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) ...
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37 votes

100 pieces 1 opportunity, choose wisely!

Actually,
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  • 33.6k
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:
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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,
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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 ...
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  • 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 ...
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  • 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. ...
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  • 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 ...
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  • 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 ...
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  • 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 (...
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  • 4,943
12 votes
Accepted

Two player Monty Hall variation

They win with probability
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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 ...
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  • 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. ...
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  • 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....
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  • 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.
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  • 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 ...
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  • 21.4k
6 votes

100 pieces 1 opportunity, choose wisely!

Since It should be clear that
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6 votes
Accepted

Simplist proof you have for the Monty Hall Problem

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  • 4,641
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:
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  • 9,301
5 votes
Accepted

The Monty Hall Arena

Let me give it a try. Attempt at the bonus
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  • 9,786
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 ...
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  • 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?
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  • 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 ...
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  • 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 ...
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  • 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 ...
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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),
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4 votes

Two player Monty Hall variation

I believe
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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:
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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 ...
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  • 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 ...
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