[Closed] Math question

Originally posted by Shevtshenko

Originally posted by jbpatzer
But the point is that it's much more interesting if there is a villain. And it's much more like poker. I'm really into this stuff! ♦

What happens if there are three colours? Or M colours?

Best argument in this thread!

I asked a friend about this (Masters in Mathematics and PHD in string theory) and he says the problem we are concerned with seems to not follow bayesian probability theory by nature.

The colour of the last ball is not determined by previous events and thus the chance of it being green is 1/(all colours out there in the spectrum).

Put simply, unless the question is worded incorrectly we cannot take worse odds than that.

Originally posted by DaPhunk
I asked a friend about this (Masters in Mathematics and PHD in string theory) and he says the problem we are concerned with seems to not follow bayesian probability theory by nature.

The colour of the last ball is not determined by previous events and thus the chance of it being green is 1/(all colours out there in the spectrum).

Put simply, unless the question is worded incorrectly we cannot take worse odds than that.

and we are not interested with the color of the last ball it's either green or not

Originally posted by roswellx

Originally posted by DaPhunk
I asked a friend about this (Masters in Mathematics and PHD in string theory) and he says the problem we are concerned with seems to not follow bayesian probability theory by nature.

The colour of the last ball is not determined by previous events and thus the chance of it being green is 1/(all colours out there in the spectrum).

Put simply, unless the question is worded incorrectly we cannot take worse odds than that.

and we are not interested with the color of the last ball it's either green or not

String theory is certainly a load of balls, so I'm sure he'd know.

Originally posted by cyzo
But jbpatzer, you are answering a different question that you made up because you knew the answer to it. The question OP posed can only be answered using information not in the question. One would have to posit the probability of details left out of the original question.

I tend to agree with this...

Originally posted by jbpatzer

Originally posted by cyzo
But jbpatzer, you are answering a different question that you made up because you knew the answer to it. The question OP posed can only be answered using information not in the question. One would have to posit the probability of details left out of the original question.

Well, I didn't know the answer when I posed the question, and the original question didn't make sense, on which point you seem to be agreeing with me (highlighted above)!

The information provided is not enough for the answer, as there is not known probability of the G vs N. If there are only 2 colors, then the answer is 50/50.

However, only 10% of jars containing N colored ball get to this state while 90% fail to have 9 consecutive G balls drawn.
Once in this state it is not relevant unless it is known the number of jars containing N ball is the same as all G jars.

For me, ordering the balls seemed to help my thinking. It doesn't seem like a fallacy to do so, as you can have situations where a different ball in the pack that you pick is Green/Not Green. There are 11 possible combinations of colours that enable us to be able to draw out 9 green balls.

GGGGGGGGGG
GGGGGGGGGN
GGGGGGGGNG
...
...
NGGGGGGGGG

For an all-Green bag, the chance of you picking 9 Greens initially is 1/1.

For a given bag with one non-Green, the chance of you picking 9 Greens initially is 1/10.

Chance of an all-Green bag: 1/11
Chance of a one-non-Green bag: 10/11

So chance of ball being Green = 1/11*1/1+10/11*9/10=10/11
Chance of ball being not Green=1/11*0/1+10/11*1/10=1/11

So if I were to bet on taking out a Green, I would only take odds that were better than lose £10 if not Green, and win £1 if Green.

Amirite? Or amianidiot.

Of course, this has to assume that the colour of one ball does not affect the colour of another ball, and the chance of it being Green is equal to that of it not being Green. If the probabilities were not 50/50 but known (say 70/30) they could be calculated in a similar way. If not known, or the events are dependent, then there is no way to calculate the answer.

Any obvious fallacies here?

why dont you just look in the bag? ♠_biggrin:

Rather than:

NG
GG

Shouldn't it be:

"
NH
GG
You f*****g lucky fish!"

*Sigh. Must...stop...trolling...*

I thought I'd sorted this one out earlier in the thread. Sigh.........

it's green or not. that makes it 50/50. did anyone say that yet?

OFC, chances that last ball is green or not is 50/50 if balls were placed in jar randomly with equal possibilities for each color. But if we want to bet for it and we don't know it, we can do some math.
We can calculate chances to take out 9 balls of 10 if there are 9 green and 1 not. According to our information - 9!/10!=0.1. So, if we have no idea how balls are placed in jar, I'd bet at least 1:9

your ball removal is not random. random selection would be with replacement.

if you drew 9 of the same color and had to guess the 10th one, if we assume all the balls have an equal chance of being pulled from the jar then we would guess that the next ball has greater than a 90% chance of being green.

No, it's exactly 90%. Imagine ten balls - 9 green and 1 other color in a row. We take them from one side and stop when we take non-green. How many places that ball could be ? It's in last place 1/10 times and other 9 times it's in some other place. So if we take 9 green balls we can say it's 90% that all balls are green.

Ignore this post. My previous equations were incorrect.

yes, it's true, but that's what we want to find out. We don't know are there 10 green balls or only 9. We know how often this happens if there are 9 green balls. If we have any idea about possibility that there are 9 or 10 green balls, our odds is that possibility. e.g. we're sure for 40% that there are 10 green balls, so odds are 4:6. I just want to say: this is our only way to get possibility if balls are placed there at total random, if we don't believe so, we shouldn't bet on it

for the probability that the last ball is green to be exactly 90% we would have to have an exactly representative sample where 90% of the balls were green.

Our sample contains 100% green balls, but we can't say that its exactly representative either because we have one unknown remaining.

Its gotta be at least a 90% chance and we might be able to decide that its more than 90 but i can't figure how much more... there might be a way.