Brain Teasers at Xmas

quote:

Originally posted by fatcat:
Winky

The odds on the contestant winning are always 1 in 2. It makes no difference if he changes or not.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2

Don't agree. State your logic.

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:
Winky

The odds on the contestant winning are always 1 in 2. It makes no difference if he changes or not.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2

Don't agree. State your logic.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2 seems logical enough.

No matter which door the contestant chooses, the host will ALWAYS eliminates one of the empty doors. It always comes down to a 1 in 2 chance. The 1 in 3 chance doesn’t exist.

quote:

Originally posted by fatcat:

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:
Winky

The odds on the contestant winning are always 1 in 2. It makes no difference if he changes or not.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2

Don't agree. State your logic.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2 seems logical enough.

No matter which door the contestant chooses, the host will ALWAYS eliminates one of the empty doors. It always comes down to a 1 in 2 chance. The 1 in 3 chance doesn’t exist.

Can you agree that a strategy that has the contestant NEVER changing doors will result in a 1 in 3 chance of them winning? The contestant makes his choice of 1 winning door out of 3 then sticks with it.

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:
Winky

The odds on the contestant winning are always 1 in 2. It makes no difference if he changes or not.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2

Don't agree. State your logic.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2 seems logical enough.

No matter which door the contestant chooses, the host will ALWAYS eliminates one of the empty doors. It always comes down to a 1 in 2 chance. The 1 in 3 chance doesn’t exist.

Can you agree that a strategy that has the contestant NEVER changing doors will result in a 1 in 3 chance of them winning? The contestant makes his choice of 1 winning door out of 3 then sticks with it.

No

The contestant has a 1 in 3 chance of CHOOSING the correct door. When the host intentionally eliminates an empty door, the contestant is left with a simple 1 in 2 chance of selecting the winning door. It doesn’t matter which of the 3 doors he originally chooses, the contestant will always be left with a 1 in 2 chance of winning.

There will only ever be a 1 in 3 chance of winning if the host doesn’t actually know which is the winning door.

quote:

Originally posted by fatcat:

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:
Winky

The odds on the contestant winning are always 1 in 2. It makes no difference if he changes or not.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2

Don't agree. State your logic.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2 seems logical enough.

No matter which door the contestant chooses, the host will ALWAYS eliminates one of the empty doors. It always comes down to a 1 in 2 chance. The 1 in 3 chance doesn’t exist.

Can you agree that a strategy that has the contestant NEVER changing doors will result in a 1 in 3 chance of them winning? The contestant makes his choice of 1 winning door out of 3 then sticks with it.

No

The contestant has a 1 in 3 chance of CHOOSING the correct door. When the host intentionally eliminates an empty door, the contestant is left with a simple 1 in 2 chance of selecting the winning door. It doesn’t matter which of the 3 doors he originally chooses, the contestant will always be left with a 1 in 2 chance of winning.

There will only ever be a 1 in 3 chance of winning if the host doesn’t actually know which is the winning door.

No, you misunderstand. I am only talking about 1 possible strategy at this stage. This particular contestant is stubborn and won't change, regardless of what the host does. He just chooses from the three, but NEVER changes in spite of being given the opportunity. How can this not give him a 1 in 3 chance of winning? The host only opens doors after the contestant's choice of 1 out of 3 doors. This cannot change anything as the contestant's choice is fixed prior to this event. If the contestant has been lucky enough to choose the prize door in the first place, he wins, if not, he loses. This strategy (of never changing) must surely have a 1 in 3 chance of success.

If I can't explain this first piece of simple logic to you, then I will stop, as the next step (the alternate, opposite and better strategy) is a little more complex and you will never get it on the basis of my obviously ineffective explanation.

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:
Winky

The odds on the contestant winning are always 1 in 2. It makes no difference if he changes or not.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2

Don't agree. State your logic.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2 seems logical enough.

No matter which door the contestant chooses, the host will ALWAYS eliminates one of the empty doors. It always comes down to a 1 in 2 chance. The 1 in 3 chance doesn’t exist.

Can you agree that a strategy that has the contestant NEVER changing doors will result in a 1 in 3 chance of them winning? The contestant makes his choice of 1 winning door out of 3 then sticks with it.

No

The contestant has a 1 in 3 chance of CHOOSING the correct door. When the host intentionally eliminates an empty door, the contestant is left with a simple 1 in 2 chance of selecting the winning door. It doesn’t matter which of the 3 doors he originally chooses, the contestant will always be left with a 1 in 2 chance of winning.

There will only ever be a 1 in 3 chance of winning if the host doesn’t actually know which is the winning door.

No, you misunderstand. I am only talking about 1 possible strategy at this stage. This particular contestant is stubborn and won't change, regardless of what the host does. He just chooses from the three, but NEVER changes in spite of being given the opportunity. How can this not give him a 1 in 3 chance of winning? The host only opens doors after the contestant's choice of 1 out of 3 doors. This cannot change anything as the contestant's choice is fixed prior to this event. If the contestant has been lucky enough to choose the prize door in the first place, he wins, if not, he loses. This strategy (of never changing) must surely have a 1 in 3 chance of success.

If I can't explain this first piece of simple logic to you, then I will stop, as the next step (the alternate, opposite and better strategy) is a little more complex and you will never get it on the basis of my obviously ineffective explanation.

If only it was logical.

There is no suggestion the contestant is stubborn, intent on not changing.

THE FACT REMAINS, IF THE HOST ELIMINATES ONE OF THE EPMTY DOORS, THE CONTESTANT IS LEFT WITH A 1 IN 2 CHANCE OF WINNING. It doesn’t matter if he swaps or not. 1 in 2 chance.

Winky, Fatcat,

Simply try it. Write down (say) 30 fairly random sets of distribution. Then write down 30 random sets of 1st door choice.

Record the results for each set of doors assuming (1) that the contestant NEVER changes his mind (2) the contestant ALWAYS changes his mind (3) the contestant ALWAYS alternates.

I wrote a little Excel programme to simulate this for 1,000 sets of doors with random location of the car and random selection of the first door.

Not a pure maths solution, but gave me sufficent confidence that my pure maths approach was right.

Cheers

Don

quote:

Originally posted by fatcat:

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:

quote:

Originally posted by winkyincanada:

quote:

Originally posted by fatcat:
Winky

The odds on the contestant winning are always 1 in 2. It makes no difference if he changes or not.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2

Don't agree. State your logic.

The fact that the host ALWAYS opens an empty door means there is NEVER a 1 in 3 chance. ALWAYS a 1 in 2 seems logical enough.

No matter which door the contestant chooses, the host will ALWAYS eliminates one of the empty doors. It always comes down to a 1 in 2 chance. The 1 in 3 chance doesn’t exist.

Can you agree that a strategy that has the contestant NEVER changing doors will result in a 1 in 3 chance of them winning? The contestant makes his choice of 1 winning door out of 3 then sticks with it.

No

The contestant has a 1 in 3 chance of CHOOSING the correct door. When the host intentionally eliminates an empty door, the contestant is left with a simple 1 in 2 chance of selecting the winning door. It doesn’t matter which of the 3 doors he originally chooses, the contestant will always be left with a 1 in 2 chance of winning.

There will only ever be a 1 in 3 chance of winning if the host doesn’t actually know which is the winning door.

No, you misunderstand. I am only talking about 1 possible strategy at this stage. This particular contestant is stubborn and won't change, regardless of what the host does. He just chooses from the three, but NEVER changes in spite of being given the opportunity. How can this not give him a 1 in 3 chance of winning? The host only opens doors after the contestant's choice of 1 out of 3 doors. This cannot change anything as the contestant's choice is fixed prior to this event. If the contestant has been lucky enough to choose the prize door in the first place, he wins, if not, he loses. This strategy (of never changing) must surely have a 1 in 3 chance of success.

If I can't explain this first piece of simple logic to you, then I will stop, as the next step (the alternate, opposite and better strategy) is a little more complex and you will never get it on the basis of my obviously ineffective explanation.

If only it was logical.

There is no suggestion the contestant is stubborn, intent on not changing.

THE FACT REMAINS, IF THE HOST ELIMINATES ONE OF THE EPMTY DOORS, THE CONTESTANT IS LEFT WITH A 1 IN 2 CHANCE OF WINNING. It doesn’t matter if he swaps or not. 1 in 2 chance.

The "never change" strategy isn't logical. I admit that. All I ask is that you accept that IF a contestant took this strategy (NEVER changing), they would have a 1 in 3 chance. The corollary is that if a contestant took the opposite strategy of ALWAYS changing, they MUST have a 2 in 3 chance of winning (as they will always get the exact opposite outcome to the person who NEVER changes). The best strategy is to swap. Not intuitive, but absolutely correct.

I'm done now. Do your research.

As an alternative, do an experiment. Use three playing cards, one of which is the "winning ace" and repeat the experiment 100 times. Don't forget that after making the first RANDOM choice, you non-randomly (as the host, you know) remove a "losing card" before swapping or not swapping. Practically, the way do do it is shuffle the cards, pick one at random as the contestant (without looking) and set it aside. Now look at the other two and remove a "losing card" (host opens a door to eliminate a losing possibility). Now look at your chosen card and decide whether "swapping doors" or "not swapping doors" would have resulted in a win. Note the winning strategy and repeat. You will see that the "Swap" strategy wins about 67% of the time and "Don't Swap" wins about 33% of the time.

As I said, this one is counter-intuitive.

Happy days, I found I could do the first three conundrums and then turned to page 2 to find that so could everybody else!!

So instead can I join the debate on question 3 ....I remember this one running and running in the newspapers between mathematics professors!

For what it is worth I too would definitely change doors. I think the confusion arises from the assumption that the two doors have an equal probability of hiding the prize. The nub of the solution is that they do not.

If you originally selected door A then your choice is "stick with A" or "select the best of door B and C". I would obviously prefer the latter?

Put it another way - you're betting on a tennis match between player A (who had a bye in the first round) or the winner of the game between B and C. Again I'd bet on the latter!

Is this a good time to confess to being a risk manager (not sure whether that counts for or against)?

......now can anybody remind me how to expand (X^3 - 1)?

quote:

......now can anybody remind me how to expand (X^3 - 1)?

I don't think you can - I presume you are looking for factors?

Cheers

Don

Fred,

Are you certain that you have provided sufficient info re the shipment of DAC555s?

If (I say IF), If the shipper knew the weight of a standard DAC555 AND the DIFFERENCE in weight of a sub-standard DAC555, then the solution would be simple - (put 1 DAC from box 1)+ (2 DACs from box 2) + (3 DACs from box 3) .....+(n DACs from box n) on the scales

Difference from expected weight, divided by diffence betwwen "ggod" DAC and "Bad" DAC will reveal number of "bad" DACs in samples on the scales and hence Box No with sub-standard DACs

But if you HAVE given us all the info we need, well!, then I will have to think agaian........and call on Dave, Winky and Ken c for help

Cheers

Don

Fred,

Are you certain that you have provide ALL the info needed for the shipper to find the sub-standard DACs?

IF (I say IF), IF the shipper knew the weight of a standard DAC AND the DIFFERENCE in weight between a standard DAC and a sub-standard DAC, then it would be easy for him to find the sub-standard box.

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)

The difference between the expected weight of (1 + 2 + 3....+ n) standard DACs, divided by the difference in weight of a sub-standard DAC will reveal the number of sub-standard DACs in the weighed sample and hence the Box No containing the rogue DACs.

If, however, you have provided ALL the info, and we don't need the weights of DACs as described above, well!, I shall have to call on Dave, Winky and Ken c to help out!!

Cheers

Don

Oooppps! I thought I'd lost the first reply, hence the "double" post

Winky,

Nice solution to the Explorer!

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........

Cheers

Don

Winky,

I like your use of playing cards in the Monty Hall teaser.

I simulated something similar in my spreadsheet simply to eliminate any "bias" in the decision making process.

Chances of winning if you NEVER change your mind are .333': If you ALWAYS "change your mind" (*) are 0.666': and if you make a 50/50 random choice when there are two boxes left, it's 0.5.

(*) At the start, you can deliberately select to open (say) Doors B & C. (this gives you a 2/3 chance of winning). You "force" the gameshow host to open both B & C by cunningly telling him that your initial choice of Door is "A".

But, as others have said, the debate amongst maths professors is still raging!

Cheers

Don

quote:

Happy days, I found I could do the first three conundrums and then turned to page 2 to find that so could everybody else!!

Happy days...........

Please join the debate on No 3, it really is a wicked little teaser.

As for No 4......? (you do need to be very careful with this one)

Cheers

Don

quote:

I'll have a go at number 2...

Good shot Dave, as Winky has confirmed.

Cheers

Don

quote:

An explorer set off on a journey. He walked a mile south, a mile east and a mile north. At this point he was back at his start. Where on earth was his starting point? OK, other than the North Pole, which is pretty obvious, where else could he have started this journey
?

Another (but not exactly precise for a purist)
Start at any point on the equator, 1mile south 1mile east (or west) 1mile north & your'e back at the equator

Any one done the equator water spiral (plughole) trick ??
20 metres north the water spiral goes clockwise
20 metres south & it turns anti-clock
On the equator it runs out without a spiral

Don .....re Q4

oh yeah he gets one chest not one coin so that means there are X^3-X coins to be distributed?

so that's (X^2-1)X/3 per person

or

(X+1)(X-1)(X/3) per person

so either X-1, X or X+1 has to be divisible by 3 ...which of course is ALWAYS true.

Therefore there will be no bloodshed

...YIPPEE do I win a quality street?

Martin C

quote:

...YIPPEE do I win a quality street?

errr, you'll have to ask George aka Fredrick all about Quality Street......

But well done!

Cheers

Don

quote:

Another (but not exactly precise for a purist)

VERY good Mike. I've never seen your solution before - and as you say, not for the purists.

But hey, this is the NAIM forum - so we must ALL be purists?

Cheers

Don

PS. It also seems to work just the same on the 49th Parallel............the 60 Deg N Lattitude......the 60 Deg S Lattitude....crikey, everywhere!!

Cheers

Don

Fatcat

If the contestant randomly (50/50) picks either of the two doors left (after the gameshow host has opened one door), his probability of winning is 0.5.

It's worth trying a reasonable run of examples eg 30 or more. It won't "prove" who's right and who's wrong, but it will reaveal a definite trend.

Cheers

Don

quote:

Originally posted by Don Atkinson:
Fatcat

If the contestant randomly (50/50) picks either of the two doors left (after the gameshow host has opened one door), his probability of winning is 0.5.

It's worth trying a reasonable run of examples eg 30 or more. It won't "prove" who's right and who's wrong, but it will reaveal a definite trend.

Cheers

Don

You're right. If the contestant's strategy is to randomly choose to switch or not, then over time they will win 50% of the time. But the best strategy is to ALWAYS switch. That way, they will win 2/3 of the time.

Monty Hall Problem

...oh and while I'm on a roll (it's not that I'm bored you understand)

distance from dealer ..I make the equation

(30)/(x-30) = (X+15)/(2x-15)

Therefore 60x = x^2 - 15x

or X = 75 miles?

......I'll get my coat....or should that be anorak

quote:

anorak

Spot on Martin, on both counts!

Cheers

Don

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.