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Two questions / challenges appropriate to this thread

Posted By: sita
Date: Monday, 29 December 2014, at 4:06 p.m.

In Response To: Two questions / challenges appropriate to this thread (Colin Owen)

Mh, I couldn't reproduce it again. :-(

I had a red, and a black die, and rolled them 201 times, so that I have 402 test cases, which is dividable by 6. 402/6 = 67. With a standard deviation of 7.472, I need 80 rolls of 6s or more, to proof that there are too many 6s with probability 95%. Here are the results for red, black, and total:

1: 39 + 28 = 67

2: 34 + 33 = 67

3: 33 + 41 = 74

4: 31 + 27 = 58

5: 32 + 31 = 63

6: 32 + 46 = 78

Even though I have too many 6s, they are not quite enough. This result could have occured with a probablity of around 7% with true random dice. Well, I'm not going to do it again with 1,000 rolls. :-)

Some things are still interesting. The red die seems to be ok, the black shows to be problematic. Measuring it alone would be good to proof the result, unfortunately that was not how the experiment was set up. Second, I had too few 4s, and 5s. Even with the black die alone. So, probably my memory was incorrect that these dice produce also too many 4s, and 5s.

I took Colins advice, and measured the dice. The red die is extremely equal, all sides seem to be 14.9mm. Maybe the 43-side is more towards 14.85mm. For the black die, sides and lengths are below:

16: 14.8mm

25: 14.85mm

34: 14.7mm

This would explain all the 3s this die produces (one would expect the shortest side to show up, because the square of the two other sides lies on the ground). But then, why does it have so few 4s? We would also expect more 4s.

Well, it seems, we know nothing. Maybe I made the mistake of measuring the dice before the experiment, and used the most equal red and black die. This would again suggest that Colin is right, and the deviations just come from the different lengths of the sides.

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