Dice roll probability

[MyUserName]

Okay, I am helping write a game and am stumped trying to figure out the probabilities when rolling dice.

I need to roll a set number of dice and count how many 1s and 2s there are (the rolls are not added together). The tricky bit is that rolling a 1 allows another die to be rolled, if that rolls a 1 another die is rolled etc.

Can anyone tell me the equation I need to calculate these probabilities?

Also what would the average number of 1s and 2s rolled if I have a varying number of starting dice?

[SimonMac]

Depends on the dice

[sasguru]

Okay, I am helping write a game and am stumped trying to figure out the probabilities when rolling dice.

I need to roll a set number of dice and count how many 1s and 2s there are (the rolls are not added together). The tricky bit is that rolling a 1 allows another die to be rolled, if that rolls a 1 another die is rolled etc.

Can anyone tell me the equation I need to calculate these probabilities?

Also what would the average number of 1s and 2s rolled if I have a varying number of starting dice?

Sorry I couldn't help earlier.
My instinct tells me this a series type problem but I'm afraid my maths is to old and rusty.

[RedSauce]

Ts pretty simple, how many sides do each die have?

[RedSauce]

Sorry I couldn't help earlier.
My instinct tells me this a series type problem but I'm afraid my maths is to old and rusty.

So is your spelling

[Ticktock]

I'm not sure I understand the question.

Are you asking what the probability is of rolling n 1s in a row, where there are multiple dice to begin with?

So, for example, you roll six dice, and three land as 1s.
So you then roll those three dice again and get one land as a 1.
So you roll that one die, get a 1.
Roll it again and get a 4, so your sequence stops.

Something like that?

[MyUserName]

Sorry I couldn't help earlier.
My instinct tells me this a series type problem but I'm afraid my maths is to old and rusty.

No probs, thanks for trying :-)

Ts pretty simple, how many sides do each die have?

6

I'm not sure I understand the question.

Are you asking what the probability is of rolling n 1s in a row, where there are multiple dice to begin with?

So, for example, you roll six dice, and three land as 1s.
So you then roll those three dice again and get one land as a 1.
So you roll that one die, get a 1.
Roll it again and get a 4, so your sequence stops.

Something like that?

Yes, pretty much. It is a skill resolution system for an RPG.

Each roll of a 1 or 2 is a 'success'. The higher your skill level the more dice you roll.

So if something has a difficulty of 4 you need 4 successes when you roll.

If you roll 6 dice and get 111256 you have 4 successes
You roll 3 more dice as you have 3 1s and get 123 you now have 4+2 successes.
You roll 1 more die as you had 1 1 and get a 2 you now have 4+2+1 = 7success and your rolling stops.

I am trying to work out the probability equation that lies behind this mechanic to answer the following type of questions:

If someone has a skill of 2 (allowing 4 dice as you get 2 per skill level) what is the chance of them getting 3 successes?

If someone rolls, say, 6 dice what is the average number of successes they roll?

[RedSauce]

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

[MyUserName]

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

But isn't that just for a known amount of dice being rolled?

A 1 causes another separate dice roll here.

[OwlHoot]

Okay, I am helping write a game and am stumped trying to figure out the probabilities when rolling dice.

I need to roll a set number of dice and count how many 1s and 2s there are (the rolls are not added together). The tricky bit is that rolling a 1 allows another die to be rolled, if that rolls a 1 another die is rolled etc.

Can anyone tell me the equation I need to calculate these probabilities?

Also what would the average number of 1s and 2s rolled if I have a varying number of starting dice?

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.