Maths Problem

[KrazyHorse]

Originally posted by Urban Ranger
Throwing 3D6 12 times is the same as throwing 36 dice all at once.

No it isn't

[Snowflake]

Yes. That's ED&D jargon, I suppose.

[KrazyHorse]

I think he's dumping the 6 lowest scores out of 12 total, not the 6 lowest dice

[Snowflake]

Originally posted by KrazyHorse
I think he's dumping the 6 lowest scores out of 12 total, not the 6 lowest dice

Right, he's dumping the scores which are the sum of the three dices (in the range of 10.5 or so).

[Urban Ranger]

Originally posted by Snowflake
If it was thrown 12 times, and the six lower ones are thrown away, then I suppose we could assume six scores are lower than the average and six scores are higher than average. So the expected value of the six remain scores would be perhaps the average of 10.5 and 18 (maximum), which is 14.25. Therefore the total score would be 6*14.25=85.5

3D6 has a normal distribution curve, not a uniform one.

[KrazyHorse]

Not quite normal. Vaguely bell-shaped, but not exactly Gaussian.

[Lul Thyme]

Originally posted by Urban Ranger


You're not done yet, you still need to throw away the 6 lowest values.

Actually I would be done...
Reread what I did, I was assuming we WERE already throwing the 6 lowest values , and keeping the 6 other one.

[Lul Thyme]

Originally posted by Snowflake
What, do we really need to complicate things this much?

The way I look at it is like this:

The expected value of score from a six side dice is:
(1+2+...+6)/6=3.5
So expected score of 3d6 would be 10.5.

If it was thrown six times, the expected score is 6*10.5=63.

If it was thrown 12 times, and the six lower ones are thrown away, then I suppose we could assume six scores are lower than the average and six scores are higher than average. So the expected value of the six remain scores would be perhaps the average of 10.5 and 18 (maximum), which is 14.25. Therefore the total score would be 6*14.25=85.5

Would that be a reasonable estimate? Hopefully a computer program would be able to confirm this.

A reasonning like that would give a reasonable estimate, but not the actual expected value, Im sorry to say.

For example, your explanation does not account at all for the fact that you may well get 12 throws that are all under the average, so even the 6 top would be etc. etc..


And also, why would you compute the mean of 10.5 and 18 when the distribution is not symmetrical at all between them (check my earlier table in my big post), so we take expected value above 10.5 we would get something close to 13. (didnt compute but would be lower than 14.25)

I bet such a reasoning would be close to the actual one, but I was trying to show what kind of computations were needed to prove the actual Expected Value.

[Lul Thyme]

Originally posted by KrazyHorse
Not quite normal. Vaguely bell-shaped, but not exactly Gaussian.

Exactly
Thats why I was saying we could probably get a pretty close approximation by using a normal, but that would not be the exact value.

UR, dont forget that a normal curve is a continuous curve.

This is a discrete variable, so already there is no way this could be EXACTLY a normal.

[Lul Thyme]

dp

[SpencerH]

Some interesting conundrums and proof positive that statisticians dont live in the real world (as I've known for some time). The odds for the monty hall problem and for the birth are 50/50. Those who answered otherwise didnt check their result experimentally and are likely to focus their studies on string theory or voodoo (whichever seems more plausible to their belief systems).

[Snowflake]

Ok, so if it is some kind of a bell shape the possibility of getting the middle value (10.5) would be (3 times?) higher than the maximum (18), in other words the expected value would be less than 14.25.

What fancinated me is that everything here is set, why is it that we would not be able to derive the expected value mathematically, in stead, we have to resume to computer simulation for a sulotion?

[Snowflake]

Originally posted by Lul Thyme

For example, your explanation does not account at all for the fact that you may well get 12 throws that are all under the average, so even the 6 top would be etc. etc..

But we should be able to compute the possibility that all 12 throws are under the average, right? And it should be rather small, I think. Also there would be the possibility that all 12 are above average, so these two cases may very well offset each other, leaving the average somewhat close to the average we originally estimated ... I don't know.

What was the principle that determines the shape of a 3d6?

[Lul Thyme]

Originally posted by Snowflake
Ok, so if it is some kind of a bell shape the possibility of getting the middle value (10.5) would be (3 times?) higher than the maximum (18), in other words the expected value would be less than 14.25.

What fancinated me is that everything here is set, why is it that we would not be able to derive the expected value mathematically, in stead, we have to resume to computer simulation for a sulotion?

Youre getting this wrong.
It IS possible to compute the exact value mathematically, i even started the computation, its just very VERY long. (at least I cant see an easy way)
Not all problems can be solved in a few lines.

[Lul Thyme]

Originally posted by Snowflake


But we should be able to compute the possibility that all 12 throws are under the average, right? And it should be rather small, I think. Also there would be the possibility that all 12 are above average, so these two cases may very well offset each other, leaving the average somewhat close to the average we originally estimated ... I don't know.

What was the principle that determines the shape of a 3d6?

Its sort of a cumulative binomial distribution.


I gave the odds for 3D6 earlier in an other post.
Your argument makes sense sorta, buts its a farcry from a mathematical proof.

I also think we could get a close estimate by using such arguments, but thats all they still are, estimates.