Brain Teaser No 1

Not sure I follow that one!

What confuses me generally, is that if the probability of the 500 being in EITHER box 1 or box 3 before the first choice is made is 1/3 + 1/3 = 2/3, then the probability of 500 being in box 2 at that time = 1/3.

If the probability of it being in 3 is set forever because of the conditions when the choice was made, why after box 1 is shown to be empty do the probabilities between 2 and 3 effectively reverse?

My wife thinks that it is a brand new question with two choices so the probablity must be 50/50, so as she is probably wrong.......;)

........yes you've convinced me. I think!!

Chris

Correct.

"If the probability of it being in 3 is set forever because of the conditions when the choice was made, why after box 1 is shown to be empty do the probabilities between 2 and 3 effectively reverse?"

They do not reverse. 3 stays the same and 2 doubles.

"My wife thinks that it is a brand new question with two choices so the probablity must be 50/50,"

No. Just because there are two doors and one prize does not imply the probability of the prize being behind a particular door is 50%. The chance of a newcomer (who hasn't seen the first part of the game) choosing right is 50%. But your chances are 67% because you are aware of the bias towards door 2.

To reduce its size, I had 'deleted' 998 lines of simulation. To recreate the original you need to highlight and 'drag' line 5 down to line 1003, where you will see the 'totals' at line 1005.

You need to do this for both sheet 1 and sheet 2.

Unfortunately, its still about 62k in size, which I think is too big to be allowed AND I HAVEN'T GOT A CLUE how to add an attachment on this Forum.

(This is a REAL brain teaser!)

If you would like a copy, e-mail me and I will attach one to my reply. Or could someone post 'step by step' instructions for adding attacments to posts? and say whether you can attach spreatdshets?

Cheers

Don

If you don't change, you'll win the 500 if your original choice was correct, which is a 1 in 3 shot.

I think that sums it up rather elegantly.... and I myself still tend towards Mr S needing to know..

Cheers

Don

After Paul has opened a door and it is empty his prior knowledge of the location of the prize is irrelevant. This is the premise of the story and so you needn't worry about whether he knew or not. This does not affect the bias of door 2.

As an aside, let's work out the overall odds of winning if Paul doesn't know (this is a deviation from the posted story):

Again, let's say you pick door 3.

p(winning) = 1 - p(losing)

p(losing) =
p(door 1 has prize) + p(door 1 empty)xp(door 3 has prize)
= 1/3 + 2/3 x 1/3 = 5/9
p(winning) = 1-5/9 = 4/9.

There will be a 4/9 chance of winning the prize if you switch. If you don't switch the chance is 2/9.

[This message was edited by bam on SUNDAY 25 November 2001 at 01:51.]

The stories premise is that Paul picks an empty door. Paul's pick is a certainty. How Paul achieved this (whether by prior knowledge or luck) is irrelevant to your odds of winning in either the switch or stick cases.

When Paul picks an empty door with certainty:

p(win | switch) = p(your pick is empty) = 2/3

p(win | stick) = 1/3

However, and this is Omer's point, the odds of winning are altered if Paul's pick is uncertain. However, this is not the premise of the puzzle.

If Paul's pick is uncertain:

p(win | switch) = p(your pick is empty) x p(Paul's pick is empty)

= 2/3 x 1/2 = 1/3

p(win | stick) = 1/3 as before

In any case, the simple rule of always switching assures you of maximizing your chances. So the answer to the puzzle is independent of Paul's prior knowledge.

BAM

Please do, I've vacillated too much on this one already.

Chris

Cheers

Don

How about

All Pauls play the game, and when the hats have been distributed they each look at the hats that the others are wearing.

Each Paul who sees two hats of the same colour votes that he is wearing a hat of the opposite colour. If they see two different colours they pass.

If all hats turn out to be the same colour they would lose, but if not they should win??

Chris

Nice puzzle. There are 8 hat permutations. It appears that the only way the team can lose with their strategy is if all their hats are the same colour. The probability of this is 2/8 so their chance of winning is 6/8. Very good.

Baron Von Stephenson kept his gold treasure (fairly got by manufacturing hi quality electronics) in a treasure house. In each room of the treasure house were as many chests as there were rooms. In each chest there were as many gold coins, as there were chests in that room. All the gold coins were of the same size and value.

When he died (ok, this is a futuristic story set in the year 2222), the Baron's will was that his favorite Forum member (Guesses on a postcard please!!) should receive one chest of gold coins. The remaining coins were to be divided equally between the Baron's three forum administrators, Paul Des, Doug G and Paul Dar .

The three administrators were proud and fierce men who would definitely resort to bloodshed if the coins could not be divided equally.

The question is simply: -

a) Was there blood shed.
b) Was there no blood shed.
c) Is there no way of knowing whether there was blood shed or not.

PS. The proof is the real requirement, not simply a one in three guess.

Cheers

Don

n^3 - n = n(n*n -1).

We now have to check this expression is divided by 3 so the 3 primadonas do not fight.

n^3-n = n(n*n-1) = n(n-1)(n+1) = (n-1)n(n+1)

The last term gives three factors of the number of coins to be divided.

It also represent three consecutive whole numbers.

Magic property of two consecutive numbers ? one must even and divisible by two

Magic property of three consecutive numbers ? one must be divisible by three

if one of the factors is divisible by three, then the original whole number must be divisible by three ?

Cheers

Don

PS 'scales' = those old fashioned ones that look a bit like the Scales of Justice !

Cheers

Don

Sort the balls into 3 sets of 4, label A, B, C

.... ....

Weight A vs B and B vs C. By noting which balance and which don't you can identify the set that contains the odd ball and whether the odd ball is heavy or light.

Now weigh the four from the set with the odd ball.

....

You now know which two contain the odd ball. Now weigh one against the other.

....

BAM

PS doesn't know
50:50

PS does know and is feeling tight
stick with your first choice

PS does know and is feeling generous
how far does generousity go, maybe switch, maybe not

PS does know and is trying to bankrupt naim
switch

Balls, Bam, how do you know which side of the balance the non-stand ball is on?

I suggest 6.

Matthew

Balls to you too!

A>B and B=C => A contains a heavy ball
A<B and B=C => A contains a light ball
A=B and B<C => C contains a heavy ball
A=B and B>C => C contains a light ball
A<B and B>C => B contains a heavy ball
A>B and B<C => B contains a light ball

A=B and B=C impossible
A<B and B<C impossible
A>B and B>C impossible

Knowing whether the odd ball is too heavy or too light you can always tell which side of the scale it is on.

BAM

[This message was edited by bam on MONDAY 26 November 2001 at 14:42.]

Yes 4 is correct. Doh!

If PS doesn't know then it is 50:50.

By PS opening 1 he affects the whole system so the probablities of both 2 and 3 change, not just 2. The probability of PS finding the 500 is 1/3, we are know moving into 2/3 probablity world and in a 2/3 probablity world 1/3 is actually 1/2.

If you ran a test on this you wuld find that the 2/3 of the time that PS didn't find the NAP the NAP would be split 50:50 between 2 and 3. So your only guide to whether to change or not is if PS does know.

Now if PS does know and doesn't want you to get the 500 he would have opened the door with it in. Therefore it can not be 2 and must be 3. If he was trying to bankrupt naim he would pick the empty door (if you had chosen incorrectly) or not at all (if you had chosen correctly), unless he was constrained by some evil gameshow rules, therefore you would pick to change, unless of course you didn't like the design of the new naim gear and felt like it would clash with your CDSII/52 setup or you couldn't afford the extra long legs for your fraim.

cheers

Matthew

However, out of curiosity, I did some surfing and found many sites with the solution. I won't spill the beans here and spoil others enjoyment, but one site cited (!) a solution based on "the finite projective plain of order 3".

So the next riddle is ... what the hell does that mean?!?!

But not with relative density guaranteed.

Matthew

There is a way but it's very elegant!

Cheers

Don

Whilst you are trying to figure out how to do it in three (and I think it's very decent of Duncan F and others to sit tight whilst yet others try to figure it out) here's a much more straightforward one.

Naim have two delivery vans that shuttle between Salisbury and their favourite dealer. (Ok, ok, in the original version it was 'two ships operate a ferry service across a river….')

Naim have two delivery vans that shuttle between Salisbury and their favourite dealer. On a particular day, one van leaves Salisbury bound for the dealer, whilst at the same time the other van leaves the dealer, bound for Salisbury. Both vans follow the same route and pass each other (giving the famous 'Naim Salute') 30 miles from the dealer's. At their initial destinations each van takes 15 minutes to load/unload. On the return journey, following the same route, the vans pass each other (giving the famous 'Naim Salute') 15 miles from Salisbury.

How far is it from Salisbury to the dealer's ?

Assume that each van travels at a constant speed and that acceleration/deceleration is allowed for in the 15 minute turn round time.

Cheers

Don

There is a way but it is very ugly....
There is a way but it's very elegant!

Ever wondered how Fermat felt ??

Cheers

Don