Puzzle #1

[isthereanybodyoutthere]

https://www.youtube.com/watch?v=cXqDIFUB7YU


Apologies, don't know how to upload actual video. If someone could PM me how, it would be appreciated. Thanks.

[cp3o]

https://www.youtube.com/watch?v=cXqDIFUB7YU


Apologies, don't know how to upload actual video. If someone could PM me how, it would be appreciated. Thanks.

in the quick reply box across the top you will see little icons ,second last on right click and inset url .

[Meursault]

http://en.wikipedia.org/wiki/Monty_Hall_problem

The solution: Always switch

[UB40]

I've woken up next to worse

The evils of drink - I presume!

[EIFII]

Brilliant love this thread! Best in I don't know how long....

[Empirical]

http://en.wikipedia.org/wiki/Monty_Hall_problem

The solution: Always switch

Now explain why you should switch in words of one syllable!

[Empirical]

Cant say I get that. There are three doors. You know nothing about what is behind any door. You pick one at random. One of the doors you didn't choose turns out to have goats behind it, so your chances of having picked the correct door have gone up from one in three to one in two. But you have learnt nothing additional about what is behind either A or C.

Therefore I suggest that as between A and C it is irrelevant that B had goats behind it. But it is also irrelevant that you previously picked A. After B has been opened, the game - effectively - starts again: you have two doors, behind one is the car behind the other is the goat. You have to choose one door at random, you have a fifty/fifty chance of being correct. Therefore choosing either A or C the second time around is equally rational.

The "obvious" answer, but...

[EIFII]

The "obvious" answer, but...

Initially I agreed with Dr Ivory until I read the Wikipedia link

You have to think about this from the point of view of the car, not the player.

When the host removes one of the "wrong" doors, the car remains behind the same door. So you can really think of the right door and one of the wrong doors as being the same door, but with a 2/3 probability of having the car behind....

[Magicman]

Don't think so. The removal of the other eight doors changes the chances of the door you have chosen being the correct one. The chances of that are now 50/50.

Probability is the ratio of favourable outcomes to total outcomes. What the deletion of the other eight doors has done is to change the total number of possible outcomes from 10 to 2. Your choice is still one, but it has gone from 1:10 to 1:2.

If I asked you to pick the ace of hearts out of a face down pack and you picked one card and left it aside unchecked...................what are the chances of it being the ace of hearts?
1 chance in 52................correct?
Now if I take the other fifty one and say............."I'll remove 50 of these cards and if the ace of hearts is here then I'll not remove it."
Now the chances of the ace of hearts being in that bundle are 51 out of 52 so there is a better chance of the card I leave being the ace of hearts than the original card you choose.While there are only two cards it is not a 50/50 chance as the two sides of the equation haven't changed
your card is still a 1 in 52 chance while the new card offered instead of 51 cards is almost a certain winner.

IF the two cards were to be mixed after the cards were removed then it would be a 50/50 chance but that is not the case.

Should you be unable to understand the above scenario then I refer you to this site for further elucidation
http://betterexplained.com/articles/...-hall-problem/

Magicman

[godeeper]

Imagine you are a contestant on a US game show. You have the chance to win a Rolls-Royce (or a Porsche).

There are three doors, and you must choose one. The Rolls is behind one of the doors; there are goats behind the other two.

You choose one door, let's call it A. but you don't open it. The host—let's call him Monty—chooses another, say B, and opens it; there is a goat there.

Now it's your turn to open a door. Monty asks if you want to stick with your original choice A, or if you want to change and open the third door, C.

So, should you go for A or C? And why?

Both A and C have an equal 50/50 chance of having the car behind them. The solution therefore is to toss a lucky coin.