Question on probability...

[LegendKiller]

Ok, a co-worker and I are discussing the following scenario

You have three cups, one has a dollar bill underneath it. You choose a cup that you think has a dollar bill. The other person opens up a cup and shows there is no money, leaving two cups. What do you do?

A. Choose another cup?

B. keep your same cup?

I maintain that you keep your cup, since the current discreet probability of you choosing the correct cup is 50/50. However, he says that you should switch cups, since the cup you didn't choose has a higher probability of being correct (50%, vs your original 33%).

Sounds like a dumb question, which we both know has several facets to it regarding probably, I understand all facets, but I still maintain that you have a 50/50 chance of choosing at that point.

 

[paulney]

Originally posted by: LegendKiller
You choose a cop that you think has a dollar bill.

 

[PurdueRy]

We discussed this exact question back in probabilistic methods class. You are correct, you must redistribute the percentages based on the new information.

But it wouldn't matter if you switched or stayed because the odds are now 50/50...

 

[FP]

Not this one again.

 

[LegendKiller]

Originally posted by: paulney

Originally posted by: LegendKiller
You choose a cop that you think has a dollar bill.

Har har

 

[LegendKiller]

Originally posted by: binister
Not this one again.

We all know that the search function is less than perfect. In fact I calculated the probability of actually finding this and...

 

[mooglekit]

Originally posted by: PurdueRy
We discussed this exact question back in probabilistic methods class. You are correct, you must redistribute the percentages based on the new information.

But it wouldn't matter if you switched or stayed because the odds are now 50/50...

winner! The third cup becomes irrelevant the moment you found out there was nothing beneath it...you're now working in a world of only 2 cups.

 

[James Bond]

Originally posted by: mooglekit

Originally posted by: PurdueRy
We discussed this exact question back in probabilistic methods class. You are correct, you must redistribute the percentages based on the new information.

But it wouldn't matter if you switched or stayed because the odds are now 50/50...

winner! The third cup becomes irrelevant the moment you found out there was nothing beneath it...you're now working in a world of only 2 cups.

Wrong.

Isn't this the same as the GOAT/FERRARI riddle?

If you stay with your original cup, your odds are 33% (because you chose 1/3). If you switch, your odds are 66% (you are choosing between 3, but you already know one is wrong, thus the percentage is 2/3)

 

[Viper GTS]

The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

 

[LegendKiller]

Originally posted by: Tizyler
Wrong.

Isn't this the same as the GOAT/FERRARI riddle?

If you stay with your original cup, your odds are 33% (because you chose 1/3). If you switch, your odds are 66% (you are choosing between 3, but you already know one is wrong, thus the percentage is 2/3)

This is only the case if you know the original cup you chose was wrong also. You cannot effectively eliminate your cup as an alternative.

 

[Vic]

The plane takes off.

 

[blackdogdeek]

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

 

[angminas]

Originally posted by: Tizyler

Wrong.

Wrong.

 

[Viper GTS]

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

This may help for some. The real issue here is that to achieve the 50% odds they think exist you would need to surrender your cup and then blindly choose again. Without that step you are always better switching, whether it's 3 or 100.

Viper GTS

 

[mooglekit]

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

From a mathematical probability theory perspecitve, you don't have the luxury of ignoring obtained information. If you start with 100 cups and remove all but 2, you can't say your odds are 1/100 for the original cup because there were 100 when I picked it, but if I switch I have a 1/2 chance. The entire problem, and the probabilities, change every time the system changes and the probabilities must be redistributed. If there are 2 cups left, and you pick one of them you will always have a 50% chance of picking the correct cup, regardless of how many cups you started with.

 

[jman19]

Originally posted by: LegendKiller

Originally posted by: Tizyler
Wrong.

Isn't this the same as the GOAT/FERRARI riddle?

If you stay with your original cup, your odds are 33% (because you chose 1/3). If you switch, your odds are 66% (you are choosing between 3, but you already know one is wrong, thus the percentage is 2/3)

This is only the case if you know the original cup you chose was wrong also. You cannot effectively eliminate your cup as an alternative.

Um, if you know your cup doesn't have the dollar, and you know which of the other cups doesn't have it... :roll:

 

[UCDAggies]

You should watch. The probability is 2/3 if you switch.

 

[angminas]

Dammit moogle, you beat me to it! Damn.

 

[LegendKiller]

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

This is a great way of putting it, after doing some research.

 

[UCDAggies]

Does the person who removed the second cup know that the dollar isn't under it, or is it random.

If the person always removes a cup with no dollar than it is 2/3, if they are picking cups at random than it 1/2.

 

[UCDAggies]

Originally posted by: mooglekit

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

From a mathematical probability theory perspecitve, you don't have the luxury of ignoring obtained information. If you start with 100 cups and remove all but 2, you can't say your odds are 1/100 for the original cup because there were 100 when I picked it, but if I switch I have a 1/2 chance. The entire problem, and the probabilities, change every time the system changes and the probabilities must be redistributed. If there are 2 cups left, and you pick one of them you will always have a 50% chance of picking the correct cup, regardless of how many cups you started with.

This is incorrect, since if the person who removed one of the cups knew it didn't have a dollar, than his choice was influenced by your choice, thus the original probability still holds.

 

[crownjules]

It can be explained best like this.

Assume that Cup 1 contains the object.

SCENARIO 1: You pick Cup 1. Cup 2 or 3 is shown. If you KEEP the cup, you WIN. If you SWITCH the cup, you LOSE.

SCENARIO 2: You pick Cup 2. Cup 3 is shown to you. If you KEEP the cup, you LOSE. If you SWITCH, you WIN.

SCENARIO 3: You pick Cup 3. Cup 2 is show to you. If you KEEP the cup, you LOSE. If you SWITCH, you WIN.

In 2 of the 3 scenarios, you win when switching and lose by keeping. That's why it's better to switch.

 

[b0mbrman]

Originally posted by: LegendKiller
Ok, a co-worker and I are discussing the following scenario

You have three cups, one has a dollar bill underneath it. You choose a cup that you think has a dollar bill. The other person opens up a cup and shows there is no money, leaving two cups. What do you do?

A. Choose another cup?

B. keep your same cup?

I maintain that you keep your cup, since the current discreet probability of you choosing the correct cup is 50/50. However, he says that you should switch cups, since the cup you didn't choose has a higher probability of being correct (50%, vs your original 33%).

Sounds like a dumb question, which we both know has several facets to it regarding probably, I understand all facets, but I still maintain that you have a 50/50 chance of choosing at that point.

Choose another

[/Marilyn vos Savant]

 

[Kev]

Originally posted by: LegendKiller

Originally posted by: blackdogdeek

Originally posted by: Viper GTS
The correct answer is switch, yet some people still haven't figured that out.

Viper GTS

this is correct.

the easiest way i've heard the solution explained so that it made sense was this:

imagine there are 100 cups. you choose #1. the person then flips over all the other cups except for cup #42. wouldn't you change to cup #42?

technically, your original cup #1 still has a 1/100 chance of having the dollar bill but cup #42 has a 99/100 chance of having the dollar bill.

This is a great way of putting it, after doing some research.

Yay, great. Please no responses after this. If this thread goes past 100 posts I will be very upset.

 

[olds]

Originally posted by: binister
Not this one again.

Get used to it.
There are only 5 original threads. They all just get recyled in one way or the other.