Problem on Mathematics - Probability

[sehsuan]

"What is the Matrix?"

sounds to me like asking that question...

sorry, a hangover from watching the Matrix Reloaded...

 

[roygoh]

Originally posted by hong2
First a few comments to Roy's QBasic codes:

The prg is correct but it's coding the wrong logics, i.e. it doesn't take care of the situation that an empty box was opened by the host. Naturally the result will be 1/3 and 2/3. If you change
prize = INT(RND(1) * 3 + 1) to
prize = INT(RND(1) * 2 + 1) , the answer will be 1/2 and 1/2 because the third box can't contain the prize.

Thanks for verifying my program.

Why should the prize location be random between 2 instead of 3? There are 3 boxes that the organizer can put the prize in randomly right?

The host will only open an empty box (after the initial choice is made). So the box he will open is determined by the box the player picks initially.

After the initial pick, the host opens an empty box, and the polayer is allowed to decide if he wants to switch. Note that at this point the key location does not change between the 2 un-opened boxes.

Originally posted by hong2
Now please allow me to give a new twist to the problem.
We still have three boxes, A, B and C but we now have 2
players X and Y. Let's say X chooses A, and Y chooses B. Supposed the host opens box C which happens to be empty (if it contains the prize, both players lose - no need for advice).

What does the "switching camp" advise our players now? X to switch to B but Y to switch to A? We have a paradox here:
1) Switch for both players mean that both will have 2/3 probability which add up to >1, obviously not allowed.
2) Prob(box to contain a prize) changes depending on who chooses it.

For box A, it is 1/3 for X but 2/3 for Y
for box B, it is 2/3 for X but 1/3 for Y
we would expect probability to be at least impartial!

How now:dunno:

The twist changes the game enough that the probablility calculations of the original game do not apply anymore.

With both players chooses different boxes, the host is left with only one box to open. Once that box is opened, the probability P(either one of the player wins) immediately collapses to either 1 (if the box host opens is empty) or 0 (if the box host opens has the key).

To look at the new game holistically:

case 1: P(player A picks the correct box initially) = 1/3
case 2: P(player B picks the correct box initially) = 1/3
case 3: P(both players picked empty boxes initially) = 1/3

So the probabilty that they will be offered to switch is P(case 1) + P(case 2) = 2/3. If they both picked empty boxes, the game is over.

That changes things alot, as now, there is no bebefit in switching. From the beginning, both player A and B has the same probability of getting the correct box. If it has been confirmed that one of them has actually got the correct box (the box the host opens is empty), they still have equal odds of winning the game, whether they switch or not.

So I don't think the twisted game scenario is a prove that the logic I presented on the original game is flawed.

Let me know what you think.

Thanks!

Roy

 

[hong2]

Hi Roy,
this is really quite interesting. To switch or not to switch and the final answer is
YES and NO depending on the assumption:

1) YES (probability 1/3 and 2/3) if the host knows where the prize is, and your reasoning is correct.
2) NO (probability 1/2 and 1/2) when the host just makes a random pick, and my reasoning is correct.

Since your assumption is that the host KNOWS, then you are correct with your argument.

Since the host knows, this is not really a probability problem (randomness) anymore. One can argue that because of the your reasoning, the host will open an empty box to tempt the player to switch. Again it is assumed that the host was instructed to help keep the prize. Of course he could just be trying to make the show interesting. Good job though.

I have a similar problem. With a population of families with 2 children, given that one of the children is a girl, what is the probability that the other is also a girl? Seems quite simple? Have fun !

 

[darkness]

The probability of having a 2nd gal is still half (in actual fact, the probability is 54%, but that's a different story altogether... ). Anyway, the probability of having a guy or gal is always 50% regardless of the other children that are ALREADY born.

 

[hong2]

Not so fast.

It can be computed to 1/3. Think about it. Just for fun.

 

[roygoh]

There are 2 children. Let's say one is older than the other.

There are 4 cases:

1. Older is Boy, Younger is Boy
2. Older is Boy, Younger is Girl
3. Older is Girl, Younger is Girl
4. Older is Girl, Younger is Boy

Given one of the children is girl, and not stating whether the girl is older or younger, this means that there are 3 possible cases, which are 2, 3, and 4.

Among these 3 cases, the probablility of the other child being a girl also is then 1/3.

The trick of the question is that most people will jump to the conclusion that it is the same as saying if the older one is a girl, then what is the probability that the younger one is a girl also.

 

[Prismatic]

How about if we look at this problem from another perspective, like looking backwards from the end results.
Since now that the choice is left to the 2 boxes, the inevitable truth is, the contestant either gets the prize or not. So it's 50-50.

I understands Roy's reasoning, I think that is that it is valid if the host takes away the 3rd box, not open it and tells you to choose between the first 2 box. Once the 3rd box is opened, the uncertainty is only limited to the 2 boxes.

 

[Jenova]

Wah... wat an old thread...

Here's what I think...

Initial...
Box A: 1/3
Box B: 1/3
Box C: 1/3

Proability of winning = 1/3
Proability of NOT winning = 2/3

Assuming the player choose A and the host opens C that is empty...
Now...

Box A: 1/2
Box B: 1/2
Box C: N/A (it is as good as having 2 boxes to start with!!!)

Proability of winning = 1/2
Proability of NOT winning = 1/2

While the proability that the key is in box B increases from 1/3 to 1/2, the proability of the key in box A also increases! We should look at the problem at the proability of winning or NOT winnning.... Since proability of winning increases from 1/3 to 1/2, and proability of NOT winning decrease from 2/3 to 1/2, the answer is clear...

If there are 100 boxes and the host opens 98 of them...

Initial
Proability of winning = 1/100
Proability of NOT winning = 99/100

After opening...
Proability of winning = 1/2
Proability of NOT winning = 1/2

 

[JillValentine]

We need grid computers to solve this. :bsmilie:

 

[Yappy]

We need grid computers to solve this. :bsmilie:

I don't know the answer.

Haha... everybody is right!

 

[gooseberry]

Wow - really old thread. Never saw it before, but now that it was brought back from the depths - provided a good brain teaser during my morning tea break.

Actually, there is no need for any computer simulations - with only 3 boxes, there are so little permutations that you can draw up all the possible scenarios. But you don't even need to do that.

If you take a look at it - 1/3 of the time, the player has chosen the correct box on the first choice. So if he switches, he gets an empty box. 2/3 of the time, the player chooses the incorrect box on the first choice, so if he switches, he gets the correct box. Therefore there is a higher chance of getting the correct box if he switches.

 

[jiaklormee]

The monty hall problem:

http://en.wikipedia.org/wiki/Monty_hall_problem

The better solution is for you to switch your choice.

My 0.02 cents

 

[ahbian]

My try.

Probability of winning by NOT switching + Probability of winning by switching = 1

Probability of winning by NOT switching = 1/3

=> Probability of winning by switching = 1- 1/3 = 2/3


=> Probability of winning by switching is higher!

 

[night86mare]

what an old question.

welcome to the monty hall problem.

 

[richardg]

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:

lets say the 3 boxes are A,B,C where A contains the prize.
strategy 1 is to stay put, strategy 2 is to switch

There is equal chance of your choosing either A, B or C.

Hence the 3 equal cases are as follows

Case 1 Choose A
Strategy 1 - You Win
Strategy 2 - You loose

Case 2 Choose B
Strategy 1 - You loose
Strategy 2 - you win ( bcos host opens empty one )

Case 3 Choose C
Strategy 1 - you loose
Strategy 2 - you win ( bcos host opens empty one )


Hence strategy 2 is better with 2 wins versus strategy 1 with 1 win.

In strategy 1 you stay put hence your probablity is 1/3
In stragegy 2 you make use of the fact that the host will reveal the empty box and that your first choice is more probably wrong hence you improve yr chances.

QED

Studied this 30 years ago , was confused
30 years later... still confused :embrass:

 

[charlesleong]

Monty Hall Problem.... been debated 100000000 times already.

Simple solution is: switch - the gameshow host is NOT allowed to open the prize.

 

[Jenova]

It is a mind game... That is IF the host knows the which is the prize...

If you pick the right one, most likely the host does the door opening to confuse you...

 

[charlesleong]

Out of the 2 chests that you did no pick, the host opened one that is empty.

Of course the host is constrained not to open the prize.

 

[rvxing]

May be there is an assumption that was wrong in the 1st place. In computer programming it also depends on the same assumption, so there is NO point of using computer programming to proof the question.

The assumption that the opened box is still in the equation may be where the flaw lies. If one know that the host is going to open one of them after he selected and that open box is for sure an empty one, then it is actually only 2 boxes and not 3 boxes to begin with.

To proof it, someone can just practically try out for 100 selections with NO change and 100 selections with change. The result will show which should be the correct answer.

Just my half cent

 

[roygoh]

May be there is an assumption that was wrong in the 1st place. In computer programming it also depends on the same assumption, so there is NO point of using computer programming to proof the question.

The assumption that the opened box is still in the equation may be where the flaw lies. If one know that the host is going to open one of them after he selected and that open box is for sure an empty one, then it is actually only 2 boxes and not 3 boxes to begin with.

To proof it, someone can just practically try out for 100 selections with NO change and 100 selections with change. The result will show which should be the correct answer.

Just my half cent

Wow....cannot believe this thread was revived!

Your second paragraph has a slight flaw in the logics because the sequence of events does matter. The host only opens a box after the guest has made a selection first.

The original situation is the same as if the guest were allowed to pick one box and then the host asked him if he/she would like to trade his/her one box for the BOTH the 2 boxes that he/she did not pick.