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Message
re: Solve this OT
Posted on 7/23/18 at 10:29 am to NewGrad1212
Posted on 7/23/18 at 10:29 am to NewGrad1212
No, you’re looking at it wrong. The question is not “what would your odds be of picking 2 gold balls?” The question is “knowing that you already picked a gold ball, what are the odds of you picking a 2nd one?”
Knowing you already picked a gold ball dramatically increased the odds of you picking a second one. It’s like a coin flip. The odds of you flipping a coin and getting heads 10 times in a row is .000967%. But if I were to tell you that you’ve flipped heads 9 times in a row already, then now your odds of getting heads 10 times in a row have dramatically increased (to 50%).
It’s the same concept here: the original premise changes the odds of the problem.
Knowing you already picked a gold ball dramatically increased the odds of you picking a second one. It’s like a coin flip. The odds of you flipping a coin and getting heads 10 times in a row is .000967%. But if I were to tell you that you’ve flipped heads 9 times in a row already, then now your odds of getting heads 10 times in a row have dramatically increased (to 50%).
It’s the same concept here: the original premise changes the odds of the problem.
2
Posted on 7/23/18 at 10:29 am to castorinho
quote:
See your problem is that you're STILL accounting for both gold balls in box one AFTER drawing.
Negative.
Each number represents a single scenario.
Posted on 7/23/18 at 10:29 am to NewGrad1212
Think of it this way people.
Instead of 2 balls, each box has 50 balls.
One has all gold, one has all silver, one has 49 silver and one gold.
If you draw a gold ball and have to pick another from the same box, the probability that it is going to be gold again is huge. Because the probability that you grabbed the one gold ball from the box of 49 silver is not good. Just because there are two boxes left, and drawing from one will definitely be gold while drawing from the other will definitely be silver, doesn't mean it's a 50/50 chance.
The same logic applies to the scenario with only two balls per box, just not on as grandiose a scale.
Instead of 2 balls, each box has 50 balls.
One has all gold, one has all silver, one has 49 silver and one gold.
If you draw a gold ball and have to pick another from the same box, the probability that it is going to be gold again is huge. Because the probability that you grabbed the one gold ball from the box of 49 silver is not good. Just because there are two boxes left, and drawing from one will definitely be gold while drawing from the other will definitely be silver, doesn't mean it's a 50/50 chance.
The same logic applies to the scenario with only two balls per box, just not on as grandiose a scale.
This post was edited on 7/23/18 at 10:31 am
Posted on 7/23/18 at 10:29 am to CptRusty
quote:
If you have only the first two boxes, and you pull a single ball from each of those boxes, at random, 100 times...what do you think the outcome distribution will look like?
How many times will you have pulled gold, how many times will you have pulled silver?
This is no way is relevant to the question in the OP ..
But you would pull a gold ball 75 times and a silver 25 times.
Posted on 7/23/18 at 10:30 am to Havoc
quote:
you either picked GG box or GS box
Thats not what it is asking. You dont know which of those 2 you picked. It is basically saying you ahve 3 balls, 2 gold and one silver. What are the odds that you reach in the box (without knowing which of the 2 boxes you started with) and pick a gold ball. 2/3
Posted on 7/23/18 at 10:31 am to Cold Drink
Ahh I got you. That's a good point. 
Posted on 7/23/18 at 10:31 am to Havoc
quote:
You either picked GG box or GS box.
But you've eliminated the 1/4 possibility of picking silver by starting with the fact that you've drawn gold.
If it were not for the fact that drawing gold first was already established, you'd be correct...but the fact that you've completely eliminated the possibility of drawing silver first changes the odds.
Posted on 7/23/18 at 10:33 am to Cold Drink
quote:
It’s like a coin flip. The odds of you flipping a coin and getting heads 10 times in a row is .000967%. But if I were to tell you that you’ve flipped heads 9 times in a row already, then now your odds of getting heads 10 times in a row have dramatically increased (to 50%).
I think some fluid leaked out of my brain. Isn’t a coin toss completely independent of the prior toss and thus 50/50 each time? It’s not as if the prior outcome affects the next unlike the present problem.
Posted on 7/23/18 at 10:33 am to castorinho
quote:You have to add the 16.67% probability difference in choosing a gold from box 1 compared box 2. You don't know which box you chose from. But originally there was a 33.33% chance of pulling gold from box 1 and a 16.67% chance of pulling it from box two. You KNOW you grabbed a gold in this single instance which was 16.67% more likely to happen from box one. So that gets added to the 50% of picking gold in the first place.
See your problem is that you're STILL accounting for both gold balls in box one AFTER drawing.
This post was edited on 7/23/18 at 10:40 am
Posted on 7/23/18 at 10:33 am to memphis tiger
quote:
It is basically saying you ahve 3 balls, 2 gold and one silver. What are the odds that you reach in the box (without knowing which of the 2 boxes you started with) and pick a gold ball. 2/3
HUH
Posted on 7/23/18 at 10:33 am to Fe_Mike
quote:
Because the probability that you grabbed the one gold ball from the box of 49 silver is not good
Most important part. It’s more likely that you’ve drawn from the box with 2 gold than the box with 1.
Posted on 7/23/18 at 10:35 am to TH03
I'm going to repost this because people are only thinking in terms of balls and not in terms of unique situations:
Let's call the balls B1G1, B1G2, B2G1, and B2S1 (box 1, gold 1, etc.) Here are the different outcomes you can possibly have when drawing a gold ball first:
1) Draw B1G1. Your remaining ball is B1G2. A gold ball is the outcome
2) Draw B1G2. Your remaining ball is B1G1. A gold ball is the outcome
3) Draw B2G1. Your remaining ball is B2S1. A silver ball is the outcome.
2/3 scenarios result in a gold ball being drawn second.
Let's call the balls B1G1, B1G2, B2G1, and B2S1 (box 1, gold 1, etc.) Here are the different outcomes you can possibly have when drawing a gold ball first:
1) Draw B1G1. Your remaining ball is B1G2. A gold ball is the outcome
2) Draw B1G2. Your remaining ball is B1G1. A gold ball is the outcome
3) Draw B2G1. Your remaining ball is B2S1. A silver ball is the outcome.
2/3 scenarios result in a gold ball being drawn second.
Posted on 7/23/18 at 10:35 am to memphis tiger
quote:
(without knowing which of the 2 boxes you started with)
So that’s the rub.
ETA: it says the same box. So you do know you’re drawing from either GG or GS. The “extra” gold ball isn’t relevant.
This post was edited on 7/23/18 at 10:37 am
Posted on 7/23/18 at 10:36 am to CptRusty
I am confused at the 66.7% answer.
After you have already drawn once, the odds change.
These two cannot both exist in the same vein on the second draw. You can either take the remaining gold ball or the remaining grey ball based on which box you got originally.
50%, what am I missing?
quote:
- Take G1 from B1
2 - Take G2 from B1
After you have already drawn once, the odds change.
These two cannot both exist in the same vein on the second draw. You can either take the remaining gold ball or the remaining grey ball based on which box you got originally.
50%, what am I missing?
Posted on 7/23/18 at 10:36 am to ell_13
quote:
You have to add the 16.75% probability difference in choosing a gold from box 1 compared box 2. You don't know which box you chose from. But originally there was a 33.33% chance of pulling gold from box 1 and a 16.67% chance of pulling it from box two. You KNOW you grabbed a gold in this single instance which was 16.67% more likely to happen from box one. So that gets added to the 50% of picking gold in the first place.
Posted on 7/23/18 at 10:37 am to CptRusty
quote:Which was my mistake initially. I focused on ignoring the 3rd box completely. But even if you started with two boxes, that answer is the same giving you picked gold as well.
but the fact that you've completely eliminated the possibility of drawing silver first changes the odds.
Posted on 7/23/18 at 10:40 am to Havoc
quote:
Isn’t a coin toss completely independent of the prior toss and thus 50/50 each time
Yes, but those odds compound if you want to find the probability of a group of coin tosses.
So the odds of flipping heads 4 times in a row = .5*.5*.5*.5
But if you've already flipped heads 3 times, then your chances of flipping heads 4 times in a row are now dramatically higher (50%) than if you've started from scratch.
Posted on 7/23/18 at 10:40 am to Displaced
quote:
After you have already drawn once, the odds change.
These two cannot both exist in the same vein on the second draw. You can either take the remaining gold ball or the remaining grey ball based on which box you got originally.
50%, what am I missing?
7 pages of this thread, and you don't get it still?This post was edited on 7/23/18 at 11:26 am
Posted on 7/23/18 at 10:40 am to 50_Tiger
2/3 bc there are 3 boxes and 2 of the boxes still have a gold ball
Posted on 7/23/18 at 10:41 am to CptRusty
Yanny.
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