Monty Hall question, but varied. I haven’t seen the option that no matter what you have a 40% chance overall to win. You always pick 1 of 3 doors, then 1 is discarded & you pick 1 of 2. So don’t you always have a 2/5 chance of being right no matter what?
In: 0
Your odds of initially picking the correct door is 1/3. This means that the odds of the prize being behind one of the other two doors is 2/3.
If the door you doesn’t have the prize then you know that one of the remaining doors has a prize and the other does not. If the door you chose has a prize then both remaining doors do not have a prize. In either case, you ABSOLUTELY KNOW that one of the remaining doors does NOT have a prize.
Monty Hall then opens one of the two remaining doors that DOES NOT have the prize. This is sometimes left as an assumption, but is essential to the calculation. If he ALWAYS picks a random door to show you then the results are completely different. You either lose when he shows you the prize, or there is a 50% chance that the prize is behind the remaining doors.
Anyways, since you know that at least one of the unpicked doors has no prize, when Monty shows you one of the doors without a prize it adds no new information–you already knew one of the doors was empty.
Since showing the door doesn’t add any information, the odds for the initial pick remain the same. There was a 2/3 chance that the prize was behind the doors you didn’t pick, so even though there is only one door it has a 2/3 chance of having the prize.
Another way to think of this is to have 100 doors. You pick one. Monty opens 98 of the other doors that do not contain a prize. Do you switch? This is the same problem with just a minor tweak. Just like before where we knew at least one of the remaining doors did not have a prize, with 100 doors we know that at least 98 of the remaining doors have no prize, so Monty can show you them without changing the information that you know. My guess is that you’d definitely switch in the 100 door case.
Latest Answers