Topic: FreeBitco.in-$200 FreeBTC🏎Win Lambo🔥0.2BTC DailyJackpot🏆$32,500 Wager Contest - page 426. (Read 407440 times)

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, 02:51:33 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: wetsuit on June 20, 2014, 02:40:33 PM

Quote from: wetsuit on June 20, 2014, 02:40:33 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: Luror on June 20, 2014, 02:38:31 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: wetsuit on June 20, 2014, 02:16:38 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).

(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).

[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: wetsuit on June 20, 2014, 02:16:38 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

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).

(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).

[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

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

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).

(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).

[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

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).

(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).

[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