# Tag Info

62

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) Person 1 has a 100/100 chance of playing, and a 1/100 chance of winning. Person 2 has a 99/100 chance of playing, and a 1/99 chance of winning. Person 3 has a 98/100 chance of ...

37

Actually,

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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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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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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 when you originally picked the car, and when you originally picked a blank separately. Originally picked the car(1/3 chance) In this case, the host can open ...

14

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 switching you will win a goat. (Guaranteed) Therefore, the chance you'll win a car is $2/3$ if you switch. Simply put, the chance you picked the car before the ...

13

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 that once they reveal it. Assuming that there is an equal chance of any of the three associates coming up and talking to you, we have a two in three chance of ...

12

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 first pick, is $\frac{1}{n}$. Monty then opens $k$ doors, where $0\leq k\leq n-2$ (he has to leave your original door and at least one other door closed). ...

12

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 (but not the one the player chose), we have  Pr(C) = \frac13\\ Pr(R|C) = 0\\ Pr(R|\bar C) = \frac12\\ Pr(C|\bar R) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+...

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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 you'd like to switch to the other door. 2/3 of the time, the door Monty Hall opened is the only other door with a goat - because 2/3 of the time, you picked a ...

9

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. But if we know he will try to trick us then we can use that against him. Unless he knows that we know, and so on and so on. In the end that means there is no ...

7

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 = 8.1k. Blindly switching to the closed door raises my chances. The original Month Hall problem teaches me that I have 2/3 chance of picking the 20k by switching ...

7

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.

6

Since It should be clear that

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You have a: EDIT: Now I have better understanding of what you've after. You have a robot:

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Let me give it a try. Attempt at the bonus

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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 answer to this question is that there is not enough information to answer it as phrased. For example, if I flip a coin outside of your view and ask what are ...

5

The question is vague, and assumptions must be made that drastically change the possible answers. EDIT: Analysis added below. Am I missing something?

4

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 probability $q$. This is the best we can do: we don't know whether the Host planned to open a door or not. Note that $p=q=1$ gives the original Monty Hall scenario. Let \$...

4

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 didn't know what was behind that one): in this case it doesn't change anything if you change or not: each door has a 1/2 chance. The host chose on purpose a ...

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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),

3

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 true. Suppose we choose Door A, and Monty Hall reveals a goat behind Door B. If Monty Hall always randomly chooses between the unchosen doors when we ...

3

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 car were placed at random, we had the following possible cases to consider at the start, each of which had equal probability: S G C S C G G S C G C S C S G C G ...

3

The other answers are correct. I just wanted to add here the "long way" of doing it. I mean all the possible cases. Case 1. Car Goat Goat Case 1.1. You choose door 1. Monty opens| switch | don't switch 2 | Lose | Win 3 | Lose | Win Case 1.2. You choose door 2. Monty opens| switch | don't switch 3 | Win | ...

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I always start with box 1. I'm assuming he picks randomly if neither box 2 or 3 contains the 20,000. If he shows the -3000 box, you're better off switching. Case 1: Probability 33% - Box 1 contains 20k, remaining box contains 7500. I win 7500 by switching. Case 2: Probability 66% - Box 1 contains 7500, remaining box contains 20k. I win 20,000 by ...

2

EDIT: to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car in the worst case, no matter what the hosts motives are. If I understood your question correctly, "in the worst case" is the key here. So the correct answer is a strategy where you maximize your winning chance over all ...

2

EDIT: to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car in the worst case, no matter what the hosts motives are. You can't sweep the host's motives under the rug like some small, unimportant detail. The player's strategy is completely dependent on the host's strategy. HOST ...

2

I will answer your edited question. In case that you have prior knowledge that the ones before you didn't find the price then the better turn is...the last. It's a little underwhelming but basically you have a 1*100/n% chance in getting the price with any number left of cake pieces (being "n" the number of pieces) so 100/1 is 100%.

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