4 ms·
You have three doors. As you note, you have a 1/3 chance of winning. Monty now reveals a losing door. At this point, your door has a 1/3 chance of still winnin
by atdrummond 5y ago
You have three doors. As you note, you have a 1/3 chance of winning.
Monty now reveals a losing door. At this point, your door has a 1/3 chance of still winning - the probability of that choice can’t change. However, as we now know one door has a 0/3 chance of winning (it’s been revealed) the remaining door must have (1-1/3) chance of winning. Thus, the remaining door has a 2/3 chance of winning.
- listenallyall 5y agoI'm going to disagree with your framing here. Probabilities DO change as more information becomes available. For example, prior to rolling a pair of dice, you have a 1/36 chance of rolling a 12. However as sooner as you see the value of the first die, the probability changes to either 0 (most likely) or 1/6. In your description, if Monte opens a door randomly -- which you don't make clear whether he does or doesn't -- then the presence of the goat DOES change your P(winning a car).
- atdrummond 5y agoHe never chooses randomly in the Monte Hall problem. He’s always going to reveal a goat. That’s why the maths are so straightforward - and yet so counterintuitive.
- listenallyall 5y ago> He never chooses randomly in the Monte Hall problem. But you never state that, which is why I took exception with your explanation. You also state that probabilities never change, which is also untrue as more information is revealed.
- atdrummond 5y agoNo. All I said is that the probability of the first door winning doesn't change. Which is true - it remains 1/3. The change in probabilities impact the other two doors, whose terminal values I listed (0 and 2/3).