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Message
re: Solve this OT
Posted on 7/23/18 at 10:35 pm to UpToPar
Posted on 7/23/18 at 10:35 pm to UpToPar
You have correctly demonstrated that the starting state determines the outcome. You are failing to see that there are three possible outcomes. The fact that two of those outcomes have the same color doesn’t make them any less distinct.
ETA: but again you have hand waved the “why”. Although the math is correct, you haven’t shown how to arrive at the 67%.
Why is there a three in the denominator, and not a two?
ETA: but again you have hand waved the “why”. Although the math is correct, you haven’t shown how to arrive at the 67%.
Why is there a three in the denominator, and not a two?
This post was edited on 7/23/18 at 10:38 pm
2
Posted on 7/23/18 at 10:38 pm to CptRusty
There are three possible outcomes that result in the first ball being gold. I agree with you there.
Posted on 7/23/18 at 10:40 pm to CptRusty
quote:
ETA: but again you have hand waved the “why”. Although the math is correct, you haven’t shown how to arrive at the 67%.
Why is there a three in the denominator, and not a two?
Because there are three gold balls. If we know that the first ball pulled is gold, then the first ball pulled is either G1B1, G2B1, or G3B2. Two of those three outcomes results in the first selection being from Box 1 and one of those outcomes results in the first selection being from Box 2. Thus, a 2/3 chance that you are pulling from box 1 and a 1/3 chance you are pulling from box 2.
Posted on 7/23/18 at 10:45 pm to UpToPar
quote:
There are three possible outcomes that result in the first ball being gold. I agree with you there.
And there are three possible outcomes for the second ball as a result.
To say there are only two options because you either chose box 1 or box 2 is illogical. If you believe that, then you must also believe there are only two options for the first gold ball, box 1 or box 2. That's the same faulty logic.
This post was edited on 7/23/18 at 10:45 pm
Posted on 7/23/18 at 11:14 pm to slackster
quote:
To say there are only two options because you either chose box 1 or box 2 is illogical. If you believe that, then you must also believe there are only two options for the first gold ball, box 1 or box 2. That's the same faulty logic.
Nope. On the first pull, there are 3 possible outcomes (G1, G2, or G3). After the first pull, there are 2 gold balls left, but at least one of those gold balls is an impossible outcome (because it’s in the other box).
Posted on 7/23/18 at 11:52 pm to UpToPar
Mathematics is the language of the universe.
The OT is fricking deaf/mute in that regard. Goddamn, you are some stupid motherfrickers. Simple counting will give you the answer.
The OT is fricking deaf/mute in that regard. Goddamn, you are some stupid motherfrickers. Simple counting will give you the answer.
This post was edited on 7/23/18 at 11:59 pm
Posted on 7/24/18 at 12:14 am to TigerstuckinMS
I understand both ways of coming up with the answer.
PearJam's way starts and ends with probability.
CptRusty's way is logic and to me more akin to common core math.
Someone else earlier in the thread explained the answer almost the exact same way Pearl did and did not get attacked for it. You have a 50/50 shot of drawing gold on the second pull. However, you have to add in the extra 16.7% chance that you pulled the box with 2 gold balls in it to begin with. So either way you look at it your odds are ~67% to pull a gold ball on the second pull.
PearJam's way starts and ends with probability.
CptRusty's way is logic and to me more akin to common core math.
Someone else earlier in the thread explained the answer almost the exact same way Pearl did and did not get attacked for it. You have a 50/50 shot of drawing gold on the second pull. However, you have to add in the extra 16.7% chance that you pulled the box with 2 gold balls in it to begin with. So either way you look at it your odds are ~67% to pull a gold ball on the second pull.
Posted on 7/24/18 at 7:50 am to CptRusty
quote:
You are failing to see how this changes the distribution of possible outcomes.
The opposite.
Once a gold ball is taken from a box it tells you all but which of two balls are left.
The single silver, or the twin of the paired gold.
Do this riddle. There are three balls in a box. Two are gold and one is silver. The odds of picking a gold ball out of the box is 2/3 or 67%.
The two riddles are completely different, they can't have the same answer.
Posted on 7/24/18 at 7:58 am to doubleb
The moment you picked gold, because there are two golds in one box and one in the other, you had a 67% chance of picking the box with two golds. Which means that’s your chance of picking a second gold. Period.
This post was edited on 7/24/18 at 7:59 am
Posted on 7/24/18 at 8:02 am to doubleb
quote:
The opposite.
Once a gold ball is taken from a box it tells you all but which of two balls are left.
The single silver, or the twin of the paired gold.
Do this riddle. There are three balls in a box. Two are gold and one is silver. The odds of picking a gold ball out of the box is 2/3 or 67%.
The two riddles are completely different, they can't have the same answer.
This has been run through computer programs that clearly show the answer is 67%.
Do it yourself.
3 possible gold balls you can chose. You have to pick a gold ball first. Two of those three gold balls will result in another gold ball being chosen.
Like I told you earlier, take it to the extreme. You have a jar with 100 gold balls. You have another jar with 1 gold ball and 99 silver balls.
You randomly select a jar then randomly select a ball that is gold. What are the odds the next ball is gold?
Your inaccurate logic would suggest its still 50/50, which is blatantly false.
Posted on 7/24/18 at 8:06 am to DannyB
quote:
Someone else earlier in the thread explained the answer almost the exact same way Pearl did and did not get attacked for it. You have a 50/50 shot of drawing gold on the second pull. However, you have to add in the extra 16.7% chance that you pulled the box with 2 gold balls in it to begin with. So either way you look at it your odds are ~67% to pull a gold ball on the second pull.
Which is the wrong way to come up with the right answer.
You have a 67% chance of pulling gold on the second pull. The two pulls are not independent events. One dictates the the other. You don't add back the 16.7% chance and all that jazz. That probability isn't lost on the second pull.
You don't have a 50/50 chance to pull gold on the second pull because there are three possible balls you can select on the second pull, and two of them are gold.
Pearl and others conflate the fact that it will either be gold or silver with only two possible solutions, when the probability suggest there are 3 possible solutions (balls) but only 2 colors.
Posted on 7/24/18 at 8:10 am to doubleb
quote:
doubleb
bless your heart
Posted on 7/24/18 at 8:19 am to slackster
Actually, mathematically, that works when accounting for the difference between each box. It’s kind of going at it backwards but the math isn’t wrong.
Posted on 7/24/18 at 8:21 am to ell_13
The fact that so many people are this stupid and stubborn in regards to easy problems on this board is scary. 
Posted on 7/24/18 at 8:21 am to DannyB
quote:
CptRusty's way is logic and to me more akin to common core math.
Jesus tap dancing Christ don't lump me in with those martians.
All I am doing is giving the (correct) explanation as to where the '2' comes from for the numerator and where the '3' comes from for the denominator.
Eddie Vedder's problem is that he is assuming the observer knows which box has been chosen once the first pull is made, which is false.
Once you've made the first pull, you still don't know which of the three initial gold balls is in your hand. This leaves the all three options still on the table, all equally likely.
Yes outcome has been determined and in reality there is only one option left, but this is again imputing knowledge to the observer which they do not have.
This post was edited on 7/24/18 at 8:23 am
Posted on 7/24/18 at 8:22 am to ell_13
quote:
Actually, mathematically, that works when accounting for the difference between each box. It’s kind of going at it backwards but the math isn’t wrong.
It's correct, but it's a dicey way to explain it that becomes more difficult with bigger numbers. Like you said, it's backwards.
Posted on 7/24/18 at 8:24 am to CptRusty
Thank you sir.
I can always use a blessing.
Uncle
I can always use a blessing.
Uncle
This post was edited on 7/24/18 at 8:33 am
Posted on 7/24/18 at 9:04 am to slackster
I always look at it as taking the separate boxes out of the equation, and put them all in one. You don't know which box you're gonna grab from next, so you only have a 2/3 chance of grabbing a gold one :dunno :
Posted on 7/24/18 at 9:08 am to slackster
How in the flying frick is this still going? 
Posted on 7/24/18 at 11:21 am to KosmoCramer
quote:
Best of luck picking the goat door with Monte Hall
That doesn't make sense in this scenario. Because Monte Hall didn't force you to pick a certain door at the onset. In the Monte Hall problem I could just as likely be holding the AS box, too. But in this scenario, I not only don't have access to the AS box, I don't even have access to the silver ball in the M box. I am required to take the gold ball
Even Bertrands box says this is a paradox. If truly allowed to select from all boxes, the probability is reduced to 1/2. You can only derive the probability of 2/3, if all boxes are available to the selector. You have told me I simply cannot be allowed to select the AS box
Therefore, you have totally changed the 'probability' of the entire problem
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