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Monty Hall Problem
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| const doors = [0, 1, 2]; | |
| const tries = 100000; | |
| let pickHasCar = 0; | |
| let theOtherHasCar = 0; | |
| for (let i = 0; i < tries; i++) { | |
| const car = doors[random(3)]; | |
| const pick = doors[random(3)]; | |
| // Open a door that doesn't have the car. | |
| const open = | |
| pick === car | |
| ? doors.filter(d => d !== pick)[random(2)] | |
| : doors.find(d => d !== pick && d !== car); | |
| const theOther = doors.find(d => d !== pick && d !== open); | |
| if (pick === car) { | |
| pickHasCar++; | |
| } | |
| if (theOther === car) { | |
| theOtherHasCar++; | |
| } | |
| } | |
| console.log({ | |
| pickHasCar, | |
| theOtherHasCar, | |
| ratio: theOtherHasCar / pickHasCar | |
| }); | |
| function random(upperBound) { | |
| return Math.floor(upperBound * Math.random()); | |
| } | |
| // When the pick has the car (1/3), the other one doesn't have the car for sure. | |
| // When the pick doesn't have the car (2/3), the other one has the car for sure. | |
| // So, the other one has the car by 2/3 chance. |
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