"Help with chances on a die roll score." Topic

A plea to those who are good at maths.

If I roll 2 x D6 and I need at least either one 1 or 2 to show up,

What are my chances (in percentages) of that happening and how do you work it out?

I can see that on 1D6 I would have a 2 in 6 chance (33%) but what does the second dice do to the odds? Double them or stay the same?

Thanks in anticipation.

The best way is to figure out the chances of NOT getting it.

So, the dice each have a 2/3rd chance of not getting a 1 or 2.

Two dice = 2/3 x 2/3 = 4/9 chance of not getting a 1 or 2.

So, 1 – 4/9 =5/9 chance of getting a 1 or 2.

anydice.com

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Follow Terrement's advice and do the table manually. It will illustrate how it works.

  1 2 3 4 5 6
1 X X X X X X
2 X X X X X X
3 X X
4 X X
5 X X
6 X X

You get 2/6 odds for one die. Each die contributes the same 2/6 (well, 12/36) to the two die roll, but there is an "overlap". Four of the rolls (specifically 1,1 1,2 2,1 and 2,2) have a 1 or 2 for both dice.

What this leads to is a progression where you start with the odds for one die and add those same odds, minus the overlap, for the next die. This works up to however many dice you want.

Or just search Google for dice probability calculator…

Or just search Google for dice probability calculator…

Personally I prefer to understand how the mathematics works – I find it more satisfying.

smallroller.exe is your friend,at least on windows:

link

I have used this at least weekly for the last 10 years for probability calculations just as you describe.

I definitely agree with the posts above about creating a 6x6 matrix and looking at all of the possible cases. With only 36 to worry about it's easily feasible, and it makes the results you would get from doing it mathematically explicit.

Doug

Thanks everyone, TMP comes up trumps – as always, Norm

Whilst either method described above will give the correct answer, the advantages of the method HLT describes are
1 – it is simple calculation you can do in your head or on a calculator
2- it works for any number of dice rather than just two like the other method

It's all a bit academic anyway. Odds only work on " infinite rolls". In real life, if I need 1,2,3,4 to save (as in last night''s Chain of Command game) I have 90% chance of getting a 5 or 6!
Result-squad picked up 2 kills and 4 shocks virtually every phase and then broke when I threw three 6s out of 5 command dice, so turn ended.
Bitter, me? Never!!!!

Could be worse – I was rolling away in one game a while back, getting low low rolls (which was bad).

Then realised that in the small handful of D6s I was using a Salute "special D6" – it was a D3 – only marked with 1, 2 and 3 – and a Salute "extreme D6" – marked with 1,1, 2, 5,6, 6 (IIRC).

Doh !

"the advantages of the method HLT describes"

is also the method described on introductory statistic courses.

It is very effective.

Dear Normsmith

As you want the possibility of a 1 or 2 on either die, then it's by simple inspection. There are 36 possible combinations of the die roll- (6 each).

Since you are not summing them, you don't have to figure out the distribution of the numbers. There are only two chances on each die that a 1 or 2 comes up. So that's four results out of 36 or 1/9 or .1111111111

I am assume you don't care if a 1 or 2 shows up on both dice, but if you are then the chance will be .55555555555

Personally I prefer to understand how the mathematics works – I find it more satisfying.

It's also better to get an understanding.

Having a formula and a calculator is nice and gives you the answer. But understanding how those tools are derived gives you the answer and the basis from which to derive tools and formulas for other situations.

Working it out manually also gives you a meta-cognitive understanding of the interactions involved in the process (rolling dice), which helps guide you to developing and understanding variants.