10-01-2005 12:41									#121
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quote: Originally posted by Urban Ranger Throwing 3D6 12 times is the same as throwing 36 dice all at once.

No it isn't

	10-01-2005 12:42									#122
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Yes. That's ED&D jargon, I suppose.

	10-01-2005 12:42									#123
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I think he's dumping the 6 lowest scores out of 12 total, not the 6 lowest dice

	10-01-2005 12:48									#124
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quote: 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).

	10-01-2005 14:48									#125
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quote: 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.

	10-01-2005 15:02									#126
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Not quite normal. Vaguely bell-shaped, but not exactly Gaussian.

	10-01-2005 20:17									#127
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quote: 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.

	10-01-2005 20:19									#128
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quote: 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.

	10-01-2005 20:21									#129
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quote: 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.

	10-01-2005 20:23									#130
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dp

	10-01-2005 20:59									#131
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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).

	10-01-2005 21:03									#132
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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?

	10-01-2005 21:09									#133
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quote: 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?

	10-01-2005 23:09									#134
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quote: 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.

	10-01-2005 23:11									#135
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quote: 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.

	10-01-2005 23:12									#136
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quote: Originally posted by 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).

Its funny that you post this in a thread where a poster (gen ludd) didnt really belive us string theorist until he went out and tested it experimentally.

Hows your numerology coming along btw?

	11-01-2005 00:19									#137
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quote: Originally posted by Lul Thyme Its funny that you post this in a thread where a poster (gen ludd) didnt really belive us string theorist until he went out and tested it experimentally. Hows your numerology coming along btw?

I'll get back to this later nonsense later but since I have an appropriate audience I thought I'd ask this question.

I'm reviewing a paper for a journal (not 'Nature' but next tier down) and I've run across a use of statistics that I think is faulty.

The comparison is of three groups using a t-test for samples of unequal sizes. The problem (as I see it) is that the group sizes are 4, 7, and 12. The comparison between the groups with 7 and 12 kids results in a significant difference p=.02.

I've always been skeptical of t-tests performed with low numbers and so have always used non-parametrics for my own studies. The last statistician I worked with reinforced my previous bias but I'm wondering if I'm wrong about this. Does anyone, or everyone, have an opinion?

	11-01-2005 00:26									#138
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quote: Originally posted by Lul Thyme Its funny that you post this in a thread where a poster (gen ludd) didnt really belive us string theorist until he went out and tested it experimentally. Hows your numerology coming along btw?

Wrong.

Gen Ludd realized that its a word game having nothing to do with the actual chances of choosing the prize. Shawnmmmmc was going to test it experimentally.

EDIT: some numerology

Instead of three cups we choose 1,000,000

You pick one then we knock over 999,998 empty cups. By the previous 'logic' you should switch to the non-picked cup since its odds of winning are now only slightly less than 100%.

	11-01-2005 06:12									#139
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quote: Originally posted by SpencerH Wrong. Gen Ludd realized that its a word game having nothing to do with the actual chances of choosing the prize. Shawnmmmmc was going to test it experimentally. EDIT: some numerology Instead of three cups we choose 1,000,000 You pick one then we knock over 999,998 empty cups. By the previous 'logic' you should switch to the non-picked cup since its odds of winning are now only slightly less than 100%.

OK let me rephrase your problem so were VERY clear.
There are 1000000 cups, only one with something under.
I point to one of 1,000,000.
You remove 999,998 other empty cups.
I can either keep my first cup, or choose the only one left over.
We repeat this many times.

In the long run, I will win WAY more often if I switch, then if I stay.

Do we both agree on this?
If not you are wrong.


By the way t-tests are far away for me, and I cant help you at all with that problem.

I would seem to think if you add the incertitude just to be safe, there is no problem...

	11-01-2005 06:18									#140
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quote: Originally posted by Lul Thyme In the long run, I will win WAY more often if I switch, then if I stay. Do we both agree on this? If not you are wrong.

You (and the others) are wrong. The odds of picking the correct cup out of 1 million are in no way linked to picking the correct cup out of two choices. They are independent events.

	11-01-2005 06:18									#141
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It's pretty easy to write a computer program to test this, Spencer, or to do it yourself.

	11-01-2005 07:05									#142
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Suppose there are 7 cups.
You get to pick 3
Of the 4 remaining I uncover 3 without the prize.
Should u keep your original 3 or switch to the one remaining.






The answer is to trade your 3 cups for the one remaining.

	11-01-2005 07:12									#143
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Seems to me that you should switch.

	11-01-2005 09:15									#144
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To convert the remaining heathen(s):

Here are a bunch of trials using a random number generator to choose which cup you guess and where the prize is, with 100 cups. In all but one of 50 trials, you win if you switch. Thus this experiment confirms that the odds should be close to 100%.

code:

Guess:  Prize:  Should you switch?
35	33        Yes         
76	24        Yes 
17	59        Yes 
88	83        Yes 
49	33        Yes 
4	36        Yes 
86	46        Yes 
24	37        Yes 
28	99        Yes 
39	86        Yes 
6	27        Yes 
69	94        Yes 
44	5        Yes 
81	10        Yes 
3	46        Yes 
78	96        Yes 
43	53        Yes 
43	8         Yes 
20	90        Yes 
84	98        Yes 
85	55        Yes 
81	41        Yes 
69	97        Yes 
94	77        Yes 
88	71        Yes 
68	80        Yes 
51	25        Yes 
59	59        No
45	55        Yes 
68	45        Yes 
8	11        Yes 
16	8          Yes 
14	50        Yes 
84	64        Yes 
31	65        Yes 
45	27        Yes 
17	37        Yes 
40	60        Yes 
53	62        Yes 
3	34        Yes 
22	32        Yes 
85	33        Yes 
56	99        Yes 
99	49        Yes 
46	97        Yes 
25	37        Yes 
2	47        Yes 
6	36        Yes 
71	81        Yes 
7	54        Yes 

	11-01-2005 09:54									#145
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Does that convince you?
Why dont you try it with say 7 cups 10 times.
That shouldnt be too long, and should be VERY convincing.

Then you can explain to us what you didnt understand.

Or you can keep saying that we live in fantasy land, while not even daring to actually try it.

	11-01-2005 10:14									#146
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quote: Originally posted by 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).

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

	11-01-2005 17:32									#147
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quote: Originally posted by KrazyHorse Too bad you don't know what the hell you're talking about.

Au contraire, I know lots about voodoo.

OK, I realized my error with the cup choice question (knowingly eliminating the empty cups positively selects for the prize) and I misread the birth question (in my defense I was reading it while also reading some sequence data).

No-one has a comment about the t-test and low n then?

	11-01-2005 17:44									#148
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I have no idea what the hell a t-test is. I'm not a statistician. I'm a pseudo-mathematician and statistical physicist.

	11-01-2005 20:29									#149
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I would also like to hear it if anybody has a concete answer to that question. I myself have been quite suspicious about statistical inferences based on low numbers. However we use it a lot at work here.

	11-01-2005 23:31									#150
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Im glad you figured out the cup and birth questions...

Well I know what a t-test is, but have not close to enough knowledge to be confident about an answer to your question.

BTW its not just with t-tests you have to be quite suspicious statistical inferences based on low numbers...

Being suspicious is always a good start.

Try computing what the confidence intervals are on your tests or something to give you and idea if it seems justifiable.