Brain teasers

[rh100]

[quote name='swee'pea99']I've heard this one before - it made no sense to me then; it makes no more now.

What's done is done. It cannot affect the situation now. The situation now is, there are two doors: one good, one bad. You can choose one, or you can choose t'other (ie, stick or change). It's 50/50. I really cannot see this any other way. Call me dumb.[/QUOTE]

Think of it this way

Your first choice statistically is most likely wrong - showing the other empty door leaves the likely correct answer.

It was on telly, someone setup about 100 lines of 3 paper cups - under one of each three he placed a toy car. He made one choice, then I think changed it to one of the others (something like that) over the 100 rows he got it right most the time.

 

[beanzontoast]

Think of it this way

Your first choice statistically is most likely wrong - showing the other empty door leaves the likely correct answer.

It was on telly, someone setup about 100 lines of 3 paper cups - under one of each three he placed a toy car. He made one choice, then I think changed it to one of the others (something like that) over the 100 rows he got it right most the time.

Once one of the 3 doors is revealed, the odds of the prize being behind either of the two remaining doors is 0.5. The odds that your first choice door conceals the prize is 0.5 - and the odds that the other remaining door conceals it is also 0.5

The 2-door scenario is independent from the 3-door scenario. It's a whole new game in which there are only 2 doors, each with an equal chance of concealing the prize. What the probabilities were when there there were 3 doors does not influence what happens now there are only 2. It is not a cumulative event. Your first choice door is now just 1 of 2 possibilities - in other words, it's just as likely to be hiding the prize as the other remaining door.

Consider tossing a coin. If someone reveals that their last five coin tosses came out heads and asks you to predict their next coin toss, surely the chance is that the next coin toss is more likely to be a tail? No, it's not. Each coin toss is an independent event. The odds of the next coin toss being a head are 0.5 - exactly the same odds of it being a tails. Statistically, if you tossed a coin 1000 times, it would be likely that heads would happen just as often as tails, so you'd end up with roughly the same number of each. In a small number of tosses, because each toss is independent, it is actually not that hard or unremarkable to get several heads in a row or several tails in a row.

 

[Over The Hill]

Once one of the 3 doors is revealed, the odds of the prize being behind either of the two remaining doors is 0.5. The odds that your first choice door conceals the prize is 0.5 - and the odds that the other remaining door conceals it is also 0.5

NO!

Consider what you know before the second door is open-
1 in three chance and there is definately at least one vacant door of the remaining two.

Consider what you know after-
No new information.
So your odds have not got better.

The remaining two doors (only) are subject to the narrowing of the odds because only those are in the pot to be removed from the options.

The original chosen door is not in that pick.

The act of reveling one vacant door only affects the odds on the doors from which the vacant door is selected (ie not including the original choice).

 

[montage]

Work is quiet………does anyone have any really good brainteasers?

Here are some to get going with.

1. Which time-piece has the most moving parts?
2. What two English word's sound the same, have different meanings and are spelt without any of the same letters.
3. You are in a house where the wall's face south, a bear walks by…what colour is the bear?
4. Nice easy one, how many months have 28 days?
5. Six eggs in a box, 6 people take one egg each, there is one egg left in the box….how?

And last of all (be warned anyone that answers this is someone to be worried about as it tests your pathological tendencies)

6.Girl attends a family funeral, meets a chap. In her emotional state she is attracted to the chap and they go to the wake together. They talk, drink, fall for each other and end up back at her place for a "fantastic night of unbridled passion". She falls totally in love with the chap. Next morning she finds that the chap has left without leaving any message, so the girl goes across town and shoots her sister in the head……….Why?

if you are really good, you'll ignore Google

1. Hmmm obvious answer would be a watch or something, but perhaps it is something moving?...such as a clock on a plane...but that is also too obvious. Maybe to do with a time zone line such as the GMT. Or flip it on it's head and go for the so-obvious-it-isn't-obvious and take a simple time piece with many moving parts. my final answer - egg timer, the ones made with sand.

2. Thought this one through for a while - I'm rubbish at english .

3. for all the walls to face south you must be on the north pole, so the answer depends on how dirty the polar bear is.

4. Too easy - all of them

5. again easy, the 6th person leaves the 6th egg in the box and takes the egg and the box.

6. First thought is that she shoots him using a camera, not a gun. This happened at a family funeral, so the man could be related, or even dead (but the riddle suggests he isn't the dead person). The sister could be dead?....but why would she take a photo of her dead sister after the funeral, unless of course "she shoots her in the head" means she takes a photo of the headstone/gravestone. I believe Wakes are also very common in Ireland?...it could be possible that the girl is a member of the IRA and she slept with the man to gain information from him, finding out that her sister was a pro-brit, therefore she killed her. I'm not so sure

 

[montage]

sorry - bit late in with my answers

 

[rh100]

Once one of the 3 doors is revealed, the odds of the prize being behind either of the two remaining doors is 0.5. The odds that your first choice door conceals the prize is 0.5 - and the odds that the other remaining door conceals it is also 0.5

The 2-door scenario is independent from the 3-door scenario. It's a whole new game in which there are only 2 doors, each with an equal chance of concealing the prize. What the probabilities were when there there were 3 doors does not influence what happens now there are only 2. It is not a cumulative event. Your first choice door is now just 1 of 2 possibilities - in other words, it's just as likely to be hiding the prize as the other remaining door.

Consider tossing a coin. If someone reveals that their last five coin tosses came out heads and asks you to predict their next coin toss, surely the chance is that the next coin toss is more likely to be a tail? No, it's not. Each coin toss is an independent event. The odds of the next coin toss being a head are 0.5 - exactly the same odds of it being a tails. Statistically, if you tossed a coin 1000 times, it would be likely that heads would happen just as often as tails, so you'd end up with roughly the same number of each. In a small number of tosses, because each toss is independent, it is actually not that hard or unremarkable to get several heads in a row or several tails in a row.

Have a watch of this - not exactly as I recalled but really worth a watch

 

[661-Pete]

She would like Fnaar and Waffles, but she wouldn't like me.

I wonder if she'd care for me. Hmmm... debatable.

OK, how about this one?
A ship is moored by the quayside, in a tidal harbour, at low tide. It is not being loaded or unloaded. There is a rope ladder hanging over the ship's side, partly immersed in the water. It has rungs spaced one foot apart. At present exactly twenty-five rungs are visible above the waterline. How many rungs will be visible at high tide? You may assume that the range between high and low tide is fifteen feet.

 

[rh100]

hmmm....25?

 

[661-Pete]

hmmm....25?

Too easy!

I rather like the light bulb one.

You are shown into a windowless room which contains a single bare light bulb hanging from the ceiling, currently off. There are no switches of any kind inside the room. Then you are taken out of this room, the door being closed behind you, and into an adjacent room which contains three ordinary wall switches. It is explained that one of the switches controls the light bulb, whilst the other switches operate other equipment which does not come into this puzzle. You cannot see from the second room into the first room.

You are allowed to operate the switches in any way you like, and then will be allowed to pay one, and only one, further visit to the first room. After that, you must state for certain which switch controls the light bulb.

 

[rh100]

 

[Bayerd]

...is now trying to work out whether following the logic through, should the contestants on Deal or no Deal always switch the final box to increase chances of winning? and is this maths part of the show's success?

 

[rh100]

Interesting point. I guess the contestant would need to be pretty switched on to work it out on the fly though.

So when 3 boxes left, guess at the highest value box - wait to see what is revealed and pick the remaining one? I sort of get this but would that work in this case? Gonna have to watch it now!

 

[Bayerd]

Thinking about it, probably not as one of the lower value boxes is not neccessarily removed, which alters the odds.

 

[Sam Kennedy]

6. If there is another funeral he might come back....

 

[661-Pete]

6. If there is another funeral he might come back....

Possibly. Perhaps the man is the undertaker?