Dice roll probability

[jamesbrown]

You'll need to express your problem better. These are independent events. The probability of rolling a 1 is independent of what was rolled before it. Thus, there is no increased probability of rolling a 1 if doing so allows another roll of the die. I think you're asking about the probability of a particular sequence. This is governed by binomial probabilities and it's simple to determine the probability of "k out of n" occurrences in that context. Also, your comment about the average doesn't really make sense. The expected number of successes is equal to the probability multiplied by the number of trials.

Binomial distribution - Wikipedia, the free encyclopedia

[MyUserName]

I know you said "allows", but if a 1 compels that dice to be rolled again, and so on, then the chance of any 1s in the end must be zero, and that means the (equal) chance of any of 2 to 6 must be 1/5. If that helps (I suspect not).

Maybe you should explain more precisely the situation whose odds you're after.

I phrased the original post poorly, rolling a 1 causes another die to be rolled. Not the original dice to be rerolled.

If I have 8 dice (all with 6 sides) and need to end up with 1 or 2 being rolled at least 4 times (included additional dice being rolled to a 1 coming up) what are the odds?

Note that each roll is separate so rolls are not added together.

[RedSauce]

I phrased the original post poorly, rolling a 1 causes another die to be rolled. Not the original dice to be rerolled.

If I have 8 dice (all with 6 sides) and need to end up with 1 or 2 being rolled at least 4 times (included additional dice being rolled to a 1 coming up) what are the odds?

Note that each roll is separate so rolls are not added together.

You cannot calculate the probability, as it could be an infinite number of dice rolls. You could get an approximation quite simply and the more rolls you do (getting more 1's, it would tend towards this number)

[sasguru]

If i have understood correctly surely the probability is:

1/s^n

where s is number of sides and n is the number of dice

EDIT: edit that is for getting the same number each time, can easily be changed for matching multiple

That was my approach but its no dice (pun intended).
I think what he wants is a pretty complicated equation:
Given x dice, you'll have a certain number of 1s (probability for that is easy to get). But then each one of those leads at another throw, each of which can lead to another and so on - leading to an infinite series summation type problem?

TBH, MUN, I'm still not sure what your requirements are.
Maybe give some scenarios?

[MyUserName]

You'll need to express your problem better. These are independent events. The probability of rolling a 1 is independent of what was rolled before it. Thus, there is no increased probability of rolling a 1 if doing so allows another roll of the die.

Agreed.

I think you're asking about the probability of a particular sequence. This is governed by binomial probabilities and it's simple to determine the probability of "k out of n" occurrences in that context. Also, your comment about the average doesn't really make sense. The expected number of successes is equal to the probability multiplied by the number of trials.

What is tripping my stats up is the additional dice roll needing to factored in. Without that then the probability would be (2/6)^x where x is the number of dice.

Binomial distribution - Wikipedia, the free encyclopedia

So it is possible to work out the mean average?

[sasguru]

You cannot calculate the probability, as it could be an infinite number of dice rolls. You could get an approximation quite simply and the more rolls you do (getting more 1's, it would tend towards this number)

That is what I was thinking ...

[MyUserName]

TBH, MUN, I'm still not sure what your requirements are.
Maybe give some scenarios?

Fair enough, maybe I am not explaining things very clearly!

If a character has, say archery skill at level 3 that would mean he gets to roll 6 dice on archery tests.
If he has a magic bow he gets an extra 2 dice.

He wants to hit a target very far away, the GM has allocated it at difficulty level 5.

Hence the archer rolls 6+2 = 8 dice hoping to end up with 5 successes (where a success is a 1 or a 2 being rolled).

The other scenario is if Sir KillsALot meets a humble peasant

Sir Kills has fight skill of 4, sword 2D and armour 3D so rolls 9 dice when he fights.
Peasant has fight skill of 1, a spear 1D and no armour so rolls 3 dice when he fights.

If they fight each other they both roll and whomever gets the most successes wins.

In order to correctly calibrate task difficulty etc. I would like the equations so I can calculate probabilities.

[RedSauce]

Fair enough, maybe I am not explaining things very clearly!

If a character has, say archery skill at level 3 that would mean he gets to roll 6 dice on archery tests.
If he has a magic bow he gets an extra 2 dice.

He wants to hit a target very far away, the GM has allocated it at difficulty level 5.

Hence the archer rolls 6+2 = 8 dice hoping to end up with 5 successes (where a success is a 1 or a 2 being rolled).

The other scenario is if Sir KillsALot meets a humble peasant

Sir Kills has fight skill of 4, sword 2D and armour 3D so rolls 9 dice when he fights.
Peasant has fight skill of 1, a spear 1D and no armour so rolls 3 dice when he fights.

If they fight each other they both roll and whomever gets the most successes wins.

In order to correctly calibrate task difficulty etc. I would like the equations so I can calculate probabilities.

I'm even more confused now. But his has got me thinking, I will have a play tonight using a Markov chain, if not could tabulate the possible outcomes with 1:N dice and attempt to work out an equation from that

[OwlHoot]

Seems like the problem is equivalent to repeated rolls of a single dice, with initially 8 allowed goes, where each time you roll a 1 you accrue an extra go.

So the tricky bit is to work out the expected roll sequence length, which is a sort of queuing theory problem almost.

That done, the number of 1s rolled will be this length less 8, and the number of 2s (or any other number) the floor (integer part) of this length less 8 and divided by 5.

[Ticktock]

I'm trying to think back to my GCSE Maths, as I think this is actually quite simple.

You want the probability of rolling a 1 or a 2 on x number of dice, when throwing y number of dice in total?