Stick or change?
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).
Comments
-
Always change according to game theory.
In real life I'm rarely offered multiple doors to open.1 -
Stick, never change your mind once you have made your decision.0
-
yep0
-
Peek through the keyhole6
-
Change - 2/3 versus 1/3 if you stick - in other words double the chance of winning if you switch.
Wasn't the Monty Hall Paradigm discussed in your other thread?4 -
Agree. This is the correct, if surprising, option and there is a mathematical basis to it.Henry Irving said:Always change according to game theory.
0 -
Yup. Sounds strange but changing doors would double your chance.0
-
Here's a better one.
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).0 -
Would you care to explain why switching is best?
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?0 -
Your first choice gave you a 1 in 3 chance of being correct - so the other two doors represent 2 out of 3 chances. They still do!!superchrissypowell said:Would you care to explain why switching is best?
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?0 -
Sponsored links:
-
Buy a sledgehammer1
-
I think it depends if you were the first child born or the second and you have a sistersuperchrissypowell said:Would you care to explain why switching is best?
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?
4 -
You have to change, if you stick then you have gone from a 1/3 chance to win, to a 1/2 chance to win. However if you change then the odds go from 1/3 to 2/3 chance of getting it right. It's basic probability maths1
-
It is said that in a group of thirty, there's an even chance two people will have the same birthday, so I'd say 31.bobmunro said:Here's a better one.
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).
0 -
I'm useless at these things.
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.0 -
I always thought it was 23 people in a room for a 50-50 chance of 2 having the same birthday.0
-
RD sign someone ffs12
-
Here's one for you.... You're in a king size bed, on your left Rachel Riley is lying naked and on the right, Kelly Brook naked. You have to make a choice.
I'm still struggling with this one, the only answer I can think of, is to toss.4 -
Can you explain how the odds go from 1/3 to 2/3, I thought there was only 1 correct answer? Therefore it's a 1/2 chance whichever door you choose.sam3110 said:You have to change, if you stick then you have gone from a 1/3 chance to win, to a 1/2 chance to win. However if you change then the odds go from 1/3 to 2/3 chance of getting it right. It's basic probability maths
0 -
You may well be right, thirty was what I remembered but there's no guarantee the person who told me that knew what they were talking about.stackitsteve said:I always thought it was 23 people in a room for a 50-50 chance of 2 having the same birthday.
0 -
Sponsored links:
-
Rachel every time, then there's a chance of some intellectual stimulation over coffee the next morning.DaveMehmet said:Here's one for you.... You're in a king size bed, on your left Rachel Riley is lying naked and on the right, Kelly Brook naked. You have to make a choice.
I'm still struggling with this one, the only answer I can think of, is to toss.1 -
Think of it this way.
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 and C.
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 and C. 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?1 -
That is roughly right. It actually works out in practice. Since schools keep classes of around 30, you generally find that more often than not, two people in that class will share a birthday.AddicksAddict said:
You may well be right, thirty was what I remembered but there's no guarantee the person who told me that knew what they were talking about.stackitsteve said:I always thought it was 23 people in a room for a 50-50 chance of 2 having the same birthday.
0 -
Not only was this covered in the boy,boy thread but there was a thread on this very puzzle a few years ago - please tell me that people that read the previous thread didn't ask for the explanation again?2
-
Yes there's a 2/3 chance of it being behind door B or C but the same odds would apply to it being behind door A or B.Fiiish said:Think of it this way.
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 and C.
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 and C. So switching is the best choice.
Think of it this way. 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 chance0 -
0 -
But we already know that door A that you picked had a 1/3 chance of being the prize. Nothing has happened to change that.MartinCAFC said:
Yes there's a 2/3 chance of it being behind door B or C but the same odds would apply to it being behind door A or B.Fiiish said:Think of it this way.
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 and C.
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 and C. So switching is the best choice.
Think of it this way. 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 chance0 -
This was part of a American game show in the 1980s. Many contestants stuck with their first choice. However, it is always best to change as your odds are twice as good if you change.0
-
It is 23 - so at a football match with 22 players and a referee on the pitch there is a better than evens chance that two share the same birthday.stackitsteve said:I always thought it was 23 people in a room for a 50-50 chance of 2 having the same birthday.
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.1 -
Not quite.MartinCAFC said:
Yes there's a 2/3 chance of it being behind door B or C but the same odds would apply to it being behind door A or B.Fiiish said:Think of it this way.
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 and C.
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 and C. So switching is the best choice.
Think of it this way. 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
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.2
