[Guest]

This is a two part question:

1. You toss a fair coin four times in a row and it lands on tails all four times. What is the probability that your next coin toss will be tails?

2. A carnie at the fair has hidden a gold ring under one of three cups, marked A, B, and C. He lets you name one cup as your choice. You pick cup A. He then lifts cup C, showing you that it's empty, and allows you to choose again. What are the odds that the ring is under cup B?

[Guest]

The reason this seems like a fallacy is because the person removing the cup knows which cup the ring is under and did not remove it. For example:

This is an excellent example. Suppose there are, as stated 100,000 lottery tickets, and one and only one will win. You have one lottery ticket.

Now a person WHO KNOWS WHICH LOTTERY TICKET IS THE WINNING ONE rips up 99,998 of the tickets, all except yours and one other. One of them has to win and, obviously, the odds of yours winning are only 1/100,000. The odds of the other one winning are 99,999/10000 because it is the only one left besides yours.

Now suppose that 99,998 tickets were RANDOMLY ripped up by someone WHO DID NOT KNOW THE WINNING TICKET.

Your probability of winning is still 1/100,000 and the probability of the other ticket winning is 1/100,000. The probability is 99,998/100,000 that one of the ripped-up tickets was the winning ticket.

The probability can change in the first example because the person KNEW WHICH WAS THE WINNING TICKET and accordingly did not rip it up.

Here is the problem, asked in a slightly different way, already on Brainden:

http://brainden.com/forum/index.php?showto...p;hl=monty+hall

EDIT: added link

i know i am being dense. i can understand how the probabilities could change since the total number of possibilities has decreased, while the number of winning tickets has not. what i don't understand is why the probability of the other ticket being the winner has changed, while the probability of your own ticket being the winner has not changed. why would they not have equal chances? i agree that if the ripped up tickets were chosen randomly, then the probability for both tickets would still be 1/100,000. but logically, if 999,998 incorrect possibilities have been eradicated, if you reevaluate the situation then you have two possibilities, one of which is the winner, giving you a 1/2 chance. how is my reasoning flawed?

[Guest]

#1: always 50/50 chance whenever coin is flipped.

#2: Monty Hall paradox. 1/3 chance of Goat. Better switch cups!

yah i agree with you on the second one

if you change cups you will double your chances

[Guest]

Sorry Phatfingers, I have to agree with the comment rossbeemer has made in his last post and say that the answer you have provided isn't actually correct.

We recently had a discussion revisitng the traditional "at least one child in a two child family is a girl - what's the probability of the other being a girl" in a new post here which is a similar kind of question too. Basically we concluded that the answer depends on how we are given information.

We have a similar case here in your second question. The way your question is worded, we do not know if the carnie knows which cup the ring is under and will always show us an empty one (in which case your answer of 2/3 is correct) or if he has just picked a cup at random and, by luck, it happened to be an empty one (in which case the answer is actually 1/2).

I haven't revisited the link rossbeamer gave, but I seem to remember this principle was discussed there and the answer of 2/3 is only correct if it is a given rule that you will always be shown an empty door(/cup).

So, either the question needs to be reworded or, as it currently stands the answer can't be determined - in order to answer you have to make an assumption and assumptions are bad when it comes to accurate math!

Edited January 13, 2009 by neida

[Guest]

i know i am being dense. i can understand how the probabilities could change since the total number of possibilities has decreased, while the number of winning tickets has not. what i don't understand is why the probability of the other ticket being the winner has changed, while the probability of your own ticket being the winner has not changed. why would they not have equal chances? i agree that if the ripped up tickets were chosen randomly, then the probability for both tickets would still be 1/100,000. but logically, if 999,998 incorrect possibilities have been eradicated, if you reevaluate the situation then you have two possibilities, one of which is the winner, giving you a 1/2 chance. how is my reasoning flawed?

Your reasoning is not flawed depending on what question you are asking.

If someone walks up to you and you hold out the 2 tickets and say "Pick which ticket will win" the probability of them doing so is 1/2. There are 2 options, and one is correct.

However, that's not really the question this riddle is asking. Call the initial ticket selected A, and the remaining ticket B. The probability of A being the correct ticket is 1/100,000 and the probability of B being correct is 99,999/100,000.

The reason is because when you initially selected ticket A, the sample space was 100,000 tickets, so obviously the probability of A winning is 1/100,000. I would say everyone agrees on that.

Now when you eliminate every other ticket except one, and you know that one of the two remaining tickets wins, the probability of the second ticket (B) being the winner is 99,999/100,000.

That can be proved in the following way:

I think everyone agrees that the probability of A winning is initially 1/100,000. Therefore, the probability is 99,999/100,000 that A will NOT win. Agreed?

Now, when you eliminate every ticket except one, consequently, 99,999 times out of 100,000, A would NOT have been the winning ticket. And the remaining ticket (B) will be the winning one.

And that last one time out of 100,000, A will indeed be the winning ticket.

But hopefully this clears it up that the probability of B being the winning ticket is 99,999/100,000.

[Guest]

Well, let me try to give the kind of non-answers that I see so often here

1. Is it a quarter you're flipping? If so, the chances of getting tails are slightly higher than the chances of heads, as the heads side of a US quarter weighs more than the tails side, and is more likely to land down. That being the case, I'd also add that the odds of heads or tails on a given coin-type can't be known without extensive testing, given that we don't know what the imprint of the images does to the flight of the coin.

(On a different note, a deterministic universe would leave you with one of two choices: you have either a 100% chance or a 0% chance of tails coming up )

2. For those having trouble with part 2...

Think of it this way: the one cup you picked has a 1/3 chance of being the right cup. The other two cups combined have a 2/3 chance of being the right cup. These odds will not change, unless the carnie repositions the prize somewhere along the way. So the odds do not change when one cup is revealed. Your cup still has a 1/3 chance, and the other two cups--the one still hidden and the revealed cup, have a combined 2/3 chance of being the right cup.

But...you can only pick the hidden cup, right? the other one is empty. So...the hidden cup you didn't already pick has a 2/3 chance. See it?

Of course, to again go along with the non-sequitur answers...as the question asks, the 'chance that the prize is under cup B', strictly speaking, involves all sorts of psychological factors about the carnie himself. Does he have a preference for putting the prize under cup B? Or is he left-handed and has a preference for placing it under cup A? Did he once receive shock treatment in therapy for an addiction which strangely left him unwilling to pick the middle of a series of anything?

All of this tells me that we have no idea what the chances are that the prize is under cup B.

[Guest]

i know i am being dense. i can understand how the probabilities could change since the total number of possibilities has decreased, while the number of winning tickets has not. what i don't understand is why the probability of the other ticket being the winner has changed, while the probability of your own ticket being the winner has not changed. why would they not have equal chances? i agree that if the ripped up tickets were chosen randomly, then the probability for both tickets would still be 1/100,000. but logically, if 999,998 incorrect possibilities have been eradicated, if you reevaluate the situation then you have two possibilities, one of which is the winner, giving you a 1/2 chance. how is my reasoning flawed?

Look at it this way:

The only way you can lose if u switch is by chosing the right cup from the start.

X Y Z cups - Y is the winner cup

1. You choose X (empty) - carnie has to open cup Z (since she knows Y is winner so the only one left is Z) then you switch and choose the remaining Y cup and win!

2. You choose Z (empty) - carnie has to open cup X (since she knows Y is winner and the only one left is X) then you switch and choose the remaining Y cup AGAIN and win! (coincidence??)

3. You choose Y (winner) - carnie opens up X or Z and you choose either X or Z and lose

If you count the possibilities above, you only lose in option 3!!! wow!!! amazing!!!! so u have 2/3 chances winning if you switch a cup!!

[Guest]

WRONG! you should know the answer to your own riddle. yes, the carnie has given you 1/3 of the possibility, but you have to spread that between the two unknowns, so the possibility of B having the ring is 1/3 plus 0.5/3 or 1.5/3. I'm pretty sure that equals 50/50 (50%). Stop thinking so much. There are only two possibilities left and you have no idea which cup the ring is under. 50% chance it's B.

Welcome to the forum! Yes, this problem took some 'splaining to me the first time I encountered it. Imagine that the carnie didn't show you the empty cup, but instead let you choose between your either initial decision or BOTH of the other cups. You clearly have a 1/3 chance of your initial decision and a 2/3 chance if you pick the other two.

You do have this to your advantage. You know that there were only three options to begin with, and two of those three options forces the carnie to leave the remaining cup with a ring in it, and one of those three options leaves an empty cup.

[Guest]

If the carnie knows where the ring is and reveals that it's not under cup C, then I would definately stick with cup A. If the ring was under cup B, then the carnie shouldn't give the option of choosing cup B instead of cup A. He would just reveal that I was wrong and take my money!

"... then clearly I can choose the cup that is not in front of me!"

I gotta say... that wasn't where I was going, but I like your answer. If the carnie knew the Monty Hall paradox, and he assumed that you knew, then he might have presented the option to entice you to change your answer, knowing you already had the ring. There's nothing stipulating that he HAD to present that option.

[Guest]

"... then clearly I can choose the cup that is not in front of me!"

I gotta say... that wasn't where I was going, but I like your answer. If the carnie knew the Monty Hall paradox, and he assumed that you knew, then he might have presented the option to entice you to change your answer, knowing you already had the ring. There's nothing stipulating that he HAD to present that option.

This is what I was getting at in my previous post. The question talks about an isolated incident and doesn't give any rule about what the carnie will or won't show, therefore we can't actually deduce anything from what we see. If we observed his trick enough times to be able to identify what rule he was using, or if we otherwise knew (e.g. he told us) then we would be able to deduce something.

The reason why it's worth stressing this is that people often just take the 2/3 answer as the truth no matter what the circumstances and this is simply wrong. I'll give the following answers depending on different reasons why we were shown the empty cup:

1. The carnie will always show us an empty cup - answer is 2/3 (several people have already explained why).

2. The carnie will pick one of the remaining cups at random. Sometimes this will result in us being shown the ring (and knowing we've lost) and sometimes this will result in us being shown an empty cup - the answer is 1/2. (I can explain maths behind this if necessary)

3. The carnie will only show us an empty cup if we have already picked the cup with the ring, otherwise he shows us nothing - answer is 0.

4. The carnie will only show us an empty cup if we have picked an empty cup, otherwise he shows us nothing - answer is 1.

5. The carnie will secretly roll a dice and if it comes up 6 he will show us an empty cup, otherwise he shows us nothing - answer gets more complicated (can't be bothered to work it out)

6. The carnie will show us an empty cup if he likes your hairstyle - answer gets more complicated

7. The carnie will try to guess whether you are good at probabilities and, if he thinks you are, will only show you an empty cup if you have already picked the correct cup in the hope you will fall into the common trap of always thinking you are better off switching - answer may be close to 0, but depends on how good the carnie is at guessing your skills.

and so on...

I hope this all makes sense.

Edited January 14, 2009 by neida

[Guest]

Good point(s) Neida. People have a tendency to just say "Oh that's Monty Hall."

Other people have a tendency to say "Oh that's Monty Hall, and the probability is 1/2." But that's a different issue.

[Guest]

5. The carnie will secretly roll a dice and if it comes up 6 he will show us an empty cup, otherwise he shows us nothing - answer gets more complicated (can't be bothered to work it out)

As much I know, in the problem it is assumed that carnie is honestly shows the empty cup after looking to the both cups. Then odds=1/3.

As in your 5th item, if he has rolled a dice, then he will sometimes (1 in 6 cases) show us an empty cup. That time I would change my cup (1/3-2/3). But if he doesn't show any cup, there is no need to change (1/2).

[Guest]

As in your 5th item, if he has rolled a dice, then he will sometimes (1 in 6 cases) show us an empty cup. That time I would change my cup (1/3-2/3). But if he doesn't show any cup, there is no need to change (1/2).

You are quite correct (except your final probability of 1/2 should be 1/3 as all three cups are unturned). I had intended to say that if he rolls a 6 he will show us a cup (i.e. at random, so it may actually be the winning cup).

It just goes to show how many possibilities there are and why it is dangerous to make assumptions!

[Guest]

So you all don't think that, given the statement of the original puzzle, this was intended to be the MH Paradox, but was rather indended to be a question in which we are required to know something about the carnie's method of choosing which cup to hide the prize under, and choosing which cup to reveal?

In that case, I think the odds are 1/14, because the carnie is receiving his instructions through a hidden earpiece, manned on alternating days by two brothers, one of whom always tells the carnie to hide the ring in his palm and not under the cup, and the other of whom always tells the carnie to hide the ring under cup A. One day out of every two weeks, both brothers get drunk and the carnie is left to his own devices. Because he's rather unimaginative, on those days he always hides the ring under cup B.

Thus, the chances that the ring is under cup B are 1/14.

Wow, that was easier than trying to do a logic puzzle.

Edited January 14, 2009 by brotherbock

[Guest]

2. For those having trouble with part 2...

Think of it this way: the one cup you picked has a 1/3 chance of being the right cup. The other two cups combined have a 2/3 chance of being the right cup. These odds will not change, unless the carnie repositions the prize somewhere along the way. So the odds do not change when one cup is revealed. Your cup still has a 1/3 chance, and the other two cups--the one still hidden and the revealed cup, have a combined 2/3 chance of being the right cup.

But...you can only pick the hidden cup, right? the other one is empty. So...the hidden cup you didn't already pick has a 2/3 chance. See it?

This is the most elegant explanation of the Monty Hall situation that I believe I've ever seen. Well done.

As far as the motivation of the Carnie, I think a reasonable assumption is that he's in it for profit.(unless you believe it's the lifestyle that is appealing?) This being the case, he would want to employ any and all strategies to give him the advantage over his mark. Knowingly influencing the choice through knowledge of the rings location would be the easiest "non-cheating" or "fair" way to accomplish this.

I think if the OP were to change "carnie" to "experienced carnie trying to maximize profit" then we could satisfy the question of prior knowledge. I do agree that, as written, it is a little open to interpretation if you're looking to nit-pick.

[Guest]

Your reasoning is not flawed depending on what question you are asking.

If someone walks up to you and you hold out the 2 tickets and say "Pick which ticket will win" the probability of them doing so is 1/2. There are 2 options, and one is correct.

However, that's not really the question this riddle is asking. Call the initial ticket selected A, and the remaining ticket B. The probability of A being the correct ticket is 1/100,000 and the probability of B being correct is 99,999/100,000.

The reason is because when you initially selected ticket A, the sample space was 100,000 tickets, so obviously the probability of A winning is 1/100,000. I would say everyone agrees on that.

Now when you eliminate every other ticket except one, and you know that one of the two remaining tickets wins, the probability of the second ticket (B) being the winner is 99,999/100,000.

That can be proved in the following way:

I think everyone agrees that the probability of A winning is initially 1/100,000. Therefore, the probability is 99,999/100,000 that A will NOT win. Agreed?

Now, when you eliminate every ticket except one, consequently, 99,999 times out of 100,000, A would NOT have been the winning ticket. And the remaining ticket (B) will be the winning one.

And that last one time out of 100,000, A will indeed be the winning ticket.

But hopefully this clears it up that the probability of B being the winning ticket is 99,999/100,000.

Well, yes and no. I will read everyone's responses as well as the wikipedia entry, etc, and I am sure they will give me the reasoning I need in order to understand the solution. I think that the reasoning you gave was flawed, for the following reason: The probability of A winning was initially 1/100,000--I agree. I also agree that this means that the initial probability of A not winning is 99,999/100,000. However, using this same logic, the initial probabilities are the same for ticket B. The probability of B not being the winner is 99,999/100,000 so using this logic the chances are still equal for A and B. As someone else pointed out, I think the reason the probabilities are different lies in the fact that the person eliminating the incorrect possibilities knows which ticket is the winner and is only eliminating non-winning tickets.

[Guest]

Look at it this way:

The only way you can lose if u switch is by chosing the right cup from the start.

X Y Z cups - Y is the winner cup

1. You choose X (empty) - carnie has to open cup Z (since she knows Y is winner so the only one left is Z) then you switch and choose the remaining Y cup and win!

2. You choose Z (empty) - carnie has to open cup X (since she knows Y is winner and the only one left is X) then you switch and choose the remaining Y cup AGAIN and win! (coincidence??)

3. You choose Y (winner) - carnie opens up X or Z and you choose either X or Z and lose

If you count the possibilities above, you only lose in option 3!!! wow!!! amazing!!!! so u have 2/3 chances winning if you switch a cup!!

Thank you! This explanation is clear and concise, and makes perfect sense!

[Guest]

Well, yes and no. I will read everyone's responses as well as the wikipedia entry, etc, and I am sure they will give me the reasoning I need in order to understand the solution. I think that the reasoning you gave was flawed, for the following reason: The probability of A winning was initially 1/100,000--I agree. I also agree that this means that the initial probability of A not winning is 99,999/100,000. However, using this same logic, the initial probabilities are the same for ticket B. The probability of B not being the winner is 99,999/100,000 so using this logic the chances are still equal for A and B. As someone else pointed out, I think the reason the probabilities are different lies in the fact that the person eliminating the incorrect possibilities knows which ticket is the winner and is only eliminating non-winning tickets.

Where it became real for me was imagining that I played it all possible times. If you play the same setup (gold ring under a fixed cup) with 100,000 cups, making all 100,000 possible choices, the host can always reduce it to the two choices, and your initial choice will only be right one out of the 100,000 times, so the second choice will be correct 99,999 times out of the 100,000.

I think you're also right that it is key knowing that only incorrect possibilities are removed. You can turn it on it's head. Going back to the one cup vs. two, if the carnie adds two empty cups instead of taking away one empty cup, have your odds changed any?

[Guest]

Well, yes and no. I will read everyone's responses as well as the wikipedia entry, etc, and I am sure they will give me the reasoning I need in order to understand the solution. I think that the reasoning you gave was flawed, for the following reason: The probability of A winning was initially 1/100,000--I agree. I also agree that this means that the initial probability of A not winning is 99,999/100,000. However, using this same logic, the initial probabilities are the same for ticket B. The probability of B not being the winner is 99,999/100,000 so using this logic the chances are still equal for A and B. As someone else pointed out, I think the reason the probabilities are different lies in the fact that the person eliminating the incorrect possibilities knows which ticket is the winner and is only eliminating non-winning tickets.

Ticket A (your ticket) has a 1/100,00 probability winning. The remaining tickets, as a group, have a 99,999/100,000 probability of winning. These are established facts.

Ticket A remains 1/100,000 because you chose it before any additional information that effects overall outcome was given.

Now, you are told that all tickets other than B are NOT winners, and asked if you would like to pick a new ticket. I think this is where the break-down in understanding comes in. When we are told that the tickets are losers, that does not disqualify them as valid choices, we could still pick them. But, since the goal is to win, and we are given information that allows us to mentally eliminate 99,998 losers from the sample, we naturally feel that we need to reevaluate the problem and we see the chance as 50/50.

The fact is, that while knowing that the tickets are losers, they are still part of the remaining 99,999/100,000 that we did not initially choose, of which ticket B is still included. We just know better than to pick from the other 99,998 losing tickets. The new knowledge does not influence the original probability, just your perspective on the problem.

This is also why the carnie doesn't need to have prior knowledge of the location of the ring. If he shows a ring under cup C, you merely lose. However, once the cup C is lifted to show nothing is under it, as long as cup C is still left as a choice, cups B and C will still have a 2/3 probability of having the ring. You will just not choose cup C, because of your new knowledge.

If cup C is removed as a choice, as in the Monty Hall "Let's Make a Deal" example, then prior knowledge of the location of the prize by the carnie game must be a given for us to make the same conclusion. The carnie removing cup C in this example, as long as WE know it is based on his knowledge of the location of the ring, is the same as you mentally removing cup C based on your new knowledge in the example above. Prior knowledge is the key, but it is situational as to who needs to have it and when.

:ducking as Neida swoops in to dismantle my arguments:

[Guest]

Well, yes and no. I will read everyone's responses as well as the wikipedia entry, etc, and I am sure they will give me the reasoning I need in order to understand the solution. I think that the reasoning you gave was flawed, for the following reason: The probability of A winning was initially 1/100,000--I agree. I also agree that this means that the initial probability of A not winning is 99,999/100,000. However, using this same logic, the initial probabilities are the same for ticket B. The probability of B not being the winner is 99,999/100,000 so using this logic the chances are still equal for A and B. As someone else pointed out, I think the reason the probabilities are different lies in the fact that the person eliminating the incorrect possibilities knows which ticket is the winner and is only eliminating non-winning tickets.

Egad. Never trust Wikipedia. If you're interested why, go google "Wikipedia and Overstock.com" or "Wikipedia and naked short selling". Those people are a cult, a cult who plays around with what they declare to be 'the truth', often for personal gain.

At any rate, you're right in that the whole ball of wax changes if the carnie is flipping cups randomly. In the traditional puzzle (and on Let's Make A Deal, where the puzzle started) this wasn't the case.

This is the most elegant explanation of the Monty Hall situation that I believe I've ever seen. Well done.

Thank you

[Guest]

Its easier to understand if you think with largers proportions. Lets change a little: If it was 100.000 cups instead of only 3. One has a ring. You choose one cup and, right now, you have only 1 possibility in 100.000 to had chossen the right one.

But, the guy that knows were the ring is removes 99.998 cups, showing that the ring was not in any of them.

Now, we have only 2 cups. The one you chose before (that has a chance of 1/100.000 of been the correct) and the only other one left in the game, that had adquired, when the guy shows the others cups impossibilities, all these cups chances of beign the correct one.

Just think whats more possible: The cup you initaly choose has the ring (1/100.00, youre a very luck guy) or the only one left, with all the rest of possibilities (99.999/100.000)?

Now can you understand it?

Best explaination I have ever heard on this !!!!

[Guest]

Best explaination I have ever heard on this !!!!

Thanks!

And wow, in this topic, Monty Hall's paradox is about to turn into "Monty Pyton's" paradox . Lot of fun reading the "if"s: what if the carnie its an alien and? And if the ring was an onion ring and the cup does not isolate it, bacterias could eat the ring so there is no ring left.

We could make a crazy movie just with the ideias posted in one topic. Braindeners, your all great!

[Guest]

Thanks!

And wow, in this topic, Monty Hall's paradox is about to turn into "Monty Pyton's" paradox . Lot of fun reading the "if"s: what if the carnie its an alien and? And if the ring was an onion ring and the cup does not isolate it, bacterias could eat the ring so there is no ring left.

We could make a crazy movie just with the ideias posted in one topic. Braindeners, your all great!

So, wait...the carnie places The One Ring under a cup, and we have to tell what the odds are that The Ring is under cup B?

If the person playing is on their way to Mordor, then The Ring will have forced the carnie to place it under whatever cup the other person was going to choose

[Guest]

I think if the OP were to change "carnie" to "experienced carnie trying to maximize profit" then we could satisfy the question of prior knowledge.

This is probably the most likely scenario actually. But in this case, if you picked a cup without the ring the carnie would probably just show you that cup and point out you've just lost. If he's showing you another cup then it must be because he's trying to get you to change your mind, so you should stick with your original choice.

This is also why the carnie doesn't need to have prior knowledge of the location of the ring. If he shows a ring under cup C, you merely lose. However, once the cup C is lifted to show nothing is under it, as long as cup C is still left as a choice, cups B and C will still have a 2/3 probability of having the ring. You will just not choose cup C, because of your new knowledge.

Sorry, this bit isn't correct. Being able to choose but knowing you won't doesn't make any difference to the probabilities than not being able to choose. Think about it like this:

Label the 3 cups A, B and C and say you pick cup A (if you pick another cup it's just a rotation of this). There are 6 possibilities, each with equal probability:

1: Ring is under cup A, carnie shows empty cup B - don't switch to win

2: Ring is under cup A, carnie shows empty cup C - don't switch to win

3: Ring is under cup B, carnie shows cup B - game over

4: Ring is under cup B, carnie shows empty cup C - switch to win

5: Ring is under cup C, carnie shows empty cup B - switch to win

6: Ring is under cup C, carnie shows sup C - game over

You see here that, as you put it, in cases 3 and 6 you are shown the ring and you "merely lose" or, as I have put it "game over". So, on the occasion that he shows you an empty cup there are only 4 possible ways this could have happened (cases 1,2,4 and 5 above, each with equal probability) and 2 cases where you should switch and 2 where you shouldn't (again each with equal probability). So the probability of cup B being the winning cup is 1/2.

Egad. Never trust Wikipedia.

I agree, but on this occasion the Wikipedia page is actually fairly good. It even has a diagram showing what I've just explained above for anyone needing to think it through some more.

:ducking as Neida swoops in to dismantle my arguments:

Happy to oblige!

[Guest]

I agree, but on this occasion the Wikipedia page is actually fairly good. It even has a diagram showing what I've just explained above for anyone needing to think it through some more.

That may be, but in order to know when to trust Wikipedia, you either have to be looking at a priori info (math, for example), or you have to already know what it is that you're looking at, right? To know that their article on probability is accurate, you have to know something about probability already. So the only way you can trust the info there is if you already know the info.

[Guest]

That may be, but in order to know when to trust Wikipedia, you either have to be looking at a priori info (math, for example), or you have to already know what it is that you're looking at, right? To know that their article on probability is accurate, you have to know something about probability already. So the only way you can trust the info there is if you already know the info.

Yes definitely, but it is accurate more times than people give it credit for. It comes under a lot of stick because it can (and quite often is) wrong. I work in the web industry so am often giving talks where Wikipedia is mentioned and I always tell people to take what is said with a pinch of salt and back it up with other sources. But it is often a good starting point to give you an idea of what to look for. The unfortunate thing is that most people just take what it says as the truth, which is why it often appears at the top of Google results, as it is one of the most trusted sites on the web...

One example where it is wrong actually is the article on brain teasers. Here, although it references the Monty Hall problem, it then goes on to discuss the two children and doesn't point out that the wording of the question leads to exactly the same issues that Monty Hall faces - it just states that the answer is definitely 1/3.

But that's a bit off topic...

Edited January 14, 2009 by neida