View Full Version : Brain teaser

Steve Axford

02-08-2011, 9:41am

Imagine a game show.

There are 3 doors, one with a major prize and the other two with a donkey behind.

You are asked to pick a door.

You pick one.

The host then opens another door - always revealing a donkey.

The host then asks you - do you want to change your pick?

Do you?

So I get either a "major prize" or a donkey? That's a win/win imo :D. But if I wanted the major prize I would change my pick.

Cheers.

Steve Axford

02-08-2011, 10:06am

Why?

p.s. The donkey is an unfriendly donkey and bites - so you don't want him.

fillum

02-08-2011, 10:20am

Why?

p.s. The donkey is an unfriendly donkey and bites - so you don't want him.

I could use him as a 'guard donkey' :D

Keep my pick.

The host opens 'another door' and 'always reveals a donkey' so has already opened both the doors with a donkey behind them.

Eeeaw, eeeaw, 'e always does that.

I assume the door you picked isn't opened. So you change for improved odds.

I'd always switch doors.

Stay with me here.

I've chosen my initial door (1) - that's a 1/3 chance of getting the prize.

The hsot reveals a losing door (door 3). Thus, Door 2 remains

IF I stick with my initial chosen door (1), I stay with a 1/3 chance of winning

IF I switch to the other door (2), I now have a 2/3 chance of winning.

Why? Because by the host revealing the donkey door, statistically speaking, there is a 2/3 chance of the prize being behind the remaining door.

Ms Monny

02-08-2011, 12:10pm

Alan, that just mashed my brain!! :D :D

I have read and re-read your answer and it still doesn't make sense! You have probably got the right answer though.

I would pick the door that is NOT next to the donkey! :o

davidd

02-08-2011, 12:21pm

I'd always switch doors.

Stay with me here.

I've chosen my initial door (1) - that's a 1/3 chance of getting the prize.

The hsot reveals a losing door (door 3). Thus, Door 2 remains

IF I stick with my initial chosen door (1), I stay with a 1/3 chance of winning

IF I switch to the other door (2), I now have a 2/3 chance of winning.

Why? Because by the host revealing the donkey door, statistically speaking, there is a 2/3 chance of the prize being behind the remaining door.

Actually I believe it is now 50% chance, there are now two doors, and one has a donkey, one has the prize, so there is an equal chance that your door or the other door has the prize.

I would probably stay with my original choice.

Ms Money, try this. There are a thousand doors and after you pick one, the host will open 998 with donkeys. Same problem, but the answer's more obvious I think.

Davidd you chose before the door was opened, so your odds haven't changed. Still one in three. They only change if you pick the other door.

davidd

02-08-2011, 12:27pm

Yeah, I wondered about that, but if you have the chance of changing your mind, you are effectively choosing again, so I would think it is now 50% each. :confused013

Steve Axford

02-08-2011, 12:30pm

My guess is that some of you have heard this before, because the answer is anything but intuitive. In fact, the more you know about probability theory, the less likely you are to get it right. I blew my mind when I first heard it, and I think in probabilities.

The doors haven't changed, but the host changed the rules by picking a door that wasn't random.

Speedway

02-08-2011, 12:35pm

The odds don't alter revealing all but 2 doors they are still 50/50 on the final decision.

Keith..

Steve Axford

02-08-2011, 12:38pm

Wrong

ameerat42

02-08-2011, 12:39pm

Look. Were any donkeys hurt in this joke? Nasty biters or not.:(

Ms Monny

02-08-2011, 12:44pm

my brain is certainly hurting, stuff the donkeys!!!!

Ms Monny

02-08-2011, 12:46pm

Yes, you change the pick because the host is giving you a clue that you have a donkey!! yes??? no??? oh, please tell us!!

Steve Axford

02-08-2011, 12:50pm

I sympathise Ms. My brain hurt when I heard this too. I made sense but didn't all at the same time. It took me a full day to make sense of it (and I mean make sense rather than calculate the correct answer).

I'll come back from Lunch and do a drawing to explain it all. :p

given the host has not opened the door that i picked there is now a 50% chance that my door has the prize...

also given that the donkey that i have decided to not go for has a bad temper there is a chance that if I dont get the prize the the donkey i get might be more friendly :D

I know mongo would just take the donkey and eat it. :p

Perhaps look at it this way...

On your first pick you have more likely selected a donkey (a 2 in 3 chance). So changing your selection after one of the donkey doors is eliminated by the host should improve your chances of getting the major prize. (I think? :confused013)

Perhaps the Opera users can clear it up for us :D...

Cheers.

Look at it this way, if the host opens two doors and reveals donkeys, what do you reckon your odds are of your door having the prize? :D

Probably not still 1 in 3.

Steve Axford

02-08-2011, 1:41pm

But if he only opens one door, what are the odds? Still 1/3 - for your pick. But not for the remaining door, because the host has added extra information to that door - but not to your choice.

Mary Anne

02-08-2011, 1:44pm

I know mongo would just take the donkey and eat it. :p

Thats the only reply here that makes sense to me David :laughing1:

My guess is that some of you have heard this before...

Ooh, not fair, not fair!

But maybe true. It did sound sort of familiar.

virgal_tracy

02-08-2011, 2:36pm

I saw it explained in the movie "21" with Kevin Spacey. Thinking makes my head hurt so thankfully didn't have to think this one through.

I will have to ask Mongo if Donkeys are good eating.

its not really that hard....

you start off with a 1 in 3 chance of getting the prize, or rather lets look at it the other way... a 2 in 3 chance of getting a donkey. remove a donkey and do not get a 1 in 3 chance of a donkey as one option is removed, there are only 2 options open to kyou now... so it becomes 50% as the only options are a donkey or the prize. at this stage it doesnt matter if you change your door as the chances are the same.

i dont know how it could be otherwise or where these other theories are coming from :confused013

Here it is in pictures and I hoipe you can make more sense out of it now. It doesn't become 50% because you started out with a 33.33% chance and that chance remains no matter what - you've always had three choices. The difference is you now have more information to increase your 33.33% chance of being correct. Albiet stupid, you can still pick the door that's been opened to you because that remains as an option. Unless you REALLY wanted the donkey.

http://img94.imageshack.us/img94/7382/step1n.png (http://imageshack.us/photo/my-images/94/step1n.png/)

http://img189.imageshack.us/img189/4856/step2w.png (http://imageshack.us/photo/my-images/189/step2w.png/)

http://img9.imageshack.us/img9/5571/step3pv.png (http://imageshack.us/photo/my-images/9/step3pv.png/)

monty hall paradox... just google it :rolleyes:

I still say the answer given is wrong.

they say picking another door changes your chance from 33% to 66% provided the host knows which door holds the prize. I find that it is irrelevant as to whether the host knows the answer or not assuming that you get what is behind the door you pick

alan, would you care to explain how door B gets to inherit door Cs odds? there are only 2 doors left and only 2 choices, therefore a 50/50 split.

Ms Monny

02-08-2011, 3:14pm

OUCH!! :eek:

That was the sound my head just made. Still doesn't make sense.....well, it does but it doesn't!!

50/50... take it from an opera user ;)

alan, would you care to explain how door B gets to inherit door Cs odds? there are only 2 doors left and only 2 choices, therefore a 50/50 split.

It's because this problem is about probability and not by process of elimination. The host KNOWS which door contains the prize, as s/he always opens a door with a donkey in it. What this does if give you additional information to calculate your probability of selecting the prize door.

Using "inheret" was probably a bad choice of words but what I was trying to convey is, if you stick with your original choice, you still only have 1/3 chance of winning the car. However, with the added information which of the other doors is a donkey, the remaining door has a 2/3 chance of being the prize because you are more likely to pick the donkey to begin with (2/3 chance) than the car (1/3). So basically what you've done is inverse your odds from a 1/3 probability to a 2/3 probability of winning by swapping.

Confusing, but it makes sense in my head :p

Xebadir

02-08-2011, 3:31pm

Lies damned lies and statistics.

Probability is a paradoxical quantity anyway, realistically the world doesn't function on two outcomes.

Richard Hall

02-08-2011, 3:33pm

50/50... take it from an opera user ;)

Nup...

if you stick with your original choice, you still only have 1/3 chance of winning the car. However, with the added information which of the other doors is a donkey, the remaining door has a 2/3 chance of being the prize because you are more likely to pick the donkey to begin with (2/3 chance) than the car (1/3). So basically what you've done is inverse your odds from a 1/3 probability to a 2/3 probability of winning by swapping.

^^^ THIS!!!

Steve Axford

02-08-2011, 3:36pm

Probability relies on information. Your first choice has no additional information, so your original 1/3 chance stays - no change because no new information - at least with regards that door. But - the other remaining door has new information added because the host did NOT select a door with a donkey at random. He knew where a donkey was and so he has added new information to the remaining door, but not to your original selection. Thus the two doors now have different odds. The remaining door was 1/3, but it is now 2/3 because the host selected one of the donkeys - deliberately - thus removing a 1/3 chance of selecting a donkey. Lay-z is, of course correct, but one suspects that he didn't figure this out for himself.

Another way of thinking of it is if your selection is 1/3 then the total probability must be 1 (by definition). So the remaining door must be 2/3, as 1 - 1/3=2/3. Unfortunately that way doesn't really explain why.

ok i see where this is going... you start with a 1/3 with 3 options. one is revealed and the 1/3 remains because there was 3 options. what doesnt make sense is that your odds improve to 2/3 by changing your mind. if you change your door there are only 2 options therefore becoming 50/50 as you obviously cant pick the door already opened.

mathematically sound but incorrect as despite you mathematics the chances of winning once the first donkey is revealed is the same regardless of whether you change doors or not.

Xebadir

02-08-2011, 3:46pm

Oh and I note, even given I switched I would still pick a Donkey. Wonder why? Because no matter the probability someone always gets screwed, and I am that lucky :P.

Ever wonder why weather forecasts are wrong? Probabilistically based.

Another way of thinking of it is if your selection is 1/3 then the total probability must be 1 (by definition). So the remaining door must be 2/3, as 1 - 1/3=2/3. Unfortunately that way doesn't really explain why.i get that totally but still think it is wrong :p

Holy moly brain hurt! :eek:

Not really because:

First Step - you pick a door.

You have a 2/3 chance of picking a donkey

You have a 1/3 chance of picking the car

Step 2 - The host knows which doors contains the donkeys, so, they open the door with the donkey.

Statistically speaking, you are twice more likely to pick a door with the donkey.

If you picked the donkey, the host is forced to open the door with the other donkey.

If you swap, you have inversed your odds from being 1/3 correct to 2/3 correct.

Here's another visualisation. Below are the possible scenarios. Let's run with the assumption you always pick the first option as your door.

Car, Donkey, Donkey

Donkey, Car, Donkey

Donkey, Donkey, Car

So, if you always picked the first option, were revealed where one of the donkeys were, then swapped, you would have twice as much chance of winning the car.

David - I can see where you are coming from with this now (which is where I can now disagree :p). From my understandting, you are adressing this hypothetical with a linear approach as opposed to factoring in all the possible outcomes. Is this so?

James T

02-08-2011, 4:01pm

That last point from Alan shows it most clearly I think.

If you switch.

You have a 1/3 chance of picking a car initially - switching and losing.

You have a 2/3 chance of picking a donkey initially - switching and winning.

If you stick.

You have a 1/3 chance of picking a car and winning.

You have a 2/3 chance of picking a donkey and losing.

ie if you are wrong to begin with (more likely) you must win by switching. Whereas if you are right to begin with (less likely) well, hard-lines.

Xebadir

02-08-2011, 4:04pm

But heres the crux of the paradox. 3 people on this forum take this challenge, all follow your advice to switch. Not every time will 2 out of the 3 win the car. You actually need an incredibly large number of samples to trend towards such a result. Take flipping a coin. Take one out now, flip it ten times, do you get 5 heads, 5 tails?

Heres another example: Weather forecasting. Statiscally the 4th day forecast has a probability of 55% of being correct. But on a day to day scale it actually only has a 50% probability, assuming the outcomes are correct or incorrect. Why? You neglect a quantity known as forecast skill of a model. Which effectively acts as a multiplier to your probability. This is more amusing when you consider the 1 day forecast has a 95% probability of being correct, but the actual potential outcomes are 50/50 before one includes model skill :P. Why do weather models get it wrong so often, because eventually the probability you get it wrong happens.

In the above case what is neglected is that no matter what someone loses :P.

Probability in a complex sense is the potential outcomes as the sample size trends to infinity and converges to the result :P.

Speedway

02-08-2011, 4:09pm

Once 1 door has been revealed then it is no longer in the equation so the odds become 50/50, it doesn’t make any difference whether you change your selection or not.

Keith.

Speedway

02-08-2011, 4:18pm

Imagine a game show.

There are 3 doors, one with a major prize and the other two with a donkey behind.

You are asked to pick a door.

You pick one.

The host then opens another door - always revealing a donkey.

The host then asks you - do you want to change your pick?

Do you?

This line changes the original odds whether you change or not.

lay-z... the door the host picked and whether or not he knows whats where is completely irrelevant. all that matters is that there are 3 doors, one with a prize 2 with donkeys. if the door the host opened (which wasnt the one you picked) was the one with the prize then the outcome is different... you lost. thats the only other outcome AFAIC up to the stage given in the original post.

Lance B

02-08-2011, 4:22pm

The flipping coin scenario given by xebadir is a good example. If you flipped a coin 10 times and received a head for all those 10 times, what is the likely hood of you getting a head on the 11th flip? Answer = 50/50 as every time you flip a coin it is always a 50/50 chance. However, the odds for getting 11 head in a row are very much higher!

Steve Axford

02-08-2011, 4:31pm

lay-z... the door the host picked and whether or not he knows whats where is completely irrelevant. all that matters is that there are 3 doors, one with a prize 2 with donkeys. if the door the host opened (which wasnt the one you picked) was the one with the prize then the outcome is different... you lost. thats the only other outcome AFAIC up to the stage given in the original post.

You have a choice of two doors. Your original pick has the information that it is a 1/3 chance. The other door has changed information because the host picked one of the donkeys (deliberately) and added new information to that and the other remaining door(but not your door because he would never pick yours. Thus the two doors are no longer random and it does matter which one you choose.

The other door has changed information because the host picked one of the donkeys (deliberately) and added new information to that and the other remaining door(but not your door because he would never pick yours. Thus the two doors are no longer random and it does matter which one you choose.how is the new information passed onto the other door? the fact still remains that it is either a donkey or the prize it doesnt become 2 donkeys or a donkey and the prize....

to quote keith:

Once 1 door has been revealed then it is no longer in the equation so the odds become 50/50, it doesn’t make any difference whether you change your selection or not.

Keith.

Steve Axford

02-08-2011, 4:53pm

The host chooses between the 2 remaining doors, thus the information is imparted to both of those doors, but not to your chosen door (since he can never choose that one). You now have different information for your door and the other remaining door.

Yes, why does the door you didn't pick inherit the other door's chances? I still think both doors will equally inherit the chance, so it becomes 50/50.

You have a choice of 2 doors after the one door is revealed, so the odds are 50/50. You can stick with your original choice (50% chance) or change to the other one, (50% chance).

Steve Axford

02-08-2011, 6:15pm

Because the host chooses only from the two remaining doors and not your pick. Either one or both have a donkey behind. It is 1/3 that both have a donkey and 2/3 that only 1 has a donkey. Therefore it is 2/3 that the remaining door is not a donkey (has the prize) and 1/3 that it is (has not the prize).

He has no influence on the door that you chose, so it is 1/3 as it always was.

This is such a beautiful thing. Our brains get totally bamboozled by it.

i am not bamboozled, you are just wrong steve! :lol:

:p

bushbikie

03-08-2011, 12:07am

OK, I'll have a bite....

I wouldn't change my pick. Who's to say that the major prize isn't a donkey?

Steve Axford

03-08-2011, 7:37am

And you're not bamboozled????? :rolleyes: Work out the maths, ving. There are only 3 doors so it can easily be done long hand.

virgal_tracy

03-08-2011, 9:41am

This is the explanation as portrayed in the movie "21"

Monty Hall Problem (http://youtu.be/cXqDIFUB7YU)

It doesn't quite explain the whys and wherefores but offers some explanation.

Try this as the fulfilling explanation

Monty Hall Explained (http://youtu.be/mhlc7peGlGg)

I hope this works.

Ms Monny

03-08-2011, 9:47am

Even after the answer was revealed....people still were talking about it and working it out! WOW, that some powerful brain teaser!! :D

thanks vince. and so you increase your chances of getting the car to 66% if you first picked a goat but if your first pick was the car and you change it is 33% according to the YT vid....

twas an enlightening video makes perfect sense... is completely illogical and IMO still wrong :p

and now i am going insane! :lol:

Speedway

03-08-2011, 10:26am

No matter which way you look at this the original Odds were 2:1. When the host opened one of the other doors to reveal a donkey your odds were still 2;1 but when he offered you a chance to change your selection the third door was removed from the equation leaving only two doors, and whether you changed or not because you were given the opportunity your odds are now 1:1 or 50/50. If you were not given the chance to change and he was just prolonging the agony the odds of you having picked the correct door are still 2:1

Keith.

PS Some people think too hard finishing up only confusing themselves.

Steve Axford

03-08-2011, 10:59am

Wrong, wrong, wrong. That's what I spent a day hurting my brain with. It does matter - the doors are now in different groups. We assume that they are all equal, because that's the way they started out. But the host changed all that when he opened a door which he knew was a donkey.

As Ms says - wonderful brain teaser because even when explained, many of us still don't believe it.

but steve, this vid explains it all... and it states it does matter if your pick is either the donkey or car as what you pick effects the outcome if kyou decide to change...

mhlc7peGlGg

ps, still 50/50 IMO :p

Lance B

03-08-2011, 11:21am

Watched it and I still think that it is 50/50. It sounds all very convincing but it is irrelevent that there is a third door as it is eliminated then it leaves two doors and therefore it makes your choice 50/50.

Steve Axford

03-08-2011, 11:22am

It doesn't change your odds for your pick (since the host can never pick that door) - it is still 1/3. But it does change the odds for the remaining door because the host tossed in more information for that door. Imagine you have picked door 1, and 2 & 3 remain. They could be

P D (P for prize and D for donkey)

D P

D D

ok?

The host then picks a donkey, so the remaining door will be

P

P

D

ok?

So that is 2/3 that it will be a prize.

Your original choice was 1/3. You add 1/3 plus 2/3 and you get 1, which it must be for any probability.

How's your brain?

fillum

03-08-2011, 11:24am

The thing to keep in mind is that there are two connected events here. If the host eliminated one of the donkey doors before you make your initial choice then the odds would be 50/50. But because you make your original choice from 3 doors you have a 1/3 chance, so the remaining door (after one eliminated) must be a 2/3 chance (zero-sum game). Obviously you could select the big prize first then swap to a donkey but that is against the odds.

Cheers.

The host then picks a donkey, so the remaining door will be

P

P

D

ok?

How's your brain?

D

D

P

sorry once the host remove a 'D' there is only 2 doors and 2 possibilities left...

I understand now how the answer works but i cant convince my brain that its right... :rolleyes:

Speedway

03-08-2011, 11:49am

It doesn't change your odds for your pick (since the host can never pick that door) - it is still 1/3. But it does change the odds for the remaining door because the host tossed in more information for that door. Imagine you have picked door 1, and 2 & 3 remain. They could be

P D (P for prize and D for donkey)

D P

D D

ok?

The host then picks a donkey, so the remaining door will be

P

P

D

ok?

So that is 2/3 that it will be a prize.

Your original choice was 1/3. You add 1/3 plus 2/3 and you get 1, which it must be for any probability.

How's your brain?

How can you have 3 choices when only two dors remain.

Keith.

This is still going on? :p

Lance B

03-08-2011, 11:53am

It doesn't change your odds for your pick (since the host can never pick that door) - it is still 1/3. But it does change the odds for the remaining door because the host tossed in more information for that door. Imagine you have picked door 1, and 2 & 3 remain. They could be

P D (P for prize and D for donkey)

D P

D D

ok?

The host then picks a donkey, so the remaining door will be

P

P

D

ok?

So that is 2/3 that it will be a prize.

Your original choice was 1/3. You add 1/3 plus 2/3 and you get 1, which it must be for any probability.

How's your brain?

Sorry, but it is irrelevent what your original pick was because the door that the host selects always has the donkey behind it and therefore leaves two doors and thus your pick is always a 50/50 bet. It sounds a little convincing in the video, but I don't buy it.

Steve Axford

03-08-2011, 12:06pm

No, the host does not choose from 3 doors. He chooses from 2 - and thus leaves 1. Your pick isn't part of the 2 doors he had. Think!!!

Lance B

03-08-2011, 12:10pm

It's irrelevent. Once it comes down to two doors, you are picking from a 50/50 bet. Remember, you are making a decision after the host has opened the other door, so at that time your decision is 50/50. Your original decision is a 66% chance of being wrong and then it goes to a 50/50 split after the host opens the other door. The video sounds all very convincing, but I am not convinced!

the fact that he chooses from 2 doors only matters in the beginning... after he has removed a donkey/goat from the equation it become irrelevant leaving 2 choices only... oh and my pick might very well have been part of the 2 doors he had. you seem to be assuming that i pick the door with the prize. :p

Steve Axford

03-08-2011, 12:16pm

But but but - the host didn't pick randomly. And he only picked from the 2 remaining doors - so the rules are different for the two groups - your original 3 doors And the 2 doors which the host used. The 3 doors were random, the two were a mix of random and not random. Try it and be amazed.

James T

03-08-2011, 12:16pm

It's irrelevent. Once it comes down to two doors, you are picking from a 50/50 bet. Remember, you are making a decision after the host has opened the other door, so at that time your decision is 50/50. Your original decision is a 66% chance of being wrong and then it goes to a 50/50 split after the host opens the other door. The video sounds all very convincing, but I am not convinced!

Your original decision influences the second stage (as has been said countless times :D) so it's not 50/50.

Steve Axford

03-08-2011, 12:18pm

David, where did I ever say that. The host picks from 2 doors - whichever 2 doors remain. There will always be a donkey behind at least one of those doors, so he picks a donkey.

The more you know about probability, the harder this one is.

fillum

03-08-2011, 12:19pm

I have a bag containing 1 white marble and 99 black marbles. You select a marble from the bag without looking. I then take the bag with the remaining 99 marbles and remove 98 black ones, leaving one marble in the bag. Where is the white marble more likely to be? Is it a 50/50 chance that the white marble is in your hand?

Your original decision influences the second stage (as has been said countless times :D) so it's not 50/50.only in that if my decision had been a door with a donkey the host wouldnt have picked that door but the other donkey door.

Steve Axford

03-08-2011, 12:23pm

If you picked a donkey (2/3), then he would pick the one remaining donkey -- therefore the one door left must be the prize. 2/3

If you picked the prize (1/3), then he could pick either door - therefore the one door left would be a donkey. 1/3

ok?????

I have a bag containing 1 white marble and 99 black marbles. You select a marble from the bag without looking. I then take the bag with the remaining 99 marbles and remove 98 black ones, leaving one marble in the bag. Where is the white marble more likely to be? Is it a 50/50 chance that the white marble is in your hand?i read this 3 times... 100 marbles, 99 black and 1 white... remove 98 black and it leaves you with 1 white and one black. 50/50 if the white marble isnt in your hand then it is in the bag.

Lance B

03-08-2011, 12:26pm

Your original decision influences the second stage (as has been said countless times :D) so it's not 50/50.

Yes, it states this in the video, but I do not agree.

It seems as though the actual outcome from using Monty Hall's techniques doesn't work. If I read this right, 50.3% of people who didn't switch, won:

http://math.ucsd.edu/~crypto/cgi-bin/MontyKnows/monty2?2+9265

If you picked a donkey (2/3), then he would pick the one remaining donkey -- therefore the one door left must be the prize. 2/3

If you picked the prize (1/3), then he could pick either door - therefore the one door left would be a donkey. 1/3

ok?????no...

http://freeemoticonsandsmileys.com/animated%20emoticons/Violent%20Animated%20Emoticons/explode%20emoticon.gif

Steve Axford

03-08-2011, 12:29pm

No. It's 1/100 that the white one is in your hand and 99/100 that it is in the bag. Your explanation assumes that you pick the marble AFTER the 98 have been removed, but the question says before.

http://phumphries.com/forums/images/smilies/savethreadkittens.jpg

No. It's 1/100 that the white one is in your hand and 99/100 that it is in the bag. Your explanation assumes that you pick the marble AFTER the 98 have been removed, but the question says before.yes the thats right... before the 98 were removed my odds were 1/100 white and 99/100 black. after the 98 black are remove my odds change as there is only one black and one white.

Steve Axford

03-08-2011, 12:33pm

Which bit didn't you understand? You have to admit that what I said makes perfect sense.

Steve Axford

03-08-2011, 12:35pm

I think that this is the most wonderful brain teaser - because it makes no difference if you explain it or not. In fact it works even better if explained.

the part how after the 98 black mables are removed from the equation the odds dont change.

doing me head in it is! http://www.smileyvault.com/albums/basic/smileyvault-bash.gif

this sir is a brilliant thread!

this I will agree on! :th3:

Steve Axford

03-08-2011, 12:38pm

Must go. Have fun kittens.

Art Vandelay

03-08-2011, 12:46pm

And in another strange turn of events, we need to keep in mind that in some countries a goat or donkey would be a more preferable prize than a car.

fillum

03-08-2011, 12:47pm

When you select the marble there is a 1% chance that the white marble is in your hand and a 99% chance the white marble is in the bag. By removing the black marbles there is still a 99% chance the white marble is in the bag. If it were now a 50% chance, I would have reduced the chance of the white marble being in the bag simply by removing black marbles, which doesn't make sense to me.

Cheers.

fillum

03-08-2011, 12:50pm

Perhaps we need to put Hitler behind one of the doors and invoke Godwin's Law (http://en.wikipedia.org/wiki/Godwin%27s_law#Corollaries_and_usage) :D

I have a bag containing 1 white marble and 99 black marbles. You select a marble from the bag without looking. I then take the bag with the remaining 99 marbles and remove 98 black ones, leaving one marble in the bag. Where is the white marble more likely to be? Is it a 50/50 chance that the white marble is in your hand?

I more or less tried this at the top of the thread, but nobody bought it.

Imagine a game show.

There are 3 doors, one with a major prize and the other two with a donkey behind.

You are asked to pick a door.

You pick one.

The host then opens another door - always revealing a donkey.

The host then asks you - do you want to change your pick?

Do you?

I like donkeys. So yes.

Milbs1

03-08-2011, 10:46pm

Fabulous thread!

I didn't get it.

Then I got it.

Then I didn't get it again.

Then I read the "marble" scenario and I totally get it now!!!

Thanks Fillum!!!

Speedway

03-08-2011, 11:31pm

Applying over complex thought to a simple mathematical problem is what scientists do to try to confuse people who over think. There are only 2 left in either situation so it becomes 50/50, simple. End of story.

Keith

Steve Axford

04-08-2011, 9:17am

Very clever - and completely wrong.

i wonder how long us mathematical luddites can hang on to our belief that the whole scenario is wrong... damn my logical brain! :p

Steve Axford

04-08-2011, 9:43am

I know the feeling, David. I twisted my brain into a knot before the penny dropped - but it took a day of hard thought. It's all been explained, so all you have to do is wipe what you already know.

Milbs1

04-08-2011, 11:07am

Its great when the penny drops though!

For me it was the marbles...

If I choose a marble from 100 (and there's only 1 white one) and then someone takes away 98 black marbles from the 99 left, logic tells me I'm unlikely to have the white one...hence even though its now a choice of 2 (50/50) I still have a 1/100 chance of having the white one!!

Well I'm now swapping, just need to enter some gameshows!!!

Steve Axford

04-08-2011, 12:12pm

Such a buzz when you do figure it out. I love it. I heard this with no explanation other than it was right to swap (which I believed because it was from a very credible source) and I couldn't stop thinking about it - then it came me - just like opening a door on a new world.

I must be easily amused!!!!!!

davidd

04-08-2011, 12:16pm

In the marble example:

To keep the procedure the same as the donkey problem, the host would have to remove 98 black marbles, so you know the white marble is either in your hand or in the bag. You then get the option of re-chosing. At this point there are only two possibilities. Either the marble is in your hand, or in the bag. Since you can re-choose, I still say it is 50/50 at this point. The previous 98 marbles have been removed from the equation. At the beginning we knew that one in the 100 was white, now we know that one in two is white, and we get to choose one.

In the marble example:

To keep the procedure the same as the donkey problem, the host would have to remove 98 black marbles, so you know the white marble is either in your hand or in the bag. You then get the option of re-chosing. At this point there are only two possibilities. Either the marble is in your hand, or in the bag. Since you can re-choose, I still say it is 50/50 at this point. The previous 98 marbles have been removed from the equation. At the beginning we knew that one in the 100 was white, now we know that one in two is white, and we get to choose one.

Yes, but you know the one in your hand was a 1 in 100 chance to be white. Sure it could be the white one, but chances are its not....

Put it another way.... if you could choose the single marble in your hand, or the other 99 marbles, which would you choose?? Essentially swapping gives you the choice of the other 99 marbles.

Steve Axford

04-08-2011, 1:16pm

The really interesting thing about this problem is why so many of us just can't get it right. Somehow our brains are just wired wrong - for this problem at least.

Somebody start a poll, with this exact same scenario.

Lets see how many people get it right, then justify their reasons why they picked it :D

ameerat42

04-08-2011, 4:55pm

Somebody start a poll, with this exact same scenario.

Lets see how many people get it right, then justify their reasons why they picked it :D

What? You mean like a donkey vote?

OK........ suddenly I get it.... makes you think tho' :o

Speedway

04-08-2011, 11:24pm

It's still 50/50 no matter what you say. Overthinking is the only problem.

Keith.

Steve Axford

05-08-2011, 7:12am

Mmmmmm. I suggest you do this 100 times and see if you get closer to 50/50 or to 33/67. Fancy a small wager?

The way I realised it was like this:

Using the marbles version, it's easier to see the difference.

You choose 1 marble. You have a 1% chance of being right at this point.

By revealing 98 black marbles, and offering you the choice to change, the host is effectively saying " Do you want the one marble you chose originally, or the 99 that were left?"

You then choose between the white marble being the one you originally chose (1% chance) or in the other 99 (99% chance).

Simple really...

Steve Axford

05-08-2011, 3:56pm

Wonderful!!! It's a buzz when you get it, ain't it?

Wonderful!!! It's a buzz when you get it, ain't it?a buzz? hmm... i dont know, it still feels a bit weird. :rolleyes:

Note: any posts made under my name before #107 above, were fraudulent, false, and the product of a confused brain.

I deny authorship of any of them..... :cool:

Imagine a game show.

There are 3 doors, one with a major prize and the other two with a donkey behind.

You are asked to pick a door.

You pick one.

The host then opens another door - always revealing a donkey.

The host then asks you - do you want to change your pick?

Do you?

NO...According to your teaser, there is 'A' donkey, not two...so I'll stick with my pick thanks...there's no more donkeys... :)

Kafter244

09-08-2011, 9:52pm

Often logic and mathematics don't go as closely together as people think! Lol. I live this problem...the other way to look at it is this...

If the host had originally said 'behind these three doors are two donkeys and a car...you can pick one door or you can pick two...' what would you do? You' take two doors obviously, because your chance of one o them bein the car is greater than if you picked one door.

When the host opens the door and gives you the chance to change your mind he's effectively giving you this option as to choose two doors would (probably) give you one donkey and one car, so the likelihood is that the car is in the other door...hence takin the swap.

Love this problem, and the thread...especially the kittens!

:th3:

Kafter244

09-08-2011, 9:53pm

...not a big fan of all those typo's though! Lol. Sorry folks...goddam

iPhones ;)

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