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Sverigevader | 1 year ago
I have three doors in front of me. behind one is a great prize. Behind the other two are bad prizes. I know which door contains the prize, and I choose that door. Before I open it the game show host eliminates one of the doors that contain the bad prize. He then asks if I'd like to switch to the other remaining door instead of the one I chose. Should I switch doors?`
Big answer:
This scenario is a classic probability puzzle known as the Monty Hall problem, named after the original host of the game show *Let's Make a Deal*. Here's how it works and what you should do:
1. **Initial Choice**: You choose one of the three doors. At this stage, the probability that you have selected the door with the great prize is \( \frac{1}{3} \), and the probability that the great prize is behind one of the other two doors is \( \frac{2}{3} \).
2. **Host's Action**: The host, who knows what is behind each door, then opens one of the remaining two doors to reveal a bad prize. This step is crucial because the host always reveals a bad prize and always chooses one of the doors you did not pick.
3. **Decision Point**: You now have the option to stick with your original choice or switch to the other unopened door. Here’s why you might consider switching:
- If your original choice (let's say you chose door 1) was correct (which has a probability of \( \frac{1}{3} \)), then the other two doors (2 and 3) both have bad prizes, and switching would cause you to lose.
- However, if your original choice was incorrect (which has a probability of \( \frac{2}{3} \)), meaning the prize is behind one of the other two doors, the host's action of opening another door (say he opens door 3, revealing a bad prize) leaves the door with the prize (door 2) still closed. Switching in this case would lead you to win.
4. **Probability of Winning**:
- If you do not switch, you win if your original choice was the door with the prize (probability \( \frac{1}{3} \)).
- If you do switch, you win if your original choice was not the door with the prize (probability \( \frac{2}{3} \)).
Therefore, **you should switch doors**. Switching doors doubles your chances of winning the great prize, from \( \frac{1}{3} \) (if you stick with your original choice) to \( \frac{2}{3} \) (if you switch). The switch works in your favor because it capitalizes on the probability shift caused by the host's action of revealing a bad prize behind one of the doors you did not initially pick.
madspindel|1 year ago
In the scenario you describe, you already know which door has the great prize and you have chosen that door initially. The game show host then eliminates one of the other two doors, which must be a door with a bad prize. At this point, you are asked whether you want to switch to the remaining unchosen door.
Since you initially chose the door with the great prize and the host's action of removing a bad prize door doesn't change the contents of the doors, there is no benefit to switching. Your initial choice was the door with the great prize, and it still contains the great prize. Switching doors would only lead you to a door with a bad prize.
Therefore, you should not switch doors. You should stick with your original choice where you already know the great prize is located.
This situation is different from the classic Monty Hall problem, where the contestant does not know where the prize is initially and the probabilities change based on the host's action of removing a bad prize door. In your case, since you know where the prize is from the start, the logic of the Monty Hall problem does not apply. Stick with your initial choice.
Sverigevader|1 year ago
dwallin|1 year ago