Nezhul wrote:I don't know what to say. It's just too stupid.
YES if we take green and red die, there are 2 total variants of dropping 2+1 (the other 1+2)
However that's if we CARE of the order they fall. tha fact is that in this game to get number 3, we are interested in the situation when numbers 2&1 can drop. We don't care wich die dropped 2, and wich dropped 1 whatsoever. We care only that ONE OF THEM dropped 2, and the OTHER dropped 1 at the same time. This doesn't add or take from the odds, because it's just the COMBINATION of numbers we are interested in, not the order. There are total 21 COMBINATIONS that can be dropped with 2 d6. Not 36 of them.
Study math and logic better, really. I'm done with explaining the simpliest shit to you.
This is old, but just for the sake of correctness,
Nezhul, you are completely wrong. Les, you are absolutely right with that big table of combinations.
If you are a programmer as you say, you could simply get strong evidence that your conviction is not correct by writing a simple program to simulate dice rolls. You could take any simple or sophisticated PRNG (such as Mersenne Twister, which is very good for simulations) or even better, truly random numbers (from random.org, for example) -- but you would need a very large amount of them.
If you have a Linux box around (or CYGWIN if you are running Windows), the following very simple one-liner (in BASH) might shock you:
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rolls=100000 && while [ $((--rolls)) -ge 0 ]; do echo $(( ((RANDOM%6) + 1) + ((RANDOM%6) + 1) )); done | sort -n | uniq -c
If you took out " | sort -n | uniq -c", you would see the actual results of the sums of the rolls. In the form presented, the script above will show you an histogram of the results. Depending on your machine, this can take a while to execute. You could reduce the value of the "rolls" variable to increase speed -- but the results might be compromised.
Running on my machine, it takes around 5 seconds to execute, and I got the following output:
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2804 2 5620 3 8519 4 11194 5 13834 6 16444 7 13873 8 11120 9 8383 10 5498 11 2711 12
The first column shows the number of occurrences of the number in the second column, among the 100000 rolls.
As you can see, the number of occurrences of the sum "3" is almost exactly the double of the occurrences of the sum "2". The same goes for the pair 12 and 11.
This experimental result coincides with Les' table.
Run the script as many times as you need to convince yourself that your analysis of the problem is incorrect.
EDIT: improved wording.