Math puzzle :P

[Nightchill]

No no no. Just follow the links Thorwyn provided.

 

[Rigga Mortice]

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.

Thorwyn's working is correct, and Vae has the key to the puzzle. Flipping a coin as above, the odds are still 50/50 due to the outcome of each flip being independent of the previous flips. This is due to the ""Lack of Memory Property" of probabilty.
The events of opening of the doors events are linked (the choice of 1st door determines whether the gameshow host has to open 1 door or can open either), though they seem not to be, hence why it seems counter-intuitive that the probabilties on the 2 remaining doors should be other than 50/50, and why it catches so many people out.

Sneaky, it's one I use on my students too (I lecture mathematics at university :touch: )

edit: though judging by my spelling, english and grammar aren't my forte

 

[Melachi]

Thorwyn is right, really to understand it you just have to look at the first choice.

Where people get caught out is on the second choice.

Gah, ive tried typing out a small explanation three times but I keep making it sound complicated.

Basically, on the first move, you have 2/3 chance of picking the WRONG door, so then, switching will win 2/3rds of the time.

 

[Nightchill]

edit: yep, follow the links Thorwyn provided or just do the quick math.

Three doors. You pick randomly, so you have a one in three chance of picking any door. Let's assume A is the prize. Pick any door. One of the wrong doors is removed. Switch doors.

As said, there are only three doors so you're picking A, B or C an equal amount of times.

You pick A (right). Either B (wrong) or C (wrong) are removed. It doesn't matter. You switch. You lose.
You pick B (wrong). C (wrong) is removed. You switch to A (right). You win.
You pick C (wrong). B (wrong) is removed. You switch to A (right). You win.

That is, you have a two thirds chance of winning if you switch.

What you're missing is that the events aren't independent. The gamehost (or whoever he is) is always removing a wrong (and not a random) door.

 

[Nightchill]

My last post here Just thought of another way to think about it - from the gamehost's point of view.

After your choice:
Two in three times he's left with 1 right and 1 wrong door.
One in three times he's left with 2 wrong doors.

He has to remove a wrong door by the rules meaning two in three times, he's left the right door. Switch!

 

[Driwen]

thing is you have 2/3rd chance of choosing the wrong door first and only 1/3rd of choosing the right one. So the chance is biggest that the other door that hasnt been opened is the right one.

Both doors after opening one has a 50% chance of being it, but the probability that you have choosen the wrong one on the first shot is 2/3rd (kinda weird).
Basically the question is asked wrong I think, both doors have a 50% chance of being the prize door. However if you choose one door first and then change on the second the probability will be 2/3rd and if you dont change than it will be 1/3rd.
So you should have asked which tactic would result in the best probability on choosing the winning door. Instead of asking which of the two doors has the highest probability of holding the prize.

 

[Thorwyn]

I`m still waiting for a solution to my envelope problem!
Comon people!

 

[Bunnytwo]

Bugger I admit it was completely wrong :worthy:

Still don't know if I would switch answer in a real game show because of the "You complete tit" factor if your first answer was right. Gutless I know but there you go.

 

[Tasslehoff]

Actually, there is no answer to this. It's all about how you calculate it. One guy says: I have the proof! And comes up with a calculatation and a solution.

Then the next guy: I have the proof! And comes up with another calculation and another solution.

So it all lies in which way you calculate it.

 

[Thorwyn]

Yeah, mathematical problems are known to have a variety of possible solutions and calculations. Hence all the "nice interpretation, A-" below your exams.
Even Pi can be described as "close to 4" or "just a little higher than 3" or "definitely not 8".

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.

nobody? :touch:
*tap tap tap*