Dice roll probability

[doodab]

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What you having for lunch?

Turkey. Not too bad.

[mudskipper]

Turkey. Not too bad.

Cold turkey for the liver ward?

[sasguru]

Turkey. Not too bad.

Better than my lunch.
Missus is at her mum's, taking the sprog because I have to work this w/e.
So I made myself a very boring and dry cheese sandwich.

[shaunbhoy]

Better than my lunch.
Missus is at her mum's, taking the sprog because I have to work this w/e.
So I made myself a very boring and dry cheese sandwich.

A sandwich that might have been custom-made for your demographic.

[MyUserName]

Okay so average number of successes (where success is rolling a 1 or a 2 on a D6) via the following system:

Roll a set number of dice (do not add them up, they are calculated individually).
For every dice which has rolled 1 roll an additional dice
Count all 1s and 2s

Can be partially solved by a negative binomial distribution but this seems to give the probability of rolling 1s and does not factor in 2s. Is this simple to factor in?


Also, for compiling tables of probabilities I would like to know the odds of rolling m successes with n initial dice.

This will be use to calculate the difficulty of tasks in the game. Someone can have a skill ranging from 0 - 5 (where each level of skill gains 2 dice). A level of 5 would be awesome, a world champion of a highly competitive sport for example.

If someone is aiming to fire an arrow into the gold at 100 yards what should the difficulty level be set at? Should it need 4 successes, 10, 15? This is what the probabilities will be used to configure.


I have knocked up a program which calculates averages over many throws. It looks like that being as every dice roll is independent I can calculate the probability of the result from 1 dice and then just multiply it by the number of dice. Is this right?

[doodab]

Okay so average number of successes (where success is rolling a 1 or a 2 on a D6) via the following system:

Roll a set number of dice (do not add them up, they are calculated individually).
For every dice which has rolled 1 roll an additional dice
Count all 1s and 2s

Can be partially solved by a negative binomial distribution but this seems to give the probability of rolling 1s and does not factor in 2s. Is this simple to factor in?


Also, for compiling tables of probabilities I would like to know the odds of rolling m successes with n initial dice.

This will be use to calculate the difficulty of tasks in the game. Someone can have a skill ranging from 0 - 5 (where each level of skill gains 2 dice). A level of 5 would be awesome, a world champion of a highly competitive sport for example.

If someone is aiming to fire an arrow into the gold at 100 yards what should the difficulty level be set at? Should it need 4 successes, 10, 15? This is what the probabilities will be used to configure.


I have knocked up a program which calculates averages over many throws. It looks like that being as every dice roll is independent I can calculate the probability of the result from 1 dice and then just multiply it by the number of dice. Is this right?

I've posted the formula for n successes with a single initial dice up above somewhere. you can use that to work out the multiple die cases as follows.

For more than one initial dice you need to add up the probabilities for all the ways N successees could happen. E.g with two dice and 3 successes you need to work out the probability of getting 3,0 2,1 1,2 and 3,0 from your initial dice and add them all together.

The odds of e.g. 3,0 can be worked out by mutiplying the odds of getting 3 successes from 1 die by the odds of getting none (this latter is 4/6, the formula is only valid for > 0)

For 3 dice you need to consider e.g. 3,0,0 2,1,0 2,0,1 1,1,1 and so on. Each term worked out by multiplying the 3 individual ones together, then add them all at the end.

I will try and write this down and simplify to a single formula in terms of D and N later but I am struggling a bit today.

Oh yes average number of successes is 2D/5 with D initial dice.