# Logic

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#41

(Original post by

sum1 please explain this? surely if m discounts one door, there are two doors left...one with and one without...is it that m says 'well done' if u r right?

**lgs98jonee**)sum1 please explain this? surely if m discounts one door, there are two doors left...one with and one without...is it that m says 'well done' if u r right?

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#42

It is the fact that Monty opens one of the remaining two doors that gives you infromation about the other - this is what increases it's prize-probability.

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#43

(Original post by

It is the fact that Monty opens one of the remaining two doors that gives you infromation about the other - this is what increases it's prize-probability.

**wanderer**)It is the fact that Monty opens one of the remaining two doors that gives you infromation about the other - this is what increases it's prize-probability.

MB

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#44

(Original post by

can someone answer the question please ask the question asked earlier. If monty opens both the other doors does that make the total probability for all three doors equal to 1/3?

MB

**musicboy**)can someone answer the question please ask the question asked earlier. If monty opens both the other doors does that make the total probability for all three doors equal to 1/3?

MB

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#45

Dead thread but thought I ought to eat humble pie (and ask a question too).

I've read the replies and accept you ought to switch.

It's been better explained by others but:

My mistake was thinking that you're only ever choosing between 1 winner and 1 loser. Of course you're not. At the start, you're choosing blindly and therefore must have a 1 in 3 chance of picking correctly. Hence, there must be a two in three chance it's in one of the other doors. Monty eliminating one of those allows you to effectively choose both of those, i.e. choose the 'other doors', i.e. choose a 2 in 3 chance.

My only question is as regards the analogy with choosing bars that I gave in my earlier post (page 1 of the thread). Is my common sense wrong and am I just being pig-headed in sticking to my guns? (Forgive crap metaphor overdose.) Or does it make a difference that my friend did not know which of the three was winner whereas Monty knew at the outset?

Summary of bars analogy:

3 bars: A,B,C

1 winner, 2 losers; you don't know which; your friend doesn't know which

You're on your way to A

Your friend calls and says it's not C

Do you continue on your way to A or about face and head for B?

My gut says stick to guns as A as likely as B. Am I wrong?

I've read the replies and accept you ought to switch.

It's been better explained by others but:

My mistake was thinking that you're only ever choosing between 1 winner and 1 loser. Of course you're not. At the start, you're choosing blindly and therefore must have a 1 in 3 chance of picking correctly. Hence, there must be a two in three chance it's in one of the other doors. Monty eliminating one of those allows you to effectively choose both of those, i.e. choose the 'other doors', i.e. choose a 2 in 3 chance.

My only question is as regards the analogy with choosing bars that I gave in my earlier post (page 1 of the thread). Is my common sense wrong and am I just being pig-headed in sticking to my guns? (Forgive crap metaphor overdose.) Or does it make a difference that my friend did not know which of the three was winner whereas Monty knew at the outset?

Summary of bars analogy:

3 bars: A,B,C

1 winner, 2 losers; you don't know which; your friend doesn't know which

You're on your way to A

Your friend calls and says it's not C

Do you continue on your way to A or about face and head for B?

My gut says stick to guns as A as likely as B. Am I wrong?

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#46

(Original post by

Dead thread but thought I ought to eat humble pie (and ask a question too).

I've read the replies and accept you ought to switch.

It's been better explained by others but:

My mistake was thinking that you're only ever choosing between 1 winner and 1 loser. Of course you're not. At the start, you're choosing blindly and therefore must have a 1 in 3 chance of picking correctly. Hence, there must be a two in three chance it's in one of the other doors. Monty eliminating one of those allows you to effectively choose both of those, i.e. choose the 'other doors', i.e. choose a 2 in 3 chance.

My only question is as regards the analogy with choosing bars that I gave in my earlier post (page 1 of the thread). Is my common sense wrong and am I just being pig-headed in sticking to my guns? (Forgive crap metaphor overdose.) Or does it make a difference that my friend did not know which of the three was winner whereas Monty knew at the outset?

Summary of bars analogy:

3 bars: A,B,C

1 winner, 2 losers; you don't know which; your friend doesn't know which

You're on your way to A

Your friend calls and says it's not C

Do you continue on your way to A or about face and head for B?

My gut says stick to guns as A as likely as B. Am I wrong?

**houdini**)Dead thread but thought I ought to eat humble pie (and ask a question too).

I've read the replies and accept you ought to switch.

It's been better explained by others but:

My mistake was thinking that you're only ever choosing between 1 winner and 1 loser. Of course you're not. At the start, you're choosing blindly and therefore must have a 1 in 3 chance of picking correctly. Hence, there must be a two in three chance it's in one of the other doors. Monty eliminating one of those allows you to effectively choose both of those, i.e. choose the 'other doors', i.e. choose a 2 in 3 chance.

My only question is as regards the analogy with choosing bars that I gave in my earlier post (page 1 of the thread). Is my common sense wrong and am I just being pig-headed in sticking to my guns? (Forgive crap metaphor overdose.) Or does it make a difference that my friend did not know which of the three was winner whereas Monty knew at the outset?

Summary of bars analogy:

3 bars: A,B,C

1 winner, 2 losers; you don't know which; your friend doesn't know which

You're on your way to A

Your friend calls and says it's not C

Do you continue on your way to A or about face and head for B?

My gut says stick to guns as A as likely as B. Am I wrong?

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#47

Just because a 'proof' exists on the internet doesn't make the apparent solution any more valid.

Consider - the proof hinges on the assumption that when one door is opened containing the dud that means that its probability of containing the prize will magically switch to another door; Changing from 1/3=1/3=1/3 to 2/3=1/3

That this is nonsense can be clearly understood when you consider that, having first chosen which door to pick at random you are now left which door to assign the 2/3 odds to also at your own whim. It sin't any more likely that you have picked the right door just because he has shown you a goat because it is unlikely that if you had picked a goat he would have shown you the goat which you picked.

Furthermore the only odds that have change in the problem are the odds of you getting a goat - they have actually DROPPED from 2/3 to 1/2 as the ratio of goats to potential goat baring doors has dropped from 2:3 to 1:3

In any probability problem like this surely the best way to consider it is to ask at each stage 'what if this was a fresh problem?' If the final round was a fresh problem and someone said to you; you must pick between two doors one of which holds a goat, the other of which holds the prize and they they asked you to assign odds you would naturally assign odds of 50-50. Because Monty Hall's act of opening a goat door doesn't actually make it any more likely that the door you picked was correct ( or vice versa ) only increases your odds of a prize door when compared to the previous round then it would be madness to claim that one door has 2/3 odds while the other 1/3.

After all, if the decision to assign 2/3 odds to one of the doors is left to you then what are the chances that you assign it to the correct door? 50-50.

Consider - the proof hinges on the assumption that when one door is opened containing the dud that means that its probability of containing the prize will magically switch to another door; Changing from 1/3=1/3=1/3 to 2/3=1/3

That this is nonsense can be clearly understood when you consider that, having first chosen which door to pick at random you are now left which door to assign the 2/3 odds to also at your own whim. It sin't any more likely that you have picked the right door just because he has shown you a goat because it is unlikely that if you had picked a goat he would have shown you the goat which you picked.

Furthermore the only odds that have change in the problem are the odds of you getting a goat - they have actually DROPPED from 2/3 to 1/2 as the ratio of goats to potential goat baring doors has dropped from 2:3 to 1:3

In any probability problem like this surely the best way to consider it is to ask at each stage 'what if this was a fresh problem?' If the final round was a fresh problem and someone said to you; you must pick between two doors one of which holds a goat, the other of which holds the prize and they they asked you to assign odds you would naturally assign odds of 50-50. Because Monty Hall's act of opening a goat door doesn't actually make it any more likely that the door you picked was correct ( or vice versa ) only increases your odds of a prize door when compared to the previous round then it would be madness to claim that one door has 2/3 odds while the other 1/3.

After all, if the decision to assign 2/3 odds to one of the doors is left to you then what are the chances that you assign it to the correct door? 50-50.

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#48

(Original post by

Just because a 'proof' exists on the internet doesn't make the apparent solution any more valid.

Consider - the proof hinges on the assumption that when one door is opened containing the dud that means that its probability of containing the prize will magically switch to another door; Changing from 1/3=1/3=1/3 to 2/3=1/3

That this is nonsense can be clearly understood when you consider that, having first chosen which door to pick at random you are now left which door to assign the 2/3 odds to also at your own whim. It sin't any more likely that you have picked the right door just because he has shown you a goat because it is unlikely that if you had picked a goat he would have shown you the goat which you picked.

Furthermore the only odds that have change in the problem are the odds of you getting a goat - they have actually DROPPED from 2/3 to 1/2 as the ratio of goats to potential goat baring doors has dropped from 2:3 to 1:3

In any probability problem like this surely the best way to consider it is to ask at each stage 'what if this was a fresh problem?' If the final round was a fresh problem and someone said to you; you must pick between two doors one of which holds a goat, the other of which holds the prize and they they asked you to assign odds you would naturally assign odds of 50-50. Because Monty Hall's act of opening a goat door doesn't actually make it any more likely that the door you picked was correct ( or vice versa ) only increases your odds of a prize door when compared to the previous round then it would be madness to claim that one door has 2/3 odds while the other 1/3.

After all, if the decision to assign 2/3 odds to one of the doors is left to you then what are the chances that you assign it to the correct door? 50-50.

**Faboba**)Just because a 'proof' exists on the internet doesn't make the apparent solution any more valid.

Consider - the proof hinges on the assumption that when one door is opened containing the dud that means that its probability of containing the prize will magically switch to another door; Changing from 1/3=1/3=1/3 to 2/3=1/3

That this is nonsense can be clearly understood when you consider that, having first chosen which door to pick at random you are now left which door to assign the 2/3 odds to also at your own whim. It sin't any more likely that you have picked the right door just because he has shown you a goat because it is unlikely that if you had picked a goat he would have shown you the goat which you picked.

Furthermore the only odds that have change in the problem are the odds of you getting a goat - they have actually DROPPED from 2/3 to 1/2 as the ratio of goats to potential goat baring doors has dropped from 2:3 to 1:3

In any probability problem like this surely the best way to consider it is to ask at each stage 'what if this was a fresh problem?' If the final round was a fresh problem and someone said to you; you must pick between two doors one of which holds a goat, the other of which holds the prize and they they asked you to assign odds you would naturally assign odds of 50-50. Because Monty Hall's act of opening a goat door doesn't actually make it any more likely that the door you picked was correct ( or vice versa ) only increases your odds of a prize door when compared to the previous round then it would be madness to claim that one door has 2/3 odds while the other 1/3.

After all, if the decision to assign 2/3 odds to one of the doors is left to you then what are the chances that you assign it to the correct door? 50-50.

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#49

right. this is the simplest possible way i can explain that the correct answer is that switching increases your chances. if you cannot understand this, try a simulation (as i have repeatedly asked of the skeptical) to convince yourself.

Suppose the money is behind door C, and there are goats behind A and B.

You can pick door A, B or C.

Suppose you pick A. Monty will open the other door with a goat behind it (door B). If you switch, you will pick the winning door (door C).

Suppose you pick B. Monty will open A. Switching wins (again)

Suppose you pick C. Monty will open A or B. Switching loses.

I count 2 out of 3 in favour of switching, and 1 against.

The above example essentially makes the point that you will only lose by switching if you pick the right door at the outset. The probability of picking the right door at the outset is 1/3. Therefore, the probability that you will lose by switching is 1/3.

The same reasoning holds when the money is behind door A or B.

Suppose the money is behind door C, and there are goats behind A and B.

You can pick door A, B or C.

Suppose you pick A. Monty will open the other door with a goat behind it (door B). If you switch, you will pick the winning door (door C).

Suppose you pick B. Monty will open A. Switching wins (again)

Suppose you pick C. Monty will open A or B. Switching loses.

I count 2 out of 3 in favour of switching, and 1 against.

The above example essentially makes the point that you will only lose by switching if you pick the right door at the outset. The probability of picking the right door at the outset is 1/3. Therefore, the probability that you will lose by switching is 1/3.

The same reasoning holds when the money is behind door A or B.

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#50

(Original post by

right. this is the simplest possible way i can explain that the correct answer is that switching increases your chances. if you cannot understand this, try a simulation (as i have repeatedly asked of the skeptical) to convince yourself.

Suppose the money is behind door C, and there are goats behind A and B.

You can pick door A, B or C.

Suppose you pick A. Monty will open the other door with a goat behind it (door B). If you switch, you will pick the winning door (door C).

Suppose you pick B. Monty will open A. Switching wins (again)

Suppose you pick C. Monty will open A or B. Switching loses.

I count 2 out of 3 in favour of switching, and 1 against.

The above example essentially makes the point that you will only lose by switching if you pick the right door at the outset. The probability of picking the right door at the outset is 1/3. Therefore, the probability that you will lose by switching is 1/3.

The same reasoning holds when the money is behind door A or B.

**irishtrep**)right. this is the simplest possible way i can explain that the correct answer is that switching increases your chances. if you cannot understand this, try a simulation (as i have repeatedly asked of the skeptical) to convince yourself.

Suppose the money is behind door C, and there are goats behind A and B.

You can pick door A, B or C.

Suppose you pick A. Monty will open the other door with a goat behind it (door B). If you switch, you will pick the winning door (door C).

Suppose you pick B. Monty will open A. Switching wins (again)

Suppose you pick C. Monty will open A or B. Switching loses.

I count 2 out of 3 in favour of switching, and 1 against.

The above example essentially makes the point that you will only lose by switching if you pick the right door at the outset. The probability of picking the right door at the outset is 1/3. Therefore, the probability that you will lose by switching is 1/3.

The same reasoning holds when the money is behind door A or B.

1: There are three doors.

2: You are to pick one of the three doors.

3: There are two goats and one pile of money. These are resting behind three seperate doors.

4: After you pick one of the three doors Monty will open a goat door.

5: By opening a goat door Monty will always reduce the number of doors to two and the number of goats to one.

6: This means that every time you play the game the final odds that you have made the right choice will be one in two as there is a 50-50 chance that you have picked a goat door.

.

.

.

.

.

The flaw behind arguing about group probabilities for a set of two doors can be explained as follows.

1: It is true that when you start there are three doors and you have one choice. Therefore you will always have a one in three chance of picking the correct door the first time.

2: Regardless of your choice however Monty will open a goat door.

3: While you can say initially that the chance the prize is behind one of the doors you didn't pick is 2/3 this does not mean that when one of them is opened as being a goat door that this magically jumps the chance of the remaining door out of the pair as being the prize door to 2/3.

4: The doors you did not pick can only be of two combinations, these being GOAT-GOAT and GOAT-PRIZE. In any case there will always be a goat door and there will always be a goat there for Monty to reveal.

5: The chance of the prize being behind one of the two unpicked doors when you started was 2/3. The chance of it being behind each unpicked door individually however was always 1/3.

6: You are insane if you think that the chance of the remaining door from the pair holding the prize is 2/3 because the other door held the goat. This is because there was no garuntee that you were going to pick the other door from the three. Therefore two group these two doors together and consider the probabilities of them combined serves no purpose as they are not and never were a singular logical entity.

7: The chance of the prize being behind either the door you picked or any of the other two doors was ALSO 2/3. Why should the remaining unpicked door jump to 2/3 when GOAT DOOR ONE is opened and not the picked door which could just as easily be grouped with the opened door.

8: This returns me to what I said above. The odds of you picking the correct door in the long run are never going to be 1/3 or anything in 3 because regardless of your initial choice there will always be one door and one goat neutralized which reduces ratio of prize to goat to door from 13 to 12 and makes the odds 50-50.

9: The fact that this happens regardless of your choice is what makes this proof valid. No ammount of testing can disprove it. Consider:

The three doors shall be called P,G1 and G2 ( Prize, Goat1 and Goat2 ) Door G1 and G2 are interchangable and are only named differently to avoid confusion.

In any game there are only two ways to make the first move. Either the player chooses P and does not choose either G1 or G2 or he chooses (G1,G2) and does not choose P or (G1,G2)

Game 1 : Chosen (P), Unchosen (G1,G2) WIN

Game 2 : Chosen (G1), Unchosen (P,G2) LOSE

Game 3 : Chosen (G2), Unchosen (P,G1) LOSE

As there is one win from three possible games this makes the odds of picking the correct door in the first round 1 in 3.

In the next round of play one door and one goat is removed. If you have picked a goat door then it will be the goat door you did not pick. If you have not picked a goat door then it will be a door chosen at random from G1 or G2.

This has reduced the possible games to two however. ( G shall stand for the goat door left.

Game 1 : Chosen (P), Unchosen (G) WIN

Game 2 : Chosen (G), Unchosen (P) LOSE

The number of games do not remain at three because there will always be a goat door which you have not picked removed and you are then allowed to chance your mind. This means there are four possible outcomes to the game:

Outcome 1: Picked P, Stuck with P. WIN

Outcome 2: Picked P, Changed to G LOSE

Outcome 3: Picked G, Changed to P WIN

Outcome 4: Picked G, Stuck with G LOSE

As there are 2 wins out of four we can say that the odds are 1 in 2 or 50-50.

I can see no room for equivocation. Unless someone can point out to me a mistake in anything that I have detailed above then it must stand accepted as being a logical application of game theory to a very simple problem. No ammount of links to other websites which make obvious logical flaws or cries of 'Try it on paper you'll see you must stick/ you must change' will change that.

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#51

(Original post by

This has reduced the possible games to two however. ( G shall stand for the goat door left.

**Faboba**)This has reduced the possible games to two however. ( G shall stand for the goat door left.

You seem quite convinced of your opinion, however, if you do know some elementary probability (Bayes theorem) you can understand the proof which is presented here: http://members.shaw.ca/ron.blond/TLE...AIN/index.html

To me, this is considerably stronger than any wordy answers or pseudo-maths.

Alaric.

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#52

It's funny how people can be so sure that all mathematicians have got the probabilities wrong! I tried explaining it today to my dad why it was 2/3, and it took a lot of convincing before he actually accepted it.

I think Fahoma's "Take each possible scenario" explanation is easiest to explain to a non-mathematician (like my dad).

I even tried experimenting with it using 3 cups and a nut. Indeed, after trying 60 times and getting my dad to change his mind each time, he won 41 times, what should be expected for a 2/3 probability.

I think Fahoma's "Take each possible scenario" explanation is easiest to explain to a non-mathematician (like my dad).

I even tried experimenting with it using 3 cups and a nut. Indeed, after trying 60 times and getting my dad to change his mind each time, he won 41 times, what should be expected for a 2/3 probability.

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#53

(Original post by

Your argument breaks there. You need to examine all the probabilities, you can't just say "oh a door is revealed and the other goat door becomes G".

You seem quite convinced of your opinion, however, if you do know some elementary probability (Bayes theorem) you can understand the proof which is presented here: http://members.shaw.ca/ron.blond/TLE...AIN/index.html

To me, this is considerably stronger than any wordy answers or pseudo-maths.

Alaric.

**Alaric**)Your argument breaks there. You need to examine all the probabilities, you can't just say "oh a door is revealed and the other goat door becomes G".

You seem quite convinced of your opinion, however, if you do know some elementary probability (Bayes theorem) you can understand the proof which is presented here: http://members.shaw.ca/ron.blond/TLE...AIN/index.html

To me, this is considerably stronger than any wordy answers or pseudo-maths.

Alaric.

Game 1; Chosen (P), Unchosen (G1) - Stick. Win.

Game 2; Chosen (P), Unchosen (G1) - Change. Lose.

Game 3; Chosen (P), Unchosen (G2) - Stick. Win.

Game 4; Chosen (P), Unchosen (G2) - Change. Lose.

Game 5; Chosen (G1), Unchosen (P) - Stick. Lose.

Game 6; Chosen (G1), Unchosen (P) - Change. Win.

Game 7; Chosen (G2), Unchosen (P) - Stick. Lose.

Game 8; Chosen (G2), Unchosen (P) - Change. Win.

All possible eight games with both goats included. This leaves odds of 4:8 or 1:2 or 50-50 for whether sticking or changing results in sucess. As I said, whether the second goat is counted as the first or not makes little difference.

I'm sorry but we'll have to agree to disagree. To my mind any argument like; 'the two doors you didn't pick had odds of 1/3 each for a collective chance of holding the prize of 2/3 and so when one of these doors is opened then that means the chance the remaining door holding the prize is 2/3' flies in the face of logic simply because it could easily be applied to the door you DID pick and one of the others to 'proove' that the chances of the prize being behind the door you DID pick are 2/3.

Applets are written by programmers. We'll continue this discussion as soon as I find a pair of goats.

{Edit2} As far as the link goes, the line;

"the probability that Monty opens door B if the prize was behind door C is p(B/C). P(B/C)=1"

seems to be in error. If nothing else because we've already established that P(B/A)=1/2. How can he be certain to do X if there is a 50% chance he might do Y. If I'm wrong about this do enlighten me but as the figure of 1 for P(B/C) is vital for prooving or disprooving my interpretation of the problem then I fear it's reasonably important.

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#54

Faboba: If you had read and understood my previous post, you will notice that I didn't use the argument that: "The prob. it is behind 2 doors is 2/3, and so if we know it's not in one of them, then the prob. it's in the other is 2/3." this is wrong, as anyone familiar with conditional probabilities will know. You are knocking down a straw man. This argument is sometimes marshalled in defence of the correct answer, although it is clearly fallacious.

I'm afraid we cannot agree to disagree, because this is a matter of fact, and not of opinion.

If you really think that the monty hall simulations on the web have been skewed by programmers trying to prove a point, then use a random number generator not written explicitly for this problem. better still, write one yourself. or throw a die. i recommend you take a sizeable sample, at least 100. if you really think you are right, you should have nothing to fear from testing your "correct" answer empirically.

the simplest possible explanation of the answer:

you will lose by switching if and only if you have picked the right door at the outset.

there is a 1/3 chance of picking the right door at the outset.

therefore, there is a 1/3 chance you will lose by switching.

(and obviously, if one doesn't lose, one wins, so there is a 2/3 chance of winning by switching)

please point out the flaw in the above argument if you wish to maintain your position. if you cannot, you must admit your error.

I'm afraid we cannot agree to disagree, because this is a matter of fact, and not of opinion.

If you really think that the monty hall simulations on the web have been skewed by programmers trying to prove a point, then use a random number generator not written explicitly for this problem. better still, write one yourself. or throw a die. i recommend you take a sizeable sample, at least 100. if you really think you are right, you should have nothing to fear from testing your "correct" answer empirically.

the simplest possible explanation of the answer:

you will lose by switching if and only if you have picked the right door at the outset.

there is a 1/3 chance of picking the right door at the outset.

therefore, there is a 1/3 chance you will lose by switching.

(and obviously, if one doesn't lose, one wins, so there is a 2/3 chance of winning by switching)

please point out the flaw in the above argument if you wish to maintain your position. if you cannot, you must admit your error.

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#55

(Original post by

4: The doors you did not pick can only be of two combinations, these being GOAT-GOAT and GOAT-PRIZE. In any case there will always be a goat door and there will always be a goat there for Monty to reveal.

**Faboba**)4: The doors you did not pick can only be of two combinations, these being GOAT-GOAT and GOAT-PRIZE. In any case there will always be a goat door and there will always be a goat there for Monty to reveal.

MGG

GMG

GGM

so assuming you always choose the first door you get.

GG

MG

GM

this means that two thirds of the time there is mone and goat behind the two remaining doors so if the goat one is open you are left with

G

M

M

meaning you have a 2/3 chance of getting money if you stop. I'm so sorry for disagreeing with all the clever people who have been saying this for weeks. I am a complete fool for not realising this earlier. I'm sure my explanation won't help. It is something you just need to think about yourself and it wwill click into place .

Musicboy

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#56

(Original post by

ok after reading this I have seen my error and now believe that you should switch from your original choice. Look at the two options. The two goats and the money can be arranged in three ways:

MGG

GMG

GGM

so assuming you always choose the first door you get.

GG

MG

GM

this means that two thirds of the time there is mone and goat behind the two remaining doors so if the goat one is open you are left with

G

M

M

meaning you have a 2/3 chance of getting money if you stop. I'm so sorry for disagreeing with all the clever people who have been saying this for weeks. I am a complete fool for not realising this earlier. I'm sure my explanation won't help. It is something you just need to think about yourself and it wwill click into place .

Musicboy

**musicboy**)ok after reading this I have seen my error and now believe that you should switch from your original choice. Look at the two options. The two goats and the money can be arranged in three ways:

MGG

GMG

GGM

so assuming you always choose the first door you get.

GG

MG

GM

this means that two thirds of the time there is mone and goat behind the two remaining doors so if the goat one is open you are left with

G

M

M

meaning you have a 2/3 chance of getting money if you stop. I'm so sorry for disagreeing with all the clever people who have been saying this for weeks. I am a complete fool for not realising this earlier. I'm sure my explanation won't help. It is something you just need to think about yourself and it wwill click into place .

Musicboy

Alaric, Irishtrep, you have my apologies.

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#57

(Original post by

You've broken my heart. You've provided an intellgible explaination for WHY I should accept two to three odds in favour of switching. I hate you and love you. If it's worth anything yours was a far better proof than http://members.shaw.ca/ron.blond/TLE...AIN/index.html

Alaric, Irishtrep, you have my apologies.

**Faboba**)You've broken my heart. You've provided an intellgible explaination for WHY I should accept two to three odds in favour of switching. I hate you and love you. If it's worth anything yours was a far better proof than http://members.shaw.ca/ron.blond/TLE...AIN/index.html

Alaric, Irishtrep, you have my apologies.

MB

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#58

At least I won't need to buy a pair of goats now.

The TRUEL:

Three men decide to settle their differences with a duel. Unfortunately as taking turns in two on two fighting will leave honour unsatisfied they decide to meet each other on a field of combat and face of simultaneously ( assume a date of around 1820 ).

Each man shall take twenty paces from a centre spot away from his opponents and take turns in aiming and firing at whomsoever he may wish.

Monseiur X has juddery hands and a wife of questionable faithfulness. He only hits his target one time out of three.

Monsieur Y was a fantastic shot but lost his middle finger when the Grande Armee retreated from Russia. As a result he hits his target only two times out of three.

Monsieur Z is a cad and a braggard. Unfortunately a combination of steady hand and eagle eyes means he will hit his target every time without fail.

Play passes from X to Y to Z to X to Y and so on for as long as more than one combatant remains alive. Play will move on when a combatant discharges his rifle.

As a personal friend of Monsieur X how would you suggest he goes about fighting the Truel so as to give himself the greatest chance of survival?

The TRUEL:

Three men decide to settle their differences with a duel. Unfortunately as taking turns in two on two fighting will leave honour unsatisfied they decide to meet each other on a field of combat and face of simultaneously ( assume a date of around 1820 ).

Each man shall take twenty paces from a centre spot away from his opponents and take turns in aiming and firing at whomsoever he may wish.

Monseiur X has juddery hands and a wife of questionable faithfulness. He only hits his target one time out of three.

Monsieur Y was a fantastic shot but lost his middle finger when the Grande Armee retreated from Russia. As a result he hits his target only two times out of three.

Monsieur Z is a cad and a braggard. Unfortunately a combination of steady hand and eagle eyes means he will hit his target every time without fail.

Play passes from X to Y to Z to X to Y and so on for as long as more than one combatant remains alive. Play will move on when a combatant discharges his rifle.

As a personal friend of Monsieur X how would you suggest he goes about fighting the Truel so as to give himself the greatest chance of survival?

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#59

(Original post by

Alaric, Irishtrep, you have my apologies.

**Faboba**)Alaric, Irishtrep, you have my apologies.

For the sake of completeness:

(Original post by

{Edit2} As far as the link goes, the line;

"the probability that Monty opens door B if the prize was behind door C is p(B/C). P(B/C)=1"

seems to be in error. If nothing else because we've already established that P(B/A)=1/2. How can he be certain to do X if there is a 50% chance he might do Y. If I'm wrong about this do enlighten me but as the figure of 1 for P(B/C) is vital for prooving or disprooving my interpretation of the problem then I fear it's reasonably important.

**Faboba**){Edit2} As far as the link goes, the line;

"the probability that Monty opens door B if the prize was behind door C is p(B/C). P(B/C)=1"

seems to be in error. If nothing else because we've already established that P(B/A)=1/2. How can he be certain to do X if there is a 50% chance he might do Y. If I'm wrong about this do enlighten me but as the figure of 1 for P(B/C) is vital for prooving or disprooving my interpretation of the problem then I fear it's reasonably important.

P(B|C) should be read as the "probability of B given C"

In this situation the door that has been picked is always A; so the probability that Monty opens B given we know the prize is behind C is evidently going to be 1. As the only other options are to open either the chosen door or the one containing the prize, neither of which Monty can do.

To be honest I think that most people get confused because they're too used to independent probabilities. Earlier on in the thread you stated that the way you should approach problems is to regard it as a fresh problem after each stage, sadly that only works with independent probabilities.

I don't think the million doors example helps people who are stuck on that either.

Alaric.

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#60

(Original post by

No problem at all, it was just a question of someone finding a way to explain it in a suitable way, I'm more of the favour of maths as the equations are more absolute and formal than English.

For the sake of completeness:

I don't want to sound patronising as I don't know how much probability you've done, but I better explain the syntax just in case.

P(B|C) should be read as the "probability of B given C"

In this situation the door that has been picked is always A; so the probability that Monty opens B given we know the prize is behind C is evidently going to be 1. As the only other options are to open either the chosen door or the one containing the prize, neither of which Monty can do.

To be honest I think that most people get confused because they're too used to independent probabilities. Earlier on in the thread you stated that the way you should approach problems is to regard it as a fresh problem after each stage, sadly that only works with independent probabilities.

I don't think the million doors example helps people who are stuck on that either.

Alaric.

**Alaric**)No problem at all, it was just a question of someone finding a way to explain it in a suitable way, I'm more of the favour of maths as the equations are more absolute and formal than English.

For the sake of completeness:

I don't want to sound patronising as I don't know how much probability you've done, but I better explain the syntax just in case.

P(B|C) should be read as the "probability of B given C"

In this situation the door that has been picked is always A; so the probability that Monty opens B given we know the prize is behind C is evidently going to be 1. As the only other options are to open either the chosen door or the one containing the prize, neither of which Monty can do.

To be honest I think that most people get confused because they're too used to independent probabilities. Earlier on in the thread you stated that the way you should approach problems is to regard it as a fresh problem after each stage, sadly that only works with independent probabilities.

I don't think the million doors example helps people who are stuck on that either.

Alaric.

Anyway... the truel dear boy, the truel.

N.B - I stole... uh... borrowed it from an example in the Simon Singh book on Fermat's Last Theorum. It's an example of a problem designed to show how results obtained through the application of Game Theory can fly in the face of intuition which I felt was an appropriate link.

There's a Montaigne quote I really should remember about being eternally indebted to those who through reason can correct our mistakes and elighten us. As it's too late to look it up now imagine that I've said it.

Good Night.

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