freebitco.in - is it even worth it?

They give away order-of-magnitude multiples of 0.2 cents with the lowest roll payout being 0.2 cents and the highest being $200. I've calculated the expected value of a roll at 2.7 cents. This assumes the top prize is exactly a 1 in 10000 chance which may not be the case.

The site is provably fair which is true to get the 8-digit hex, but then what they do with it, that's where the devil is in the details, which in this case is in Point 4 below. The rules used to say the integer was rounded "UP" to the nearest whole number and now and it says the integer rounded "OFF" to the nearest whole number so that is a huge difference. Here is the explanation of the rules from their site:

Code:

Two strings are created : 
1. STRING1 = "[NONCE]:[SERVER SEED]:[NONCE]" 
    STRING2 = "[NONCE]:[CLIENT SEED]:[NONCE]"
    For your last roll : 
    STRING1 = "76:XY5UiZdUVqxX:76" 
    STRING2 = "76:t8GqWftzShE2yotF:76"
2. Then HMAC-SHA512 is used to hash STRING1 with STRING2 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.

If you convert an 8-digit hex to a decimal you need to get "ffffffff" in order to convert to 4294967295, which divided by 429496.7295, is exactly 10,000. You have exactly an 16^8, or 1 in 42,94,967,296, chance of rolling this exact number. So the question becomes what does "round off" mean. If they mean rounding up any fraction then there is really a 1 in 10,000 chance of winning the top prize. If they actually round any fraction down then the top prize is an unwinnable canard, 22 times harder to win than the Powerball! Or maybe the site means its a true mathematical rounding, so if you roll a hex that converts to 9999.5000 or higher you win, and then at least you can win the top prize 1 in 20,000 times. But in any event it would be interesting to see how many people win what prize tier and see if it ties mathematically to the number of attempts.

Here is the expected value per roll based on the three different scenarios, given the site continues to maintains $200 for the top prize and divides each tier down by 10 with the same probability for winning each tier:

Round Up: 2.71 cents
Based on what "Round Off" means:
Round Down: 0.71 cents
Mathematical round: 1.71 cents

You can see how much the top tier affects the total EV.

If anyone has actually rolled a 10,000 that would essentially eliminate the "round down" theory so post if you have. Then it's a question of whether you can win 1 in 10k or 1 in 20k times. Accordingly, if anyone has rolled a 0000 then hat would eliminate the round up theory, and we'd be left with the mathematical rounding theory (albeit, discounting the fact someone may have beat the 1 in 42 billion odds for either of those exact hex rolls).