Quote from: Don Pedro Dinero on November 28, 2017, 04:44:36 PM

Quote from: TheQuin on November 28, 2017, 03:10:26 PM

This is why you don't understand. They are displayed as whole numbers but are calculated from decimals.

From the provably fair information on the site:

Quote

How are rolls calculated?

1 Two strings are created :
   STRING1 = "[NONCE]:[SERVER SEED]:[NONCE]"
   STRING2 = "[NONCE]:[CLIENT SEED]:[NONCE]"
   For your last roll :
   STRING1 = "187642:dd2d1349d29ccb2b154633ffe1083fab798480781a2ae5f85d302e21731aad22:187642"
   STRING2 = "187642:vXqTuIUiZsRsIqSW:187642"
2 Then HMAC-SHA512 is used to hash STRING1 with STRING2 as the secret key, giving us a 128 character hex string.
3 The first 8 characters of the hex string are taken and converted to a decimal.
4 This decimal is then divided by 429496.7295 and rounded off to the nearest whole number.
5 This whole number is used as your roll, with the maximum possible value being 10,000.

The part I bolded explains where the decimal is introduced and that in turn when rounding up will reduce the chance to 20,000:1.

Over in the thread on the Micro Earnings board, someone kindly wrote a little C++ program to work it out.
https://bitcointalk.org/index.php?topic=319540.msg25099985#msg25099985

Thanks for having replied. I knew that freebitco.in had a provably fair system, as well as many other sites but I dont know anything about it. I guess Ill have to learn.

However, provably fair seems deceiving to me in this way because, yes, it is proved to be fair, but you have less likelihood of winning than the actual whole number shown represent.

From what you said before:

Quote from: TheQuin on November 27, 2017, 09:01:13 AM

Nothing has been tweaked and you can work out the probability of hitting each prize level from the numbers displayed. So level 2 is 9886 - 9985 giving 99 chances in 10,000 possible results.

I take it that all numbers would have 1 in 10.001 chances of appearing, apart from number 10.000 which would have 0,5 in 10.001 chances, and, I guess, number 0 which would have 1,5 in 10.001.

Regards.

No, 0 number will have only 1:20000 chances to appear, as well as 10000.
A number n with decimals is calculated with the two strings and the method explained on the website. Range is [0,10000], 0 will be drawn if 0<= n < 0,5 that's 1/20000 of the whole [0,10000] segment.
The same for 10000 that will be drawn if 9999,5 < n <= 10000 here too, 1/20000 of the whole segment.
Every other number p has a 1:10000 chance to be drawn, as [p-0,5, p+0,5] is 1/10000 of the whole range.