Some puzzles

[mfgoode]

No.1: 23 [spoiler]
Mallly

[biwema]

5:

The Chance of wonning in the normal goat-problem is n-1/n.

But due to the fact that the candidate has to change the door every round.
Every time, when one door is opened, the chance that this door contained the prize is spread among the other door (except that on, which the candidate has chosen.) so the probabilities differ between the doors.

The candidate has now the goal, that he keeps the porbability of one (or few) door as low as possible, that the last time he has to switch, he can jump to a door with a high probability.

If the game leader knows that, he can always choose the door with the lowest probability and so bring the chance of winning down to 50%

edit(forgot to write):
... independent from the initial number or doors.

[rogue]

Originally Posted by biwema

If the game leader knows that, he can always choose the door with the lowest probability and so bring the chance of winning down to 50%

I disagree. Remember, when there are three doors left you choose a door and one is opened. You are forced to switch. Since your last choice gave a 1 in 3 chance of winning, the switch changes that to 1 - 1/3 or 2/3.

[fetofs]

Originally Posted by biwema

5:

The Chance of wonning in the normal goat-problem is n-1/n.

But due to the fact that the candidate has to change the door every round.
Every time, when one door is opened, the chance that this door contained the prize is spread among the other door (except that on, which the candidate has chosen.) so the probabilities differ between the doors.

The candidate has now the goal, that he keeps the porbability of one (or few) door as low as possible, that the last time he has to switch, he can jump to a door with a high probability.

If the game leader knows that, he can always choose the door with the lowest probability and so bring the chance of winning down to 50%

edit(forgot to write):
... independent from the initial number or doors.

But isn't the chance (with four doors) equal to .625? I think the answer must be strictly in terms of n.

[biwema]

It always depends on the view:

I went through all possble cases:

example for 3 doors
(Prize is in 3)
candidate: 1 moderator: 2 candidate: 3 - W
candidate: 2 moderator: 1 candidate: 3 - W
candidate: 3 moderator: 2 candidate: 1 - L
candidate: 3 moderator: 1 candidate: 2 - L

50 %

if you equalize the first choices of the candidate, on the other hand, you get 2/3.

to cut down the whole a little bit: I chance of winning is equal the chance of choosing a goat at the second last step.

[fetofs]

Originally Posted by biwema

It always depends on the view:

I went through all possble cases:

example for 3 doors
(Prize is in 3)
candidate: 1 moderator: 2 candidate: 3 - W
candidate: 2 moderator: 1 candidate: 3 - W
candidate: 3 moderator: 2 candidate: 1 - L
candidate: 3 moderator: 1 candidate: 2 - L

50 %

if you equalize the first choices of the candidate, on the other hand, you get 2/3.

to cut down the whole a little bit: I chance of winning is equal the chance of choosing a goat at the second last step.

I can see that, but we are talking about the candidate side here. With choices 1 and 2 he won, and with 3 he lost (I actually equalized the last ones, as the candidate made the same choice in both)

[rogue]

Originally Posted by biwema

It always depends on the view:

I went through all possble cases:

example for 3 doors
(Prize is in 3)
candidate: 1 moderator: 2 candidate: 3 - W
candidate: 2 moderator: 1 candidate: 3 - W
candidate: 3 moderator: 2 candidate: 1 - L
candidate: 3 moderator: 1 candidate: 2 - L

50 %

if you equalize the first choices of the candidate, on the other hand, you get 2/3.

to cut down the whole a little bit: I chance of winning is equal the chance of choosing a goat at the second last step.

In the first two possibilities, the moderator has one choice. The last two possibilities are really the same possibility except that the moderator has two choices. They would be weighted as 1, 1, .5, .5, which is the same as (1+1)/(1+1+.5+.5) or 2/3.

It should be easy to write a program to demonstrate this. Although I'm not 100% certain for cases for more than 3 doors, I feel confident until someone can persuade me otherwise.

[fetofs]

Originally Posted by rogue

It should be easy to write a program to demonstrate this. Although I'm not 100% certain for cases for more than 3 doors, I feel confident until someone can persuade me otherwise.

For 4 doors, assume two possibilities:

1)The guest picks the right door at the start, and consequently at the end (.25)
2)The guest picks the wrong door first (.75), and misses again (.5) to get the car. So, the chance is .75*.5=.375

.375+.25=.625

[rogue]

Originally Posted by fetofs

For 4 doors, assume two possibilities:

1)The guest picks the right door at the start, and consequently at the end (.25)
2)The guest picks the wrong door first (.75), and misses again (.5) to get the car. So, the chance is .75*.5=.375

.375+.25=.625

2/3 is incorrect for > 3 doors because when there are four doors left and Monty opens a door, you can only choose from two of the remaining three doors and not all three of them (because the third is the one you had to switch from). I have a function in mind, but I won't post it yet. I want to work out the answer for n = 5. If someone posts it before I have worked it out, then I can see if my function is correct.

[biwema]

n=5:
P=1/2
19 cases pro and 19 cases contra

[rogue]

Originally Posted by biwema

n=5:
P=1/2
19 cases pro and 19 cases contra

Could you please show the cases? I haven't had time to work it out, so I would to see the different cases.