nextDiscs = 4 * blue + 3 * discs - 3;
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blue = nextBlue;
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discs = nextDiscs;
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}
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return blue;
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}
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```
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