Math puzzle :P

[Melachi]

Ok, I'll try word it as best as I can to avoid any confusion.

Your on a gameshow.

There are three closed doors, and behind one of them is a Prize.
You get to choose which door you want, so you point one out.
Now the gameshow host knows which doors have no prizes and which does.
The gameshow host now opens one of the doors you didnt pick which contains no prize.

So now there is only two closed doors, one with the prize and one without.

You now have the choice of sticking with your first choice, or choosing the other door.

Which door has a higher probability of having the prize behind it?
The door you chose first off?
The door which you didnt choose?
Or equal chances for both doors?

Note: This isnt a trick question, and its not word play, and the relevance of the gameshow is just to make it easier to understand the question.

 

[Tinneth]

equal imho

 

[Aoln]

I'd say equal, could be door you choose first but wording would have to be clearer tbh. :|

 

[Melachi]

Redoing it with some diagrams

3 Doors, A, B & C

Behind one, lies a prize (its pretty too!). But the doors are locked!

<Door A>
<Door B>
<Door C>

We dont know which one the prize is behind, so we have a guess at <Door B>. So we point at it, but dont open it.

Now, the gameshow host knows which door has the prize, and so removes option by removing one of the incorrect doors (it cant be the one we picked ).

So now were left with 2 options.

<Door A>
<Door B> <- this was our initial guess
<Door C> <- Gameshow host has opened this door so we cannot choose it, and it doesnt contain the prize.

Now we have our second choice. Stick with our first guess <Door B> or, switch to <Door A>.

Which door has the higher probablity of having the prize behind it?
Door A?
Door B?
Or is it 50-50 chance?

 

[Xeanor]

the door you chose has a higher chance of being the one with the prize, because if you chose one without the prize then the gamehost might have opened the one with the prize

 

[Melachi]

the door you chose has a higher chance of being the one with the prize, because if you chose one without the prize then the gamehost might have opened the one with the prize

Sorry you have a mistake here and its my fault because of more wording problems, so much easier to say these things than to type em, anyway..

The door that the gameshow host opens, cannot be the one with the prize in it, and it also cant be the one you chose at the begining.


i.e.
prize beind this door (C)
<Door A> <Door B> <Door C>

if we choose <Door C> the host removes <Door A> or <Door B>
if we choose <Door A> the host removes <Door B>
if we choose <Door B> the host removes <Door A>

Sorry for the misunderstandings ;/

 

[Moriaana]

Xeanor is actually right, but for kinda the wrong reasons ^^. If you stick with your original choice, you have a higher probability of winning the prize. If you want all the math...ask someone else. It's too early and I am too tired to type it all

Basically, if you stick to your first choice, you have almost double the chance of winning. Sounds weird yes? But it's true

Hmm let me try doing the simple version for you:

First choice: 1 in 2. (trust me ^^)

Second choice: 1 in 2.

By removing one of the 2 empty doors/boxes/whatevers from your 2nd choice, the guy actually helped you if you stick with your first choice by promoting it to a 50% chance of winning outright instead of a 25% chance if you swap

I think thats right, but it's still too early to check properly hehe

Try it out on people if you don't believe me. Theres loads of silly things like that.... ask your math teacher about the 'Birthday Trick'.

 

[Tasslehoff]

I think there is an equal chance... It just has to be, why shouldn't there be?

There are 3 doors.

You chose door number 2. There's no price in door number 3.

Therefore there must be a prize between door number 2 or 1, an equal chance.

 

[[GOA]Erivoss]

ok the gameshow host knows where the prize is and will ALWAYS choose a door that doesn't have the prize behind. So, the first choice you make doesn't matter at all as in the end you'll always be left with 2 doors, one with the prize behind and one with nothing.

2 doors = 50/50 chance as the first step doesn't affect those odds in any way shape or form.

that's assuming that the gameshow host will always pick a door without the prize behind...


... or smt...

 

[Powerslave]

someone needs to get layed

j/k xD

 

[Alan]

Which door has a higher probability of having the prize behind it?
The door you chose first off?
The door which you didnt choose?
Or equal chances for both doors?

Note: This isnt a trick question, and its not word play, and the relevance of the gameshow is just to make it easier to understand the question.

There are now only two doors remaining, of which one has the prize and one doesnt, this is therefore a 50/50 chance. The fact there were 3 doors, and one was rulled out are there to cloud the issue.

 

[Thorwyn]

Changing door increases the chance.

If you stick to your initial decision, the probability is 0,33.

If you change, there are 3 options

(A) (B) (C) lets assume prize is behind (A)

you chose (A), talkmaster opens (B) or (C), you switch to (B) or (C) = LOSE

you chose (B), talkmaster opens (C), you switch to (A) = WIN

you chose (C), talkmaster opens (B), you switch to (A) = WIN

So if your first pick was wrong, you´ll automatically end up with the correct door, which is a probability of 0,66. If you want to illustrate the system a bit more, imagine it´s not just 3 doors, but 100. You chose one, the talkmaster opens 98 other doors, now you have the chance to stick to your first pick or change. Not a hard decision is it?

 

[Thorwyn]

While we´re at it... this one is even harder:

You´re in a gameshow again and the talkmaster (who´s tired of the good old door problem) has two envelopes in his hands. Each of those envelopes contains a cheque with an unknown ammount of money. One of the cheques is worth exactly twice the ammount the other cheque.
You may now chose one of those envelopes. Now, after a short commercial break, the talkmaster asks you if you want to stick to your initial choice, or if you want to switch envelopes.
The math to calculate the expected win is simple. (0,5 * x/2)+(0,5*2x) = 5x/4
That´s a pretty good percentage, better chances than any other lottery game has to offer. If you switch, you win more (when you win) than you lose (when you lose). So you switch envelopes.
After a short commercial break, the talkmaster ask you, if you want to switch envelopes again. Our calculation above is still correct, so you switch again....
and after a short commercial break... etc.etc.etc.

What´s wrong here? How can you gain more (theoretically) than you lose while switching between the two envelopes?

Warning: This is a REAL tough one.

 

[Vae]

While your brain and logic might make you think it's 50/50 the higher probability is that the prize is behind door A i.e. we changed choice of door once one had been removed.

This is due to a trick in statistics/probability theory. If you stay with the first door your chance of winning is 1/3. If you change doors your chance of winning is 2/3.

Easy explanation:
Your first choice (door B) has a probability of 1/3. After the host opens door C we can exclude that door (i.e. the probability of that door having the prize is 0 ). Therefore door A must be the remainder of the probability i.e. 2/3

 

[Archeon]

This is the only type of Math I understand

What Makes 100%? What does it mean to give MORE than 100%? Ever wonder about those people who say they are giving more than 100%? We have all been to those meetings where someone wants you to give over 100%. How about achieving 103%? What makes up 100% in life?

Here's a little mathematical formula that might help you answer these questions:

If:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z is represented as:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.

Then:

H-A-R-D-W-O-R-K
8+1+18+4+23+15+18+11 = 98%

and

K-N-O-W-L-E-D-G-E
11+14+15+23+12+5+4+7+5 = 96%

But,

A-T-T-I-T-U-D-E
1+20+20+9+20+21+4+5 = 100%

And,

B-U-L-L-S-H-I-T
2+21+12+12+19+8+9+20 = 103%

AND, look how far ass kissing will take you

A-S-S-K-I-S-S-I-N-G
1+19+19+11+9+19+19+9+14+7 = 118%

So, one can conclude with mathematical certainty that While Hard work and Knowledge will get you close, and Attitude will get you there, it's the Bullshit and Ass kissing that will put you over the top.

 

[Nightchill]

Vae that doesn't hold up. At the start all doors have a 1/3rd chance. You're saying that the other door goes from 1/3rd to 2/3rd chance. i.e. you're recalculating. If you're doing that for the other door, you should be doing it for your door (resulting in a 50/50 chance)?

 

[Thorwyn]

Nightchill, there are 2 wrong doors (=2/3) and 1 correct door (=1/3). if you chose a wrong door and decide to switch, you´ll end up at the correct door. Therefor, your initial 2/3 probability to chose the wrong door will lead to a 2/3 probability to end up at the correct door.

 

[sibanac]

While your brain and logic might make you think it's 50/50 the higher probability is that the prize is behind door A i.e. we changed choice of door once one had been removed.

This is due to a trick in statistics/probability theory. If you stay with the first door your chance of winning is 1/3. If you change doors your chance of winning is 2/3.

Easy explanation:
Your first choice (door B) has a probability of 1/3. After the host opens door C we can exclude that door (i.e. the probability of that door having the prize is 0 ). Therefore door A must be the remainder of the probability i.e. 2/3

nope since you remove an option (door c) you need to re calculate the chances, since the inital conditions are no longer the same

 

[Chronictank]

how can one of 2 options have 2/3 probability
imo its 1/2
since you eliminated 1 door you have to recalculate, in which case each door has 50% chance.

 

[Nightchill]

Thorwyn's right. 2/3rds chance of winning if you switch doors after one has been removed. Silly me

Easy to see... Three doors A B C. Let's say A is right but you're switching doors once one is removed.

You pick A, end up at B or C (whichever wasn't removed). You lose.
You pick B, end up at A (since C was removed). You win.
You pick C, end up at A (since B was removed). You win

 

[Urme the Legend]

I had this class on university.. but didn't pay enough attention I guess.. but Thorwyn[B&Q] got the right theory I guess ..

To obvious that it would be 50/50.. why else have a course "Probability" on the University

 

[sibanac]

Thorwyn's right. 2/3rds chance of winning if you switch doors after one has been removed. Silly me

Easy to see... Three doors A B C. Let's say A is right but you're switching doors once one is removed.

You pick A, end up at B or C (whichever wasn't removed). You lose.
You pick B, end up at A (since C was removed). You win.
You pick C, end up at A (since B was removed). You win

you wrote it down wrong:
you pick A, end up at B. (C removed) you lose
you pick A, end up at C. (B removed) you lose
You pick B, end up at A (since C was removed). You win.
You pick C, end up at A (since B was removed). You win



look at it this way after one is removed you get a choise between 2 things one is winning one is loseing 50/50 since you dont know what is what

 

[Nightchill]

No sibanac.

We're only picking A a third of the time. In your case you're picking it half the time. Think about it.

 

[Thorwyn]

Read here (for example):
http://www.remote.org/frederik/projects/ziege/index.html.en


And if you don´t trust the math, you can try statistics here
http://www.phiba.de/Divers/goat.htm

 

[Ormorof]

i like archeons logic

 

[Bunnytwo]

50/50 not really any different from the flip a coin 6 times and it lands heads up 6 times, whats the odds on it landing heads up a seventh time . . 50/50.

 

[Nightchill]

Did you read the thread Bunnytwo? It's not 50/50

 

[Vae]

50/50 not really any different from the flip a coin 6 times and it lands heads up 6 times, whats the odds on it landing heads up a seventh time . . 50/50.

That's because each coin flip is an independent event. The 3 door puzzle involves 2 events (the 2 choices) which are not independent.

 

[Moriath]

thats crap of the remaining doors u have a 50/50 chance of getting the right one .. of the original doors u have a 1/3 chance of getting the right one even if one has been opened as your never gonna chose it but its still in the equation therefore it still takes its 1/3 ..

either that or its 1/3 is transfered equally between the 2 remaining doors to go back to 50/50

how can one door possibly have more of a chance than the other of having a prize behind it when u only have one prize and 2 doors left.

 

[Bunnytwo]

Did you read the thread Bunnytwo? It's not 50/50

Yep it is after first door has been removed there are 2 options left, the odds are no longer out of 3, but out of 2.

Otherwise you are saying that if you put a bullet in a revolver and choose to have the 5th pull of the trigger carried out with the gun at your head, the 4 previous chambers turn out to have no bullet, the chance of the fifth chamber having a bullet is still 1/6, as it was at the start even though there are only two chambers left. The sixth chamber somehow has a 5/6 chance of having it.