Impeccable logic - part 2

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Sorry to open a new thread on this but when I tried to post a reply to the original I found it had been locked.

For anyone interested in the maths and not critisizing others because they dont agree...

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The odds on the Monty Hall scenario are 50-50.

Axiom: (from the website listed in this post) It is important to note here that Monty would NOT open the door that concealed the car.

Proof:

Three doors - A, B and C. Let us suppose the Car is behind A, and goats behind B and C.

Scenario 1 (you choose the car first time)

Let us suppose the contestant chooses A. Monty Hall opens B (or C) and asks if you want to swap A for C (or B). You still dont know what is behind any door and you have a straight choice between keeping A or swapping for C - a car is behind one, a goat behind the other but you dont know. 50-50.

Scenario 2 (you choose a goat first time)

Let us suppose the contestant chooses B (or C). Monty Hall opens C (or B) and asks if you want to swap B (or C) for A. You dont know what is behind B (or C) and A, so again its a straight choice between a car and a goat. 50-50.

Confusion:

The website poses the question:

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the other doors, opens another door, say No. 3, which has a goat. He then says to you, 'Do you want to pick door No. 2?' Is it to your advantage to take the switch?"

But then subtly changes the problem:

"Omitting one phrase in the statement of this problem changes the answer completely and this might explain why many people have the wrong intuition about the solution. If the host (Monty Hall) does not know where the car is behind the other two doors, then the answer to the question is "IT DOESN'T MATTER IF THE CONTESTANT SWITCHES."

However the initial statement, as quoted on the same website says "Monty would NOT open the door that concealed the car."
But he can only do this if he knows where the car is!

So there is an inconsistency - who knows what the actual question is that the website author thinks they are answering (with such apparent authority) but its not the initial question.
 
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I was going to post too.

I agree that the overall odds on MH scenario are 50:50 but It looks like swapping really does buck the odds.


The contestant has only six permutations to choose from so suppose the car is behind door 1.

Pick 1 and stick = win.
Pick 2 and stick = lose.
Pick 3 and stick = lose.
Pick 1 and swap = lose.
Pick 2 and swap = win.
Pick 3 and swap = win.

Six permutations, three win and three lose, that's 50:50.
But whatever is picked first, when it comes to the decision to stick or swap, then two out of the three 'swap' permutations will win.
:eek:
 
Ans finally Kes has come to the same conclusion i did yesterday.

I wonder if Softus will now have a rethink! :LOL:
 
Pick 1 and stick = win.
Pick 2 and stick = lose.
Pick 3 and stick = lose.
Pick 1 and swap = lose.
Pick 2 and swap = win.
Pick 3 and swap = win.

Errrmmm ...

What about Pick 2 and swap 3 = Lose.
and Pick 3 and swap 2 = Lose

So ...

Pick 1 and stick = win
Pick 1 and swap = lose
Pick 2 and swap 3 = Lose
Pick 2 and swap 1 = win
Pick 3 and swap 2 = Lose
Pick 3 and swap 1 = win

Which is ... 50:50 :rolleyes:
 
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What about this.

At 1st pick 1/22 chance for either picking max or min prized box from 23.

Following the pick, assuming the choice being one or t'other, the chance against picking the opposite at the next choice is 21/22, if this pick is not the opposite then next up 20/21 and so on until at the final pick 1/2

The overall chance of finding the opposite becomes 21/22 x 20/21 x 19/20 ...... x 1/2.
mathematically (21 !) ÷ (22 !) where for example 4 factorial, usually written 4!, is equal to 24 (1 × 2 × 3 × 4 = 24. NB 2x3x4 = 24 also ;) )

(21 !) ÷ (22 !) = 0.04545...45... = 1/22 Therefore the overall chance of finding the opposite to the 1st pick is .... The same 1/22 !!
Therefore 50/50, can see no reason for a swap.
-0-
 
Probably not helpful but odds and statistics always make me smile.

For instance a man is terrified of being on the same flight as armed hijackers. He researches the subject and finds the odds on him being on the same plane as an armed man are 10,000 to 1.

So to imporove his odds, he buys himself a gun and smuggles it on board his next trip. The odds of there being TWO guns on the same flight are 100,000 to 1.
 
There are always exceptions to probability but, if you play the odds for long enough, they hold true.

This is why, occasionally, some lucky punter has a big win on (say) a 7 way accumulator ... 99.9% lose their shirt in line with the odds though ;)

In DoND, when faced with stick or swap for the big prize, you take your chances on a 50:50 probability ... Do you feel lucky punk ... Make my day :LOL:

MW
 
Pick 1 and stick = win.
Pick 2 and stick = lose.
Pick 3 and stick = lose.
Pick 1 and swap = lose.
Pick 2 and swap = win.
Pick 3 and swap = win.

Errrmmm ...

What about Pick 2 and swap 3 = Lose.
and Pick 3 and swap 2 = Lose

So ...

Pick 1 and stick = win
Pick 1 and swap = lose
Pick 2 and swap 3 = Lose
Pick 2 and swap 1 = win
Pick 3 and swap 2 = Lose
Pick 3 and swap 1 = win

Which is ... 50:50 :rolleyes:

Let's try that again. The car is behind door 1 and there are six permutations for the contestant.

The six permutations I gave are correct if a little succinct.
MW some of yours wouldn't be possible. You won't have a choice what to swap 2 or 3 with if you've picked a goat first because Monty will always open the door with the other goat.

Three of the six permutations will win and three will lose. That's 50:50 ok, but the contestant has a choice to stick or swap and two of the three 'swap' permutations will win. I find that rather interesting.
 
The Monty Hall scenario is quite simple, and the odds can easily be followed......

1) Their are 3 doors, your chance of picking the correct door with your first choice is...1 in 3 Therefore you will only win the car 33 times out of 99 attempts, if you stick with your original choice
( if you think carefully about the above sentence, its all you need to know )

2) So what is the advantage in switching, well we already know that in 2 out of 3 Attempts, the car will be behind the other 2 doors, that you did not choose.
Monty now removes the goat for you........ :D
Thus the remaining door contains the car 66 times out of 99.

Its a must switch, thanks to Monty removing the goat
 
I think that says it all, Trazor, there's really nothing else to argue about is there? There's certainly no paradox.
 
Trazor you are right. As said before in the previous thread. Your getting 2 goes of picking as you will never get the car first time. So its a 2 in 3 chance. Better swapping.

But the Deal or No deal scenario is diiferent. No one knows where the money is there there will never be a scenario like that as in Montys experiment you can pick 1 of three boxes. In deal or no deal, the contetsant has a box and they can only choose to eliminate 1 of the 2 other boxes.
So this is where the ' no one knows ' bit comes in as no one is actually taking away the wrong box first time round in deal or no deal.

Hence the 50:50 chance
 
Correct. No where's Softus when you want him?
 
joe-90 said:
There's certainly no paradox.
There certainly is. Here it is, within one single post:

Firstly trazor said:
( if you think carefully about the above sentence, its all you need to know )
Then trazor contradicted him/herself when he/she said:
Its a must switch, thanks to Monty removing the goat
If Monty offers the swap, then him having exposed a goat made no difference, other than to confuse people.

Regarding the debate about whether or not Kes' first post on the other topic represented a valid MH scenario, my offer still stands, despite the action of a bored/officious/fascist* moderator.
_____

(*delete as applicable.)
 
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