11-01-2005 23:49									#151
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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.

	12-01-2005 04:40									#152
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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?

	12-01-2005 09:22									#153
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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.

	12-01-2005 09:33									#154
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Sounds reasonable, because the colour with more marbles will be selected more often, thus getting trimmed down to size.

	12-01-2005 10:55									#155
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they key thing in the algorithm is the fact that when you draw opposite color, it gets put back in the bag I think...

	13-01-2005 03:44									#156
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1/2 is correct. I found this proof online:

quote: 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.

	13-01-2005 04:39									#157
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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.

	13-01-2005 05:51									#158
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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.

	13-01-2005 06:20									#159
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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.

	13-01-2005 07:51									#160
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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...

	13-01-2005 08:02									#161
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quote: 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"

	13-01-2005 08:17									#162
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quote: 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.

	13-01-2005 09:03									#163
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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?

	13-01-2005 09:04									#164
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Wow! That's cool!

	13-01-2005 09:07									#165
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Oh and that factor is the difference between b and c (the two larger ones), apparently.

I solved it, never mind.

	13-01-2005 09:14									#166
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Do tell.

	13-01-2005 09:14									#167
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This is pretty cool though for one to memorize the triplets. For example if we want to see what is the triplets that has 9 as the shortest side. We then get 9²=81=40+41. We then know the triplet is 9, 40, 41, without having to check if it fits a²+b²=c².

	13-01-2005 09:16									#168
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quote: Originally posted by Kuciwalker Do tell.

a²=c²-b²=(b+c)(c-b)
Therefore a²/(c-b)=b+c

It's very simple, actually, I just have never thought of it this way.

	13-01-2005 09:16									#169
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quote: Originally posted by Snowflake This is pretty cool though for one to memorize the triplets. For example if we want to see what is the triplets that has 9 as the shortest side. We then get 92=81=40+41. We then know the triplet is 9, 40, 41, without having to check if it fits a2+b2=c2.

Yeah, but many pairs of integers add up to 81.

	13-01-2005 09:18									#170
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You need to make it b+c=81 and b-c=1 though.

	13-01-2005 09:19									#171
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quote: Originally posted by Snowflake You need to make it b+c=81 and b-c=1 though.

Yes.

Is this a special case or the general case? IOW, is b-c always 1?

	13-01-2005 09:26									#172
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No, but this is the easy one to get. For example, if a=11, you get 121=60+61. If a=13, you get 169=84+85. If a = 15, you get 225=112+113 ...
So your triplets would be 11, 60, 61; 13, 84,85; 15, 112, 113 ...

For even numbers it seems you need to make b-c=2 or something ...

	13-01-2005 09:32									#173
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Ok for even numbers you cut the square into half before you figure the b and c out. For example, if a=10, you get 100/2=50=24+26, so the triplet is 10, 24,26. If a=12, you get 144/2=72=35+37, so the triplets is 12, 35, 37. If a =14, you get 196/2=98=48+50, and the triplets is 14, 48, 50 ...

	13-01-2005 09:39									#174
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quote: Originally posted by 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?

Yes. Duh.

(n+1)^2 - n^2 = 2n + 1 = n + (n+1)

Simple algebra...

	13-01-2005 09:45									#175
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In fact, it's quite easy to show that up to a scale factor all pythagorean triplets are equivalent to a = (2n + 1), b = (2n^2 + 2n), c = (2n^2 + 2n + 1)

	13-01-2005 09:45									#176
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quote: Originally posted by KrazyHorse Yes. Duh. (n+1)^2 - n^2 = 2n + 1 = n + (n+1) Simple algebra...

You are assuming the two longer sides are continous integers, which may or may not be the case, as shown above. However you are right the algebra is very simple. I'm just happy that I found a way to get the Pythagorean triplets for any number with ease.

	13-01-2005 09:51									#177
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It is the case up to a scale factor, as I've said. For all intents and purposes 3 4 5, 5 12 13, 7 24 25 etc. are the only pythagorean triplets. All others are direct multiples of these.

	13-01-2005 09:54									#178
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quote: Originally posted by KrazyHorse In fact, it's quite easy to show that up to a scale factor all pythagorean triplets are equivalent to a = (2n + 1), b = (2n^2 + 2n), c = (2n^2 + 2n + 1)

Good thinking. If you got an even number you can simply scale it back to an odd number before you get the triplets. Although this would actually give you different solutions sometimes ... Hmmm

For example for 12, I would get 12,35,37, and you would get 12,16, 20 ...

	13-01-2005 09:57									#179
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In view of your last post, the one I'd get may be a pythagorean triplet and the one you'd get would be a direct multiple.

The thing is for your method it is easy to get a real pythagorean if a is an odd number, but not if it is an even number.

	13-01-2005 10:02									#180
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Looks to me if a is an odd number, or multiples of four, then you'd be able to get a real pythagorean triplet, otherwise they would only be multiples ...