Work Out How Big The Fourtune Jack Dice Jackpot Has To Be For profitable play

[bitedge]

This thread

https://bitcointalk.org/index.php?topic=1217368.0

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 0.01฿

If you win you get 80% of the jackpot (50% if you bet 0.0001 and 20% if you bet 0.000001)

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



Thanks to BTF user ndnhc this is what we come up with
Bet amount : 0.01฿

Probability of rolling 1x.xx : 0.1
Probability of hitting the jackpot : 0.1^7

Number of rolls needed on average : 1/0.1^7 = 10000000
฿ wagered required on average : 100000

House edge : 1000฿

To be positive EV, the jackpot should be more than 1000/0.8 = 1250฿
Bet amount : 0.0001฿

Probability of hitting the jackpot : 0.1^7

Number of rolls needed on average : 1/0.1^7 = 10000000
฿ wagered required on average : 1000

House edge : 10฿

To be positive EV, the jackpot should be more than 10/0.5 = 20฿
Bet amount : 0.000001฿

Probability of hitting the jackpot : 0.1^7

Number of rolls needed on average : 1/0.1^7 = 10000000
฿ wagered required on average : 10

House edge : 0.1฿

To be positive EV, the jackpot should be more than 0.1/0.2 = 0.5฿.
The last one appears to be the most attractive option. What do yall think did we miss something? The first case of 1250฿ will probably never happen!

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.

[1713432019]

1713432019

[1713432019]

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[joksim299]

Bitsler also has dice progressive Jackpot

Code:

To win the jackpot you must hit the roll 77.77 and your two last digits of your ID bet must be 77 i.e : xx,xxx,x77

Odds to hit 77.77 1/10000 and tow last digits of bet id 77.77 1/100 = 1/1000000

Code:

A % of the jackpot will be won according to the bet amount
% = (bet_amount*100)/0.005 (BTC)

So Jackpot should be 50BTC to be EV+

[bitedge]

My friend who is a math whiz but not a BTF member suggests we need to multiply this by 7 because it takes 7 rolls to try to hit the jackpot. So if you hit 1 in every 10000000 tries the amount of rolls you need to hit is 70000000 and that is what you should base the expected loss on. This kind of makes sense in that any series of 6 rolls (or less) has a 0 % chance of hitting.

With his theory the +EV point is multiplied by 7 like this in the case of the BTC0.01 bet for 80% of the jackpot.

jackpot > (7 * betPerRoll * houseEdge) / (P(jackpot) * share)

jackpot > (7 * .01 * 0.01) / (0.0000001 * 0.8 )

jackpot >  0.0007/ 0.00000008

jackpot > 8750

exactly 7 times what he had without the 7 roles per try element.

What do yall think

[ndnh]

For reference:

Deleting the other post and moving it here:

Bet amount : 0.01BTC

Probability of rolling 1x.xx : 0.1
Probability of hitting the jackpot : 0.1^7

Number of rolls needed on average : 1/0.1^7 = 10000000
BTC wagered required on average : 100000

House edge : 1000BTC

To be positive EV, the jackpot should be more than 1000/0.8 = 1250BTC. Yeah.


Bet amount : 0.0001BTC

Probability of hitting the jackpot : 0.1^7

Number of rolls needed on average : 1/0.1^7 = 10000000
BTC wagered required on average : 1000

House edge : 10BTC

To be positive EV, the jackpot should be more than 10/0.5 = 20BTC.


Bet amount : 0.000001BTC

Probability of hitting the jackpot : 0.1^7

Number of rolls needed on average : 1/0.1^7 = 10000000
BTC wagered required on average : 10

House edge : 0.1BTC

To be positive EV, the jackpot should be more than 0.1/0.2 = 0.5BTC.

which appears to be the most attractive option.



Edit: Just noticed this was in the OP lol. Never mind. Shouldn't have deleted the other post.

[ndnh]

My friend who is a math whiz but not a BTF member suggests we need to multiply this by 7 because it takes 7 rolls to try to hit the jackpot. So if you hit 1 in every 10000000 tries the amount of rolls you need to hit is 70000000 and that is what you should base the expected loss on. This kind of makes sense in that any series of 6 rolls (or less) has a 0 % chance of hitting.

To me, only the last statement makes sense.

The number of rolls needed (average) would be 10000006 in the very first attempt.
In all other cases it will be 10000000.
When we take the average of 10000006 and 10000000 and 10000000 and 10000000 and so on, we get to 10000000.

That is,
1st roll : 0
2nd roll : 0
...
6th roll : 0
7th roll : 1/10000000
8th roll : 1/10000000
...
100000000000000000th roll : 1/10000000
...
infini'th roll : 1/10000000
Average : 1/10000000


So, the Number of rolls needed on average : 1/0.1^7 = 10000000



1/10000000 is the chance of hitting jackpot per roll. (roll > 6)
The 7 is accounted for while calculating the per roll chance.


ofc I could be wrong.