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Dice Rolling--10 dice, 10 sides, probabilities of winning?


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#1 sn0wie   Members   -  Reputation: 112

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Posted 02 September 2012 - 07:59 PM

Very new here..not sure if this is the right place, but would gladly appreciate some input. I know of someone advertising a game and am curious of your thoughts. I cannot see how he gets his numbers.

"Chance of throwing a losing roll below 47%.

//roll-dice10-sides10
We will then add up the sum of all the dice values. You get paid based on what the final value is when all the dice are added together.

Payout:
10 : win 32x
11 : win 28x
12 : win 25x
13-24: win 10x
25-29: win 5x
30-35: win 3x
36-43: win 2x
44-47: free reroll
48-62: you lose
63-66: free reroll
67-75: win 2x
76-80: win 3x
81-85: win 5x
86-97: win 10x
98: win 25x
99: win 28x
100: win 32x
Special Rolls:
LUCKY 7's (roll all 7's): win 30x
THE COUNTER (Roll 1 2 3 4 5 6 7 8 9 10 or 10 9 8 7 6 5 4 3 2 1): win 30x
Staggered (roll any 2 numbers in a staggered fashion aka 2 3 2 3 2 3 2 3 2 3): Win 20x
The Buildup (every number must be equal or larger than the last. Example - 1 1 2 3 4 4 5 6 7 10): Win 12x
The number problem ( 3 8 1 6 5 4 7 2 9 10): win 31x (Starting from left to right each number is divisible by it's sum of digits exception*- the 10 counts as a 0 digit. So 3 is divisible by 1, 38 is divisible by 2, 381 is divisble by 3, 3816 is divisble by 4 and so on)
Fibonnacci (1 1 2 3 5 8 1 3 2 1): Win 31x
Pyramid scheme (each of the first four numbers builds up to the 5th number and then gets smaller again example 1 1 3 4 10 9 9 8 6 5 ): Win 4x
Chain Rolls:
If you roll a number twice that is not a winner then it becomes a winner for the next roll (or until you stop rolling it): win 2x (this compounds so if you roll it after this it is a 4x winner then an 8x winner then a 16x until I go broke or you stop rolling it)
If you roll 2 consecutive numbers the next consecutive number will become a winner for the next roll (example, if you roll a 56 and a 57 then a 58 is a winner on your next roll)l: win 2x
UPDATE!!!! From now on the 2nd roll for your chain will become a push rather than a loss. This means if you roll a 55 you have turned 54, 55, and 56 all into pushes. This in turn reduces your chance of throwing a losing roll all the way to 46.9%!!!!"

I am very interested in Math and studied Econ & Finance in school, but I believe that this game is very, very unfairly priced, and I get something closer to 23% win chance.

Regardless, thanks! Definitely appreciate your time

Sponsor:

#2 Bregma   Crossbones+   -  Reputation: 5939

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Posted 02 September 2012 - 08:15 PM

Gambling works by suckering money out of fools who are willing to give it up. If the odds were not in favour of the house and if the rubes didn't always lose there wouldn't be much point in running the game, would there?
Stephen M. Webb
Professional Free Software Developer

#3 Álvaro   Crossbones+   -  Reputation: 15525

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Posted 02 September 2012 - 09:07 PM

The easiest way to compute those probabilities is by taking the polynomial P = 0.1*x + 0.1*x^2 + ... + 0.1*x^10, and computing the coefficients of P^10.

I'll try to figure out the odds if I get some time tomorrow. I have some tennis to watch now. :)

#4 Brother Bob   Moderators   -  Reputation: 9289

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Posted 03 September 2012 - 01:44 AM

The easiest way to compute those probabilities is by taking the polynomial P = 0.1*x + 0.1*x^2 + ... + 0.1*x^10, and computing the coefficients of P^10.

I'll try to figure out the odds if I get some time tomorrow. I have some tennis to watch now. Posted Image

Here are the exact odds using Matlab:
Spoiler

The vector of probabilities is such that the first element is the probability to roll 10, second to roll 11, and so on.

#5 LorenzoGatti   Crossbones+   -  Reputation: 3081

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Posted 04 September 2012 - 09:46 AM

Special Rolls:
LUCKY 7's (roll all 7's): win 30x
THE COUNTER (Roll 1 2 3 4 5 6 7 8 9 10 or 10 9 8 7 6 5 4 3 2 1): win 30x
Staggered (roll any 2 numbers in a staggered fashion aka 2 3 2 3 2 3 2 3 2 3): Win 20x
The Buildup (every number must be equal or larger than the last. Example - 1 1 2 3 4 4 5 6 7 10): Win 12x
The number problem ( 3 8 1 6 5 4 7 2 9 10): win 31x (Starting from left to right each number is divisible by it's sum of digits exception*- the 10 counts as a 0 digit. So 3 is divisible by 1, 38 is divisible by 2, 381 is divisble by 3, 3816 is divisble by 4 and so on)
Fibonnacci (1 1 2 3 5 8 1 3 2 1): Win 31x
Pyramid scheme (each of the first four numbers builds up to the 5th number and then gets smaller again example 1 1 3 4 10 9 9 8 6 5 ): Win 4x

All these are terribly rare (1 in 10 billions apart from the "pyramid"); I wouldn't bother.
Maybe shorter combinations could work: for example a subsequence of 4,5,...10 increasing or decreasing values at any point could give an extra payout.
I'm also worried that a game where you just run 10 dice sequentially, without any betting choice except for the trivial initial one, might be too trivial to be fun. Sequential rolls lend themselves to adjusting bets as you see dice, maybe even betting against other players.

Omae Wa Mou Shindeiru





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