Question

In the 'Who wants to be a millionaire' show, for each question, one has to choose 1 out of 4 possible answers the pass to the next level. Bob, who's a pretty good contestant, is at the final stage of the game - the million dollar question. When the question showed up, he had the shock of his life - the entire question was written in Latin, as if the host was trolling. Bob never studied Latin, has no idea what the question means, thinks his death had arrived. As every choice has equal chance of winning, he just randomly picks A and decides to give A as the final answer...

Answer 1

However, he suddenly realizes he can still use the 50/50, which lets the machine take off 2 wrong answers! The machine then took off B and C... now he has A and D left. Without a hesitation, he chose D and won the 1M$ prize. And chilled for the rest of his life.

Explain why he chose D.

Answer 2

because he said a and he didnt win

Answer 3

Well if there's still 2 answers left, it would be still a 50%-50%..

Answer 4

A/D :|

Answer 5

no he said a final answer and he didnt get it so he immediately switched to D lol

Answer 6

he didn't say A, he was going to say A then suddenly remembered he still has the "50/50" left

Answer 7

"he just randomly picks A and decides to give A as the final answer..."
A = final answer

Answer 8

Re-read the question countonme123

Answer 9

I'm pretty sure this is this problem
http://en.wikipedia.org/wiki/Monty_Hall_problem

Answer 10

Actually, he did get some extra useful knowledge out of pure luck. Ignore what I said above.

Answer 11

there is no reason for him to switch to D then

Answer 12

If the computer had chosen A as an incorrect answer, then he would have had a 50-50 chance. However, Bob got lucky, and by the monty hall principle (unofficially), changing to D gave him a 75% chance to win.

Answer 13

By switching to D, he improves his chances from 25% to 75%

Answer 14

lol i guess i am not familiar with millionare's computer

Answer 15

Then Bob goes back to time, and instead of A, he chose D, does A still has 75% ?

Answer 16

So in 2/3 cases where he uses the 50/50 Bob gets unlucky and the computer picks A, and he has a 50-50 shot. Still better than a 25% chance, but not great.

Since it didn't choose the door he privately chose, it can be modeled in the same way as if he had actually chosen the door. Hence, his chances improved from 25% to 75%.

Answer 17

As for if he went back in time, we can't say what chance he has. If the computer happened to eliminate B and C again, then yes, we would indeed have a 75% chance of switching.

Answer 18

Yes, the computer took off B and C again, but now we have a paradox:
If began with A, then switched to D --> D = 75%
If began with D, then switched to A --> A = 75%

It's the same question, same answer. How can this happen?

Answer 19

I guess that's why I hate probabilities.. ;(

Answer 20

It's a different scenario. We're focusing on long term events. If he chose D every time, and whenever it eliminated B and C he would switch to A, then he would win 75% of the time by switching.

If he chose A every time, and whenever the computer eliminated B and C he would switch to D, he would also win 75% of the time.

Answer 21

It's two entirely different cases where we don't know what's behind the door until it's revealed to us.

Answer 22

However, if the real answer is always A.

First, he mentally chose A. Then after B and C are eliminated, he chooses D. Well he'll lose, always...

Answer 23

If the real answer is always A, it's no longer a probability problem. However, the probability that A is the answer for 10 times in a row is \[1\over4^{10}\]This is very small, and it only gets smaller as time goes on.

Once again, in probabilities, we only really care about long term events, and this would be factored into any probabilities where things aren't fixed.

Answer 24

When Bob chose A first time, D will be 75%.

But about the 'go back to time' situation, when Bob mentally chose D the first time, instead of choosing A the first time...well, A then will be 75%.

We can't avoid the fact that the question and the answer remain THE SAME.

Is math really saying that, in the same question of the same answer. Choosing A has 75% and choosing D ALSO HAS 75% ?

Answer 25

Or let's put it this way.

Answer 26

The problem assumes that the correct door is already set. In probability, this isn't really the case.

Answer 27

Another contestant, Jake, is sitting beside Bob and doing the exact same mental process as Bob, except he mentally chose D the first time. He will think he'll win 75% of time.

But remember that the question & answer is the same.

Bob, who chooses the opposite of everything Jake does, will think he'll win 75% of time too.

Contradiction.

Answer 28

This is a singular case, and can't be used as a probability.

Answer 29

At this point, I guess that would turn into a guessing 'problem'.

Answer 30

If it's already decided that D is the correct door, the computer opens B and C, and one person choose A, then they have 100% chance if they switch. You're trying to turn the problem into something that has solutions already set in stone, and probability doesn't work that way.

Answer 31

This is true.

Answer 32

Very very good

Answer 33

Thank you :)

Answer 34

Mindblown Dx

Answer 35

Are you looking at

P(A wins|B and C are randomly open)
?

Answer 36

More like

P(A wins|Bob privately chooses A, and B and C are then randomly opened)

Answer 37

same difference...

'Bob privately chooses A' doesn't change the probability

Answer 38

It can be though of as an extension of the monty hall problem, where choosing A does actually change the probability if B and C are opened.

Answer 39

B and C are randomly chosen...they are not methodically chosen as in the Monty hall problem

Answer 40

But we're given that they have been/will be chosen.

Answer 41

the choosing in the MH problem isn't random at all if the person initially picks the wrong 'box'

Answer 42

Overall in the problem, I completely agree with you.

Answer 43

It's just that we're looking at a small portion of all the probabilities where we're looking for the probability

P(A wins|Bob privately chooses A, and B and C are then randomly opened)

In most cases, B and C won't be the doors opened, but in the few cases where they are, Bob does get some useful information.

Answer 44

no he doesn't...that is if B and C are chosen randomly. Now if the selection is done as in the MH problem then we do get new information.

Answer 45

work out the problem using bayes' theorem

Answer 46

if B and C are random...you will get 1/2

if B and C are chosen as in the MH case, then it is 1/4

Answer 47

B and C aren't completely randomly opened. They're randomly opened among the three doors that don't contain anything behind them.

Answer 48

So the machine makes sure to not get rid of the correct option.

Answer 49

correct

Answer 50

that is true in this case and the MH case...though they are still different

Answer 51

Perhaps I should have written

P(A wins|Bob privately chooses A, B and C are incorrect, B and C are then randomly opened)

Answer 52

if B and C are randomly open then obviously B and C are not winners..so 'B and C are incorrect' is not needed

Answer 53

But shouldn't the order factor in somehow? B and C are incorrect and THEN B and C are opened?

Answer 54

if you want you can include that they are not winners...(though it is implied in the fact that they were open) ...it doesn't change anything

Answer 55

the monty hall problem requires the person running the game to know what the original choice of the contestant is so that they can methodically open doors as to not reveal the winner and to not open your door. In this case (1) you are not telling the host your choice (or are you...it doesn't matter since... ) (2) the non-winners are chosen to be reveled at random and not with the frame of mind that I don't want to open your original choice even if it is a loser.

Answer 56

I'm sorry, but I'm still confused. Could you please write out some of the probabilities you would need for Bayes' formula?

Answer 57

cool monty hall thingy *_* never knew that! time to join WWTBAM

Answer 58

Let A= win with A
B= B is a loser and is open
C=C is a loser and is open
D= win wth D
P=pick A in the beginning

we want P(A|B,C,P)
Fist assume random


\[P(A|B,C,P)=\frac{P(B,C|A,P)P(A|P)}{P(B,C,P)}\]
\[=\frac{P(B,C|A,P)P(A|P)}{P(B,C|A,P)P(A|P)+P(B,C|D,P)P(D|P)}\]

I skipped to parts on the bottom since those have probability 0 and it made the expression too long to put on the screen

\[=\frac{\frac{1}{3}\frac{1}{4}}{\frac{1}{3}\frac{1}{4}+\frac{1}{3}\frac{1}{4}}=\frac{1}{2}\]


if we had open doors based on the MH idea then we get

\[=\frac{\frac{1}{3}\frac{1}{4}}{\frac{1}{3}\frac{1}{4}+1\frac{1}{4}}=\frac{1}{4}\]


since for P(B,C|D,P)

D is the winner and you chose A therefore I can't open A ...I must open B and C...thus this probability is 1

Answer 59

I see now. Thanks for actually writing that out.