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Author Topic: Stat question
Napolean
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Icon 1 posted April 13, 2006 12:10      Profile for Napolean     Send New Private Message       Edit/Delete Post   Reply With Quote 
Part 1
5,000 cups, one has a ball under it. What is your chance of finding the ball?

Part 2
If Monty takes away 4998 empty(bad) ones, what is your probability of finding the ball?

Posts: 3 | From: Ohio | Registered: Apr 2006  |  IP: Logged
Sxeptomaniac

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Icon 1 posted April 13, 2006 12:18      Profile for Sxeptomaniac   Author's Homepage     Send New Private Message       Edit/Delete Post   Reply With Quote 
If you ask a forum of geeks to do your homework for you on your first post, what is the probability that they will give you your answer?

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Let's pray that the human race never escapes from Earth to spread its iniquity elsewhere. - C. S. Lewis

Posts: 1590 | From: Fresno, CA | Registered: Mar 2005  |  IP: Logged
Napolean
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Icon 1 posted April 13, 2006 12:24      Profile for Napolean     Send New Private Message       Edit/Delete Post   Reply With Quote 
Naa, this is not homework. It a disagreement amonst friends.
Posts: 3 | From: Ohio | Registered: Apr 2006  |  IP: Logged
Stereo

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Icon 1 posted April 13, 2006 12:35      Profile for Stereo     Send New Private Message       Edit/Delete Post   Reply With Quote 
For question two, is the cup chosen before or after the 4998 others are removed? Are you allowed to change your choice? That makes the difference between 1/2, 1/5000 and 4999/5000. I'll let you choose which probability fits which scenario. (Given, of course, that I understand correctly that part of statistical games, but I think I do. I even think I can explain it cleanly.)

As for question 1, it is really worth answering? It is really so evident... 1 ball out of 5000 cups... [Roll Eyes]

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Eppur, si muove!

Galileo Galilei

Posts: 2289 | From: Gatineau, Quebec, Canada | Registered: Apr 2001  |  IP: Logged
Napolean
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Icon 1 posted April 13, 2006 13:02      Profile for Napolean     Send New Private Message       Edit/Delete Post   Reply With Quote 
Let me use the exact words of the person who I disagree with.

Make this Part 2 instead..

I will now remove 4,998 different cups from the game. This means that there are 2 cups remaining in the game. One of them is the cup that you selected in step 1. The other cup is from the original 4,999 cups. One of these cups contains the ball, the other cup does not. What are the odds that the cup you selected in step 1 contains the hidden ball?

Posts: 3 | From: Ohio | Registered: Apr 2006  |  IP: Logged
drunkennewfiemidget
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Icon 1 posted April 13, 2006 13:04      Profile for drunkennewfiemidget     Send New Private Message       Edit/Delete Post   Reply With Quote 
quote:
Originally posted by Napolean:
Let me use the exact words of the person who I disagree with.

Make this Part 2 instead..

I will now remove 4,998 different cups from the game. This means that there are 2 cups remaining in the game. One of them is the cup that you selected in step 1. The other cup is from the original 4,999 cups. One of these cups contains the ball, the other cup does not. What are the odds that the cup you selected in step 1 contains the hidden ball?

1 in 5000
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GrumpySteen

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Icon 1 posted April 14, 2006 17:30      Profile for GrumpySteen     Send New Private Message       Edit/Delete Post   Reply With Quote 
It's trick question based on semantic nitpicking.

The chances of picking a cup with the ball in it can only be calculated at the time the choice is made. When the choice is made it was 1 in 5000. Even if you remove all the other cups, the odds of the ball being in the cup you picked haven't changed. It was still a 1 in 5000 chance, you just happen to know that the results of the choice now.

If you get to choose again, however, then you get a new set of odds to go with the new choice which, in the example given, will be 1 in 2. The odds on the original choice have not changed.

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Worst. Celibate. Ever.

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Stereo

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Icon 1 posted April 14, 2006 20:09      Profile for Stereo     Send New Private Message       Edit/Delete Post   Reply With Quote 
quote:
Originally posted by Steen:
It's trick question based on semantic nitpicking.

The chances of picking a cup with the ball in it can only be calculated at the time the choice is made. When the choice is made it was 1 in 5000. Even if you remove all the other cups, the odds of the ball being in the cup you picked haven't changed. It was still a 1 in 5000 chance, you just happen to know that the results of the choice now.

If you get to choose again, however, then you get a new set of odds to go with the new choice which, in the example given, will be 1 in 2. The odds on the original choice have not changed.

Just to be sure, if you can only choose after 4998 bad cups are removed, yes it's a 1/2. If you pick one first, then 4998 are removed, and then you get to choose again, you have 1/5000 chance if you pick the same, and 4999/5000 if you pick the other. (That is of course, if the cups aren't mixed up in between, as this would reset the odds.)

Think of it this way. When there are 5000 cups on the table, you are asked to pick one. The chance to pick the good is 1 out of 5000. Then the "trick master" regroup all the other cups, and ask you if you think the ball is in the cup you choose at first, or one of the 4999 other cups. Then obviously, you have 4999 chances out of 5000 that the ball is in the group of cups. That the trick master removes 4998 bad cups is irrelevant in the probability count, as he can only remove bad ones, and there is only one good cup. So unless the probability is reset in some way, the chances are 1/5000 (the one chosen at first) vs. 4999/5000 (the others), for a total of 1/1. (Unless it really is a trick and the ball was put under none, and the trick master pretends it was under the one you didn't pick, but that's a whole other story.)

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Eppur, si muove!

Galileo Galilei

Posts: 2289 | From: Gatineau, Quebec, Canada | Registered: Apr 2001  |  IP: Logged


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