I don't see how you can increase your chances of winning if there is only one winning door.
You've got a 1/3 chance.
Then it becomes 50/50.
Whether the first one you picked was the winning one or the other door when you get the 50/50. You can't increase your chances of winning when it's 50/50.
You can increase your chances of selecting the winning door at some point but if your original choice was the million quid then you change it to the incorrect door, you don't keep the money right? So I can't see a logical reason for changing the door.
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).
This is exactly the kind of nonsense that Mowgli gets his rocks off in some bed-sit in Belgium. Stop it everyone. Now. Someone put up a picture of a train.
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
Assume C is the winning Door.
The possible routes to victory are, in order of pick, reveal, then choice are...
A, B is revealed, switch to C B, A is revealed, switch to C C, A or B is revealed, stick with C
The routes to losing are:
A, B is revealed, stick with A B, A is revealed, stick with B C, A or B is revealed, switch to A or B
2 thirds of winning scenarios involve switching. 2 thirds of losing scenarios involve sticking.
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
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.
People above have explained it far better than I can, but it's a classic mathematical problem, it was a 1/3 chance, but now one of the doors has been shown to be empty it's a 2/3 chance. There is always 3 doors so the the probability will always be /3
I don't see how you can increase your chances of winning if there is only one winning door.
You've got a 1/3 chance.
Then it becomes 50/50.
Whether the first one you picked was the winning one or the other door when you get the 50/50. You can't increase your chances of winning when it's 50/50.
You can increase your chances of selecting the winning door at some point but if your original choice was the million quid then you change it to the incorrect door, you don't keep the money right? So I can't see a logical reason for changing the door.
Maybe I'm just thick.
You're thinking of it as a separate event once the first door is opened. As is shown on the last page if you change your pick after the first you'll win more times than you'll lose.
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.
If I was in bed with those two the only choice I would have to make is which one of my kids gets the house because I would die from over exertion.
This was originally posited by a mathematician, right? I read it in a book once. This woman got dog's abuse from dozens of mathematicians and professors, who refused to accept her logic. But she was eventually proved right I think? You switch.
This was originally posited by a mathematician, right? I read it in a book once. This woman got dog's abuse from dozens of mathematicians and professors, who refused to accept her logic. But she was eventually proved right I think? You switch.
Correct - and learned mathematicians refused to believe the proof!!
I've used the Monty Hall Paradigm for about 20 years when training traders in probability - there are people who attended courses I ran all those years ago who still don't get it!!
I don't see how you can increase your chances of winning if there is only one winning door.
You've got a 1/3 chance.
Then it becomes 50/50.
Whether the first one you picked was the winning one or the other door when you get the 50/50. You can't increase your chances of winning when it's 50/50.
You can increase your chances of selecting the winning door at some point but if your original choice was the million quid then you change it to the incorrect door, you don't keep the money right? So I can't see a logical reason for changing the door.
Maybe I'm just thick.
Look at it this way - when you first chose, you were probably wrong. So when you get a chance to choose again, don't stick with the door that was probably wrong.
This was originally posited by a mathematician, right? I read it in a book once. This woman got dog's abuse from dozens of mathematicians and professors, who refused to accept her logic. But she was eventually proved right I think? You switch.
Correct - and learned mathematicians refused to believe the proof!!
I've used the Monty Hall Paradigm for about 20 years when training traders in probability - there are people who attended courses I ran all those years ago who still don't get it!!
Does it translate on Deal or No Deal? I can't get my head around that bit. I think it was in a Derren Brown book that I read this and he said Deal or No Deal doesn't quite work that way.
The easiest way to prove it is assume that there are 100 doors and after you choose one, 98 of the other 99 are taken away. What are the chances of you guessing correct in the first instance? 1 in 100. So the most sensible choice is to switch.
If that doesn't work for you try 1 billion doors.
If that doesn't work go back to threading your wicker basket.
Think of it this way. [...] 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.
Wrong, you've missed an option out!
If it's behind A and you choose A, regardless of whether it isBor Copened, you would LOSE by switching If it's behind A and you choose A, regardless of whether it is B orC opened, 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 isBor Copened, you would LOSE by switching If it's behind B and you choose B, regardless of whether it is B orC opened, 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 isBor Copened, you would LOSE by switching If it's behind C and you choose C, regardless of whether it is B orC opened, 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 12 possible outcomes and 6 would be winners by switching and 6 would be losers.
The easiest way to prove it is assume that there are 100 doors and after you choose one, 98 of the other 99 are taken away. What are the chances of you guessing correct in the first instance? 1 in 100. So the most sensible choice is to switch.
Wrong, switching would ALSO give you a door which originally had odds of 1/100 - much like the very first door you chose.
The easiest way to prove it is assume that there are 100 doors and after you choose one, 98 of the other 99 are taken away. What are the chances of you guessing correct in the first instance? 1 in 100. So the most sensible choice is to switch.
Wrong, switching would also giving you a door which originally had odds of 1/100 - much like the first door you chose.
OK I'll have a bet with you. You put £10 in and I'll put £500 in. All the money will be put in a single pot and put behind one out of a hundred doors. You pick one door at random, then 98 of the other doors will be opened and I get what's behind the 99th door. If it's truly 50/50 then you have far more to gain from betting than I do. I'll even offer you double or nothing if you lose the first time.
If it's behind A and you choose A, regardless of whether it isBor Copened, you would LOSE by switching If it's behind A and you choose A, regardless of whether it is B orC opened, 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 isBor Copened, you would LOSE by switching If it's behind B and you choose B, regardless of whether it is B orC opened, 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 isBor Copened, you would LOSE by switching If it's behind C and you choose C, regardless of whether it is B orC opened, 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 12 possible outcomes and 6 would be winners by switching and 6 would be losers.
The easiest way to prove it is assume that there are 100 doors and after you choose one, 98 of the other 99 are taken away. What are the chances of you guessing correct in the first instance? 1 in 100. So the most sensible choice is to switch.
Wrong, switching would also giving you a door which originally had odds of 1/100 - much like the first door you chose.
Why have you doubled up on the same options? (in bold). You are wrong but you'll get it eventually.
And on the 100 door example - you are wrong again. Yes you would choose a door that was originally 1/100 by switching - but it aint 1/100 now because some nice kind person has removed 98 losers!
And on the 100 door example - you are wrong again. Yes you would choose a door that was originally 1/100 by switching - but it aint 1/100 now because some nice kind person has removed 98 losers!
What you're saying is that my originally chosen door is 1/100 - but now that all the other doors are removed, the one I'd switch to somehow has better odds - it doesn't.
Odds before doors removed: Both 1/100 Odds after doors are removed: Both 1/2.
OK I'll have a bet with you. You put £10 in and I'll put £500 in. All the money will be put in a single pot and put behind one out of a hundred doors.
You pick one door at random, then 98 of the other doors will be opened and I get what's behind the 99th door. If it's truly 50/50 then you have far more to gain from betting than I do. I'll even offer you double or nothing if you lose the first time.
The bolded section doesn't change anything to the original scenario (which you repeated underneath). Why are we putting in different amounts? You used amounts of money put in to justify door probabilities - they're not related. The unbolded part is still 50/50. I'll take you up on that bet!!
And on the 100 door example - you are wrong again. Yes you would choose a door that was originally 1/100 by switching - but it aint 1/100 now because some nice kind person has removed 98 losers!
What you're saying is that my originally chosen door is 1/100 - but now that all the other doors are removed, the one I'd switch to somehow has better odds - it doesn't.
Odds before doors removed: Both 1/100 Odds after doors are removed: Both 1/2.
What if I had a deck of 52 playing cards. You are trying to pick out the Ace of Spades, so you pick out one at random, face down. We set aside the facedown card to one side - 1/52 chance of it being the Ace of Spades. Therefore it is 51/52 chance of the Ace of Spades being in the rest of the deck I'm still holding.
I then take the other 51 cards, look at them, then remove 50 of the cards which are not the Ace of Spades. I have one facedown card left.
Which one are you going to think is the Ace of Spades? A card you have chosen randomly out of a deck of 52, or a card which has been filtered from a set of 51 cards which had a 51/52 chance of containing an Ace of Spades, or a 1/52 chance of not containing the Ace of Spades?
The easiest way to prove it is assume that there are 100 doors and after you choose one, 98 of the other 99 are taken away. What are the chances of you guessing correct in the first instance? 1 in 100. So the most sensible choice is to switch.
Wrong, switching would also giving you a door which originally had odds of 1/100 - much like the first door you chose.
OK I'll have a bet with you. You put £10 in and I'll put £500 in. All the money will be put in a single pot and put behind one out of a hundred doors. You pick one door at random, then 98 of the other doors will be opened and I get what's behind the 99th door. If it's truly 50/50 then you have far more to gain from betting than I do. I'll even offer you double or nothing if you lose the first time.
This is the sapiential version of "meet you at Makros to sort it out"
And on the 100 door example - you are wrong again. Yes you would choose a door that was originally 1/100 by switching - but it aint 1/100 now because some nice kind person has removed 98 losers!
What you're saying is that my originally chosen door is 1/100 - but now that all the other doors are removed, the one I'd switch to somehow has better odds - it doesn't.
Odds before doors removed: Both 1/100 Odds after doors are removed: Both 1/2.
OK I'll have a bet with you. You put £10 in and I'll put £500 in. All the money will be put in a single pot and put behind one out of a hundred doors.
You pick one door at random, then 98 of the other doors will be opened and I get what's behind the 99th door. If it's truly 50/50 then you have far more to gain from betting than I do. I'll even offer you double or nothing if you lose the first time.
The bolded section doesn't change anything to the original scenario (which you repeated underneath). Why are we putting in different amounts? You used amounts of money put in to justify door probabilities - they're not related. The unbolded part is still 50/50. I'll take you up on that bet!!
As I said, you'll get it eventually.
One last go at convincing you.
The purpose of the game is to choose an Ace.
Take three playing cards let's say two Jacks and an Ace. Lay those three cards at random face down and choose one. Now look at the other two and I'm sure you will agree that the Ace will be in the other two 2/3 of the time. When the 1/3 of the time it isn't there you would lose if you switched because taking one away will always leave a Jack. However the 2/3 of the time the Ace is there, taking one away (or revealing one) will always leave the Ace (you can't remove or reveal the Ace). So by switching on those 2/3 occasions you would win.
Seriously take three cards and do it time and time again and you will approach a situation where you will bag the Ace 2/3 of the time when you switch.
The easiest way to prove it is assume that there are 100 doors and after you choose one, 98 of the other 99 are taken away. What are the chances of you guessing correct in the first instance? 1 in 100. So the most sensible choice is to switch.
Wrong, switching would also giving you a door which originally had odds of 1/100 - much like the first door you chose.
OK I'll have a bet with you. You put £10 in and I'll put £500 in. All the money will be put in a single pot and put behind one out of a hundred doors. You pick one door at random, then 98 of the other doors will be opened and I get what's behind the 99th door. If it's truly 50/50 then you have far more to gain from betting than I do. I'll even offer you double or nothing if you lose the first time.
This is the sapiential version of "meet you at Makros to sort it out"
I should open an SE7 Door Riddle Casino, judging by the responses in this thread I'd be raking it in.
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.
I'd go for Brook with Riley touching herself whilst solving (out loud) complicated maths.
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.
The best way to prove it to the disbelievers that switching is better is to play the game 20 times with someone. It will soon become clear. If still unconvinced, play for money.
If it's behind A and you choose A, regardless of whether it isBor Copened, you would LOSE by switching If it's behind A and you choose A, regardless of whether it is B orC opened, 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 isBor Copened, you would LOSE by switching If it's behind B and you choose B, regardless of whether it is B orC opened, 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 isBor Copened, you would LOSE by switching If it's behind C and you choose C, regardless of whether it is B orC opened, 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 12 possible outcomes and 6 would be winners by switching and 6 would be losers.
The easiest way to prove it is assume that there are 100 doors and after you choose one, 98 of the other 99 are taken away. What are the chances of you guessing correct in the first instance? 1 in 100. So the most sensible choice is to switch.
Wrong, switching would also giving you a door which originally had odds of 1/100 - much like the first door you chose.
Why have you doubled up on the same options? (in bold). You are wrong but you'll get it eventually.
And on the 100 door example - you are wrong again. Yes you would choose a door that was originally 1/100 by switching - but it aint 1/100 now because some nice kind person has removed 98 losers!
Nope. False logic. There are three places you can choose and three places it can be. That's 9 combinations not 12. The odds are skewed because the door that is opened is not random. It's a known empty one. It did hurt brain when I first heard it 30 years ago and I was much sharper then.
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 B/Switch
Comments
You've got a 1/3 chance.
Then it becomes 50/50.
Whether the first one you picked was the winning one or the other door when you get the 50/50. You can't increase your chances of winning when it's 50/50.
You can increase your chances of selecting the winning door at some point but if your original choice was the million quid then you change it to the incorrect door, you don't keep the money right? So I can't see a logical reason for changing the door.
Maybe I'm just thick.
Stop it everyone. Now.
Someone put up a picture of a train.
The possible routes to victory are, in order of pick, reveal, then choice are...
A, B is revealed, switch to C
B, A is revealed, switch to C
C, A or B is revealed, stick with C
The routes to losing are:
A, B is revealed, stick with A
B, A is revealed, stick with B
C, A or B is revealed, switch to A or B
2 thirds of winning scenarios involve switching. 2 thirds of losing scenarios involve sticking.
I've used the Monty Hall Paradigm for about 20 years when training traders in probability - there are people who attended courses I ran all those years ago who still don't get it!!
If that doesn't work for you try 1 billion doors.
If that doesn't work go back to threading your wicker basket.
If it's behind A and you choose A,
regardless of whether it isBor Copened, you would LOSE by switchingIf it's behind A and you choose A,
regardless of whether it is B orC opened, you would LOSE by switchingIf 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 isBor Copened, you would LOSE by switchingIf it's behind B and you choose B,
regardless of whether it is B orC opened, you would LOSE by switchingIf 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 isBor Copened, you would LOSE by switchingIf it's behind C and you choose C,
regardless of whether it is B orC opened, you would LOSE by switchingIf 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 12 possible outcomes and 6 would be winners by switching and 6 would be losers.
Therefore: 50/50 odds Wrong, switching would ALSO give you a door which originally had odds of 1/100 - much like the very first door you chose.
Therefore: 50/50 odds
And on the 100 door example - you are wrong again. Yes you would choose a door that was originally 1/100 by switching - but it aint 1/100 now because some nice kind person has removed 98 losers!
Odds before doors removed: Both 1/100
Odds after doors are removed: Both 1/2. The bolded section doesn't change anything to the original scenario (which you repeated underneath). Why are we putting in different amounts? You used amounts of money put in to justify door probabilities - they're not related. The unbolded part is still 50/50. I'll take you up on that bet!!
I then take the other 51 cards, look at them, then remove 50 of the cards which are not the Ace of Spades. I have one facedown card left.
Which one are you going to think is the Ace of Spades? A card you have chosen randomly out of a deck of 52, or a card which has been filtered from a set of 51 cards which had a 51/52 chance of containing an Ace of Spades, or a 1/52 chance of not containing the Ace of Spades?
One last go at convincing you.
The purpose of the game is to choose an Ace.
Take three playing cards let's say two Jacks and an Ace. Lay those three cards at random face down and choose one. Now look at the other two and I'm sure you will agree that the Ace will be in the other two 2/3 of the time. When the 1/3 of the time it isn't there you would lose if you switched because taking one away will always leave a Jack. However the 2/3 of the time the Ace is there, taking one away (or revealing one) will always leave the Ace (you can't remove or reveal the Ace). So by switching on those 2/3 occasions you would win.
Seriously take three cards and do it time and time again and you will approach a situation where you will bag the Ace 2/3 of the time when you switch.
Now do you get it?
Shlong division.
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%
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 B/Switch