Help Required: Dice Statistics (for table game)

Prime · December 2013

I'm having a talk with my mate regarding a table top game (warhammer in this case) and how much he will be relying on a small % chance of something happening. Statistics are not my friend so I thought some smart bugger can settle this.

What is the probability of the following happening at least once, assuming all dice are 6 sided.

Start by rolling 4 dice, keep any number of them that show 2 or more, discard any 1s.
Roll the left over dice, keep any number of them that show a 6, re roll any 1s that are showing applying this rule to it once more but ignoring the reroll if a further 1 is rolled part.
Take any left over dice and roll them, any number between 1 and 4 show up keep those.

Finish.

So yeah what's the chance of all of that happening at least once? My mind melts after the first stage.

Zombie Hero · December 2013

Sometimes its best to start at the probability of one die completing those possible tasks, and then just expand everything for six four dice. Since die are independent, its pretty easy for one dice to do this.

So P(one dice completing all tasks):

P(not a 1)*P(6 with possible reroll)*P(between 1 and 4)

The first and the last probabilities are trivial.
The second is P(6 w/ reroll) = P(6) or P(1 and then 6)

These are disjoint events so: P(6 w/ reroll) = 1/6 + 1/6*1/6 = 7/36

So

P(completing task) = 5/6*7/36*4/6 = .1080 , or 10.80%

The probability of at least one dice out of four completing task is the same as 1 minus P(no dice making it), so

1 - P(no success in four dice) = 1 - (1-.1080)^4 = .3667 or 36.70%

lessthanpi · December 2013

What final outcome are you looking for? Or just all the possibilities? This is a huge problem, albeit not impossible to solve. I need a bit more info to solve it for you.

Zombie Hero · December 2013

And thanks to our friend the binomial distribution. (heavy rounding)

P(0 success) = .6331
P(1 success) = .3066
P(2 success) = .0557
P(3 success) = .0045
P(4 success) = .0001

Prime · December 2013

lessthanpi: I think Zombie Hero nailed it unless his statistic knowledge is wrong (which im in NO position to question), but im after the chance of AT least 1 success (a success being a roll of a 2+, 6+, 4 or less sequentially).

Zombie Hero, thanks for the working more then anything. Means I can reproduce it in similar but different situations!

tinwhiskers · December 2013

In the second step with the 6's what do you mean by keep?
Like:
Step 1, remove any dice that are 1s from the pool
Step 2a, roll remaining dice, Put 6s aside
Step 2b, reroll any dice from step 2a that were 1s, if 6s Put aside?
Step 3, reroll any remaining dice, do what with dice 1-4?

Like are we looking for a sum after step 3? Including the 6s set aside in step 2? Or just the number of dice that are between 1-4 in step 3? Or that number + the number of 6s from step 2?

Prime · December 2013

tinwhiskers wrote: »

In the second step with the 6's what do you mean by keep?
Like:
Step 1, remove any dice that are 1s from the pool
Step 2a, roll remaining dice, Put 6s aside
Step 2b, reroll any dice from step 2a that were 1s, if 6s Put aside?
Step 3, reroll any remaining dice, do what with dice 1-4?

Like are we looking for a sum after step 3? Including the 6s set aside in step 2? Or just the number of dice that are between 1-4 in step 3? Or that number + the number of 6s from step 2?

Step 1, remove any dice that are 1s from the pool
Step 2a, roll remaining dice, remove any dice that are not 6s from the pool
Step 2b, reroll any dice from step 2a that were 1s, remove any dice that are not 6s from the pool
Step 3, reroll any remaining dice, values of 1-4 are successes

Does that make it clearer? Im after the prob of at least 1 success.

Bliss 101 · December 2013

Let's see, I've got a bit of a splitting headache but I think this is right. First, calculate the probability for rolling just 1 die through the trials:

Step 1: 5/6 = 83,3%

Step 2: 1/6 + 1/36 = 19.4%

Step 3: 4/6 = 66.7%

Total: 0.833 * 0.194 * 0.667 = 10.7%

It follows that the probability of failure with just 1 die would be 89.3%. Now, let's look at the 4 dice. The probability of at least one of them passing the trials is (1 - the probability that none of them passes):

P(total) = 1 - 0.893^4 = 36.4%

So that's the probability that you get to keep at least one of the die rolls in the end.

edit: beat'd.

JohnnyCache · December 2013

What test in 40k is this

Prime · December 2013

Im afraid its the square base version

Its the chance of a Bret Lord HKB a chaos demon prince with a certain load out.

tarnok · December 2013

If Zombie's numbers are correct then all you need to do for "at least one" is subtract the probability for "none" from 1. Which is about 36% from his numbers, agreeing with an earlier poster. (And a later poster now that I scroll back up.)

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