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

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#1
Members - Reputation: **112**

Posted 02 September 2012 - 07:59 PM

"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

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#2
Crossbones+ - Reputation: **5651**

Posted 02 September 2012 - 08:15 PM

*Stephen M. Webb*

Professional Free Software Developer

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#3
Crossbones+ - Reputation: **14188**

Posted 02 September 2012 - 09:07 PM

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

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#4
Moderators - Reputation: **8713**

Posted 03 September 2012 - 01:44 AM

Here are the exact odds using Matlab: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.

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

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#5
Crossbones+ - Reputation: **2807**

Posted 04 September 2012 - 09:46 AM

All these are terribly rare (1 in 10 billions apart from the "pyramid"); I wouldn't bother.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

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.