Bertrand's Box Paradox and Monty Hall Problem

In summary, the conversation discussed two different probability paradoxes: Bertrand's Box Paradox and the Monty Hall Problem. In the first, there is a 1/3 chance of drawing a gold coin from three drawers, but if one of the gold coins is revealed, the chance of the other being gold increases to 2/3. In the second, with 10 doors, there is a 1/10 chance of choosing the correct door, but after 8 doors are revealed, the chance of the remaining closed door being correct increases to 9/10.
  • #1
Quincy
228
0
This thing is making me pull my hair out: http://en.wikipedia.org/wiki/Bertrand's_box_paradox Can someone give me a good explanation of this?

As for the Monty Hall Problem, I think I understand it. This is how I think of it: There is a 1/3 possibility of picking the correct door and 2/3 possibility of picking the wrong door. And after one of the empty doors has been revealed, then by switching your choice, that 2/3 possibility of picking the wrong door becomes a 2/3 possibility of picking the correct door. -- Is this a legitimate way of understanding it?
 
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  • #2
Quincy said:
This thing is making me pull my hair out: http://en.wikipedia.org/wiki/Bertrand's_box_paradox Can someone give me a good explanation of this?

All the explanation was trying to say was that when you drew a gold coin, there is a 1/3 chance of drawing each gold coin of the three gold coins. 2 of these gold coins when checked with its partner will give another gold since both are in the same drawer whereas the last gold coin is with a silver coin in a drawer. Therefore, the chance that the partner coin is gold is 2/3.
 
  • #3
Monty Hall Problem: it´s easy to understand it with 10 doors. There is a 1/10 possibility of picking the correct door and 9/10 possibility of picking the wrong door. You make your choice. And after 8 of the empty doors has been revealed, you choice still 1/10 possibility of picking the correct door and the other closed door is 9/10 possibility of picking the correct door.
 

1. What is Bertrand's Box Paradox?

Bertrand's Box Paradox is a thought experiment that involves three boxes, each containing a different amount of money. The paradox arises when you are given the option to switch your choice after being shown the contents of one of the boxes. The question is whether it is advantageous to switch or keep your initial choice.

2. What is the Monty Hall Problem?

The Monty Hall Problem is a famous puzzle that was made popular by the game show "Let's Make a Deal." It involves three doors, one of which hides a prize. After choosing a door, the host opens one of the other doors to reveal that it does not contain the prize. The question is whether it is advantageous to switch your choice to the remaining unopened door.

3. Are Bertrand's Box Paradox and the Monty Hall Problem related?

Yes, both Bertrand's Box Paradox and the Monty Hall Problem involve a similar scenario where you are given the option to switch your choice after being shown new information. However, the two problems have different probabilities and outcomes, and the optimal strategy for each is different.

4. What is the optimal strategy for Bertrand's Box Paradox?

The optimal strategy for Bertrand's Box Paradox depends on the probabilities assigned to each box. In general, if the probabilities are equal, then it does not matter whether you switch or keep your initial choice. However, if one box has a higher probability, it is more advantageous to switch to that box after being shown the contents of one of the other boxes.

5. What is the optimal strategy for the Monty Hall Problem?

The optimal strategy for the Monty Hall Problem is to always switch your initial choice. This is because, by switching, you have a 2/3 chance of choosing the correct door, compared to a 1/3 chance if you stick with your initial choice. This may seem counterintuitive, but it has been mathematically proven to be the best strategy.

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