What are the odds of winning?

[nomadking]

With the 3 doors scenario, if you pick correctly first time, then the host has TWO doors to pick from. If you pick one of the 2 the wrong ones, then only ONE to pick from. Overall there are 4 possible end results.
There is a prize door and your initial choice of door. There is a 50/50 chance they are one in the same. That is the question you are being asked, no matter how many doors you start with.

The mistake has been made in treating all non-prize doors as a single outcome.
Prize is behind A.
Combination 1 : Pick A, Host opens B, stick wins
Combination 2 : Pick A, Host opens C, stick wins
Combination 3 : Pick B, Host opens C, swap wins
Combination 4 : Pick C, Host opens B, swap wins
If you're going to treat 1&2 as a single combination, then you have to treat 3&4 as a single combination. That still brings you around to a 50/50 chance.
Now 4 starting doors, prize is behind A
Comb 1 : Pick A, opens B&C, stick wins
Comb 2 : Pick A, opens B&D, stick wins
Comb 3 : Pick A, opens C&D, stick wins
Comb 4 : Pick B, opens C&D. swap wins
Comb 5 : Pick C, opens B&D. swap wins
Comb 6 : Pick D, opens B&C. swap wins
50/50 again.
Would you when throwing a pair of dice, treat all non-double six combinations as single combination?

[daveeb]

Originally Posted by nomadking

With the 3 doors scenario, if you pick correctly first time, then the host has TWO doors to pick from. If you pick one of the 2 the wrong ones, then only ONE to pick from. Overall there are 4 possible end results.
There is a prize door and your initial choice of door. There is a 50/50 chance they are one in the same. That is the question you are being asked, no matter how many doors you start with.

The mistake has been made in treating all non-prize doors as a single outcome.
Prize is behind A.
Combination 1 : Pick A, Host opens B, stick wins
Combination 2 : Pick A, Host opens C, stick wins
Combination 3 : Pick B, Host opens C, swap wins
Combination 4 : Pick C, Host opens B, swap wins
If you're going to treat 1&2 as a single combination, then you have to treat 3&4 as a single combination. That still brings you around to a 50/50 chance.
Now 4 starting doors, prize is behind A
Comb 1 : Pick A, opens B&C, stick wins
Comb 2 : Pick A, opens B&D, stick wins
Comb 3 : Pick A, opens C&D, stick wins
Comb 4 : Pick B, opens C&D. swap wins
Comb 5 : Pick C, opens B&D. swap wins
Comb 6 : Pick D, opens B&C. swap wins
50/50 again.
Would you when throwing a pair of dice, treat all non-double six combinations as single combination?

It's already been explained several times, please don't get lost down that rabbit hole

[spiderplant]

Originally Posted by nomadking

Combination 1 : Pick A, Host opens B, stick wins
Combination 2 : Pick A, Host opens C, stick wins
Combination 3 : Pick B, Host opens C, swap wins
Combination 4 : Pick C, Host opens B, swap wins

But you're twice as likely to pick either B or C than you are to pick A. So swap wins twice as often.

The odds are always in favour of swapping.
If you dont understand that after the example I gave, you never will.

[ and no, of course its not a real game show, its just a maths teaser, no one would run it as a real show because the odds are in favour of the participant ]