## View Poll Results: Do you think the player should switch?

Voters
48. You may not vote on this poll
• Yes

20 41.67%
• No

13 27.08%
• Don't know

4 8.33%
• No difference

11 22.92%

# Thread: Problem on Mathematics - Probability

1. Originally posted by James_T

P ( prize in A | prize not in C ) = P( prize in A and prize not in C) / P (prize not in C) = (1/3) / (2/3) = 1/2
I disagree with this because P(Not in C) is 1.

This is because the choice is not being made by you-the game show contestent. The choice is made by the host and it is always correct.

This is to me sounds more like a trick question as there P(Not in C) is always 1.

Hence the choice has always been between A or B. Which is 1/2.

2. P(prize not in C) = 1 only because it is now known that with probability 1, the prize is not in C.

Mathematically, it is correct to say:
P(Choose correct box given prize not in C) = 1/2

To change or not to change is really up to the contestant, cos given prize not in C, you have 50-50 chance tat your original choice was correct.

3. Originally posted by Zerstorer
I disagree with this because P(C) is 1.

This is because the choice is not being made by you-the game show contestent. The choice is made by the host and it is always correct.

This is to me sounds more like a trick question as there P(C) is always 1.

Hence the choice has always been between A or B. Which is 1/2.
The choice is definitely between A, B and C in the beginning. It is not always between A or B.

You have overlooked the fact that the host only opens the empty box AFTER the player has made the initial choice.

If after the player has made the intital choice and the host opens an empty chest from the 2 chests not selected, but he did not offer the trade. Does it mean that the player's chances of winning has increased from 1/3 to 1/2 automatically?

No right?

4. ## Included a poll now

I have used my moderator capabilities to add a poll in this thread for you to cast your vote.

Please remember to stay civil should you choose to participate in the discussion on this problem.

Even mathematical professors have debated heatedly about this puzzle. I do not want this discussion here to end up in a flame war.

We each have the freedom to believe in our own reasoning. In this case I feel that who's right and who's wrong is less important than having fun and interlectual stimulation out of this discussion.

Thanks.

Roy

5. Originally posted by roygoh

If after the player has made the intital choice and the host opens an empty chest from the 2 chests not selected, but he did not offer the trade. Does it mean that the player's chances of winning has increased from 1/3 to 1/2 automatically?

No right?
IMO, still technically, the computed probability is 1/3 but now the player is facing a 1/2 chance of winning.

6. ok i posted a "yes" for the poll.

at first i thought no, but after reading the explanations, i decided on the yes.

the "proof" by induction got me convinced.

7. Originally posted by roygoh
The choice is definitely between A, B and C in the beginning. It is not always between A or B.

You have overlooked the fact that the host only opens the empty box AFTER the player has made the initial choice.
The thing is that he is allowed to change his choice now. Whatever choice he made doesn't affect the choice he makes now. The opening of Box C is inconsequential, it will always be empty and it doesn't affect the decision in anyway.

If after the player has made the initial choice and the host opens an empty chest from the 2 chests not selected, but he did not offer the trade. Does it mean that the player's chances of winning has increased from 1/3 to 1/2 automatically?

No right?
Nope. Because the computation should be made at the branch when he is faced with only 2 boxes.

So if he isn't allowed to change, the probability remains 1/3.
Since he is allowed to change his choice, the decision actually occurs at the point when there is only 2 boxes left so its 1/2.

8. To bring this back to the original question as stated.

"Should he change now?"

Simply because the chance of his choice being right at this juncture is already 1/2. Whether he changes or not doesn't matter.

9. heres my 2cents:

the intital probability of picking the right chest with no re-choosing is 33.3%. but with this classic problem of selection without replacement and conditional probability, the actual probability of chossing the right chest at the onset is actually 1/3*1/2, i.e. 16.6%, not 33% - survivorship bias. opening 1 empty chest does not change that initial probability.

but the situation here is that you get to choose again and not simply leaving it to chance, so your probability of choosing the 1 out of two is then rises dramatically from 16.6% to 50%.

Therefore, you should choose again!

10. but the situation here is that you get to choose again and not simply leaving it to chance, so your probability of choosing the 1 out of two is then rises dramatically from 16.6% to 50%.

Therefore, you should choose again!
let me rephrase that, the probability rises to 50% whether or not you rechosse or not, so its up to you whether you re-choose or not......

11. So the key to this quesiton is, does the fact that the host offeres the player to choose again wipes out entirely the probabilities carried over from previous actions, thus resulting in a 50-50 chance of winning now?

I am leaning towards saying that it does not wipe out the "carry overs".

The proposition that the chances of wining is 50/50 is only true if the player decides to toss a coin to decide if he should switch or not. This is the only way to wipe out the probability "carry overs", because the coin toss is then totally independent of prior events, and when applied to the case of 2 chests with a grand prize in one of them, can truly represent the 50/50 case.

Bear in mind that the question is whether the player should switch or not, instead of asking if the player should make another random choice among the 2 chests.

So, my position is still, if the player decides not to switch, his chances of winning is 1/3. If he makes a totally random choice after being offered the opportunity of switch (such as basing the decision on a coin toss), his chances of winning is 1/2. If he switches, then his chances of winning is 2/3.

- Roy

12. Originally posted by James_T

Using conditional prob

P ( prize in A | prize not in C ) = P( prize in A and prize not in C) / P (prize not in C) = (1/3) / (2/3) = 1/2
This works for me. P(prize not in C)= 2/3 is correct rather than 1. The probabilities should be calculated before the point where C is shown to have no prize.

Maths shouldn't be contradicting logic right?

13. mmm... i feel that the chance is now 1/2 and not 1/3 anymore. when presented with 3 boxes, the prob is 1/3. however, after the host has revealed a box that is empty, the contestant effectively have to choose from 2 boxes. Hence his probability is increased to 1/2.

Though not likely, if the organisers of the gameshow do not want contestants to win, then, if the contestant chose the correct box when presented with 3 boxes, the host will offer to open up an empty box. Cos that effectively increase the chance for the contestant NOT to win the car to 50%.

14. To add to my earlier post, I think we can consider these two events as mutually exclusive. The first event is when presented with 3 boxes. The second event is when presented with only 2 boxes. These two events should be independent.

15. Originally posted by toasty
...snip
The difficulty in understanding is that the probability for box A has not changed. It is still 1/3 after the host has opened B. That is because your decision to choose A was made before the game show host did anything, and you cannot go back and undo your choice which was made before the extra information was presented.
mmm... arent the contestant allowed to rechoose again (in a way, undo his choice) If the contestant is not allowed to rechoose, then his chance is still 1/3. If he is allowed to rechoose, he effectively only have to choose from 2 boxes. The third box which is opened by the host should no longer be considered in the event.

16. I think the probablity attached to the box containing the key and the probablity of the person winning are 2 separate computations.

17. I will not base my decision on the probability attached to the box. Therefore, I base my decision on the chance of winning. I say the chances of winning are 50-50.

18. Originally posted by trendmatrix
To add to my earlier post, I think we can consider these two events as mutually exclusive. The first event is when presented with 3 boxes. The second event is when presented with only 2 boxes. These two events should be independent.
This is the key part to this puzzle as it has to be treated that the 2 choices are mutually exclusive.

At the first choice, he has a 1/3 probability of getting the correct chest.

At the second choice, he will have a 1/2 probability of getting the correct chest.

"What is the probability that his original choice was correct?"
The answer is 3C1 x 2C1 =1/6

"What is the probability that his original choice was incorrect and the remaining chest is correct?"
The answer is 3C2 x 2C1 x 2C1 = 1/6
The first 2C1 refers to the choice made by the host. If he is fully aware of the actual location of the correct chest, then this is no longer a random probability problem and all this calculation does not come into play.

Because it is no longer random, then we have to start the second part with only 2 chests namely the one of his original choice and the one which is not his choice. Whether he change his mind or not, the probability this time round will be 1/2 as there are only 2 chests to choose from.

Just my 3 cents' worth to this thread..........

19. Originally posted by roygoh

So, my position is still, if the player decides not to switch, his chances of winning is 1/3. If he makes a totally random choice after being offered the opportunity of switch (such as basing the decision on a coin toss), his chances of winning is 1/2. If he switches, then his chances of winning is 2/3.

- Roy
Since there is such a huge gap in not switching and switching, i.e. 33% and 66%, would it be possible to write a computer program that does 1 million+ iterations of this and see the result? i.e., is the person who switches more likely to win than the person who does not?

20. Originally posted by erwinx
Since there is such a huge gap in not switching and switching, i.e. 33% and 66%, would it be possible to write a computer program that does 1 million+ iterations of this and see the result? i.e., is the person who switches more likely to win than the person who does not?
I am sure someone has already done that.

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