Maths Problem

[Snowflake]

Confidence intervals will not give you much more info in this case, I believe. He has good statistical inferences (low t-score, high r square, etc.), the problem is that with only a few observations in the sample, how much could we trust those statistical inferences.

[Mysterious Gene]

Here's one:

I have 741 black marbles and 247 white marbles. I will draw one at random and discard it. I will keep drawing and discarding marbles as long as they match the color of the first. If I draw one of a different color, I will put it back into the back and shuffle the marbles. Then the process will begin again.

Example:
I draw a black marble, and discard it.
I draw another black marble, and discard it.
I draw a white marble, and shuffle it back into the bag.
I draw a black marble, and discard it.
I draw a white marble, and shuffle it back into the bag.
I draw a white marble, and discard it.
I draw another white marble, and discard it.
I draw another white marble, and discard it.
I draw a black marble, and shuffle it back into the bag.

Eventually, only one marble is left. What is the probability of it being black?

[Lul Thyme]

I havnt made a complete proof, but Im going to guess that the probabilty is one half for any starting number of marbles as long as there is at least one of each.

[Urban Ranger]

Sounds reasonable, because the colour with more marbles will be selected more often, thus getting trimmed down to size.

[Lul Thyme]

they key thing in the algorithm is the fact that when you draw opposite color, it gets put back in the bag I think...

[Mysterious Gene]

1/2 is correct. I found this proof online:

The probability is exactly 0.5 that the marble is black.

The situation can be scaled down to the case where there are 3 black and 1 white balls. The probabilities for the are eight possible combinations are listed below. Four end in black:
WB,BBB P=(1/4)=6/24
BW,WB,BB P=(3/4)(1/3)(1/3)=2/24
BW,BW,WB,B P=(3/4)(1/3)(2/3)(1/2)(1/2)=1/24
BBW,WB,B P=(3/4)(2/3)(1/2)(1/2)=3/24

Four end in white:
BW,BW,BW,W P=(3/4)(1/3)(2/3)(1/2)(1/2)=1/24
BW,BBW,W P=(3/4)(1/3)(2/3)(1/2)=2/24
BBW,BW,W P=(3/4)(2/3)(1/2)(1/2)=3/24
BBBW,W P=(3/4)(2/3)(1/2)=6/24

You can see that the probabilities will sum to 0.5 for each ball

I also made a probability table for 4 black and 1 white marble, which also came out to a 50% chance (I don't feel like typing it up). I'm pretty sure that yes, it will be 1/2 for any number of marbles using the given rules.

[Lul Thyme]

Why scale it down to 3 and 1 though tahts harldy a proof.
I think the best would be to do 1 and 1 which is trivial.
Then n and 1 by induction, and then n and m by induction.

I sort of started that other time (yesterday) but couldnt be arsed to finish it.

[Mysterious Gene]

I solved it the same way you did; just thinking about it logically, without any math. Because it's 50/50 with 2 black and 1 white, 3 black 1 white, and 4 black 1 white, I'm going to go out on a limb and make the early assumption that it will be 50/50 with any combination of marbles. I could be wrong, but it seems like it would work no matter what.

[Lul Thyme]

Yes, but what Im saying is realy math is about proving such statements, and not go out on a limb, so you dont have to worry about being wrong anymore thats all , dont take offense.

[Mysterious Gene]

I didn't invent the problem though, only repost it. The quote I posted was the "solution" on the site where I found the problem. It's meant as a "challenge", anyways...

[KrazyHorse]

Originally posted by Lul Thyme
Why scale it down to 3 and 1 though tahts harldy a proof.
I think the best would be to do 1 and 1 which is trivial.
Then n and 1 by induction, and then n and m by induction.

I sort of started that other time (yesterday) but couldnt be arsed to finish it.

That's what I figured was a decent idea for the proof.

3 and 1 is "engineer's induction"

[Jaguar]

Originally posted by KrazyHorse


Too bad you don't know what the hell you're talking about.

It's understandable to be a heathen for a few minutes before seeing the light, but this is too much.

[Snowflake]

Ok this is another strange thing that I never thought about before. Remember some of the Pythagorean triplets? 3,4,5; 5,12,13; 7,24,25 ... Well obviously we know that for them a²+b²=c². However, it seems that there's something else that are true: 3²=4+5, 5²=12+13, 7²=24+25 ...

I wonder if this is merely coincidental, or is there something there that could be proved. I also tried the set 8,15,17, while 8²is not equal to 15+17, if you cut them by half, 4²=15/2+17/2 ... Can we prove that for each of the triplets there always exist a factor where we could make this relationship to be true?

[Jaguar]

Wow! That's cool!

[Snowflake]

Oh and that factor is the difference between b and c (the two larger ones), apparently.

I solved it, never mind.