Brain Teasers at Xmas

Posted by: Don Atkinson on 23 December 2010

Brain Teasers at Xmas

Xmas time - and many of us will be off on holiday or meeting up with friends and family. For those with a bit of time on their hands, I thought I would post a dozen (12 days of Xmas ?) brain teasers to help keep your minds in gear.

All are taken from the Brain Teaser thread from a few years back and credit is given to the original poster where this is other than me. Ken c still posts here, but BAM and Matthew T….??

None of these teasers are “trick” questions, but you will need to read them carefully in order to understand some of them. Discuss at leisure......

I know it’s not as exciting as today’s hi-tech electronic games, but I hope it helps to keep some of you amused over Xmas, and as Ken c would say – “enjoy”

Cheers

Don
Posted on: 01 January 2011 by winkyincanada
quote:
Originally posted by fatcat:
The Monty Hall wilkipedia information seems convincing in places but not so in others.

Can anybody explain this

Contestant chooses door 3. The chance the car is behind door 3 or door 2 is 2 in 3. The chance it is behind door 1 is 1 in 3. The host opens door 2 to reveal a goat. Therefore the chance the car is behind door 3 is 2 in 3. This indicates he should not swap.


Interesting. But I think your logic misses the point that the host couldn't have chosen to open door 3, as that has already been chosen by the contestant. The host is knowledgeably selecting between doors 1 and 2.

The chance that the car is behind door 3 is always 1 in 3 and that doesn't change. What the host does when he opens door 2 is collapse the probability that it is behind doors 1 or 2 (2 in 3) to just door 1. The probability that the car is behind door 1 is now 2 in 3. Don't forget, the host knows the car isn't behind 2 (he doesn't choose door 2 at random), and when he shares this information with the contestant, it should be used for future decisions.
Posted on: 01 January 2011 by fatcat
quote:
Originally posted by winkyincanada:

Interesting. But I think your logic misses the point that the host couldn't have chosen to open door 3, as that has already been chosen by the contestant. The host is knowledgeably selecting between doors 1 and 2.


In your logic the host can not open door 3.
Posted on: 01 January 2011 by winkyincanada
quote:
Originally posted by fatcat:
quote:
Originally posted by winkyincanada:

Interesting. But I think your logic misses the point that the host couldn't have chosen to open door 3, as that has already been chosen by the contestant. The host is knowledgeably selecting between doors 1 and 2.


In your logic the host can not open door 3.


That's right. He can't open the door that the contestant has chosen. Don't forget in this puzzle, the host always opens an empty door AFTER the contestant has chosen their door. And the host always offers the contestant a chance to swap.
Posted on: 01 January 2011 by fatcat
quote:
Originally posted by winkyincanada:
quote:
Originally posted by fatcat:
quote:
Originally posted by winkyincanada:

Interesting. But I think your logic misses the point that the host couldn't have chosen to open door 3, as that has already been chosen by the contestant. The host is knowledgeably selecting between doors 1 and 2.


In your logic the host can not open door 3.


That's right. He can't open the door that the contestant has chosen. Don't forget in this puzzle, the host always opens an empty door AFTER the contestant has chosen their door. And the host always offers the contestant a chance to swap.


Contestant chooses door 3. The chance the car is behind door 3 or door 2 is 2 in 3. The chance it is behind door 1 is 1 in 3. The host opens door 2 to reveal a goat. Therefore the chance the car is behind door 3 is 2 in 3. This indicates he should not swap.

In the above, the host chooses door 2 after the contestant chooses door 3. He has the option to swap. But the odds show he should not swap.
Posted on: 01 January 2011 by Martin_C
quote:
..great to see the debate still bouncing back and forth ....Don you've got a lot to answer for!!

Fatcat: I fear your logic is flawed because, having claimed the odds of 2 in 3 for it being behind either door 2 and 3 you then only consider the scenario of the prize not being behind door 2....this is a statistical no no..... You must also include the scenario of the prize being behind door 2 (in which case the host will open door 1) and therefore the odds of it being behind door three is zero for that specific case...adding the two cases together gives you average odds of 1 in 3 as argued by better minds than mine! (though this is a convoluted way of working things out!)

I'm afraid I know what I mean but probably didn't describe it very well. As always with statistics it doesn't mean you can't win by sticking just that, on average, you are better off changing.

The other way at this is to do a simple event tree. Say I always pick door 1 and always stick

Case 1: prize is behind door 1. I win. (It doesn't matter which door the host opens because they both have a goat)

Case 2: prize is behind door 2. I loose. (The host opens door 3)

Case 3: prize is behind door 3. I loose. (The host opens door 2)

There are no other options and each of the cases are equally likely so on average I win 1 time in 3.

Now instead I always swap, having chosen door 1 initially.

Case 1. Prize is behind door 1. I loose. (I swapped to whichever door the host didnt open)

Case 2. Prize is behind door 2. I win. (Host opened door 3 and I swapped to door 2)

Case 3. Prize is behind door 3. I win. (Host opened door 2 and I swapped to door 3)

I won 2 times in 3 using the swapping strategy, therefore this is definitely preferable.

...isn't this fun and Happy New Year
Posted on: 01 January 2011 by Don Atkinson
quote:
Can anybody explain this

Contestant chooses door 3. The chance the car is behind door 3 or door 2 is 2 in 3. The chance it is behind door 1 is 1 in 3. The host opens door 2 to reveal a goat.Therefore the chance the car is behind door 3 is 2 in 3. This indicates he should not swap.


I have highlighted the false statement in bold.

The host actually opens EITHER door 1 OR door 2.

You are correct in stateing the probability of the car being behind either Door 2 or Door 3 as 0.666'
The only way to benefit from this probability is to FORCE doors 2 and 3 to be opened.
If the gameshow host can open either Door 1 or Door 2, you can't be certain he will open Door 2
If he were to open Door 1.........

Fatcat, you are asking very reasonable questions. However, explaining why your various hypotheses are wrong, is very difficult. It's much easier to re-iterate why "changing your mind" is the best option. Have you tried Winky's suggestion of using cards? or my suggestion of writing down (say) 30 random sets of Doors and 30 random choices of initial door?

Cheers

Don
Posted on: 01 January 2011 by Derry
My mate Mike does not understand this, nor do I really...
Posted on: 01 January 2011 by Derry
Nor does Teddy Novaks...
Posted on: 02 January 2011 by Don Atkinson
quote:
My mate Mike does not understand this, nor do I really...


Welcome to the Club Derry.....I don't think any of us understands it really!!

Cheers

Don
Posted on: 02 January 2011 by Don Atkinson
quote:
As you say, he could start anywhere around the line of latitude approximately (1 + 1/6) miles north of the South Pole.

There are a few other places as well, just as obvious once you've "seen" them........


Winky, or anybody else.....

Any further thoughts?

Cheers

Don
Posted on: 02 January 2011 by fatcat
quote:
Originally posted by Don Atkinson:

Fatcat, you are asking very reasonable questions. However, explaining why your various hypotheses are wrong, is very difficult. It's much easier to re-iterate why "changing your mind" is the best option. Have you tried Winky's suggestion of using cards? or my suggestion of writing down (say) 30 random sets of Doors and 30 random choices of initial door?

Cheers

Don


I've just performed a few random tests. Out of seven tests, it would have been benifitial to swap is six. Red Face
Posted on: 02 January 2011 by fatcat
quote:
Originally posted by Don Atkinson:
quote:
As you say, he could start anywhere around the line of latitude approximately (1 + 1/6) miles north of the South Pole.

There are a few other places as well, just as obvious once you've "seen" them........


Winky, or anybody else.....

Any further thoughts?

Cheers

Don


1 + 1/12 and 1 + 1/24 miles north of south pole.
Posted on: 02 January 2011 by winkyincanada
quote:
Originally posted by fatcat:
quote:
Originally posted by Don Atkinson:
quote:
As you say, he could start anywhere around the line of latitude approximately (1 + 1/6) miles north of the South Pole.

There are a few other places as well, just as obvious once you've "seen" them........


Winky, or anybody else.....

Any further thoughts?

Cheers

Don


1 + 1/12 and 1 + 1/24 miles north of south pole.


Yes, anywhere that you end up on a latitude that lets you lap the pole an exact number of times so you return back up your original path. Clever.
Posted on: 02 January 2011 by Don Atkinson
quote:
I've just performed a few random tests. Out of seven tests, it would have been benifitial to swap is six.


Convincing hey!

All you need now, is to invent the maths to predict it!!!!

happy new year

Don
Posted on: 02 January 2011 by Don Atkinson
quote:
anywhere that you end up on a latitude that lets you lap the pole an exact number of times so you return back up your original path. Clever.


Brilliant, neat eh!

Cheers

Don
Posted on: 02 January 2011 by Don Atkinson
So, five down, only seven to go!!

............. and the 12 days of Xmas is far from over!

Cheers

Don
Posted on: 02 January 2011 by Don Atkinson
quote:
1 + 1/12 and 1 + 1/24 miles north of south pole.


Brilliant, fatcat. Almost unlimited number of places other than the North Pole!!

Cheers

Don
Posted on: 02 January 2011 by Don Atkinson
quote:
#5 Odd Ball?


For the avoidance of doubt....the balls can be put on either side of a balance-scale eg 3 x balls on the left hand side and 3 x balls on the right hand side. If the scales balance, then we know the "odd" ball isn't one of the six on the scales.

This would also represent "one" weighing.

Hope this makes Teaser No 5 clearer.

Cheers

Don
Posted on: 02 January 2011 by fatcat
Don

In Q8. (Ladder)

Presumably there are 2 distances the ladder can be placed from the edge of the crate.
Posted on: 02 January 2011 by Don Atkinson
quote:
Presumably there are 2 distances the ladder can be placed from the edge of the crate.


Correct! the ladder could be (sort of) horizontal or, more practically, sort of vertical - well, (1 + "d")m off vertical, where "d" is the unknown.

Cheers

Don
Posted on: 04 January 2011 by Fred Mulder
quote:
Originally posted by Don Atkinson:
IF the shipper..
He could take and weigh (1 DAC from box 1) + (2 DACs from box 2) + (3 DACs from box 3).....+(n DACs from box n)


Et voila, that's it!

It was a riddle once used in an episode of Columbo. The goal is How to determine which are the fakes, given the limited information (not to solve it by calling an exact weight).

Using made-up weights your approach will solve it: ie real= 10 KG, fake= 9.9KG
1x10
2x10
3x9.9 (minus 0.1)
4x10
totals 99.7 = container 3
(0.3 difference devided with 0.1)

When I can remember one which is harder to solve, I will post it! let me think... something about an egg.. a chicken.. but what was the question.. ;-)

Cheers, Fred
Posted on: 08 January 2011 by Don Atkinson
fatcat,

any progress with the ladder?

cheers

Don
Posted on: 14 January 2011 by Don Atkinson
Just to help out in case anybody is stuck.....

The answer to #5 (odd ball) is 3.

Now you just need the method of weighing.

Cheers

Don
Posted on: 16 January 2011 by Fred Mulder
quote:
Originally posted by Don Atkinson:
The answer to #5 (odd ball) is 3.
Now you just need the method of weighing.


You didn't like the answer on page one?
the 'a,b,c,x' are ball types, with 'x' being the odd one

Now please don't tell me it's wrong, I was happy with the result

Fred
Posted on: 16 January 2011 by Ian G.
The answer to Q7 is 36 pi (I think).

Ian