Dice Games - Provably Fair - ELI5 please?

[whohackedme]

I understand the server seed and the client seed, both are suppose to be secret.  I dont need all the technical details, I am just looking for a general/basic explanation.

(1)  How are the future OVER/UNDER results calculated?  (I know the results are calculated in advance, but how if one does not know the others secret seed?)

(2)  If the client must have the server seed to know results in advance, how does one result get displayed while keeping the others secret?

Thank you!

[FinneysTrueVision]

The client seed is not kept secret. The casino has to know this in order to generate a result.

You have control over the client seed in order to make it fair.

[ReiMomo]

This has been already discussed here and there's a good explanation of how provably fair works by the author @webtricks.
https://bitcointalk.org/index.php?topic=5227525.0

It is fully detailed and I guess that is what you are looking for, just read and don't hesitate to ask again.

The client seed is not kept secret. The casino has to know this in order to generate a result.

You have control over the client seed in order to make it fair.

Exactly, you have a right to verify this as a user.

[whohackedme]

Thanks everyone.  I will read the links.

So, in theory, If you are playing a strategy  (exactly the same without deviation) that the site can figure out,  The site "COULD" generate a bunch of games until they see results that would cause you to loose-- IF you keep betting your strategy?

This seems like what is happening, I just did not think it was possible.

[whohackedme]

This has been already discussed here and there's a good explanation of how provably fair works by the author @webtricks.
https://bitcointalk.org/index.php?topic=5227525.0

Webtricks recommends changing BOTH Client and Server seed every bet.  Would just the client side be sufficient?  I mean, that eliminates any risk, correct?  They still have to generate new set of rolls.  I do not understand the risk by just changing client seed but not server seed?


Also, if someone would like to give me a reality check to confirm my statistics on OVER/UNDER  on Dice:

games or flips to generate consecutive OVER or UNDER score (house = 0.00)

4 games for 2 consecutive
8 games for 3
16 games for 4 etc.

[whohackedme]

I dont want to blast any dice site, This seems sketchy to me.
Below is a chart of my balance after each bet.
I changed my client seed/server seed about every 50 bets.
That is a bot, about 1000 bets/hour.

To me, Blue looks normal or random, Red does not.
Red is about 3000 bets.

(I have all the server/client seed nonce info for each bet)

Anyone want to share their thoughts?

[FinneysTrueVision]

I dont want to blast any dice site, This seems sketchy to me.
Below is a chart of my balance after each bet.
I changed my client seed/server seed about every 50 bets.
That is a bot, about 1000 bets/hour.

To me, Blue looks normal or random, Red does not.
Red is about 3000 bets.

(I have all the server/client seed nonce info for each bet)

Anyone want to share their thoughts?

There is not a lot of information to go by in this screenshot. Depending on the strategy you were using it is entirely possible to produce these results. If you have client/server seeds and nonce then you can use a third party verifier to see if the results match the same results displayed by the casino.

Thanks everyone.  I will read the links.

So, in theory, If you are playing a strategy  (exactly the same without deviation) that the site can figure out,  The site "COULD" generate a bunch of games until they see results that would cause you to loose-- IF you keep betting your strategy?

This seems like what is happening, I just did not think it was possible.

They would need to have a really flawed provably fair system if they were able to manipulate the results against you. The only scenarios where I see this being a possibility is if you used the default client seed or you re-used a client seed when the server seed was changed.

[whohackedme]

Thanks for your thoughts.  Yes, not much info there, but the random trendline went out of whack.  When I start zooming in on the details, I can not see the shift.  It is probably just random.  I know there is some numerical difference there, I was just curious if anyone here thought that was an odd shift to be random or?  It is just a long term trend shift, like I changed my betting strategy.  Most strategies have a linear trendline, then you get wrecked by a large (random) negative losses that wipe out all your wins Quickly. (But not typically would you expect FIVE 1 in 250 chance events in 100 games/bets)   I guess the thing about random it is random  (I guess computer generated is true random, nature is not random  - Benford's law)  


Here is the scenario I was imagining:

Lets say I am playing simple opposite of a run and increasing bets after each loss. If two OVER in a row, I am betting UNDER and increasing my bet on each loss.  We can not do this forever, we would run out of money.

Say #5 bet in a row is always my largest and last bet (If there are 5 in a row, and I have been betting to break the streak each time, but I never bet to stop a 6th consecutive score.)  Very easy for a site to see this strategy.

Lets say I always get a new client seed after the #5 loss.

All a site would have to do is generate 2^6 (64 sets of scores) evaluate which one has 6 or more consecutive runs very early and use that set of scores.  They can prove it was "fair".

OR, lets say I change both the Server and Client seed every single bet.

All a site would have to do is on average generate 2 sets of scores and they could make the next bet be the one that is the score they want and then choose to use that set.  Still it can be proven as fair.

I am not so sure that changing the seeds all the time is a good option for security.   I think that changing that often allows the sites to change the next bet everytime.  Maybe the best option is to keep track of the seeds and prove they did not change the results.

[whohackedme]

So, after some further research, what I was worried about is not possible.

One Server Seed + One Client Seed  Generate ONE set of scores.  Not many combinations of different sets of scores.

[o_e_l_e_o]

Have a look at this site: https://dicesites.com/provably-fair

It provides a good explanation of how rolls are made provably fair, and also provides verifiers for a variety of dice sites. If the site you are using is on there you can plug your numbers in and check that they match. There are also a number of verifiers provided here: https://www.btcgosu.com/tools/provably-fair-verifier/

One Server Seed + One Client Seed  Generate ONE set of scores.  Not many combinations of different sets of scores.

Provided you are also including the nonce, then yes. You can use one server seed and one client seed to play thousands of games, by simply incrementing the nonce by 1 after each bet, which sites will do automatically. Even such a minor change will result in a completely different and completely unpredictable hash output, will will result in a completely random and completely unpredictable dice roll.

If at any point you think something is not random (as in the case of your red rolls in the image you linked), then you can reveal the server seed and check everything is fair.

[whohackedme]

Thank You.   I have a statistical problem I would love to learn an explanation.

Statistically to get two in a row (OVER or  UNDER) it takes 4 games
3 in a row takes 8 games
4 in a row takes 16

For example, when I evaluate the data from a dice site.

3 in a row averages just under 7 games
4 in a row averages just under 14 games

I have confirmed this with 20,000 games to sets of 4,000 games.

IF I generate in excel or python script,  combine random four numbers (0-9) and divide by 100 to generate the score, (and generate the same data qty) I get the statistical average.

I have only tested one dice site,  Does anyone know why this happens and does it happen on all sites?

I want to speculate that this "Slightly" increases the number of longer runs to upset any statistical based betting strategies.

Looking forward to everyones comments.

[o_e_l_e_o]

Statistically to get two in a row (OVER or  UNDER) it takes 4 games
3 in a row takes 8 games
4 in a row takes 16

How did you arrive at these numbers? They are not correct.

Assuming a 50/50 chance (so flipping a coin, for example), then the equation for calculating the expected coins flips required, x, to produce n consecutive identical results (so n heads in a row, for example), is as follows:

x = 2⁽ⁿ⁺¹⁾-2

So, to flip 2 heads in a row, it will statistically take 6 flips.
To flip 3 heads in a row, it will statistically take 14 flips.
To flip 4 heads in a row, it will statistically take 30 flips.

Now, if we are looking at flipping n in a row of EITHER heads or tails, then you simply add one flip to the number before. For example, I want to flip either 3 heads or 3 tails in a row, then statistically I flip once (which will start me down the path of either heads or tails), and then I need 2 in a row of whichever path I am on.

So, to flip 3 either heads or tails in a row, it will statistically take 1+6 = 7 flips.
To flip 4 either heads or tails in a row, it will statistically take 1+14 = 15 flips.

[whohackedme]

Statistically to get two in a row (OVER or  UNDER) it takes 4 games
3 in a row takes 8 games
4 in a row takes 16

How did you arrive at these numbers? They are not correct.

Assuming a 50/50 chance (so flipping a coin, for example), then the equation for calculating the expected coins flips required, x, to produce n consecutive identical results (so n heads in a row, for example), is as follows:

x = 2⁽ⁿ⁺¹⁾-2

So, to flip 2 heads in a row, it will statistically take 6 flips.
To flip 3 heads in a row, it will statistically take 14 flips.
To flip 4 heads in a row, it will statistically take 30 flips.

Now, if we are looking at flipping n in a row of EITHER heads or tails, then you simply add one flip to the number before. For example, I want to flip either 3 heads or 3 tails in a row, then statistically I flip once (which will start me down the path of either heads or tails), and then I need 2 in a row of whichever path I am on.

So, to flip 3 either heads or tails in a row, it will statistically take 1+6 = 7 flips.
To flip 4 either heads or tails in a row, it will statistically take 1+14 = 15 flips.

You are correct, Thank you 

I just did basic 2^n, for 3 flips I was calculating 2*2*2=8. 

I had the same thought as what you explained, but I was not sure which was correct, 7 or 8 flips,  so I created a test to confirm.

I created a spreadsheet with 10,000 rows, and hit F9 over and over, the average of doing this 50 times
3 in a row = 8
4 in a row = 16

After doing that I felt that the basic 2^n was correct.

I apologize for not having a better understanding, do you mind explaining the difference I see in excel example?

This is how I calc the random scores in excel
I have 4 cells, each one =RANDBETWEEN(0,9), I CONCATENATE those 4 cells and divide by 100 to get the score
I copy this for 10,000 rows.
at the change of OVER/UNDER result = 1st flip in looking for the same result. 
(if current result was same as previous it incremented 1, if it was different, it reset to 1)
I gave the house all results between 49.5-50.5
For the average # of flips for 3 in a row, I counted how many times "3" occurred and 10,000/that number = 8.
Same for 4, 5, etc in a row.

Thanks again.

[shield132]

Thanks everyone.  I will read the links.

So, in theory, If you are playing a strategy  (exactly the same without deviation) that the site can figure out,  The site "COULD" generate a bunch of games until they see results that would cause you to loose-- IF you keep betting your strategy?

This seems like what is happening, I just did not think it was possible.

That's just an old popular opinion. Right now, trustworthy casinos like PrimeDice, Bitsler, crypto-games, etc have 100% provably fair games, they don't cheat/lie. They don't even need to build an AI app that will analyze the player's behaviour and act according to it, no. Game is provably fair but you miss one thing, it has a house edge, 1% in most cases. This house edge changes the whole situation here. Mathematically, because of this house edge, casinos have an opportunity over you and on the long term, they are guaranteed winners.

Just think about this: I have 51 apples and you - 49 and we have to eat the same amount but the winner will be the one left with more apples. When I and you eat 49, I'll still have 2 apples. There is no way you can change it (only if you push me or steal from me).

Idk if it's a good example but tried to show you the reason why the casino is the winner in the long term.

[o_e_l_e_o]

I apologize for not having a better understanding, do you mind explaining the difference I see in excel example?

I'm not sure I completely follow your process, but I suspect the issue might be here:

For the average # of flips for 3 in a row, I counted how many times "3" occurred and 10,000/that number = 8.

Are you only counting runs of exactly 3 in a row, or are you counting runs of at least 3 in a row? That means you would need to count how many times "3" occurred, and "4", and "5", and so on, add them all up, and then divide 10,000 by the total.

[whohackedme]

Thanks everyone.  I will read the links.

So, in theory, If you are playing a strategy  (exactly the same without deviation) that the site can figure out,  The site "COULD" generate a bunch of games until they see results that would cause you to loose-- IF you keep betting your strategy?

This seems like what is happening, I just did not think it was possible.

That's just an old popular opinion. Right now, trustworthy casinos like PrimeDice, Bitsler, crypto-games, etc have 100% provably fair games, they don't cheat/lie. They don't even need to build an AI app that will analyze the player's behaviour and act according to it, no. Game is provably fair but you miss one thing, it has a house edge, 1% in most cases. This house edge changes the whole situation here. Mathematically, because of this house edge, casinos have an opportunity over you and on the long term, they are guaranteed winners.

Just think about this: I have 51 apples and you - 49 and we have to eat the same amount but the winner will be the one left with more apples. When I and you eat 49, I'll still have 2 apples. There is no way you can change it (only if you push me or steal from me).

Idk if it's a good example but tried to show you the reason why the casino is the winner in the long term.

Yes, thank you.

[whohackedme]

I apologize for not having a better understanding, do you mind explaining the difference I see in excel example?

I'm not sure I completely follow your process, but I suspect the issue might be here:

For the average # of flips for 3 in a row, I counted how many times "3" occurred and 10,000/that number = 8.

Are you only counting runs of exactly 3 in a row, or are you counting runs of at least 3 in a row? That means you would need to count how many times "3" occurred, and "4", and "5", and so on, add them all up, and then divide 10,000 by the total.

I am counting every time a 3 occurred.  if it was only 3 in a row or 10 in a row, I would count both.

So I would take 10,000 divided by the total number of times 3 occurred, the long term average would = 8.

ONLY 3 in a row would be the total - count>3.  (correct?)

[o_e_l_e_o]

Let's take rolling either 2 heads or 2 tails in a row. Statistically, this will take us 3 flips. Let's look at all possible combinations of 3 flips:

Code:

HHH
HHT
HTH
HTT
TTT
TTH
THT
THH

There are 8 combinations, 6 of which will give you at least either 2 heads or 2 tails in a row. 24 flips for 6 "success", and 24/6 = 4, which is where your answer comes from.

You are calculating the number of successes in x rolls, as opposed to calculating the number of rolls required for a success. They two are not synonymous.

[whohackedme]

Let's take rolling either 2 heads or 2 tails in a row. Statistically, this will take us 3 flips. Let's look at all possible combinations of 3 flips:

Code:

HHH
HHT
HTH
HTT
TTT
TTH
THT
THH

There are 8 combinations, 6 of which will give you at least either 2 heads or 2 tails in a row. 24 flips for 6 "success", and 24/6 = 4, which is where your answer comes from.

You are calculating the number of successes in x rolls, as opposed to calculating the number of rolls required for a success. They two are not synonymous.

Thank you for your answer.  (and patience lol)

I think I finally understand.  It only takes one flip to get halfway there.  You only need one more flip to get 2 in a row, you have a 50/50 chance, so on average it will take both of those flips to get 2 in a row = 3 flips.

So the equation is 2^n  -1  like you said.  (It would be 4 if we wanted specific Heads or Tails in a row, 3 if we do not care which one)  Thanks!

[o_e_l_e_o]

I think I finally understand.  It only takes one flip to get halfway there.  You only need one more flip to get 2 in a row, you have a 50/50 chance, so on average it will take both of those flips to get 2 in a row = 3 flips.

This is correct.

So the equation is 2^n  -1  like you said.  (It would be 4 if we wanted specific Heads or Tails in a row, 3 if we do not care which one)  Thanks!

It would actually be 6 if we wanted specifically 2 heads in a row. I'll try to explain why below.

Let the number of flips required to get 2 heads in a row be x. On any single flip, we have a probability of 0.5 of flipping heads, and a probability of 0.5 of flipping tails.

If we flip tails on the first flip, then we still need to flip x more times to get 2 heads, but we have already flipped once, so it becomes (x+1). Since the probability of this is 0.5, it becomes 0.5(x+1).
If we flip heads on the first flip and then tails on the second flip, again we are back where we started needing to flip x more times to get 2 heads, but we already flipped twice, so it becomes (x+2). The probability of flipping a heads then a tails is 0.25, so it becomes 0.25(x+2).
If we flip heads on the first flip and then heads on the second flip, we have succeeded. We have flipped twice, with a probability of 0.25, which becomes (0.25)2.

So, of all the possible outcomes:

x = 0.5(x+1) + 0.25(x+2) + (0.25)2
x = 0.5x + 0.5 + 0.25x + 0.5 + 0.5
x = 0.75x + 1.5
0.25x = 1.5
x = 6