Dice roll probability

[sasguru]

Problem is you can't simply work out one die and multiply by the number of dice.
Let's ignore getting a two for now.
If you have 1 dice the probability of success is 1/6.
If you have 2 dice the probability of success is NOT 2x1/6=2/6, but 3/36.

Think of the sample space with two dice. There are 36 possibilities.
For success, you could have

1, Number other than 1
Number other than 1, 1
1, 1

So that's 3/36. An additional complication is that for the first 2 possibilities on that list (a single 1) you would throw a further die.
But for the last possibility (2 ones) you would throw 2 further dice, if I understand correctly.

I think simulation with a program is the way to go - and I don't think that's a trivial task either. If only I didn't have all this work to do before Christmas, sigh

[sasguru]

This is all taken into account. Only rolling a 1 results in a further roll. That is where the series 1 + 1/6 + 1/36 + ... comes from.



So why keep telling me my formula for the average number of successes per die is wrong then? Especially when you keep saying you can't be arsed to work out the answer properly.

I'll try and sort the answer to the odds problem out in the morning, I'm a bit tired now, been a long day.

Glad to see you're well enough to give the wrong answer Doodab
How are things?

[d000hg]

I couldn't be arsed reading all the posts but when rolling D dice, you expect to get D/6 ones and D/6 twos. Then you roll a further D/6 dice and expect to get an additional D/36 ones and D/36 twos. Roll another D/36 dice and expect to get D/216 ones and twos...

It's gives you nice infinite series, number of one or two dice

n12 = D/3 + (D/6)/3 + (D/6^2)/3 + (D/6^3)/3 + ... + (D/6^n)/3

Which is the infinite sum of terms D/(3*6^n) from n=0...∞

I don't know if that summation has a solution BUT even if it does, is it applicable to a discrete system? It seems anyway that unless D is very large only the first couple of elements in the series are useful anyway so you could say for small D, n12 = D*7/18... But this makes me think it is too early for maths and I screwed up.

[sasguru]

I couldn't be arsed reading all the posts but when rolling D dice, you expect to get D/6 ones and D/6 twos. .

I stopped reading at that point.

[d000hg]

I stopped reading at that point.

got a bit over your head?

[MyUserName]

I couldn't be arsed reading all the posts but when rolling D dice, you expect to get D/6 ones and D/6 twos. Then you roll a further D/6 dice and expect to get an additional D/36 ones and D/36 twos. Roll another D/36 dice and expect to get D/216 ones and twos...

But you only roll a further D6 for each dice where you have rolled a 1.

[sasguru]

got a bit over your head?

That or you're talking bollux. Which do you think it is and why?

Proverbs 17:28

[NotAllThere]

The approach of writing down the probabilities for each n is not the way to go. This is because there are infinitely many possible values for n, as you could keep rolling a 1 for ever.

You need to write it as a geometric series like what doodab did.

Having more than one die doesn't complicate things. If you can work out the expected number of successes if you have one die (which is 0.4), the if you have two die you simply double it.

For any given "n" you can calculate the probability of scoring n. The fact that you can get infinitely many values for n is irrelevant. As n gets large, the probability of scoring n dwindles towards zero - quite rapidly.

[doodab]

Glad to see you're well enough to give the wrong answer Doodab
How are things?

I'm pretty sure that's the right answer, each initial die's contribution to the number of extra dice rolled is independent, it doesn't change depending on how many other dice there are. And obviously the expected successes is 2/6 the expected number of dice rolled.

It's more complex when working out the odds of N successes with D dice cos of the permutations and having to weight them accordingly.

I'm off the codeine for a bit so will either crack it shortly or become delirious like the old blokes in the nappies I'm surrounded by.

[No2politics]

Problem is you can't simply work out one die and multiply by the number of dice.
Let's ignore getting a two for now.
If you have 1 dice the probability of success is 1/6.
If you have 2 dice the probability of success is NOT 2x1/6=2/6, but 3/36.

Think of the sample space with two dice. There are 36 possibilities.
For success, you could have

1, Number other than 1
Number other than 1, 1
1, 1

So that's 3/36. An additional complication is that for the first 2 possibilities on that list (a single 1) you would throw a further die.
But for the last possibility (2 ones) you would throw 2 further dice, if I understand correctly.

I think simulation with a program is the way to go - and I don't think that's a trivial task either. If only I didn't have all this work to do before Christmas, sigh

When I say work out the number of successes for one die and multiply by two, I'm referring to the number of STARTING Die (excuse capitals).

We have to start thinking in terms of expected number of successes (ie how many 1,s and 2,s are you going to roll in total assuming you have one starting die), rather than the probability of success. I agree that you can't just double probabilities. However you CAN double expectations.

So, assuming you have one starting die and also lets assume the expected number of successes is 0.4. Then to get the expected number of successes for two starting die that is simply double ie 0.8.

To calculate the expected number of successes for one starting die, you need to calculate as doodab did.