63
votes
Accepted
100 pieces 1 opportunity, choose wisely!
What you're missing here is the chance of playing at all, given that the game ends when someone finds the prize. (or, chance of finding a prize goes to 0, which is the same thing)
...
- 7,154
37
votes
22
votes
100 pieces 1 opportunity, choose wisely!
Other people have already given correct answers, but I wanted to suggest a different way of thinking about the question that involves less calculation:
- 111k
21
votes
Monty Hall variation
A difference between the Monty Hall scenario and the train platform scenario is that
After you remember that the train will definitely not be leaving from platform 2,
- 111k
19
votes
Accepted
The Monty Hall problem
You switch your door, because you have double the chance of getting the car
Explanation:
When you pick a door, there is a 1 in 3 chance that the door contains the car.
Let's examine that case of ...
- 669
14
votes
The Monty Hall problem
Switch.
There's a $2/3$ chance that you'll choose a goat, and a $1/3$ chance you'll choose the car.
If you chose a goat, after switching you will win a car. (Guaranteed)
If you chose a car, after ...
- 864
13
votes
Accepted
4 Doors and a Half Value prize
Note: I assume the host knew door #1 would be empty. This is analogous to the original Monty Hall problem. If the host randomly opens a door, the door-opening is moot and can be left out completely.
...
- 851
13
votes
Accepted
The Old Millionaire
For simplicity, I'm going to refer to the associates as knight, knave, and joker. Also, I'm assuming that they won't reveal the location of the money. If they did, then obviously you would switch to ...
- 5,767
12
votes
The Monty Hall problem
I've answered this over on Math.SE, so I'll just quote most of that.
Suppose we have $n$ doors, with a car behind $1$ of them. The probability of choosing the door with the car behind it on your ...
- 4,956
12
votes
Accepted
Monty Hall all over again?
Here's the formal expression of what we seek. Let $C$ be the event "chose the door with the car" and $R$ be the event "revealed the door with the car". Now, assuming that the door opened is random (...
- 4,943
12
votes
Accepted
10
votes
Monty Hall variation
Here's how the original Monty Hall worked:
You pick a door. (2/3 chance of it being a goat, 1/3 chance of it being a car)
Monty Hall opens a door that he knows has a goat in it, and then asks you if ...
- 8,947
9
votes
Monty Hall all over again?
Without assuming anything more? No, we have absolutely no idea whether it is beneficial to switch or not.
If the host knows what is behind each door and is malicious, then he will try to trick us. ...
- 199
7
votes
Accepted
The 3 Boxes Show!
If I want, I can always win my money back!
Not switching is academic: I have a 1/3 chance of picking the 20k, 1/3 of picking the -3k and 1/3 of picking the +7.5k. My expected win is 6.6k-1k+2.5k = 8....
- 851
7
votes
Monty Hall all over again?
I will stay with the door I have. I am banking on a malicious host who takes advantage of my Monty Hall familiarity and only offers me a switch when I am pointing to the car.
- 1,136
7
votes
Two player Monty Hall variation
This solution was derived independently
but the result is equivalent to
the answer already
given by Mark Tilford
and the confident tone of
a suggestion by Greg Martin
led to the realization that this ...
- 21.4k
6
votes
6
votes
Accepted
6
votes
Simplist proof you have for the Monty Hall Problem
You have a:
EDIT: Now I have better understanding of what you've after.
You have a robot:
- 9,301
5
votes
Accepted
5
votes
Monty Hall all over again?
The accepted answer conflates "the host's knowledge and motivations are unknown" with "the host's actions are uniform and random", which is at best a debatable assumption. I actually think the correct ...
- 196
5
votes
Monty Hall all over again?
The question is vague, and assumptions must be made that drastically change the possible answers.
EDIT: Analysis added below. Am I missing something?
- 15k
4
votes
Monty Hall all over again?
Let's assume that the Host randomly decides whether to reveal a goat and offer a switch: if your first door has a car, he reveals with probability $p$, and if it was first a goat, then with ...
- 31.4k
4
votes
Monty Hall all over again?
Well in this case the game becomes more or a Poker game because you don't know if the host followed a rule or not. There are two (already well-known) cases:
The host chose the other door randomly (he ...
- 141
4
votes
Monty Hall variation
When doing Bayesian updating on new information, we should change our probability assignment for a hypothesis only if the probability of seeing the new information depends on whether the hypothesis is ...
- 978
4
votes
Variation of Monty Hall problem for gambling on who is the champion soccer team
Original answer (now Part 1)
If we assume each team has an equal chance of winning any particular match and that you know A plays B and C plays D in the first round.
Then
Second answer (now Part 2),
- 14.8k
4
votes
4
votes
Two player Monty Hall variation
I believe the answer is:
Start with all possible permutations:
Test the different strategies:
Strategy 1:
Strategy 2:
Strategy 3:
Strategy 4:
- 1,742
4
votes
Two player Monty Hall variation
Alice and Bob start by assuming that the Goat is behind a specific door. If the doors are numbered 1, 2, and 3 then they might, for example, decide to assume the Goat is behind door 3. With this ...
- 41
3
votes
Monty Hall all over again?
If he always opens one of the doors you did not pick at random, then the door he opens should contain a car 1/3 of the time, a sheep 1/3 of the time, and a goat 1/3 of the time.
If the animals and ...
- 1,191
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