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You are presented with three, identical doors. Behind one of the doors is a million pounds. Behind each of the other two doors is a losing lottery ticket.

If you choose the door with the money behind it, you win the dough.

However, after you choose a door, and before it's opened, one of the remaining doors opens, to reveal a losing lottery ticket.

You're then given a chance to stick with your original choice of door, or change your mind.

What should you do? Stick? Change? Or does it make no difference?

(Acknowledgements to Monty Hall for this one).

If you choose the door with the money behind it, you win the dough.

However, after you choose a door, and before it's opened, one of the remaining doors opens, to reveal a losing lottery ticket.

You're then given a chance to stick with your original choice of door, or change your mind.

What should you do? Stick? Change? Or does it make no difference?

(Acknowledgements to Monty Hall for this one).

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## Comments

In real life I'm rarely offered multiple doors to open.

Wasn't the Monty Hall Paradigm discussed in your other thread?

How many people would you need in a room to have more chance than not that two of them share the same birth anniversary (not year, just day and month).

Initially you have a 1 in 3 chance of getting the right door.

You make it past the first stage and so you have the option of two doors (as you can completely rule out the door that has been opened showing a losing ticket), 1 with the winning ticket and other with a losing ticket. Thus your chances of winning at that point are now 1 in 2. Surely it doesnt matter if you change your mind or not?

Sponsored links:I'd think it's still the same chance of being the money.

One door containing a losing ticket will be opened leaving 50/50 chance of you being right whether you've pick a losing one or the one with the bucks.

I'm still struggling with this one, the only answer I can think of, is to toss.

Sponsored links:There are doors A, B and C. You are told the winning ticket is behind one of these 3 doors, each equally likely (1/3).

Say you pick door A. That means you have a 1/3 chance of getting it right, or there is a 2/3 chance of it being behind doors B

andC.The host then reveals Door B to be a losing door. The odds still haven't changed though, there is a 1/3 chance you pick the correct door, and a 2/3 chances of it being behind B

andC. So switching is the best choice.Similarly, say there are two coats, one has one pocket and one has two pockets. You are told that one of the pockets has £100 in it, and you pick one of the two coats and get to keep the money found in any of the pockets. Would you pick the coat with one pocket or two pockets? Even if you definitely knew that at least one of the pockets in the two pocket coat were empty?

The coat pockets explanation makes sense and I agree with because the odds were in it's favour with two pockets to begin with but in this case each door is an equal 1/3 chance

It starts to go up pretty quickly after that - with 75 people it's 99.9% that two (or more) will share the same birthday.

If it's behind A and you choose A, regardless of whether it is B or C, you would LOSE by switching

If it's behind A and you choose B, C would have to be opened, so by switching you would WIN

If it's behind A and you choose C, B would have to be opened, so be switching you would WIN

If it's behind B and you choose B, regardless of whether it is A or C, you would LOSE by switching

If it's behind B and you choose A, C would have to be opened, so by switching you would WIN

If it's behind B and you choose C, A would have to be opened, so be switching you would WIN

If it's behind C and you choose C, regardless of whether it is A or B, you would LOSE by switching

If it's behind C and you choose A, B would have to be opened, so by switching you would WIN

If it's behind C and you choose B, A would have to be opened, so be switching you would WIN

Therefore, there are 9 possible outcomes and 6 would be winners by switching and 3 would be losers.