FreeBitco.in-$200 FreeBTC🏎Win Lambo🔥0.2BTC DailyJackpot🏆$32,500 Wager Contest - page 416.

natri

newbie

Activity: 7

Merit: 0

Hey guys, I'm back!

I've gotta say I spent quite some time working things out, and this time I've manually obtained numbers from exactly 1000 rolls! Yes, 1k rolls, every single one recorded and written down in my excel file for later analysis. And I've gotta say the results don't surprise. Running chisquare test gave again a quite confident rejection of the null. Numbers probabilistically don't seem random.

Some info you might wonder (my test size of 1000 rolls):
MAX value: 9986
MIN value: 3
Median: 4815
Mean: 4832,296

9056 values never occurred, 885 singles, 53 duplicates and 3 (!!) triplicates.

Some numbers are even complete obvious duplicates of what I got in my past rolls. I am VERY strongly inclined to believe the probability of certain values is much MUCH higher than some others (here we esp. wonder about 9998,9999 and 10000 - who has ever even seen those?!).

Quote from: wetsuit on June 21, 2014, 01:50:38 AM

The javascript based roll verifier is now online at http://roll-verifier.s3-website-us-east-1.amazonaws.com/. I shall link it via the website shortly.

You indeed did not change the roll outcome and you used the same server and client seeds (all correct), however, even with completely random inputs (client seed, server seed) the final outcome is most likely not a random number ranging from 0-10000, I am quite confident in that.

arieq

sr. member

Activity: 364

Merit: 254

Quote from: wetsuit on June 22, 2014, 03:52:48 AM

Quote from: arieq on June 22, 2014, 02:49:16 AM

Quote from: wetsuit on June 21, 2014, 01:50:38 AM

The javascript based roll verifier is now online at http://roll-verifier.s3-website-us-east-1.amazonaws.com/. I shall link it via the website shortly.

It would be great if all fields from the above URL are automatically filled with previous rolls details when someone clicks on CLICK HERE TO VERIFY YOUR ROLLS link, so that he only needs to click VERIFY button to verify the rolled number and no need to copy paste one by one

done!

great, it works.

wetsuit

legendary

Activity: 981

Merit: 1026

Quote from: arieq on June 22, 2014, 02:49:16 AM

Quote from: wetsuit on June 21, 2014, 01:50:38 AM

The javascript based roll verifier is now online at http://roll-verifier.s3-website-us-east-1.amazonaws.com/. I shall link it via the website shortly.

It would be great if all fields from the above URL are automatically filled with previous rolls details when someone clicks on CLICK HERE TO VERIFY YOUR ROLLS link, so that he only needs to click VERIFY button to verify the rolled number and no need to copy paste one by one

done!

wetsuit

legendary

Activity: 981

Merit: 1026

Quote from: arieq on June 22, 2014, 02:49:16 AM

Quote from: wetsuit on June 21, 2014, 01:50:38 AM

The javascript based roll verifier is now online at http://roll-verifier.s3-website-us-east-1.amazonaws.com/. I shall link it via the website shortly.

It would be great if all fields from the above URL are automatically filled with previous rolls details when someone clicks on CLICK HERE TO VERIFY YOUR ROLLS link, so that he only needs to click VERIFY button to verify the rolled number and no need to copy paste one by one

Will do shortly.

arieq

sr. member

Activity: 364

Merit: 254

Quote from: wetsuit on June 21, 2014, 01:50:38 AM

The javascript based roll verifier is now online at http://roll-verifier.s3-website-us-east-1.amazonaws.com/. I shall link it via the website shortly.

It would be great if all fields from the above URL are automatically filled with previous rolls details when someone clicks on CLICK HERE TO VERIFY YOUR ROLLS link, so that he only needs to click VERIFY button to verify the rolled number and no need to copy paste one by one

wetsuit

legendary

Activity: 981

Merit: 1026

The javascript based roll verifier is now online at http://roll-verifier.s3-website-us-east-1.amazonaws.com/. I shall link it via the website shortly.

natri

newbie

Activity: 7

Merit: 0

Quote from: Luror on June 20, 2014, 05:05:56 PM

Quote from: natri on June 20, 2014, 03:59:06 PM

I will check more thoroughly into it when I have some spare time. Gonna post my results here.

I have an idea for you: try to do it in various browsers, as it's possible that Javascript math.random implementation is to blame. Which one did you use for your test, by the way?

Google chrome.

Luror

full member

Activity: 215

Merit: 100

Quote from: natri on June 20, 2014, 03:59:06 PM

I will check more thoroughly into it when I have some spare time. Gonna post my results here.

I have an idea for you: try to do it in various browsers, as it's possible that Javascript math.random implementation is to blame. Which one did you use for your test, by the way?

natri

newbie

Activity: 7

Merit: 0

Quote from: Luror on June 20, 2014, 03:56:23 PM

Quote from: natri on June 20, 2014, 03:36:44 PM

Quote from: Luror on June 20, 2014, 03:12:02 PM

But you said you made only 500 rolls? Or did you mean some bigger test? You need more rolls to get 3rd and 4th prizes (at least 1250-5000).

That last statement ("I've never rolled more than 9993 on any of my multiplier or regular rolls on neither faucet") I referred to all of my plays on either freebitco.in and freedoge.co.in which total way over 2000 rolls so by calculating 1-(9993/10000)^2000+ the probability should be 75%+.

You are forgetting there's also possible to roll "0" (probability is also 1/20000), so it's a little less 1-(1/20000+9993/10000)^2000 ~ 73%

Doesn't change the number much though

Well, yeah. When it comes to large numbers, statistics shall prevail

I will check more thoroughly into it when I have some spare time. Gonna post my results here.

Luror

full member

Activity: 215

Merit: 100

Quote from: natri on June 20, 2014, 03:36:44 PM

Quote from: Luror on June 20, 2014, 03:12:02 PM

But you said you made only 500 rolls? Or did you mean some bigger test? You need more rolls to get 3rd and 4th prizes (at least 1250-5000).

That last statement ("I've never rolled more than 9993 on any of my multiplier or regular rolls on neither faucet") I referred to all of my plays on either freebitco.in and freedoge.co.in which total way over 2000 rolls so by calculating 1-(9993/10000)^2000+ the probability should be 75%+.

You are forgetting there's also possible to roll "0" (probability is also 1/20000), so it's a little less 1-(1/20000+9993/10000)^2000 ~ 73%

Doesn't change the number much though

wetsuit

legendary

Activity: 981

Merit: 1026

Quote from: natri on June 20, 2014, 03:51:33 PM

Quote from: wetsuit on June 20, 2014, 03:40:33 PM

Quote from: natri on June 20, 2014, 03:36:44 PM

Quote from: Luror on June 20, 2014, 03:12:02 PM

But you said you made only 500 rolls? Or did you mean some bigger test? You need more rolls to get 3rd and 4th prizes (at least 1250-5000).

That last statement ("I've never rolled more than 9993 on any of my multiplier or regular rolls on neither faucet") I referred to all of my plays on either freebitco.in and freedoge.co.in which total way over 2000 rolls so by calculating 1-(9993/10000)^2000+ the probability should be 75%+.

Ok, lets assume prob for rolling 10k is 1/20000. But by simply checking the freebitco.in WEBSITE STATS, there was currently 538,271,966 played and 394.52490422 BTC earned, if we consider that you can ONLY win top roll price (huge undervaluation) and probability is 1/20000 to earn ONLY 0.1 BTC (undervalued) -> the expected earned value of BTC should be 2700 BTC (way more than 394).

And if we say that all ppl were extremely unlucky we can check;
variance=npq=538271996*1/20000*19999/20000
sqrt(var)=164

99.7% rule (3 std deviations): Still users should earn at LEAST: 2700-3*164= 2200 BTC, and this is with only top win of 0.1 btc.

Those rolls include both free and multiply rolls, and multiply rolls are way more in number.

Alright, that explains it sufficiently enough then

But why is there such a huge discrepancy when generating a roll. Notice the frequency of occurrences inside the 5001-6000 bracket. I am not trying to accuse or anything but the randomness procedure might well be flawed.

Posting results:

Number   Expected   Observed   (O-E)^2/E
1-1000 50 45 0,5
1001-2000   50   42   1,28
2001-3000   50   50   0
3001-4000   50   39   2,42
4001-5000   50   52   0,08
5001-6000   50   77   14,58
6001-7000   50   51   0,02
7001-8000   50   41   1,62
8001-9000   50   42   1,28
9001-10000   50   61   2,42

I am not sure to be honest, maybe you need a larger sample size ? I shall look into ways to make the client and server seed generation more random.

natri

newbie

Activity: 7

Merit: 0

Quote from: wetsuit on June 20, 2014, 03:40:33 PM

Quote from: natri on June 20, 2014, 03:36:44 PM

Quote from: Luror on June 20, 2014, 03:12:02 PM

But you said you made only 500 rolls? Or did you mean some bigger test? You need more rolls to get 3rd and 4th prizes (at least 1250-5000).

That last statement ("I've never rolled more than 9993 on any of my multiplier or regular rolls on neither faucet") I referred to all of my plays on either freebitco.in and freedoge.co.in which total way over 2000 rolls so by calculating 1-(9993/10000)^2000+ the probability should be 75%+.

Ok, lets assume prob for rolling 10k is 1/20000. But by simply checking the freebitco.in WEBSITE STATS, there was currently 538,271,966 played and 394.52490422 BTC earned, if we consider that you can ONLY win top roll price (huge undervaluation) and probability is 1/20000 to earn ONLY 0.1 BTC (undervalued) -> the expected earned value of BTC should be 2700 BTC (way more than 394).

And if we say that all ppl were extremely unlucky we can check;
variance=npq=538271996*1/20000*19999/20000
sqrt(var)=164

99.7% rule (3 std deviations): Still users should earn at LEAST: 2700-3*164= 2200 BTC, and this is with only top win of 0.1 btc.

Those rolls include both free and multiply rolls, and multiply rolls are way more in number.

Alright, that explains it sufficiently enough then

But why is there such a huge discrepancy when generating a roll. Notice the frequency of occurrences inside the 5001-6000 bracket. I am not trying to accuse or anything but the randomness procedure might well be flawed.

Posting results:

Number   Expected   Observed   (O-E)^2/E
1-1000 50 45 0,5
1001-2000   50   42   1,28
2001-3000   50   50   0
3001-4000   50   39   2,42
4001-5000   50   52   0,08
5001-6000   50   77   14,58
6001-7000   50   51   0,02
7001-8000   50   41   1,62
8001-9000   50   42   1,28
9001-10000   50   61   2,42

EDIT: No proof or whatsoever, but I somehow also got a gut feeling that some individual numbers appear with way higher frequency than the others...

wetsuit

legendary

Activity: 981

Merit: 1026

Quote from: Luror on June 20, 2014, 03:38:31 PM

Quote from: wetsuit on June 20, 2014, 03:16:38 PM

Quote from: Luror on June 20, 2014, 03:12:02 PM

Quote from: natri on June 20, 2014, 01:00:44 PM

Anyways, I've actually tested the randomness of rolls and the result was that rolls ARE NOT RANDOM, p=0,004 (sample 500, chi-squared).
This low p means it's almost impossible that I made error of type I, so rolls actually aren't random at all...
More info regarding my testing procedure here: http://www.reddit.com/r/Bitcoin/comments/28dtaa/wondering_freebitcoin_provably_unfair/

This is interesting, it is possible after all that pseudo-random numbers (for seeds) are not quite random. Why didn't you share full test results? It would be nice for someone to repeat it (though it's quite time consuming: 500 hours).

Quote from: natri on June 20, 2014, 01:00:44 PM

(btw, I've never rolled more than 9993 on any of my multiplier or regular rolls on neither faucet, probability of which should be over 75%...)

But you said you made only 500 rolls? Or did you mean some bigger test? You need more rolls to get 3rd and 4th prizes (at least 1250-5000).

Quote from: Luror on March 31, 2014, 11:13:56 AM

[0 - 9885]: ~ 1/20000 + 9885/10000 = 0.98855
[9886 - 9985] ~ (9985-9886+1)/10000 = 0.01 (~ once per 100 rolls)
[9986 - 9993] ~ (9993-9986+1)/10000 = 0.0008 (~ once per 1250 rolls)
[9994 - 9997] ~ (9997-9994+1)/10000 = 0.0004 (~ once per 2500 rolls)
[9998 - 9999] ~ (9999-9998+1)/10000 = 0.0002 (~ once per 5000 rolls)
[10000] ~ 1/20000 = 0.00005 (~ once per 20 000 rolls)

sanity check: 0.98855+0.01+0.0008+0.0004+0.0002+0.00005 = 1

The client seeds are actually generated using javascript entirely on the client side. This is the part of the Javascript code that generates the client seed (can be found in the JS file http://static4.freebitco.in/min/main1403284658.js now but the name of the file changes when I make updates) :

Code:

charSet = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
var randomString = '';
for (var i = 0; i < 16; i++)
{
var randomPoz = Math.floor(Math.random() * charSet.length);
randomString += charSet.substring(randomPoz,randomPoz+1);
}
$('#next_client_seed').val(randomString);

Ok, I don't know much about Javascript programming, but I googled a little, and it seems Math.random() is not so good for some things, still it is unclear whether it can give effects like natri got:

https://stackoverflow.com/questions/5651789/is-math-random-cryptographically-secure
https://stackoverflow.com/questions/578700/how-trustworthy-is-javascripts-random-implementation-in-various-browsers
https://security.stackexchange.com/questions/20029/generate-cryptographically-strong-pseudorandom-numbers-in-javascript

The best answer seems to be this: http://stackoverflow.com/a/18507748

"Recent browsers expose window.crypto.getRandomValues() which is cryptographically strong" <--- can you use this?

EDIT : Looked into that and seems like it is experimental so users might run into compatibility issues. I shall try to replace the current random string generator with a stronger one.

wetsuit

legendary

Activity: 981

Merit: 1026

Quote from: natri on June 20, 2014, 03:36:44 PM

Quote from: Luror on June 20, 2014, 03:12:02 PM

But you said you made only 500 rolls? Or did you mean some bigger test? You need more rolls to get 3rd and 4th prizes (at least 1250-5000).

That last statement ("I've never rolled more than 9993 on any of my multiplier or regular rolls on neither faucet") I referred to all of my plays on either freebitco.in and freedoge.co.in which total way over 2000 rolls so by calculating 1-(9993/10000)^2000+ the probability should be 75%+.

Ok, lets assume prob for rolling 10k is 1/20000. But by simply checking the freebitco.in WEBSITE STATS, there was currently 538,271,966 played and 394.52490422 BTC earned, if we consider that you can ONLY win top roll price (huge undervaluation) and probability is 1/20000 to earn ONLY 0.1 BTC (undervalued) -> the expected earned value of BTC should be 2700 BTC (way more than 394).

And if we say that all ppl were extremely unlucky we can check;
variance=npq=538271996*1/20000*19999/20000
sqrt(var)=164

99.7% rule (3 std deviations): Still users should earn at LEAST: 2700-3*164= 2200 BTC, and this is with only top win of 0.1 btc.

Those rolls include both free and multiply rolls, and multiply rolls are way more in number.

Luror

full member

Activity: 215

Merit: 100

Quote from: wetsuit on June 20, 2014, 03:16:38 PM

Quote from: Luror on June 20, 2014, 03:12:02 PM

Quote from: natri on June 20, 2014, 01:00:44 PM

Anyways, I've actually tested the randomness of rolls and the result was that rolls ARE NOT RANDOM, p=0,004 (sample 500, chi-squared).
This low p means it's almost impossible that I made error of type I, so rolls actually aren't random at all...
More info regarding my testing procedure here: http://www.reddit.com/r/Bitcoin/comments/28dtaa/wondering_freebitcoin_provably_unfair/

This is interesting, it is possible after all that pseudo-random numbers (for seeds) are not quite random. Why didn't you share full test results? It would be nice for someone to repeat it (though it's quite time consuming: 500 hours).

Quote from: natri on June 20, 2014, 01:00:44 PM

(btw, I've never rolled more than 9993 on any of my multiplier or regular rolls on neither faucet, probability of which should be over 75%...)

But you said you made only 500 rolls? Or did you mean some bigger test? You need more rolls to get 3rd and 4th prizes (at least 1250-5000).

Quote from: Luror on March 31, 2014, 11:13:56 AM

[0 - 9885]: ~ 1/20000 + 9885/10000 = 0.98855
[9886 - 9985] ~ (9985-9886+1)/10000 = 0.01 (~ once per 100 rolls)
[9986 - 9993] ~ (9993-9986+1)/10000 = 0.0008 (~ once per 1250 rolls)
[9994 - 9997] ~ (9997-9994+1)/10000 = 0.0004 (~ once per 2500 rolls)
[9998 - 9999] ~ (9999-9998+1)/10000 = 0.0002 (~ once per 5000 rolls)
[10000] ~ 1/20000 = 0.00005 (~ once per 20 000 rolls)

sanity check: 0.98855+0.01+0.0008+0.0004+0.0002+0.00005 = 1

The client seeds are actually generated using javascript entirely on the client side. This is the part of the Javascript code that generates the client seed (can be found in the JS file http://static4.freebitco.in/min/main1403284658.js now but the name of the file changes when I make updates) :

Code:

charSet = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
var randomString = '';
for (var i = 0; i < 16; i++)
{
var randomPoz = Math.floor(Math.random() * charSet.length);
randomString += charSet.substring(randomPoz,randomPoz+1);
}
$('#next_client_seed').val(randomString);

Ok, I don't know much about Javascript programming, but I googled a little, and it seems Math.random() is not so good for some things, still it is unclear whether it can give effects like natri got:

https://stackoverflow.com/questions/5651789/is-math-random-cryptographically-secure
https://stackoverflow.com/questions/578700/how-trustworthy-is-javascripts-random-implementation-in-various-browsers
https://security.stackexchange.com/questions/20029/generate-cryptographically-strong-pseudorandom-numbers-in-javascript

The best answer seems to be this: http://stackoverflow.com/a/18507748

"Recent browsers expose window.crypto.getRandomValues() which is cryptographically strong" <--- can you use this?

natri

newbie

Activity: 7

Merit: 0

Quote from: Luror on June 20, 2014, 03:12:02 PM

But you said you made only 500 rolls? Or did you mean some bigger test? You need more rolls to get 3rd and 4th prizes (at least 1250-5000).

That last statement ("I've never rolled more than 9993 on any of my multiplier or regular rolls on neither faucet") I referred to all of my plays on either freebitco.in and freedoge.co.in which total way over 2000 rolls so by calculating 1-(9993/10000)^2000+ the probability should be 75%+.

Ok, lets assume prob for rolling 10k is 1/20000. But by simply checking the freebitco.in WEBSITE STATS, there was currently 538,271,966 played and 394.52490422 BTC earned, if we consider that you can ONLY win top roll price (huge undervaluation) and probability is 1/20000 to earn ONLY 0.1 BTC (undervalued) -> the expected earned value of BTC should be 2700 BTC (way more than 394).

And if we say that all ppl were extremely unlucky we can check;
variance=npq=538271996*1/20000*19999/20000
sqrt(var)=164

99.7% rule (3 std deviations): Still users should earn at LEAST: 2700-3*164= 2200 BTC, and this is with only top win of 0.1 btc.

wetsuit

legendary

Activity: 981

Merit: 1026

Quote from: Luror on June 20, 2014, 03:12:02 PM

Quote from: natri on June 20, 2014, 01:00:44 PM

Anyways, I've actually tested the randomness of rolls and the result was that rolls ARE NOT RANDOM, p=0,004 (sample 500, chi-squared).
This low p means it's almost impossible that I made error of type I, so rolls actually aren't random at all...
More info regarding my testing procedure here: http://www.reddit.com/r/Bitcoin/comments/28dtaa/wondering_freebitcoin_provably_unfair/

This is interesting, it is possible after all that pseudo-random numbers (for seeds) are not quite random. Why didn't you share full test results? It would be nice for someone to repeat it (though it's quite time consuming: 500 hours).

Quote from: natri on June 20, 2014, 01:00:44 PM

(btw, I've never rolled more than 9993 on any of my multiplier or regular rolls on neither faucet, probability of which should be over 75%...)

But you said you made only 500 rolls? Or did you mean some bigger test? You need more rolls to get 3rd and 4th prizes (at least 1250-5000).

Quote from: Luror on March 31, 2014, 11:13:56 AM

[0 - 9885]: ~ 1/20000 + 9885/10000 = 0.98855
[9886 - 9985] ~ (9985-9886+1)/10000 = 0.01 (~ once per 100 rolls)
[9986 - 9993] ~ (9993-9986+1)/10000 = 0.0008 (~ once per 1250 rolls)
[9994 - 9997] ~ (9997-9994+1)/10000 = 0.0004 (~ once per 2500 rolls)
[9998 - 9999] ~ (9999-9998+1)/10000 = 0.0002 (~ once per 5000 rolls)
[10000] ~ 1/20000 = 0.00005 (~ once per 20 000 rolls)

sanity check: 0.98855+0.01+0.0008+0.0004+0.0002+0.00005 = 1

The client seeds are actually generated using javascript entirely on the client side. This is the part of the Javascript code that generates the client seed (can be found in the JS file http://static4.freebitco.in/min/main1403284658.js now but the name of the file changes when I make updates) :

Code:

charSet = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
var randomString = '';
for (var i = 0; i < 16; i++)
{
var randomPoz = Math.floor(Math.random() * charSet.length);
randomString += charSet.substring(randomPoz,randomPoz+1);
}
$('#next_client_seed').val(randomString);

Luror

full member

Activity: 215

Merit: 100

Quote from: natri on June 20, 2014, 01:00:44 PM

Anyways, I've actually tested the randomness of rolls and the result was that rolls ARE NOT RANDOM, p=0,004 (sample 500, chi-squared).
This low p means it's almost impossible that I made error of type I, so rolls actually aren't random at all...
More info regarding my testing procedure here: http://www.reddit.com/r/Bitcoin/comments/28dtaa/wondering_freebitcoin_provably_unfair/

This is interesting, it is possible after all that pseudo-random numbers (for seeds) are not quite random. Why didn't you share full test results? It would be nice for someone to repeat it (though it's quite time consuming: 500 hours).

Quote from: natri on June 20, 2014, 01:00:44 PM

(btw, I've never rolled more than 9993 on any of my multiplier or regular rolls on neither faucet, probability of which should be over 75%...)

But you said you made only 500 rolls? Or did you mean some bigger test? You need more rolls to get 3rd and 4th prizes (at least 1250-5000).

Quote from: Luror on March 31, 2014, 11:13:56 AM

[0 - 9885]: ~ 1/20000 + 9885/10000 = 0.98855
[9886 - 9985] ~ (9985-9886+1)/10000 = 0.01 (~ once per 100 rolls)
[9986 - 9993] ~ (9993-9986+1)/10000 = 0.0008 (~ once per 1250 rolls)
[9994 - 9997] ~ (9997-9994+1)/10000 = 0.0004 (~ once per 2500 rolls)
[9998 - 9999] ~ (9999-9998+1)/10000 = 0.0002 (~ once per 5000 rolls)
[10000] ~ 1/20000 = 0.00005 (~ once per 20 000 rolls)

sanity check: 0.98855+0.01+0.0008+0.0004+0.0002+0.00005 = 1

wetsuit

legendary

Activity: 981

Merit: 1026

I shall set up a javascript based roll verifier within a couple of days to hopefully stop all the "your game is not fair" accusations.

Luror

full member

Activity: 215

Merit: 100

Quote from: natri on June 20, 2014, 01:00:44 PM

Hey guys. I love this faucet however I think it is deceiving in terms of payoff and it's probability of high winnings.
Rolling a number between 1-10000 would imply a 1/10000 probability of hitting 10000, however, I think it is much lower.

The probability is actually 1/20000, take a look at my earlier post:

Quote from: Luror on March 03, 2014, 11:27:06 AM

Quote from: ingrownpocket on March 02, 2014, 06:31:48 PM

With 2889 referrals, I find it odd that I've never gotten a 50% commission from a 0.22500000 BTC payout (10000 lucky number).

I'm in doubts the probability of rolling "10000" is 0.0001 with current algorithm, it would be interesting if someone would do an independent test (run an implementation of current algorithm 1,000,000 times, look at the amount of 10000's).

upd Ok, I looked at algorithm closer, and if everything is properly randomized, we can calculate the probability theoretically:

"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 up to the nearest whole number.
5. This whole number is used as your roll, with the maximum possible value being 10,000."

We have hex strings from "00000000" to "ffffffff" (0 to 4294967295), 4294967296 in total. "ffffffff" converts to 4294967295/429496.7295 = 10000.

The biggest number which will be rounded to 10000 is 9999.5, back converting it (*429496.7295), we'll get 4294752546.63525.

So the numbers from the interval [4294752547, 4294967295] will be converted to "10000", the quantity of them = 4294967295-4294752547+1 = 214749.

The probability of getting "10000" = 214749/4294967296 = 0.00005000014789402484893798828125 (~ 1/20000)

P.S. After 40,000 rolls one should get 86% probability of rolling "10000" at least once

(proof: https://www.wolframalpha.com/input/?i=1-%281-0.00005000014789402484893798828125%29^40000)

Topic: FreeBitco.in-$200 FreeBTC🏎Win Lambo🔥0.2BTC DailyJackpot🏆$32,500 Wager Contest - page 416. (Read 400708 times)