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Red Marble?

Name: Anonymous 2007-11-19 21:06

You have a bag with 7 marbles, 6 are blue and 1 is red. 7 people will choose, one at a time, trying to avoid the red marble. Is there any advantage to when you choose? (1st , last or other) Some say last since that allows everyone else to take their 1 in 7 chance before you, therefore increasing the chances one of them will get it before you. Others feel it makes no difference, since the odds are 1 in 7 for every one.

Name: Anonymous 2007-11-19 21:11

No advantage, unless the whole thing is stopped as soon as the red marble is drawn, in which case go last.

Name: Anonymous 2007-11-19 21:18

Assuming the marbles aren't replaced, then no all spots have equal (1 in 7) chance of losing.  If the marbles ARE replaced and the game ends once someone draws red, then it's better to be last, since the chance of the nth person losing is (6/7)^(n-1)*(1/7)  (this scenario doesn't guarantee that someone loses in the first 7 draws, however)

Name: Anonymous 2007-11-19 21:25

You are confusing the question.

It is asking for the odds in choice of spot, not the actual pick.
Before any picks are made out of the bag, what is the best option to pick? Should you pick out of the bag first, second, or third and so on?
If you go first you have a 1/7 shot
Second would have a 1/6 but only if #1 didn't pick

You have to add in the probability to each pick that the red marble isn't picked.

So wouldn't all slots be 1/7?

Name: Anonymous 2007-11-19 22:13

So wouldn't all slots be 1/7?
Yes. As >>2 and >>3 said.

Name: Anonymous 2007-11-19 22:24

even if the marbles arent replaced the odds are still 1/7 for each person that goes. if you go first you have a 1 in 7 shot and if you go 2nd you have a 1/6 shot if the guy before you didnt get the marble...however if lets say your the 7nth and everyone else has picked and noone has the marble there is a 100% chance that you will get it basically allowing others to dictate your faith for you

Name: Anonymous 2007-11-20 18:53

It doesn't matter what position you take.  Everyone, no matter what position, has a 1/7 chance to draw the red marble.  Think about it like this: If you go last, and the first six choose before you, but no one reveals what color they have, what are the odds that you draw the red marble on the final draw?  1/7.  You don't know if anyone got it first.  But going last makes it like this:  6 draws are made before yours.  Your draw is only 1 draw.  there's a 6/7 chance that ANYONE ELSE will get the red marble.  There's a 1/7 chance that YOU will get the red marble.  6/7 + 1/7 = 100% chance of SOMEBODY getting the red marble.

Just like on Deal or No Deal, where there's a constant 1/26 chance that the contestant has the million-dollar case, no matter how many have been eliminated.  The odds that they flash on the screen from time to time are bullshit that's dumbed down so that the American part of the audience feels like they understand it.

Name: Anonymous 2007-11-20 18:55

If the marbles are taken out of the bag, the person with the best chance is the first one to draw.

Name: Anonymous 2007-11-20 20:28

>>8
Not actually true. Read the rest of the thread, faggot.

Name: Squirrel King 2007-11-25 4:13

Assuming that everyone keeps the marble as they draw, the math would be as such:

First person draws - 7 marbles

1/7 chance

Second person draws - 6 marbles

1/6 + 1/7 = 13/42 chance

Third

1/5 + 1/6 + 1/7 = 9/105

Fourth
      (1/5 + 1/6 + 1/7)
1/4 + 9/105 = 13/420

Fifth

1/3 + 13/420 = 4/315

Sixth

1/2 + 4/315 = 1/105

I don't know though, I'm terrible with probability.

Name: Anonymous 2007-11-25 4:17

I don't know though, I'm terrible with probability.
Ain't that the truth.

Name: Anonymous 2007-11-25 6:03

You are in nth position, n=1,2...7

There are 7-(n-1) balls left when you pick.

P(You get red) = P(Red is left on your go)*P(You pick red given that it's still there)

P(Red is left on your go) = 1 if you're first, 6/7 if you're second, 5/7 if you're third etc = [7-(n-1)]/7

P(You pick it if it is left) = 1/7 if you're first, 1/6 if you're second etc = 1/[7-(n-1)]
 
P(You get red) = [7-(n-1)]/7 * 1/[7-(n-1)] = 1/7

Fuck yeah, Bayes.

>>7
About DoND, they don't give probabilities over here in Britain (at least they didn't last time I saw it) but what really annoys me is when they talk about tactics, there are no fucking tactics, you're just picking boxes at random!

Name: Anonymous 2007-11-29 22:54

>>7

You're confusing DonD with the Monty Hall game.  The difference is that in Monty Hall, the door that is revealed is opened by someone who knows that the prize is not behind that door.  In DonD, the cases are opened by someone with no such information.  The difference is that in DonD, the odds that they report on the show really are the correct ones.  The odds of having the million (or whatever the highest amount is on a given day) changes as the game progresses.  It is only in a Monty Hall-like scenario that the probability of the initial choice being a winner stays constant.

Name: Anonymous 2007-11-30 13:28

It has already been decided that one of those people will draw the red ball, that person has 100% chance of getting the red ball as he is the only one who can draw it.

Name: Anonymous 2007-12-04 18:56

There would also be no advantage/disadvantage if everyone just simultaneously grabbed a marble. The same exact probability occurs when you go last and when you go first.

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