Dice roll probability

[MyUserName]

Great, thanks. I did not realise it was a straight binomial probability.

My analysis of what you've written is that you have 12 1:1 fights (trials) and the first fighter is weighted to the tune of 9/12 in winning these fights.

Sorry if these are stupid questions but I have not studied stats, ever!

Is this done purely on the number of dice I mentioned? So if the knight had 10 dice the equation would be 10/13?

In that case, p=0.75 (9/12), n=12 and k= 7 successes or more from the 12 trials.

Sorry where does 7 come from?


Does this also work for the other scenario:

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).

[sasguru]

One way to do this (and quite possibly the only way as my first impression is there isn't a closed form solution- ie no equation can describe it) is by simulation.

You would need a random number generator and a set of rules/code to do this. The random number is between 0 and 1. You would say if random number < (1/6) then dice roll=1. You would keep a count of the number of success etc, and write the code to cover all outcomes. Run the simulation 10000 times and calculate the average number of successes.

That was indeed a possibility I suggested to MUN in my initial conversation.
Write some code.

[scooterscot]

I've already given you the equation for computing binomial probabilities. I think the problem you're having is expressing this in the right format. Most of the difficulty with maths is analysis of the problem.

WHS - the question is all backwards.

Once again, if it's not simply explained it's poorly understood.

[scooterscot]

This is not relevant as it misses the complicating factor of there being an unknown number of dice involved after the initial allocation.

Each dice has a 1/6 chance of bringing in another die and repeat. Logically this causes some kind of series to form of 1/6^X or something but I cannot nail it.

This game will be fun for about one minute is my guess!

In that case the probability that this series occurs for n times is equivalent to sum <sigma> (with limits between one & infinity) where Prb_of_1st_die . Prb_of_2nd_die and so no n...

[MyUserName]

That was indeed a possibility I suggested to MUN in my initial conversation.
Write some code.

This is my fall back position. I was hoping to learn a bit about probability and stats by solving it myself and when that failed asking someone to explain it to me. It is not relevant to anything I am doing, I would just like to know.

I can write programs like this in my sleep, I will not learn anything by doing that so I will only do it like that if I run out of time.

[scooterscot]

What is it you sell those poor Germans?

What makes you think they're poor? They're minted man. Loads of money over here.

Want in?

[jamesbrown]

Great, thanks. I did not realise it was a straight binomial probability.



Sorry if these are stupid questions but I have not studied stats, ever!

Is this done purely on the number of dice I mentioned? So if the knight had 10 dice the equation would be 10/13?



Sorry where does 7 come from?


Does this also work for the other scenario:

Not stupid questions at all, but the starting point is to analyse and express the problem clearly. I think you just need to express it as the probability if success (relative to the other fighter), rather than the number of throws. If a fighter is stronger, they have a greater p and the other fighter has 1-p chance to win each fight.

The 7 is the number of successes required to win overall from 12 trials, i.e. 6 would be a tie, 7 an overall win.

[MyUserName]

In that case the probability that this series occurs for n times is equivalent to sum <sigma> (with limits between one & infinity) where Prb_of_1st_die . Prb_of_2nd_die and so no n...

Not sure this is it. A success is a 1 or a 2 but only a 1 allows another die to be rolled. Does Prb_of_Xth_die above relate to the probability of a 1 or probability of a 1 or 2?

Not sure why you think it is poorly explained. A success is a 1 or a 2 and rolling a 1 also allows another die to be rolled.

What is the equation for rolling X amount of successes given Y amount of starting dice?

[No2politics]

Actually I think there might be a closed form!

It's a geometric progression.

http://en.m.wikipedia.org/wiki/Geometric_progression

With a=2/6 and r=2/6

If u first assume you only have one roll and work out the expected number of successes for that, and then multiply by the number of die you actually have.

So the expected number of successes is (2/6) + (2/6) squared +(2/6) cubed etc

[No2politics]

Depends whether there is a equation for the sum of a geometric progression- good luck. I'm outside waiting for train freezing my fingers off!!