Brain Teaser No 1

I still prefer the glacier solution.

OK, providing you have a couple of years to spare!! When I was in Alaska last year, the glaciers were moving at 8 inches per day!!

On the other hand, 'Scott's' tours of the Antarctic do seem to take for ever!!

Cheers,

Don

Fire away, and we'll give it our best shot.

I've got a couple more, but they can wait.

Cheers

Don

[This message was edited by Don Atkinson on TUESDAY 20 November 2001 at 22:35.]

Late one cold, rainy night in November after a few pints and a rather dubious curry you flop down onto your favourite listening couch and turn on your Naim system. Ah...that's better. But even the PRaT isn't enough to keep your eyes from closing and you soon fall into a deep, deep slumber and start to dream...

Congratulations! You have been selected as a participant in a very popular national TV gameshow. This is your first time on TV and on such a show and you are overwhelmed with excitement and nervousness. The gameshow host is none other than Paul Stephenson. Dressed in a gold lame suit with sequens, standing centre-stage on a Fraim. The grand prize: a Naim NAP500 and installaton in your home by Roy George himself. Your heart pounds with excitement.

You find the first rounds easy and before long the other contestants have been eliminated and you have made it to the final challenge. The audience go wild! You try hard to look at them through the bright spot lights and recognize them as your chums from the Naim Forum - all clapping and cheering you on. What a night!

Now the final challenge begins. The lights dim and the cheers of the crowd die down. Paul shows you three doors numbered 1, 2 and 3. He says that the NAP500 is hidden behind one of the doors. You must choose which one. If you choose correctly you win the NAP500. If not you win Paul's hideous suit. The tension mounts. The audience can't contain themselves: shouts begin..."door 1"..."no door 2"...etc. You hesitate. Paul urges you to make a choice. Finally you choose door number 3 (after your existing Nait 3 system). A green light comes on over door 3. Oh no...did I choose wisely...

A hush comes over the theatre as Paul walks over to door number 1 and motions to open it. The atmosphere can be cut with a knife. Slowly, Paul opens door 1 to reveal nothing! Phew that was lucky. The crowd bursts into a relieved cheer. But soon Paul has gripped the handle of door 2 and the crowd go silent. You can hear a pin drop. The tension is unbearable...

Suddenly Paul stops turning the handle of door 2 and turns towards you. He says he is in a good mood tonight and wants to help you as much as he can. He offers to do you a favour and allow you to change your selection. You can now stick with your original choice...door 3 or switch to door 2. It is up to you and you have just 10 seconds to decide. The audience erupts! "door 2 go for door 2"..."no, no stick with door 3"...

You have 5 seconds left. You are sweating. Everyone is hanging on your next words. Do you stick with door 3 or switch to door 2?

Presumably the NAP500 etc is worth more than the gold lame suit?

After carefully pondering your puzzle, I think the answer depends on whether Mr Paul KNOWS which door hides the NAP500.

If Mr Paul DOESN'T know which door hides the 500 (which is the way I have read the puzzle), then AFTER Door 1 has been opened and revealed nothing, then each of Doors 2 and 3 have an equal probability of hiding the 500 ie each probability is now 1/2. It therfore doesn't matter which you chose.

If, however Mr Paul DOES know where the 500 is, (and, given this a regular GAME SHOW where the RULES and SEQUENCE of events are well know in advance this seems likely, and assuming that in order to build up the THRILL Mr Paul (or the usual host) always opens an EMPTY Door), then the choice is simple, go for Door 2 (ie the unopened door that you didn't initially select). The probability of this being the right door is */*, whereas the probability of the your first choice being right (Door 3) is only */*.

I will hold back on quoting probabilities or explaining my reasoning until we have sorted out Mr S's state of mind!

Cheers

Don

I e-mailed you the probabilities missing above, but my e-mail connection seemed to 'glitch'.

Did you recieve them ?

Cheers

Don

Can't summon up my wits right now to solve the puzzle, but if you don't mind, I think I have to use your post in my next translation class. That would be great fun, I think.

Thomas

Don,
Got your email, have replied. You are half right.
Re the price of the suit: you don't care because of the ridicule you'll receive on the forum the next day

Regarding Mr P's state of mind...who among us would claim to be able to fathom such a thing?
Has anyone else got an opinion about whether Mr P's state of mind affects your choice of doors?

quote:

---

Is my use of English so bad it needs translating?

---

Oh no, not at all. The scenario you've created just lends itself perfectly to vocab expansion and translation practice, particularly with collocations such as "the tension mounts", and it's a good read at that. I teach general translation from English into German at the local language occupations college (80% of the students are young ladies - it's a tough job).

Thomas

The state of Mr S's mind is completely unimportant.

Change to Door 2 and your chances are 2/3

Stick with Door 3 and your chances are 1/3

I'm still willing to take a chance on the gold lame cloak if its worth more than the 500

BTW I will post my 'logic' for the above chances later.

Cheers

Don

quote:

---

This reasoning doesn't apply to our current choice however, as the elimination of Door 1 means it's become a straight 50-50 chance between door 2 and door 3.

---

Nope. The elimination of door one means that 100% of the 2/3rds probability of doors 1 or 2 holding the 500 now rests with door 2.

Before it was 1/3rd each door. Alternatively stated: 1/3 my door, 2/3rds "not my door". It's still 2/3rds "not my door" - the 500 hasn't moved - but having eliminated door 1, that 2/3rds now rests farily and squarely with door 2. So go for it.

quote:

---

Before it was 1/3rd each door. Alternatively stated: 1/3 my door, 2/3rds "not my door". It's still 2/3rds "not my door" - the 500 hasn't moved - but having eliminated door 1, that 2/3rds now rests fairly and squarely with door 2. So go for it.

---

I disagree...the probablities are *not* conditional, although the way the problem is presented makes it appear as though they are. You have free option to pick a door in both samples; so each sampling is a single event in which the outcome is not dependant on previous trial. In which case the second trial is not a case of '2/3rds "not my door" ' since that door has been removed from the trial.

First sample: pick a door, any door: probablilty that the 500 lies behind is 1/3 for each door.

Second sample: pick a door, any door: probability is that the 500 lies behind one of two doors = 1/2 for each door.

Although you *know* the 500 hasn't moved, that alone tells you no more about where it is. Your chance of winning has increased though, it's gone up from 1 in 3 to 1 in 2 ...

Cheers,
MC

It took me a while before I accepted the (acknowledged) correct answer, which is that you SHOULD change your mind and swap.

Bearing in mind that the presenter knows which door is the right one, imagine that there are 1 million doors to start with. You pick, say doorA ; your chances are (literally) 1 in a million.
The presenter now opens 999,998 doors leaving only two remaining - your original choice DoorA, and one other. The chance that you were originally correct is still only 1 in a million. Therefore you should definitely swap.
The more doors there are to start with, the more you should swap. Most people's intuition is the opposite - ie the more doors which become "eliminated", the more they believe their first choice was right!

When I saw this thread on puzzles, I was going to post this very same "doors" puzzle, but someone beat me to it.
It is probably my favourite puzzle, simply because it causes so much debate (nearly as much as Mana v Quadraspire!).

If you search on the web using "three,doors,puzzle" you will see that there are a number of explanations of it.

Steve D

[This message was edited by steved on FRIDAY 23 November 2001 at 17:11.]

So far I have created two sheets, one where the contestant 'sticks' with his original choice, the other where he always changes his mind. In both cases, Mr S always opens an empty box. (sometimes Mr S has a choice about this box, sometimes he doesn't). In other words, Mr S KNOWS where the prize is.

Running this simulation 1,000 times always gives a probability close to .333 if you don't change your mind and .667 if you always do change your mind. (Excel hides the boxes, picks the first choice, does Mr S's thinking for him - BRILLIANT - etc, etc)

The results of 10 runs of each spreadsheet are given below

.336; .322; .342; .355; .346; .344; .330; .360; .332; .363; Avg = .343
.682; .672; .704; .614; .648; .684; .669; .679; .685; .664; Avg= .670

I intend to extend the spreadsheet to investigate the results assuming Mr S DOESN'T know where the prize is, ie he opens box 1 or 2 at random. This means that from time to time he will reveal the prize - much to the contestant's dismay! What I'm not sure about, is whether the contestant is still better off changing his mind (when he gets this chance). At present, I am inclined towards...................???

Cheers

Don

At the start of the show, all boxes have equal probability of containing the 500.
The probability of the 500 being in any particular box is 1/3
Probability of 500 being in EITHER box 1 or box 2 is 1/3 + 1/3 = 2/3
Probability of 500 being in box 3 = 1/3
(At this point the probability of 500 being in box 1 CAN be expressed as 0.5x2/3 and likewise the probability of 500 being in box 2 can ALSO be expressed as 0.5x2/3)

Revealing that box 1 is empty changes the probability of box 1 from 0.5x2/3 to 0x2/3 which = 0
and also changes the probability of box 2 from 0.5x2/3 to 1x2/3 which = 2/3

Hence it's best to 'change your mind'

Cheers

Don

Brain teaser no.3: It's 5.29 Friday afternoon; do I worry about this some more or go to the pub...?

Cheers,
MC

ME ? i'm on three nights of flying in a row....don't want to be thinking about probabilities on the approaches... just certainties

Cheers

Don

quote:

---

Probability of 500 being in EITHER box 1 or box 2 is 1/3 + 1/3 = 2/3
Probability of 500 being in box 3 = 1/3

---

So the probability of it being in EITHER box 1 or box 3 is ???

Chris

For the record, Don and Duncan are correct and Steve D has given the explanation that I was going to give. I'm impressed that Don has actually proved it analytically using Excell! Thorough or what?

But let me add some more reasoning to help with the 50-50 argument. If there were two doors to start with and the NAP500 was placed behind one of them at random then the chances of it being behind a particular door would be 50%. But because of the method used to arrive at 2 doors the 500 has not, in effect, been placed randomly. This is the real crux of the matter.

Your original choice was made when there were 3 doors so your chance of being right was 33%. This chance CANNOT be changed by any actions affecting any other doors (apart from revealing the 500 behind another door in which case the probability becomes zero). It is set forever because of the conditions WHEN THE CHOICE WAS MADE. It is as if Mr S., knowing which door you have chosen, has biased the game by placing the 500 behind the remaining door twice as often as behind your door. The way Mr S eliminates a door effectively biases the two-door situation against your original choice. The placement of the 500 between the two remaining doors IS NOT RANDOM.

In other words, just because there end up being two doors and one prize does not mean that there is an equal chance of the prize being behind either door. It depends on the bias used in the placement of the prize. It's like rolling a loaded dice.

Note that these two statements are BOTH true at the time that two doors remain:
a) The chance of finding the prize by randomly selecting one of the two remaining doors is 50% (this is the reasoning used by Martin, Declan and Paul).
b) The chance of the prize being behind door 2 is 67% and door 3 is 33% (as reasoned by Don and Duncan).

Most confusion about this puzzle arises when a person believes these statements conflict with one another. But they don't!

Here's why:
In a) the chance of RANDOMLY selecting the prize door equals the chance of choosing door 2 times the probability that the prize is behind it + the chance of choosing door 3 x the probability of the prize being behind it, or:

( 0.5 x 0.667 ) + ( 0.5 x 0.333 ) = 0.5

But because you, the contestant, have witness the biasing process you have extra information and you needn't make a random selection. So your odds of choosing the right door have increased to 0.667.

BAM
(takes another shot of novocaine in the frontal lobe to ease the pain)

[This message was edited by bam on SATURDAY 24 November 2001 at 01:03.]

Once again,
The fact that the chance of finding the prize by selecting one of the two remaining doors at random is 50% DOES NOT IMPLY that the chance of the prize being behind a particular door is 50%.

Chris

Someone who had not seen the first part of the game and simply came across two doors would only have a 50% chance of winning the prize.

Door 3 is reserved by the contestant

Mr S is not allowed to open door 3 and must choose between doors 1 and 2. Their combined probability is 67%.

Case A: If Mr S opens door 1 and it is empty
door: 1 2 3
prob: 0% 67% 33%

Case B: If Mr S opens door 1 and finds the prize
door: 1 2 3
prob: 100% 0% 0%

The odds are not affected by whether Mr S knows door 1 to be empty in advance or not. The story states it was empty and that is all that matters. In a real gameshow Mr S. would probably know in advance and choose Case A to increase the tension and keep the ratings up.

[This message was edited by bam on SATURDAY 24 November 2001 at 12:13.]

A few posts back you ask So the probability of it being in EITHER box 1 or box 3 is ???

At the start of the show the answer is .667 and you would 'chose' Box 2, knowing full well that when Mr S opens Box 1 or 3 (YOU have now forced him to do this) you will then open the other box (ie 3 or 1, if you see what I mean?)

In other words, IF (if, if, if)the NAP500 IS in box 1 or 3 YOU are DEFINITLEY (absolutely, no doubt, 100%)going to get it!!!!(assuming Mr S knows where it is AND opens the empty box)

Well, at least I think there is a slight possibility that this is the case..........what a brain teaser!!!

Cheers

Don

If you switch doors your chance of winning the prize is 100% if you first select an empty door and 0% if you first select the prize door.

p(winning given that you switched) =
p(selecting an empty door at start) x 100%
+
p(selecting prize door at start) x 0%
=
67% x 100% + 33% x 0% = 67%

If you don't switch doors, the chance of winning is 100% if you select the prize at first and 0% if you didn't

p(winning given that you stuck) =
p(selecting prize door at start) x 100%
+
p(selecting empty door at start) x 0%
=
33% x 100% + 67% x 0% = 33%

Therefore you have 67% chance of winning if you always switch and 33% if you always stick.

[This message was edited by bam on SATURDAY 24 November 2001 at 12:34.]