Work out how big the Nitrogen Dice Jackpot has to be for play to be profitable

[bitedge]

Edit: summary and conclusions of this thread

Nitrogen’s dice game has an accumulating jackpot which is won by rolling a number with a 7 in it 12 times in a row. In the long run you will make money by betting when the jackpot is over ฿3.65 because the chance of winning the jackpot is greater than the money you expect to lose trying to win it.

Details and calculations here https://bitroll.com/bitcoin-dice-guide/mathematically-profitable-dice-betting/



The conditions of the Nitrogen Jackpot are that

• You roll a number with a 7 13 times in a row
• Each bet is at least BTC0.001

So we need to work out what is the expected loss of rolling BTC0.001 however many times you need to roll in order to expect to hit the jackpot. Whenever the jackpot is higher than that number it is profitable to roll.

For this we need to know at least

• House edge/expected value
  Each roll has a 1% house edge, 99% expected value.
  
• probability of rolling a 7
  each roll has 4 numbers ##.## and each number has a 10% chance of being a 7 so each roll has a 40% chance of including a 7
  
• probability of rolling 13 consecutive 7s
  40% x .4 x .4 x .4 x .4 x .4 x .4 x .4  x .4 x .4 x .4 x .4 x .4 = 0.00067108864%

100/0.00067108864 = 149011

This means we need to roll 149011 times to expect to roll 13 consecutive 7s.

We can expect a jackpot for every BTC149.011 bet which has an expected loss of 1%, that is BTC1.49011. As such we should be rolling any time the jackpot is more than BTC1.4911. Usually the jackpot is more than that, the last time it went off it was for BTC5.9.

Right now it's BTC3.9 so we should all be betting there with the auto betting bot that lets you set it to stop rolling if the jackpot is hit, that is what I am doing right now.

I am not 100% confident in my calculations and assumptions, corrections and input are welcome.

If you did not know Nitrogen now have dice here is a review http://www.bitroll.co/bitcoin-dice-reviews/#nitrogen

[joksim299]

I think that your calculations are not correct, Coef calculated chance to hit jackpot and NLNico has added EV calculation.

What are our odds to hit jackpot at the moment? 12 times in a row does not sound realistic to me.

The chance for getting a 7 in any of the 4 digits in one bet is 1 - 0.9 ^ 4 = 0.3439.
So the chance to get it done 12 times in a row should be 0.3439 ^ 12 = 0.000002736 or 0.0002736% or 1 in 365438.

Didn't realize Coef calculated it too, lol.

But yeh, jackpot needs to be at least: 1/((3439/10000)^12)*0.001*0.01 = BTC3.65. If the JP is that amount, then making 0.001 bets is EV+ Smiley

[BigMac]

• probability of rolling a 7
  each roll has 4 numbers ##.## and each number has a 10% chance of being a 7 so each roll has a 40% chance of including a 7
  

This part is just plain wrong. Think about this: If you have a 10-digit number, is the chance of having a 7 somewhere in it 100%? I am afraid not.

[bitedge]

First thing to note is that the Jackpot has sometimes been set to 12 consecutive 7s and sometimes at 13 consecutive 7s, I am not sure what Nitrogen are doing with this or if other numbers are possible but if we work out the calculation we can substitute any number in there.

Someone much smarter than me sent me this

If each roll has 4 numbers and you need only one of them to be 7 you calculate the probability like this:

P(any number is 7) = 1 - P(no number is 7)
                   = 1 - P(first not 7) * P(second not 7) * P(third not 7 ) * P(forth not 7)
                   = 1 - 0.9 ^ 4
                   = 0.3439

Then you will want:

P(13 rolls have a 7) * jackpot  > 13 * bet_cost * house_edge
jackpot > 13 * 0.001 * 0.01 / (0.3439 ^ 13)
        > 138.14

And if I substitute a 12 for the 13 I get

12 * 0.001 * 0.01 / (0.3439 ^ 13) = 127.515

Then NL Nico recons

1/((3439/10000)^12)*0.001*0.01 = 3.65

And If I substitute the 12 for an 13 I get

1/((3439/10000)^13)*0.001*0.01 = 10.626

Can anyone defend either of the above over the other one? NL Nicos result look closer to right...

[NLNico]

First of all

And if I substitute a 12 for the 13 I get

12 * 0.001 * 0.01 / (0.3439 ^ 13) = 127.515

If you change the first 13 to 12, you have to change the second one too :p which would make it BTC43.85





Both agree it's 0.3439 probability to hit a 7 in one roll. So for a 12 streak it's 0.3439^12 = 0.00000273644

So you should hit once in every 1/0.00000273644 = 365438.155453 rolls. To calculate how much you will spend on that just multiply by roll amount and house edge:
365438.155453*0.001*0.01 = BTC3.65

1/(0.3439^13) = 1062629.12315 rolls.
1062629.12315 *0.001*0.01 = BTC10.626


I think your first explanation shouldn't multiply by 12 and 13. (guess what 138.14 divided by 13 is and 43.85 divided by 12 )



Also I assume nitro just makes it to 13 whenever the jackpot is BTC3.65 which makes the jackpot never EV+. The advantage for nitro is that there is a much bigger chance the jackpot will increase to really nice numbers as it gets less likely to hit when the streak-number increases.

edit: actually according to this post, nitro doesn't do that atm. So I am not sure on which amount they make it 13.

[ndnh]

• House edge/expected value
  Each roll has a 1% house edge, 99% expected value.
  
• probability of rolling a 7
  each roll has 4 numbers ##.## and each number has a 10% chance of being a 7 so each roll has a 40% chance of including a 7
  
• probability of rolling 13 consecutive 7s
  40% x .4 x .4 x .4 x .4 x .4 x .4 x .4  x .4 x .4 x .4 x .4 x .4 = 0.00067108864%

Probability of having a 7 in the result of a 4 digit dice roll.
=> 1 - chance of 7 not appearing in 1 place ^ number of digits
= 1-(0.9^4) = 0.3439 or 34.39%.

Probability for the event to occur 13 times consecutively = 0.3439^13 = 9.4106210549937454718076051037294e-7 = 0.00000094106210549937454718076051037294.

That is 1 in 1062629.12315 rolls. (1/(0.3439^13) directly)

That is 1 bet in 1062.62912315BTC wagered.
House edge would be 10.6262912315BTC.

That is, if the number of required consecutive rolls with 7 to win the jackpot is 13, the jackpot should be more than 10.6262912315BTC for the player to have a positive EV.

Edit: Looks like it has been changed to 12 rolls.
Jackpot > 3.6543815545BTC in that case.

[1/(1-0.9^4)^12]*0.001*0.01 = 3.65438155453BTC

[Haggisahh]

It should also be noted that rolls below 10 only have 3 digits, making it even more difficult.

[ndnh]

It should also be noted that rolls below 10 only have 3 digits, making it even more difficult.

Not really. it is exactly the same. 4.12 is 04.12, you see?


Edit:
Taking numbers from 0.00 to 99.99, we have,
number of times 7 appears in tens place = 10^3 = 1000 (70.00 to 79.99 => 7999-7000+1)
ones place = 1000
tenths place = 1000
hundredths place = 1000
Total = 4000. (40%)

Now, you notice that some numbers were counted more than once. for instance, a 77.77 has been counted four times, and a 7.77 or 07.77 has been counted 3 times.
So we reduce that.
3*1 +  2*9*4 +  1*9*9*6 = 561 or 5.61%.

4000-.61 = 3439/10000 or 34.39%.

[Haggisahh]

Oh, you're right :-)
Either way it seems like people are rolling for this way way too early.

[bitedge]

Nitrogens dice jackpot is now so big it is mathematically profitable to bet because the Jackpot is more than the amount you would expect to lose by rolling enough that you would expect to win it Details and calculations here

https://bitroll.com/bitcoin-dice-guide/mathematically-profitable-dice-betting/

[bitedge]

This thread did a great job in finding out the +EV point of the Nitrogen dice jackpot. Now we have a similar challenge to work out the same thing for FortuneJack's new dice jackpot. It works differently.

The house edge is the same 1%

The min bet is 10 mBTC (all bitcoin figures will be in mBTC)

If you win you get 80% of the jackpot

In order to win you have to roll a four figure number starting with a 1, then starting with a 2, then starting with a 3, through to 7. For example

17.65
27.65
37.65
47.65
57.65
67.65
77.65

Here is how FortuneJack explains it



So my first and probably wrong guess is something like.

P(num is ok) = 1 / 10
∴
P_Jackpot = 0.1^7

which is .0000001 indicating we expect to hit the jackpot once in 10,000,000 rolls.

min bet of 10 makes that 100,000,000 that we need to bet

house edge of 1% means we expect to lose

100,000,000 * 0.01 = 1,000,000

So we expect to lose 1,000,000 betting enough to win the jackpot

we only get 80% of the jackpot

1,250,000 * .8 = 1,000,000

therefore the jackpot needs to be more that 1,250,000, which is ฿1250 for betting to be profitable? that can't be right, I must have set the probability of hitting to high or something...

What do yall think?

FWIW I used to find FortuneJack slightly shady but they have kept on improving and running an honest operation, I quite like them now. Here is a full review.

https://bitroll.com/bitcoin-dice-reviews/fortunejack-dice-review/

Roll good.

[bitedge]

Thanks alot ndhc,

This old thread is not getting viewed because old. I should have kept this for discussing the Nitrogen Jackpot and made a sperate thread about the FortuneJack one, so that's what I did

https://bitcointalk.org/index.php?topic=1502112

[bob123]

Im definetly gonna look into this.
Thanks for your time calculating this.

[Hatcher]

• House edge/expected value
  Each roll has a 1% house edge, 99% expected value.
  
• probability of rolling a 7
  each roll has 4 numbers ##.## and each number has a 10% chance of being a 7 so each roll has a 40% chance of including a 7
  
• probability of rolling 13 consecutive 7s
  40% x .4 x .4 x .4 x .4 x .4 x .4 x .4  x .4 x .4 x .4 x .4 x .4 = 0.00067108864%

Probability of having a 7 in the result of a 4 digit dice roll.
=> 1 - chance of 7 not appearing in 1 place ^ number of digits
= 1-(0.9^4) = 0.3439 or 34.39%.

Probability for the event to occur 13 times consecutively = 0.3439^13 = 9.4106210549937454718076051037294e-7 = 0.00000094106210549937454718076051037294.

That is 1 in 1062629.12315 rolls. (1/(0.3439^13) directly)

That is 1 bet in 1062.62912315BTC wagered.
House edge would be 10.6262912315BTC.

That is, if the number of required consecutive rolls with 7 to win the jackpot is 13, the jackpot should be more than 10.6262912315BTC for the player to have a positive EV.

Edit: Looks like it has been changed to 12 rolls.
Jackpot > 3.6543815545BTC in that case.

[1/(1-0.9^4)^12]*0.001*0.01 = 3.65438155453BTC

I had to necro this thread because I've been thinking about this a lot lately, and I'm not actually sure that the math above is sufficient to calculate the EV for this. I'd like to get some input on my thought process here.

The thing is, the bet isn't a one-off. You're not betting 1000 satoshis for a chance to multiply your bet by 1062629.12315x all at once. It would be great if it were that simple and the math in this thread would suffice for that. However, here you're going for a sequence of 13 rolls, each with the odds of 34.39% to qualify, but also with each bet costing you 1000 satoshis. By the time you've rolled a jackpot hitting sequence, that particular winning sequence would have cost you 13000 satoshis. It complicates things a bit.

The question is: considering you have to pay 1000 satoshis a spin, and need 13 consecutive spins of 34.39%, what is your total investment before you are mathematically expected to hit the jackpot? Are you expected to see a specific 13 number streak within a sequence of 1062629 rolls? If so, you can consider your total investment for those rolls to cost $1,062,629 and as long as the prize is worth more than that you are good to go. But it's more complicated than that, evidenced by the fact that you would rather pay $1 with a million sided dice to hit $1M at once, rather than pay $1 for each 10 sided dice you roll trying to get 6 1s in a row (same problem, easier to visualize).

[actmyname]

But it's more complicated than that, evidenced by the fact that you would rather pay $1 with a million sided dice to hit $1M at once, rather than pay $1 for each 10 sided dice you roll trying to get 6 1s in a row (same problem, easier to visualize).

Those two have the exact same probability.
For the former, it's 1 out of 1 million, obviously.
For the latter, it's 1/10^6 = 1 out of 1 million.

You're concerned about the price per roll however ndnh already factored in the house edge. That considers all possible roll percentages and gives you the required ev.

Consider this very simple proposal.

1/10000 chance to hit the jackpot.

One option is to pick a 10000-sided die. You are still betting on the outcome of this dice. Say, >=5000 for 2x. EV of rolling the dice without factoring jackpot: 0. EV of rolling the dice with factoring in the jackpot: jackpot/10000.

The other option is to roll 100-sided dice. The rolls are independent but having two 100's in a row means that you will hit the jackpot. Say you bet >=50 for 2x. EV of rolling the dice without factoring jackpot: 0. EV of rolling the dice with factoring in the jackpot: jackpot/10000.

Now if you factor in house edge you can see the similarities.

[Hatcher]

But it's more complicated than that, evidenced by the fact that you would rather pay $1 with a million sided dice to hit $1M at once, rather than pay $1 for each 10 sided dice you roll trying to get 6 1s in a row (same problem, easier to visualize).

Those two have the exact same probability.
For the former, it's 1 out of 1 million, obviously.
For the latter, it's 1/10^6 = 1 out of 1 million.

You're concerned about the price per roll however ndnh already factored in the house edge. That considers all possible roll percentages and gives you the required ev.

Consider this very simple proposal.

1/10000 chance to hit the jackpot.

One option is to pick a 10000-sided die. You are still betting on the outcome of this dice. Say, >=5000 for 2x. EV of rolling the dice without factoring jackpot: 0. EV of rolling the dice with factoring in the jackpot: jackpot/10000.

The other option is to roll 100-sided dice. The rolls are independent but having two 100's in a row means that you will hit the jackpot. Say you bet >=50 for 2x. EV of rolling the dice without factoring jackpot: 0. EV of rolling the dice with factoring in the jackpot: jackpot/10000.

Now if you factor in house edge you can see the similarities.

We're beyond factoring the house edge in - I have already done that. The jackpot bets that I am referring to solely concern the house edge. The house edge for a qualifying bet (.001 BTC) is 1000 satoshis. That's what you're paying the house each roll when you are trying to hit the jackpot.

I know that the two options have the probability of 1 million, that's why I used it as an example. But you would much rather do it as a 1 off (1000000x your 1000 satoshi bet with a million sided dice) rather than pay 1000 satoshis for each time you roll 6 individual dice to hit the same jackpot.

1 1 million sided dice - you paid 1000 satoshis to win 10 BTC
6 10 sided dice - you paid 6000 satoshis to win 10 BTC

[actmyname]

1 1 million sided dice - you paid 1000 satoshis to win 10 BTC
6 10 sided dice - you paid 6000 satoshis to win 10 BTC

I see what's happening here. I'm not sure if this follows a more complicated binomial distribution (since you have non-independent probabilities) but shouldn't you then expect to pay N * jackpot house edge to win the amount, where N is the number of dice? Given E(X) = 1/10^6 * 6(-1000) + (1-1/10^6) * 0

So then we would have (in your example) a breakeven point of 60 BTC.

[Hatcher]

1 1 million sided dice - you paid 1000 satoshis to win 10 BTC
6 10 sided dice - you paid 6000 satoshis to win 10 BTC

I see what's happening here. I'm not sure if this follows a more complicated binomial distribution (since you have non-independent probabilities) but shouldn't you then expect to pay N * jackpot house edge to win the amount, where N is the number of dice? Given E(X) = 1/10^6 * 6(-1000) + (1-1/10^6) * 0

So then we would have (in your example) a breakeven point of 60 BTC.

That would indeed be the case if we were forced to buy the rolls in groups of 6. However as soon as we encounter a failure in our streak we start again from the beginning.

Example of buying the rolls in groups of 6: [4,0,5,7,7,7], [7,7,7,9,8,3] Though you hit 6 specific numbers in a row, you would have lost 12,000 satoshis here because of the fact that you had to buy them in groups of 6. In this set up, you would need 60 BTC to break even since you need to hit the 6 specific numbers as soon as you are buying a new group of 6 rolls.

In the actual way this is set-up, that restriction does not exist. [4,0,5,7,7,7,7,7,7,9,8,3]] is a winner after spending 9,000 satoshis.

So, I guess the way of figuring this out is by calculating the average "cost per attempt". 90% of the time you are only paying 1 unit, as you won't hit the 7 and you'll be starting again. However once you do hit the 7, then you need to make a distribution of costs in accordance to the odds of hitting a particular streak (the highest being a 6 streak, with a 1 in 1 million shot to win, but having cost you 6 units once you hit it).

[actmyname]

So, I guess the way of figuring this out is by calculating the average "cost per attempt"

How's this?

Let J = jackpot, p = chance of getting required number, n = amount of numbers required for jackpot
E(X) = (J-n(1000)) * pⁿ + (1-pⁿ)(-1000)

My math could be completely wrong but this seems right to me. Any time you don't run into the n-in-a-row lucky numbers, you lose 1000. Once you do, you win the jackpot and lose n*1000.

I believe(?) this calculation is similar to a negative binomial distribution

Thus we have to find 0 = (J - 12(1000)) * 0.3439^12 + (1-0.03439^12)(-1000)
And J = 3.6545 x 10^8 meaning the jackpot must exceed 3.6545 BTC. This is only very slightly higher than ndnh's prediction.

[Hatcher]

So, I guess the way of figuring this out is by calculating the average "cost per attempt"

How's this?

Let J = jackpot, p = chance of getting required number, n = amount of numbers required for jackpot
E(X) = (J-n(1000)) * pⁿ + (1-pⁿ)(-1000)

My math could be completely wrong but this seems right to me. Any time you don't run into the n-in-a-row lucky numbers, you lose 1000. Once you do, you win the jackpot and lose n*1000.

I believe(?) this calculation is similar to a negative binomial distribution

No, because each number in the streak costs another unit. The "cost per attempt" isn't restricted to a single unit and the amount of numbers required to hit the jackpot, but everything in between.

For the 6 10 sided dice example: If you roll 2 of the specific numbers in a row and lose, you lose 2000 satoshis. If your roll 5 of the specific numbers in a row and lose, you lose 5000 satoshis. The failed attempts have different costs, and the various probabilities for hitting each one must be factored in to determine the average cost per attempt.