Problem on Mathematics - Probability

Do you think the player should switch?


Results are only viewable after voting.

Status
Not open for further replies.

roygoh

Senior Member
Jan 18, 2002
5,011
0
0
Northwest
Visit site
#1
Imagine you are on a TV game show, and you managed to rule out all other contestants and now face your final challenge.

You are presented with 3 treasure chests, and you are told that one of the chests contains the big prize of a sports car. The other 2 chests are empty.

You are given 1 choice. If you pick the chest with the keys to the sports car, you get to drive it home the very same day. If you pick one of the empty chests, you go home with nothing.

After some consideration, you picked one of the chests. Before opening the chest to check if you have won the big prize, the game show hosts suddenly presented you with another challenge. Out of the 2 chests that you did no pick, the host opened one that is empty.

So now there are 2 chests left. The one you have picked, and the on that you did not pick and the host did not open.

The host then tells you that you now have a choice wether you want to change your pick to the other unopened chest.

Would you change? And why?

:gbounce:
 

Fizzy

New Member
Mar 2, 2003
352
0
0
#2
Not very confident of my answer but will give it a try:

Do not change. The probability of winning is the same i.e. 1/3.

Before the host opened one of the box, the probability of winning was 1/3, after the host opened the box, using conditional probability, the probability of winning would still work out to be equal to 1/3 (because of 1/2 * 2/3).
 

kingpin

Senior Member
Sep 17, 2002
2,256
0
36
Visit site
#3
At that point the host opened the empty chest and gives u the option of making a choice, the probability becomes 1 out of 2 ie 1/2.
 

U

UStime

Guest
#6
Originally posted by Fizzy
Not very confident of my answer but will give it a try:

Do not change. The probability of winning is the same i.e. 1/3.

Before the host opened one of the box, the probability of winning was 1/3, after the host opened the box, using conditional probability, the probability of winning would still work out to be equal to 1/3 (because of 1/2 * 2/3).
I don't think this is correct. Before the host opens the box, the probability of getting the prize is 1/3. Conversely, the prob of not getting the prize is 2/3.

After one of the 3 boxes has been opened and it's found to be empty, the prob of you winning the prize has now increased to 1/2. And the prob of you not winning it is reduced to 1/2 also.

At this stage, it's up to you (based on gut feeling, last minute prayers, etc?!) whether you want to change or not. The chances are 50-50.
 

toasty

New Member
Apr 8, 2003
400
0
0
Singapore, near NUS
Visit site
#7
This is a probability puzzle that has been debated repeatedly and I guess that the same debate is going to take place here. :D

The correct choice is to choose the other box that the game show host did not open, because the probability of getting the car keys is 2/3 if you take that box. Let's say that the 3 boxes are A,B and C. Let's say that you choose A. (If you choose another box, then we just change the labels). Let's say the game host opens box B leaving only 2 possible boxes, A and C. The probability of the prize being in A is 1/3, the probability of the prize being in C is 2/3. It is therefore beneficial for you to choose box C.

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.

If you have 1 million boxes and you choose 1, you have a 0.000001 chance of choosing the correct box. If the host opens 999998 boxes and shows them all to be empty, your 1 box is still 0.000001 chance of having the prize. The other remaining box has a 0.999999 chance of having the prize. You should change to the other box.
 

roygoh

Senior Member
Jan 18, 2002
5,011
0
0
Northwest
Visit site
#8
I heard that this puzzle has actually got a lot of mathematical professors in heated debates.

I am convinced that the player should make the switch.

Here's how to look at it:

The scenario is the same as the host asking the player if he would trade his selected chest with BOTH the 2 chests that he did not select. If the host did not open the empty chest and offered to trade BOTH the 2 chests with the one that the player picked, the obvious answer is to take the offer, right?

So what difference does it make if we know that the host is only going to open the chest that is empty? The probability that the gran prize is within the 2 that was not picked initially is 2/3, and remains at 2/3 even after the one that is empty has been opened.

The chances that the player has picked the chest with the grand prize is only 1/3. And by switching, he will increase his odds to 2/3.

Another way to look at this is by induction. Say if the number of chests is increased to 100, with still only 1 chest containing the grand prize while the rest are empty. The player picks 1, and his probability of winning is 1/100. The probability that the grand prize is in one of the other 99 chests becomes 99/100.

Now out of the 99 that the player did not pick, the hosts opens 98 chests that are empty and offer the player for a trade. If you are the player would you trade now?

The logic is the same for both cases, even though the increase in chances of winning is more obvious in the second case.

Let me know if you are not convinced, and we can discuss further.

- Roy
 

roygoh

Senior Member
Jan 18, 2002
5,011
0
0
Northwest
Visit site
#9
Originally posted by toasty
This is a probability puzzle that has been debated repeatedly and I guess that the same debate is going to take place here. :D

The correct choice is to choose the other box that the game show host did not open, because the probability of getting the car keys is 2/3 if you take that box. Let's say that the 3 boxes are A,B and C. Let's say that you choose A. (If you choose another box, then we just change the labels). Let's say the game host opens box B leaving only 2 possible boxes, A and C. The probability of the prize being in A is 1/3, the probability of the prize being in C is 2/3. It is therefore beneficial for you to choose box C.

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.

If you have 1 million boxes and you choose 1, you have a 0.000001 chance of choosing the correct box. If the host opens 999998 boxes and shows them all to be empty, your 1 box is still 0.000001 chance of having the prize. The other remaining box has a 0.999999 chance of having the prize. You should change to the other box.
Ah....you posted the same argument as I did and beat me to it.
 

DanMan

Senior Member
Mar 27, 2003
644
0
16
Singapore, Pasir Ris
www.photo.net
#11
Wah... chim! The probability the sponsors of the game show didn't want the contestant is high (save money). Instructions have been given to the host to throw the contestant off if he chose the right one. So if I were the contestant, I would stay with my choice. :D
 

#12
Originally posted by roygoh
I heard that this puzzle has actually got a lot of mathematical professors in heated debates.

I am convinced that the player should make the switch.

Here's how to look at it:

The scenario is the same as the host asking the player if he would trade his selected chest with BOTH the 2 chests that he did not select. If the host did not open the empty chest and offered to trade BOTH the 2 chests with the one that the player picked, the obvious answer is to take the offer, right?

So what difference does it make if we know that the host is only going to open the chest that is empty? The probability that the gran prize is within the 2 that was not picked initially is 2/3, and remains at 2/3 even after the one that is empty has been opened.

The chances that the player has picked the chest with the grand prize is only 1/3. And by switching, he will increase his odds to 2/3.

Another way to look at this is by induction. Say if the number of chests is increased to 100, with still only 1 chest containing the grand prize while the rest are empty. The player picks 1, and his probability of winning is 1/100. The probability that the grand prize is in one of the other 99 chests becomes 99/100.

Now out of the 99 that the player did not pick, the hosts opens 98 chests that are empty and offer the player for a trade. If you are the player would you trade now?

The logic is the same for both cases, even though the increase in chances of winning is more obvious in the second case.

Let me know if you are not convinced, and we can discuss further.

- Roy
I am not convinced, but I am lousy at math too.... though it kind of makes sense. Isn't it that when you have 3 chests : A, B, C.

Probability is A: 0.3, B:0.3, C:0.3.

Say you pick A, the host pick C, C is empty.

So you can now eliminate C, and the prize is now between A and B. So why doesn't the probability increase to 0.5, 0.5? That the host opened one that is empty (or even 998 that is empty) doesn't increase the chance of that other box being the prize right?

:dunno:

Regards
CK
 

Yongliang

New Member
Apr 27, 2002
167
0
0
Singapore, East
#13
Wow!

Interesting problem and solutions.
Well, I'm no maths freak, but theoretically, I guess Roygoh and Toasty are correct.

However in reality, simply put, the situation has changed.
You are now faced with 2 chests, and you need to choose one.
Of course, the probability is still 1/2. It doesn't make practical sense that one chest shd have a probability of 1/3, and the other 2/3.

Just my personal views. =)
 

roygoh

Senior Member
Jan 18, 2002
5,011
0
0
Northwest
Visit site
#14
Originally posted by ckiang
I am not convinced, but I am lousy at math too.... though it kind of makes sense. Isn't it that when you have 3 chests : A, B, C.

Probability is A: 0.3, B:0.3, C:0.3.

Say you pick A, the host pick C, C is empty.

So you can now eliminate C, and the prize is now between A and B. So why doesn't the probability increase to 0.5, 0.5? That the host opened one that is empty (or even 998 that is empty) doesn't increase the chance of that other box being the prize right?

:dunno:

Regards
CK
There are a few other ways to look at this if the scenarios I presented earlier does not convince you.

In you statement "Say you pick A, the host pick C, C is empty." it is important to note that the host already know which box is empty and he intentioanlly pick and open one that is empty. He does not pick the box by chance.

So no matter which box the player picked up front, his chances of getting it is 1/3 and it does not change even if additional information is presented later.

Let me try another way of explaining this.

Say the player is aware that the host is going to offer the trade before the game starts and he decides that he will make the switch when offered.

He will win if he picks an empty chest initially. He will lose of he picks the correct chest initially. So his chances of winning is 2/3.

Say the player is aware that the host is going to offer the trade before the game starts and he decide not to switch when offered.

He will win if he picks the correct chest initially, and he will loose if he picks an empty chest initially. So his chances of winning is 1/3.

So his chances of winning is 2/3 if he switches.

Convinced now?

- Roy
 

Marx

New Member
Apr 2, 2003
379
0
0
East
Visit site
#15
My idea is though technically 1/3 odd is obviously smaller than the 2/3, may I suggest that this 2/3 is ths sum of the 2 individual 1/3 odds they carry. For me, choosing say box A, than the remaining box C for that matter (if B is opened by the host), will not increase my chances of winning. My thinking is it's the same 1/2 odds now, in reality, you're faced with two unopened boxes! Same goes if the choices are 100s.
Just my 2¢.
:confused:
 

roygoh

Senior Member
Jan 18, 2002
5,011
0
0
Northwest
Visit site
#16
Originally posted by Yongliang
Wow!

Interesting problem and solutions.
Well, I'm no maths freak, but theoretically, I guess Roygoh and Toasty are correct.

However in reality, simply put, the situation has changed.
You are now faced with 2 chests, and you need to choose one.
Of course, the probability is still 1/2. It doesn't make practical sense that one chest shd have a probability of 1/3, and the other 2/3.

Just my personal views. =)
The situation only changed after the intial choice is made, so it cannot change the probability that the intial choice is right or wrong.

If, before the player made his initial choice, the host opens an empty chest and let the player choose from the remaining 2, then yes, the probability of winning is an obvious 0.5.

The twist is that the empty chest is opened after the player has made the initial choice. So it cannot change the probability of winning from 1/3 to 1/2 automatically.

Here's yet another way to look at this.

Say the player is allowed to pick 2 out of 3 chests. What is his chances of winning?

2/3, right?

So after he has picked the 2 chests, the host opens one of the 2 that he picked and it is empty. Does it increase his chances of winning? No. Because we already know that at least one of the chests his picked will be empty, and the host knows exactly which one is empty. The host merely picks the one that he already know is empty and opens it. The player's chances of winning is still 2/3, and it does not drop to 1/2 because now there are only 2 unopened chests.

In this case, if the hosts offers the player a chance to switch with the box he did not pic initially, is it more obvious that he should not switch?

Reversing the logic back to the original scenario, where the player picks one chest, his chances of winning is 1/3 if he does not switch, and his chances does not automatically increase from 1/3 to 1/2 after the host opens an empty chest from the two that he did not pick.

Convinced yet?
 

HOCL

New Member
Jan 30, 2002
492
0
0
NE
Visit site
#17
Very interesting.

I think I am mathematically convinced.

Just ask yourself: given a choice, would you choose
A) 1 out of 3 or
B) 2 out of 3?

Which probability is higher? Obviously B.

HAHA! However, you must keep in mind that we are only talking abt probability here. In the end, the prize may still end up in the choice with lower probability!
 

James_T

New Member
May 14, 2003
2
0
0
Visit site
#18
Couldn't resist giving my 2cents worth.

1/3 or 1/2 all depends on when you are computing.

If prior to the event, you are told that the host would open one of the 2 unpicked boxes and allow you to trade, then probability is 1/3 as explained very painstakingly by Roy.

If you fast-forward to the point where the host opens one box, and you are allowed to switch, the probability is 1/2 , since it is either this or that box.

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
 

Zerstorer

Senior Member
Jul 8, 2002
3,437
0
0
#19
I feel that whatever choice the person made initially is irrelevant because he is allowed to switch.

Hence the computation is only made at the point when he is allowed to choose between 2 boxes.

The fact that the host opened a empty box doesn't change anything because the probability of that is 1.(He knows which is empty).
 

Status
Not open for further replies.
Top Bottom