- My Forums
- Tiger Rant
- LSU Recruiting
- SEC Rant
- Saints Talk
- Pelicans Talk
- More Sports Board
- Fantasy Sports
- Golf Board
- Soccer Board
- O-T Lounge
- Tech Board
- Home/Garden Board
- Outdoor Board
- Health/Fitness Board
- Movie/TV Board
- Book Board
- Music Board
- Political Talk
- Money Talk
- Fark Board
- Gaming Board
- Travel Board
- Food/Drink Board
- Ticket Exchange
- TD Help Board
Customize My Forums- View All Forums
- Show Left Links
- Topic Sort Options
- Trending Topics
- Recent Topics
- Active Topics
Started By
Message
0
Posted on 7/23/18 at 9:49 pm to doubleb
quote:
Envelope 1 has two aces.
Envelope 2 has an ace and a queen
Envelope 3 has two queens.
The dealer reached into a random envelope and pulls out an ace.
The odds that the remaining card is an ace or a queen is 50/50
Umm, no.
Forgot the envelope without an ace, it's irrelevant. Add all of the non-aces to envelope 2.
One envelope with 2 aces, another with 1 ace and 48 other cards. You reach in and pull out an ace.
What are the chances the next card is an ace? I'm legitimately asking you.
Posted on 7/23/18 at 9:53 pm to slackster
quote:50/50 because it is either an ace or not an ace.
What are the chances the next card is an ace?
There are only two options.
Posted on 7/23/18 at 9:54 pm to Scruffy
Et tu, Scruffe
Posted on 7/23/18 at 9:55 pm to Scruffy
Exactly. Thread/
Posted on 7/23/18 at 10:00 pm to slackster
I'd love to wager with all the 50/50 folks on this game.
I'd be like taking money from an idiot.
I'd be like taking money from an idiot.
Posted on 7/23/18 at 10:01 pm to slackster
(no message)
This post was edited on 7/23/18 at 10:09 pm
Posted on 7/23/18 at 10:05 pm to CptRusty
quote:
Lets say the balls are hollow. Inside each ball is paper with a unique indentifier for that ball, G1B1, G2B1, G1B2, S1B2.
When you pull the first gold ball, what are the odds of each respective identifier being in that ball? 1/3...
Once this pull is made, but before you look at the identifier inside...what are the odds for each of the following identifiers being found in the next ball:
G1B1 - 1/3
G2B1 - 1/3
S1B2 - 1/3
You can't account for the fact that if you selected box 2 then G1B2 is no longer an option while not ALSO accounting for the fact that G1B1 and G2B1 are each an option, but they cannot BOTH be options. It has to be one or the other.
Posted on 7/23/18 at 10:09 pm to GregMaddux
quote:
It's pointless to think about this problem like that in the practical world. What happens when you change the question to the following.
Same boxes as described by the OP. You select a box at random and pick a ball. Whats the probability that the second ball you choose from the same box is the same color (either color) as the first ball you chose?
Obviously it's 67%. Very easy to see how it's not 50%. Two out of three boxes have the same color balls. Given that question, before picking boxes starts the answer is 67%.
Anyone claiming the OP's problem's answer is equal to that (67%) is being foolish
Mind blown by stupidity
Posted on 7/23/18 at 10:09 pm to UpToPar
Thank you
Posted on 7/23/18 at 10:11 pm to UpToPar
What will the second ball be if you pull the following:
G1B1?
G2B1?
G1B2?
Hopefully you can agree that each of those is equally likely to be your starting pull. If you can agree to that much, then the distribution of possible outcomes should be obvious.
G1B1?
G2B1?
G1B2?
Hopefully you can agree that each of those is equally likely to be your starting pull. If you can agree to that much, then the distribution of possible outcomes should be obvious.
Posted on 7/23/18 at 10:13 pm to UpToPar
quote:
You can't account for the fact that if you selected box 2 then G1B2 is no longer an option while not ALSO accounting for the fact that G1B1 and G2B1 are each an option, but they cannot BOTH be options. It has to be one or the other.
As I told PearlJam, you guys should be worried that the 50/50 crew agrees with your logic.
Posted on 7/23/18 at 10:21 pm to CptRusty
quote:
What will the second ball be if you pull the following:
G1B1?
G2B1?
G1B2?
1/3
1/3
1/3
We are getting to the same answer, but we are getting there using different logic.
I'm saying that the reason the answer is 67% is because once we know that the first selection is a gold ball, there is a 67% chance you picked the box with the two gold balls and a 33% chance you selected the box with the mixed gold and silver balls. At this point, there are only two possible outcomes, a selection of the "other gold ball" or a selection of the silver ball. Because there's a 67% chance you selected the box with 2 gold balls, theres a 67% chance your second selection will be the "other gold ball."
You're looking at the probability, from the initial selection, of selecting two gold balls if we know that the first ball pulled is gold. Under your reasoning, there are three possible combinations of pulls prior to the first ball being pulled that result in the first pull being a gold ball: G1/G2, G2/G1, and G3/S1. 2 of these 3 result in two gold balls being pulled, thus, the probability of the second ball being gold is 67%
Both reasonings get to the correct answer, they just differ in how they get there.
Posted on 7/23/18 at 10:23 pm to slackster
quote:
As I told PearlJam, you guys should be worried that the 50/50 crew agrees with your logic.
There's a reason this is so. It's because their logic would be correct if they had taken into account the increased odds that the initial selection came from the box with 2 gold balls.
Posted on 7/23/18 at 10:27 pm to UpToPar
Nope
Posted on 7/23/18 at 10:27 pm to UpToPar
quote:
1/3 1/3 1/3
That is not what I asked.
quote:
'm saying that the reason the answer is 67% is because once we know that the first selection is a gold ball, there is a 67% chance you picked the box with the two gold balls and a 33% chance you selected the box with the mixed gold and silver balls.
This is circular logic. You literally just said “the reason it’s 67% is because it’s 67%”
Posted on 7/23/18 at 10:27 pm to UpToPar
quote:
There's a reason this is so. It's because their logic would be correct if they had taken into account the increased odds that the initial selection came from the box with 2 gold balls.
Yeah, I'm fine with the way you're explaining it, but pearl and I have gone back on forth over the semantics of it.
I'm focusing on what's left, as that's the question. The next ball can be one of three unique balls, two of which are gold, and one of which is silver.
Posted on 7/23/18 at 10:30 pm to UpToPar
There was no initial selection.
You picked a gold ball at random. You had a 50/50 chance of picking one, but that's not the question.
We were given a set of circumstances. A box produced s gold ball.
Now the odds that the other ball is gold I'd 50/50. That's been demonstrated.
You picked a gold ball at random. You had a 50/50 chance of picking one, but that's not the question.
We were given a set of circumstances. A box produced s gold ball.
Now the odds that the other ball is gold I'd 50/50. That's been demonstrated.
Posted on 7/23/18 at 10:32 pm to doubleb
quote:
A box produced s gold ball.
You are failing to see how this changes the distribution of possible outcomes.
Posted on 7/23/18 at 10:33 pm to CptRusty
quote:
This is circular logic. You literally just said “the reason it’s 67% is because it’s 67%”
Follow me here:
Let's label the boxes:
Box 1- 2 gold
Box 2- 1 gold 1 silver
Box 3- 2 silver.
If I tell you that you pull the first ball and its gold, what are the chances you selected the first ball from Box 1? 67%.
What are the chances you selected the 1 gold ball in Box 2? 33%.
If you selected the first ball from Box 1 then the probability of the second ball being gold is 100%.
If you selected the first ball from Box 2 then the probability of the second ball being gold is 0%.
Thus, (67% * 1) + (33% * 0)= 67%.
Page 31 of 39
Popular
Back to top
Follow TigerDroppings for LSU Football News