I just gave you my response and I am not going to wade through your clear misunderstanding of Lewis's argument. — Jeremiah
But if your selected door does not have the car behind it then it is very relevant information, because it means the car is behind the remaining closed door that you did not select. — andrewk
Yes. Your initially chosen door has 1/3 probability of containing the prize. The other two doors have a total of 2/3 probability. When the host shows you that one of those doors is empty, its probability goes to 0. — Andrew M
Yes, if you switch you are effectively choosing two doors, which leads to the accepted solution that it is optimal to switch. That's what I said above.The Monty Hall problem and its analysis is identical to choosing 1 door, or choosing 2 doors. — tom
We have P(C) = 1/3. If you retain your selection then the probability of a win is:
1/3 x P(win | C & not switch) + 2/3 x P(Win | ~C & not switch) = 1/3 x 1 + 2/3 x 0 = 1/3.
On the other hand if you switch the probability of a win is:
1/3 x P(win | C & switch) + 2/3 x P(Win | ~C & switch) = 1/3 x 0 + 2/3 x 1 = 2/3. — andrewk
num.trials <- 10000 car.location <- 1 + trunc(3 * runif(num.trials)) # assume contestant chooses door 1 at first first.pick <- 1 monty.opens <- ifelse(car.location == first.pick, 2 + trunc(2 * runif(num.trials)), 5 - car.location) # Monty picks randomly between 2 and 3 if car is behind door 1, # else he chooses the door (2 or 3) with no car behind it new.pick <- 5 - monty.opens # pick door 2 if Monty opened 3, else pick 3 print(paste("probability of winning if not switched is", sum(first.pick == car.location) / num.trials)) print(paste("probability of winning if switched is", sum(new.pick == car.location) / num.trials))
> print(paste("probability of winning if not switched is", sum(first.pick == car.location) / num.trials)) [1] "probability of winning if not switched is 0.3277" > print(paste("probability of winning if switched is", sum(new.pick == car.location) / num.trials)) [1] "probability of winning if switched is 0.6723"
One thought here is that you could use a similar tree structure to argue for a quarter probability for heads. The first two branches represent the first and second awakenings with 1/2 probability each. The first awakening branch splits into heads and tails branches with 1/2 probability each. The second awakening branch extends to a tails branch with probability 1. Multiplying, we get first awakening + heads = 1/4, first awakening + tails = 1/4, second awakening + tails = 1/2. — Andrew M
Now those first two branch probabilities don't seem particularly plausible since there are twice as many first awakening states. But the same can be said for the halfer view since there are twice as many tail states. — Andrew M
P(Heads|Awake) = (P(Heads) * P(Awake|Heads)) / P(Awake) = (1/2 * 1/2) / 3/4 = 1/3
There are four equally probable states in the experiment, three awake states and one sleep state, so P(Awake) = 3/4 and P(Awake|Heads) = 1/2. — Andrew M
But that doesn't mean that there was a 1/3 chance of getting a red ball. — Michael
This is why I suggested the alternative experiment where we don't talk about days at all and just say that if it's heads then we'll wake her once (and then end the experiment) and if it's tails then we'll wake her twice (and then end the experiment). There aren't four equally probable states in the experiment. — Michael
I’m saying that if it’s heads then only I’m asked and if it’s tails then I’m asked and 1,000 other people are asked. — Michael
Sleeping Beauty knows she has 1/2 chance of going through any possibility. You have 1/2 chance of going through Monday+heads on 1/4 chance of going through each tails outcome. Those situations are not comparable. — BlueBanana
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