EGPRC

5 ms·

1.
by EGPRC 3y ago
I believe the confusion with "ignoring the car picked cases" comes due to thinking that they are additional games to the original ones, instead of being part of those that constitute the 2/3 in which the player starts picking
2.
by EGPRC 3y ago
That's not correct, because if you agree that one kind of result tends to happen more than the other, then it is easier that the single game that you are currently playing belongs to the larger group. When you play once, it is like if
3.
by EGPRC 4y ago
The Monty Fall problem asks you about the case when the host has managed to reveal a goat. That only includes a subset of the total attempts that you would start selecting a door, not all. Once a goat is revealed, you could only be in the 1
4.
by EGPRC 4y ago
Moreover, to understand why with a random revelation of the goat the chances of each option are 50%, you first need to understand the real reason why they are not 50% in standard Monty Hall problem. It is because when the player has picked
5.
by EGPRC 4y ago
You can get that the answer must be 50% once the host just reveals a goat by chance using reductio ad absurdum. Notice that since the host is doing it randomly, it would be the same if you (the contestant) were who made the revelation inste
6.
by EGPRC 4y ago
Yes, you could still win 2/3 of the time by switching in all games, only that with those rules the probability for each remaining door once a goat is revealed would be 1/2, as you stated. So, you could also win 2/3 of the tim
7.
by EGPRC 4y ago
This only occurs because the host knows the locations of the contents and cannot reveal the player's door and neither which has the car. So the other that he leaves closed is always the best possible one that could be found among the r
8.
by EGPRC 5y ago
In Monty Hall Game you improve your odds to be 2/3, not 1/2. If you think the staying door wins 1/3 of the time and the switching door 2/3, where is the car located in the 1/6 missing?