A fascinating mathematical problem
Posted by: Minky on 18 April 2004
I've just finished reading "The curious incident of the dog in the night-time" by Mark Haddon. The book has a brief passage on something called "The Monty Hall" problem. Marilyn vos Savant, the smartest person in the world by virtue of her Guinness record IQ, was asked the following question (my words) :
In a game show there are three doors. Behind two of them are goats. Behind the third is a car. You are a contestant trying to win the car. You choose a door and the game show host opens one of the two remaining doors to reveal a goat. Which door is most likely to have the car behind it (should you stick or switch) ?
If you have never encountered this problem before, have a think about it.
When Marilyn vos Savant gave her answer she was quite soundly chastised by some very heavy-hitting brainiacs, and this is at least as fascinating as the problem itself, because (as I have also found) this baby seems to have an almost supernatural ability to turn otherwise mild-mannered people into red-eyed monsters
In a game show there are three doors. Behind two of them are goats. Behind the third is a car. You are a contestant trying to win the car. You choose a door and the game show host opens one of the two remaining doors to reveal a goat. Which door is most likely to have the car behind it (should you stick or switch) ?
If you have never encountered this problem before, have a think about it.
When Marilyn vos Savant gave her answer she was quite soundly chastised by some very heavy-hitting brainiacs, and this is at least as fascinating as the problem itself, because (as I have also found) this baby seems to have an almost supernatural ability to turn otherwise mild-mannered people into red-eyed monsters
Posted on: 20 April 2004 by Haroon
quote:
Originally posted by Vuk's son:
Haroon,
Crucial to understand is that - the other door left underwent 999,998 experiments where each time it wasn't chosen to be open. Each success in such experiment of not being chosen to open increases its chances of being the one.
Hmm this is where I disagree with this. After 999,998 experiments you say one is more likely to be the prize than the other? why? Each time a door is opened the prob of which one is the prize is divided equally between the remaining doors including the one chosen, so when you get to two its 50-50!
start with a 1,000,000, so 1 in 1,000,000
we chose the first as our door, we open another door,
we now have 999,998 not open + our door + 1 open not the prize
Prob is 1 in 999,999 (999,998 + our door)
we open another door,
we now have 999,997 not open + our door + 2 open not the prize
Prob is 1 in 999,998 (999,997 doors left + our door)
we open another door,
we now have 999,996 not open + our door + 3 open not the prize
Prob is 1 in 999,997 (999,996 doors left + our door)
So on until...
we open another door,
we now have 1 not open + our door + 999,998 open not the prize
Prob is 1 in 2 (1 door left + our door)
quote:
On the other hand, the door you chose didn't go under any experiment, as the rules state the host opens one door but never the door you chose.
Omer.
Like shown above our door although not open has the same chances as the remaining door that was subjected to the test.
Posted on: 20 April 2004 by jayd
So, it seems nobody has any problem with my approach, which is simply to outline all possible outcomes and then add the results to see which strategy is best.
That was a lot easier than fractions.
That was a lot easier than fractions.
Posted on: 20 April 2004 by Markus S
quote:
Originally posted by Minky: Pick one of the three doors and stick with it. Forget about opening either of the other doors. Do this 1000 times. Approximately how many cars will you have in your driveway ?
NOW. Pick one of the three doors and stick with it. Open the first of the other doors to reveal a goat, then the second to reveal either a goat or a car. Do this 1000 times. How many cars will you have in your driveway ?
Once you understand that the chance of winning a car doesn't switch to 50:50 after the first door is opened you are home and hosed.
A combination of this post by Minky and an earlier post of Two-Sheds have convinced me. Switch it is. Doh!
Posted on: 20 April 2004 by long-time-dead
quote:
Originally posted by Minky:
............. the game show host opens one of the two remaining doors to reveal a goat.
The original problem had a fixed outcome to the host' door - a goat.
Given that fixed variable then the probability of any door having a car MUST be 1 in 2. Switching or sticking has therefore the same chance of winning.
Should the host have the option of opening a door in a truly random fashion and the possibility of spoiling the game by revealing a car, then the probability will vary accordingly and switching is the only option.
Read the original post again !!!!
Posted on: 20 April 2004 by matthewr
Honestly, if THIRTY MILLION ITERATIONS isn't enough to convince people I don't know what is.
"Given that fixed variable then the probability of any door having a car MUST be 1 in 2"
Except the host doesn't pick the goat door until you picked your first door -- meaning its probability of having a car (ie 0) is subtracted from the probabiltiy of the switch door (which therefore remains at 2/3).
"Read the original post again !!!!"
Quite.
Matthew
"Given that fixed variable then the probability of any door having a car MUST be 1 in 2"
Except the host doesn't pick the goat door until you picked your first door -- meaning its probability of having a car (ie 0) is subtracted from the probabiltiy of the switch door (which therefore remains at 2/3).
"Read the original post again !!!!"
Quite.
Matthew
Posted on: 20 April 2004 by matthewr
To the remaining descentors:
I will play an independently adjudicated version of this game with anyone.
We will play a minimum of 10,000 iterations where you stick with your first pick and I get the switch door.
If you win 4500 times or more you win £10,000.
Less than 4500 times and I win £10,000.
We can play for more or less if you wish.
If you really beleive its 50:50 you simply cannot lose.
Matthew
I will play an independently adjudicated version of this game with anyone.
We will play a minimum of 10,000 iterations where you stick with your first pick and I get the switch door.
If you win 4500 times or more you win £10,000.
Less than 4500 times and I win £10,000.
We can play for more or less if you wish.
If you really beleive its 50:50 you simply cannot lose.
Matthew
Posted on: 20 April 2004 by Haroon
quote:
Originally posted by Vuk's son:
Haroonquote:
Hmm this is where I disagree with this. After 999,998 experiments you say one is more likely to be the prize than the other? why? Each time a door is opened the prob of which one is the prize is divided equally between the remaining doors including the one chosen, so when you get to two its 50-50!
Please read this carefully.
Your sentence above is exactly the issue. The probability of the opened goat door is not divided between all the left doors, but only between the doors that participated in that event. The chosen door didn't participate, it couldn't be exposed as goat, and no pain no gain, so it also doesn't benefit from the divided probability
Omer.
What so the chosen door which didnt participate has zero chances of having the prize? I think you need to read again.
Posted on: 20 April 2004 by Haroon
quote:
Originally posted by Matthew Robinson:
"Given that fixed variable then the probability of any door having a car MUST be 1 in 2"
Except the host doesn't pick the goat door until you picked your first door -- meaning its probability of having a car (ie 0) is subtracted from the probabiltiy of the switch door (which therefore remains at 2/3).
Matthew
The host always picks a goat door! there two of them so no matter what you choose he will take out one.
Posted on: 20 April 2004 by Minky
Not sure about you lot but I'm going to plead insanity on this one. I'm off to find me a couple of goats.
Posted on: 20 April 2004 by Haroon
The switching methos relies on believing that on the first round you have a 1/3 chance of having the car and the other two choices represent 2/3.
Yeah?
So on the second round your current choice is likely to be 1/3 goat and now that one door has been revealed as goat the other one has 2/3 chances of being car.
I have shown that this logic is not correct, by the host revealing a goat you can no longer assume your first choice was 2/3 goat. New information has been added the scenario is changes, we need to change our assumptions, new bets on the table! Another way to look at it
We choose C this has 1/3 Car 2/3 chance of being goat
But both B and A have the same probs attached to them each
- car -goat
A 1/3 -2/3
B 1/3 -2/3
C 1/3 -2/3
Having choosen C you assume that it only has 1/3 chance of car and 2/3 goat, and A and B each have 1/3 put together 2/3, right? Then you must also acknowledge that A+C has 2/3 chance of car too and as do B+C. chances of any two pair have car are always 2/3 BEFORE elimination.
Let say we elim A
We chose C which had 1/3 chance
and A+B had 2/3 of car, but as explained above so to do A+C have 2/3 of car
Whether we stick or switch does not matter
A+B = A+C = 2/3;
A=0,
B=C
We have choice bwetween B and C that is equal, also given that also B+C = 2/3, then B=1/3=C
Just because we choose C it doesnt exclude itself from the probs with A, its only after we choose that the host decides to elim A and we become aware that it is a goat. There is a second goat and because A+B = A+C both options have the same chance of having the goat.
A+B = 2/3 chance of Car, what is A+B chance of being goat?
[This message was edited by Haroon on Wed 21 April 2004 at 1:03.]
Yeah?
So on the second round your current choice is likely to be 1/3 goat and now that one door has been revealed as goat the other one has 2/3 chances of being car.
I have shown that this logic is not correct, by the host revealing a goat you can no longer assume your first choice was 2/3 goat. New information has been added the scenario is changes, we need to change our assumptions, new bets on the table! Another way to look at it
We choose C this has 1/3 Car 2/3 chance of being goat
But both B and A have the same probs attached to them each
- car -goat
A 1/3 -2/3
B 1/3 -2/3
C 1/3 -2/3
Having choosen C you assume that it only has 1/3 chance of car and 2/3 goat, and A and B each have 1/3 put together 2/3, right? Then you must also acknowledge that A+C has 2/3 chance of car too and as do B+C. chances of any two pair have car are always 2/3 BEFORE elimination.
Let say we elim A
We chose C which had 1/3 chance
and A+B had 2/3 of car, but as explained above so to do A+C have 2/3 of car
Whether we stick or switch does not matter
A+B = A+C = 2/3;
A=0,
B=C
We have choice bwetween B and C that is equal, also given that also B+C = 2/3, then B=1/3=C
Just because we choose C it doesnt exclude itself from the probs with A, its only after we choose that the host decides to elim A and we become aware that it is a goat. There is a second goat and because A+B = A+C both options have the same chance of having the goat.
A+B = 2/3 chance of Car, what is A+B chance of being goat?
[This message was edited by Haroon on Wed 21 April 2004 at 1:03.]
Posted on: 20 April 2004 by Haroon
quote:
Originally posted by Minky:
Not sure about you lot but I'm going to plead insanity on this one. I'm off to find me a couple of goats.
Ive got nowt better to do
Posted on: 20 April 2004 by Two-Sheds
ok so I've just got back from tha pub watching the big game between the maple leafs and the senators and I'll have one final go at explaining this thing and convincing one more person before taking me goat to bed.
right then:
1=goat 2=goat 3=car
you pick 1. host opens 1. you lose.
you pick 1. host opens 2. you switch and win.
you pick 1. host opens 3. you lose.
you pick 2. host opens 1. you switch and win.
you pick 2. host opens 2. you lose.
you pick 2. host opens 3. you lose.
you pick 3. host opens 1. you switch and lose.
you pick 3. host opens 2. you switch and lose.
you pick 3. host opens 3. you win
again the ones in italics can never happen because the host will never open a door to reveal the car 1st or your door first.
To understand the problem we have to see it as a complete problem. So we will take it that we always switch and above are all the possible outcomes assuming we always switch.
The main point to 'getting' this is to merge the final two possibilities I've listed into one:
you pick 3. host opens 1. you switch and lose.
you pick 3. host opens 2. you switch and lose.
The only important aspect of this game is if you win or lose. As we can see from these last 2 cases no matter what we do if we pick 3 we lose, so the probability of losing if we pick 3 is 1 (or the chance of winning is 0). On the other hand, if we pick door 1 or 2:
you pick 1. host opens 2. you switch and win.
you pick 2. host opens 1. you switch and win.
we can see if we pick 1 or 2 then the chance of winning is 1 (chance of losing is 0).
So combining these if we can pick 1 or 2 we win, if we pick 3 we lose. At our first choice we don't know which door is which so we have a 2/3 chance of picking 1 or 2 which would give us a win and a 1/3 chance of picking 3 which would give us a 3.
If we were to include all the lines in italics above then indeed if he opened a door to reveal a goat by chance then we would have a 50-50 of winning the car, but the whole point is that there is no chance involved in revealing the first goat.
right then:
1=goat 2=goat 3=car
you pick 1. host opens 1. you lose.
you pick 1. host opens 2. you switch and win.
you pick 1. host opens 3. you lose.
you pick 2. host opens 1. you switch and win.
you pick 2. host opens 2. you lose.
you pick 2. host opens 3. you lose.
you pick 3. host opens 1. you switch and lose.
you pick 3. host opens 2. you switch and lose.
you pick 3. host opens 3. you win
again the ones in italics can never happen because the host will never open a door to reveal the car 1st or your door first.
To understand the problem we have to see it as a complete problem. So we will take it that we always switch and above are all the possible outcomes assuming we always switch.
The main point to 'getting' this is to merge the final two possibilities I've listed into one:
you pick 3. host opens 1. you switch and lose.
you pick 3. host opens 2. you switch and lose.
The only important aspect of this game is if you win or lose. As we can see from these last 2 cases no matter what we do if we pick 3 we lose, so the probability of losing if we pick 3 is 1 (or the chance of winning is 0). On the other hand, if we pick door 1 or 2:
you pick 1. host opens 2. you switch and win.
you pick 2. host opens 1. you switch and win.
we can see if we pick 1 or 2 then the chance of winning is 1 (chance of losing is 0).
So combining these if we can pick 1 or 2 we win, if we pick 3 we lose. At our first choice we don't know which door is which so we have a 2/3 chance of picking 1 or 2 which would give us a win and a 1/3 chance of picking 3 which would give us a 3.
If we were to include all the lines in italics above then indeed if he opened a door to reveal a goat by chance then we would have a 50-50 of winning the car, but the whole point is that there is no chance involved in revealing the first goat.
Posted on: 21 April 2004 by John Sheridan
quote:
Originally posted by Haroon:quote:
Originally posted by Minky:
Not sure about you lot but I'm going to plead insanity on this one. I'm off to find me a couple of goats.
Ive got nowt better to do
Haroon, rather than continually spouting the same flawed theories, why don't you just do the experiment yourself (as obviously the other 30M done by people in this thread aren't enough for you)?
Posted on: 21 April 2004 by matthewr
Haroon -- Just think what you could do with £10k! It's sitting there on the table just waiting for your bet.
Two Sheds -- I understood the problem before you started to explain it! Which brings me back neatly to my original point which is the correct answer to this is the intuitive answer and people seem to stuggle to understand or to explain it by making it way too complicated.
Matthew
Two Sheds -- I understood the problem before you started to explain it! Which brings me back neatly to my original point which is the correct answer to this is the intuitive answer and people seem to stuggle to understand or to explain it by making it way too complicated.
Matthew
Posted on: 21 April 2004 by Paul Ranson
If the probability of a win if you stick is 'x' then the probability of a win if you switch is 1-x.
So what's the probability of a win if you stick?
Clearly it is 1/3. It doesn't matter in what order the host opens the other doors, or whether you ever find out what was behind them. All that matters is your door and you will win 1/3 of the time. Therefore switching must win the other 2/3 of the time, there's no possibility of a half-win/half-lose state.
Does Haroon disagree?
Paul
So what's the probability of a win if you stick?
Clearly it is 1/3. It doesn't matter in what order the host opens the other doors, or whether you ever find out what was behind them. All that matters is your door and you will win 1/3 of the time. Therefore switching must win the other 2/3 of the time, there's no possibility of a half-win/half-lose state.
Does Haroon disagree?
Paul
Posted on: 21 April 2004 by count.d
As another perspective and to clear the mind of goats, if the probability of throwing a double-6 with two dice is 1/36, then the more times we throw the dice, the closer, in proportion, will be the number of double-6s thrown to of the total number of throws. This is what in everyday language is known as the law of averages. The overlooking of the vital words "in proportion" in the above definition leads to much misunderstanding among gamblers. The gambler's fallacy lies in the idea that in the long run, chances will even out. Thus if a coin has been spun 100 times, and has landed 60 times head uppermost and 40 times tails, many gamblers will state that tails are now due for a run to get even. The belief is that the law of averages really is a law which states that in the longest of long runs the totals of both heads and tails will eventually become equal.
In fact, the opposite is really the case. As the number of tosses gets larger, the probability is that the percentage of heads or tails thrown gets nearer to 50%, but that the difference between the actual number of heads or tails thrown and the number representing 50% gets larger.
So if you did another 100 spins and result in 56 heads and 44 tails. The corrective has set in, as the percentage of heads has now dropped from 60 per cent to 58 per cent. But there are now 32 more heads than tails, where there were only 20 before. The law of averages follower who backed tails is 12 more tosses to the bad.
The host will always eliminate the incorrect door which leaves it to heads or tails.
So basically the more you carry out the experiment, you would end up either with a large garage or a farm.
In fact, the opposite is really the case. As the number of tosses gets larger, the probability is that the percentage of heads or tails thrown gets nearer to 50%, but that the difference between the actual number of heads or tails thrown and the number representing 50% gets larger.
So if you did another 100 spins and result in 56 heads and 44 tails. The corrective has set in, as the percentage of heads has now dropped from 60 per cent to 58 per cent. But there are now 32 more heads than tails, where there were only 20 before. The law of averages follower who backed tails is 12 more tosses to the bad.
The host will always eliminate the incorrect door which leaves it to heads or tails.
So basically the more you carry out the experiment, you would end up either with a large garage or a farm.
Posted on: 21 April 2004 by Rasher
Good morning...It's Groundhog day..!! 
Posted on: 21 April 2004 by greeny
O I do like this!
For Haroon. A slightly different tack.
1 car 2 goats 3 doors
You choose door 1.
You agree that your chance of picking the Car is 1 in 3.
The host SAYS to you "Behind one of the doors you didn't pick is a goat"
You agree that this has not changed your changes of being right - still 1/3
The host says " I know which door that you didn't pick has the goat behind it"
Again because the host knows where the items are does not change your chances of being right. - still 1/3
Now the host knows that either door 2 or door 3 has a goat behind it. The host will only show you which door has the goat, so he will open door 2 or door 3 to reveal a goat. If the car was behind door 3 he shows you door 2 and via verca. Hence the door the host shows has Zero chance of having the car.
If the host had chosen randomly then we have a totally different scenario as the host would also have a 1/3 chance of picking the car, but this isn't so.
For Haroon. A slightly different tack.
1 car 2 goats 3 doors
You choose door 1.
You agree that your chance of picking the Car is 1 in 3.
The host SAYS to you "Behind one of the doors you didn't pick is a goat"
You agree that this has not changed your changes of being right - still 1/3
The host says " I know which door that you didn't pick has the goat behind it"
Again because the host knows where the items are does not change your chances of being right. - still 1/3
Now the host knows that either door 2 or door 3 has a goat behind it. The host will only show you which door has the goat, so he will open door 2 or door 3 to reveal a goat. If the car was behind door 3 he shows you door 2 and via verca. Hence the door the host shows has Zero chance of having the car.
If the host had chosen randomly then we have a totally different scenario as the host would also have a 1/3 chance of picking the car, but this isn't so.
Posted on: 21 April 2004 by matthewr
count.d,
"The gambler's fallacy lies in the idea that in the long run, chances will even out"
In the long run they will and the only question is how long is the long run. This is why Poker is a game of skill rather than gambling becuase eventually, if we play long enough, we will all get the same cards.
"So if you did another 100 spins and result in 56 heads and 44 tails. The corrective has set in, as the percentage of heads has now dropped from 60 per cent to 58 per cent. But there are now 32 more heads than tails, where there were only 20 before"
Hmm, I can sort of see what you mean but I am not sure that its terribly useful to know.
Gamblers (as opposed to idiots who gamble) generally think in terms of odds. If the odds are 4/1 and for a £1 bet I can win £5 then this is a bet I make as it has "positive expectation" (+EV). If would only win £3 then the bet would have negative expectation (-EV) and I shouldn't make it. Ultimately this is all that matters in the long run someone who consistently makes +EV bets will win and someone who doesn't will lose.
What you are talking about is, I think, variance which is usually expressed as standard deviation. I am a little unclear as to exactly how this is calculated although I know that my Poker playing has a SD of about 13 "Big Bets" (in my case $26US) a number which AFAICT expresses the usual +/- deviation about my average win rate I can expect.
[Interestingly Poker generally rewards aggressive play -- as long as it's correct -- which makes your average win rate rise but also increases your variance/SD. Hence the old adage about "To win a lot you have to be prepared to lose a lot"]
The other stat related to this is "Risk of Ruin" -- which expresses how bad a run of bad luck can be for a paricular game. In the case of Limit Texas Hold Em this is 300BB (I have no idea how this is worked out) and so one must have a bankroll of 300 times the big bet in order to avoid running out of money before the "law of averages" kicks in and you start winning (assuming correct play).
So, as I understand it, working out SD and RoR or your coin flip example would be the way to understand the issues you raise.
Matthew
"The gambler's fallacy lies in the idea that in the long run, chances will even out"
In the long run they will and the only question is how long is the long run. This is why Poker is a game of skill rather than gambling becuase eventually, if we play long enough, we will all get the same cards.
"So if you did another 100 spins and result in 56 heads and 44 tails. The corrective has set in, as the percentage of heads has now dropped from 60 per cent to 58 per cent. But there are now 32 more heads than tails, where there were only 20 before"
Hmm, I can sort of see what you mean but I am not sure that its terribly useful to know.
Gamblers (as opposed to idiots who gamble) generally think in terms of odds. If the odds are 4/1 and for a £1 bet I can win £5 then this is a bet I make as it has "positive expectation" (+EV). If would only win £3 then the bet would have negative expectation (-EV) and I shouldn't make it. Ultimately this is all that matters in the long run someone who consistently makes +EV bets will win and someone who doesn't will lose.
What you are talking about is, I think, variance which is usually expressed as standard deviation. I am a little unclear as to exactly how this is calculated although I know that my Poker playing has a SD of about 13 "Big Bets" (in my case $26US) a number which AFAICT expresses the usual +/- deviation about my average win rate I can expect.
[Interestingly Poker generally rewards aggressive play -- as long as it's correct -- which makes your average win rate rise but also increases your variance/SD. Hence the old adage about "To win a lot you have to be prepared to lose a lot"]
The other stat related to this is "Risk of Ruin" -- which expresses how bad a run of bad luck can be for a paricular game. In the case of Limit Texas Hold Em this is 300BB (I have no idea how this is worked out) and so one must have a bankroll of 300 times the big bet in order to avoid running out of money before the "law of averages" kicks in and you start winning (assuming correct play).
So, as I understand it, working out SD and RoR or your coin flip example would be the way to understand the issues you raise.
Matthew
Posted on: 21 April 2004 by Simon Perry
Oh my god I cannot believe we are still on the 2 goats problem! I'd seen that the replies to this thread were going up but I'd assumed that we were onto other mathematical problems.
Matthew, you have the patience of a saint.
Simon
PS - Haroon, please see sense!
Matthew, you have the patience of a saint.
Simon
PS - Haroon, please see sense!
Posted on: 21 April 2004 by Haroon
John and Simon
Sorry to be boring you, but if you are happy that switching is right why come back to read the thread?
If you want other problems to solve why not start a NEW topic hey? Or is that too much of a problem for you to solve?
Look here i'm arguing my position out of interest and debate not because i have something personal agaist those who thinking switching is correct. If you find it tiresome thats up to you, you go and do something else and leave us be on this thread.
Im gonna stop here now because carrying on is useless when you get people resonding like you both have. I've put my views forward no one has actually explained why its wrong, just keep tellin' me another way and that pointless. I have tried to put my self in the view of the switching theory and found for me that it doesnt seem to work - explanation given in posts above. Now if you think my theories are wrong then why not try to actually explain to me why and how MY view is wrong rather than just telling me it is flawed.
Like I say now end of. Matthew thanks for at least trying to explain your position. As far as im concerned we can just put this aside.
Sorry to be boring you, but if you are happy that switching is right why come back to read the thread?
If you want other problems to solve why not start a NEW topic hey? Or is that too much of a problem for you to solve?
Look here i'm arguing my position out of interest and debate not because i have something personal agaist those who thinking switching is correct. If you find it tiresome thats up to you, you go and do something else and leave us be on this thread.
Im gonna stop here now because carrying on is useless when you get people resonding like you both have. I've put my views forward no one has actually explained why its wrong, just keep tellin' me another way and that pointless. I have tried to put my self in the view of the switching theory and found for me that it doesnt seem to work - explanation given in posts above. Now if you think my theories are wrong then why not try to actually explain to me why and how MY view is wrong rather than just telling me it is flawed.
Like I say now end of. Matthew thanks for at least trying to explain your position. As far as im concerned we can just put this aside.
Posted on: 21 April 2004 by John Sheridan
quote:
Originally posted by Haroon:
John and Simon
Sorry to be boring you, but if you are happy that switching is right why come back to read the thread?
If you want other problems to solve why not start a NEW topic hey? Or is that too much of a problem for you to solve?
Look here i'm arguing my position out of interest and debate not because i have something personal agaist those who thinking switching is correct. If you find it tiresome thats up to you, you go and do something else and leave us be on this thread.
Im gonna stop here now because carrying on is useless when you get people resonding like you both have. I've put my views forward no one has actually explained why its wrong, just keep tellin' me another way and that pointless. I have tried to put my self in the view of the switching theory and found for me that it doesnt seem to work - explanation given in posts above. Now if you think my theories are wrong then why not try to actually explain to me why and how MY view is wrong rather than just telling me it is flawed.
Like I say now end of. Matthew thanks for at least trying to explain your position. As far as im concerned we can just put this aside.
Haroon,
the host's choices are cg, gg, gc. By removing a goat he's left with c, g, or c.
You, on the other hand, still have g, g or c - your original 1/3 choice.
Now if you 'switch' to the SAME door you originally picked it can't suddenly have just g or c behind it, it MUST still have the original options g, g or c.
If you switch to the hosts 2nd door then it has either c, g or c behind it. ie you have 2/3 chance of finding the car.
But like I said before, if this isn't working for you, get your significant other to play host and see for yourself. Better yet, take up Matthews offer of £10K as if we're wrong and you're right then you can't possibly lose.
Posted on: 21 April 2004 by matthewr
Haroon,
I find your arguments very difficult to follow so it is difficult to point out exactly where you are going wrong.
But, if i read you correctly, when you say...
"Let say we elim A
We chose C which had 1/3 chance
and A+B had 2/3 of car, but as explained above so to do A+C have 2/3 of car"
...you don't really have 2/3 of a chance as you are sticking with C whatever happens so the odds of A+C are irrelevant to your choice.
In contrast the switchers get to choose *both* A+B (2/3) and then decide which one they really want *after* the host has told them which of A or B has a goat.
Your argument requries that we transfer probability to your choice after you have made it -- which would mean that is you just got the host to reveal goats behind both doors this would transfer a 100% probability to your original choice was and you'd always get the car.
Matthew
I find your arguments very difficult to follow so it is difficult to point out exactly where you are going wrong.
But, if i read you correctly, when you say...
"Let say we elim A
We chose C which had 1/3 chance
and A+B had 2/3 of car, but as explained above so to do A+C have 2/3 of car"
...you don't really have 2/3 of a chance as you are sticking with C whatever happens so the odds of A+C are irrelevant to your choice.
In contrast the switchers get to choose *both* A+B (2/3) and then decide which one they really want *after* the host has told them which of A or B has a goat.
Your argument requries that we transfer probability to your choice after you have made it -- which would mean that is you just got the host to reveal goats behind both doors this would transfer a 100% probability to your original choice was and you'd always get the car.
Matthew
Posted on: 21 April 2004 by John Sheridan
ok, looking at this:
A+B = A+C = 2/3;
A=0,
B=C
for starters you missed out B+C which also = 2/3
A+B = A+C = B+C = 2/3
if a=0 then continuing with this logic
B = C = B+C = 2/3
so now you've got an obvious problem.
What actually happens is that you pick C = 1/3 and the host then picks A+B = 2/3.
He then eliminates (A OR B). So you still have C=1/3 and he STILL has A OR B = 2/3. What you don't have is A+C or B+C because you've only picked 1 door. You get A+C or B+C by switching doors. If you switch to the SAME door then you would have C+C but we know that's not a valid option.
A+B = A+C = 2/3;
A=0,
B=C
for starters you missed out B+C which also = 2/3
A+B = A+C = B+C = 2/3
if a=0 then continuing with this logic
B = C = B+C = 2/3
so now you've got an obvious problem.
What actually happens is that you pick C = 1/3 and the host then picks A+B = 2/3.
He then eliminates (A OR B). So you still have C=1/3 and he STILL has A OR B = 2/3. What you don't have is A+C or B+C because you've only picked 1 door. You get A+C or B+C by switching doors. If you switch to the SAME door then you would have C+C but we know that's not a valid option.
Posted on: 21 April 2004 by kevinrt
Difficult one to fathom. I've been swayed by both sides, but have decided the sensible thing to do is switch.
Either your chances go up or stay the same, they don't go down.
Or is there a 3rd arm to this discussion?
Either your chances go up or stay the same, they don't go down.
Or is there a 3rd arm to this discussion?