odds increase if the person opening the door knows the location of the winning ticket and intentionally reveals a losing ticket
OP does not contain enough information for us to know whether this happened
if the door was opened at random then odds remain at 50%
I originally thought this was the case but actually it isn't.
Assume that the game ends as soon as the prize is revealed and the revealed door is picked from random from one of the two doors you do not pick, and that door C is the winning door. Here are the winning scenarios (switches are bolded) in order of Pick/Reveal/Stick or Switch
A/B/Switch B/A/Switch C/A or B/Stick
Losing Scenarios
A/B or C/Stick or Game Over B/A or C/Stick or Game Over C/A or C/Switch
nope, it's 50/50 if person opening the door does not know what is where
odds increase if the person opening the door knows the location of the winning ticket and intentionally reveals a losing ticket
OP does not contain enough information for us to know whether this happened
if the door was opened at random then odds remain at 50%
I originally thought this was the case but actually it isn't.
Assume that the game ends as soon as the prize is revealed and the revealed door is picked from random from one of the two doors you do not pick, and that door C is the winning door. Here are the winning scenarios (switches are bolded) in order of Pick/Reveal/Stick or Switch
A/B/Switch B/A/Switch C/A or B/Stick
Losing Scenarios
A/B or C/Stick or Game Over B/A or C/Stick or Game Over C/A or C/Switch
nope, it's 50/50 if person opening the door does not know what is where
proof is here
That scenario specifically excludes scenarios where the outcome could be the winning door being opened at random, unlike my proof. Probability theory does not work on excluding certain scenarios.
that outcome has already been exluded in the OP. @Chizz is asking what we should do in the event of a losing door being opened
True. Your proof has an error though (although I imagine you found it on the Internet and you did not write it). It gives each of the cases where Monty opens the door you picked a probability of 1/9, even though both the proof and the OP state that Monty will not do this, he will always open a remaining door.
I do agree with you that if it were truly random which door he picked, then it would be 50/50, but if you exclude the possibilities that the proof excludes, then it is not random in the true sense of the word.
that outcome has already been exluded in the OP. @Chizz is asking what we should do in the event of a losing door being opened
If the door opened reveals the $1million then the game is over because the player doesn't have the option to swap.
Whether or not the person opening the door to a losing ticket knows or not is irrelevant - IF the door opened reveals a losing ticket then swapping gives you a 2 in 3 chance of finding the prize.
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?
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?
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.
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
Not quite.
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.
that outcome has already been exluded in the OP. @Chizz is asking what we should do in the event of a losing door being opened
True. Your proof has an error though (although I imagine you found it on the Internet and you did not write it). It gives each of the cases where Monty opens the door you picked a probability of 1/9, even though both the proof and the OP state that Monty will not do this, he will always open a remaining door.
I do agree with you that if it were truly random which door he picked, then it would be 50/50, but if you exclude the possibilities that the proof excludes, then it is not random in the true sense of the word.
tbh I didn't even read that proof, just assumed what it was
but you get my point- OP does not give us enough info to know if switching will improve our chances. we do however know 100% that it won't reduce our chances
Imagine I have 10 envelopes, 9 empty and 1 with £100. You can pick 1 of the envelopes and I keep the other 9 We can all agree you have a 10% chance of the money, while I have 90%
If I then look in my envelopes and throw away 8 of the empty ones, we now have 1 each. Even though it seems its now 50-50, there is still 90% chance I have the money.
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nope, it's 50/50 if person opening the door does not know what is where
proof is here
I do agree with you that if it were truly random which door he picked, then it would be 50/50, but if you exclude the possibilities that the proof excludes, then it is not random in the true sense of the word.
Whether or not the person opening the door to a losing ticket knows or not is irrelevant - IF the door opened reveals a losing ticket then swapping gives you a 2 in 3 chance of finding the prize.
...except for spelling 'by' incorrectly thrice.
but you get my point- OP does not give us enough info to know if switching will improve our chances. we do however know 100% that it won't reduce our chances
;-)
You can pick 1 of the envelopes and I keep the other 9
We can all agree you have a 10% chance of the money, while I have 90%
If I then look in my envelopes and throw away 8 of the empty ones, we now have 1 each.
Even though it seems its now 50-50, there is still 90% chance I have the money.