# Odds calculations

17 replies [Last post]
questccg Offline
Joined: 04/16/2011

Hi all,

I was wondering if someone could calculate the odds for my dice mechanic:

-1d12 (Dice #1) vs. 1d8 (Dice #2)

Dice #1 must be HIGHER than Dice #2.

I wanted to know what the odds would be that this occurs...

questccg Offline
Joined: 04/16/2011
Hmm...

questccg wrote:
I wanted to know what the odds would be that this occurs...

So I have calculated there are 60 favorable rolls, 36 unfavorable rolls for a total rolls of 96.

60 out of 96 = 62.5%

36 out of 96 = 37.5%

Is this correct? I'm not certain I did it right...

questccg Offline
Joined: 04/16/2011
And the odds would be:

5:8 (Favorable) and 3:8 (Unfavorable).

MarkKreitler Offline
Joined: 11/12/2008
Yup

That's what I get.

M.

questccg Offline
Joined: 04/16/2011
I was not sure...

MarkKreitler wrote:
That's what I get.

I was not sure because when I use the same calculations with 2d6 (1d6 vs. another 1d6), I expected to get the odds of 50:50... But it turns out 5:12 favorable and 7:12 unfavorable...

Am I correct with these odds???

The Chaz Offline
Joined: 11/20/2012
??

Why would you expect 50-50??

The probability that die A is higher than die B is the same as
the probability that die B is higher than die A.

Hopefully that symmetry is obvious. Then there's the probability that they're equal. 11,22,33,44,55,66 --- six ways
6/36 = 1/6.
A > B half of the remaining 30/36 times.

questccg Offline
Joined: 04/16/2011
...?

The Chaz wrote:
Why would you expect 50-50??

Just because the dice were the same, I thought the odds would be 50:50... But I understand that when the dice rolls are EQUAL, they are not favorable rolls. And therefore they need to be subtracted (6 out of 36). That means there are 30 out of 36, halved 15 out of 36 (5:12).

MarkKreitler Offline
Joined: 11/12/2008
Yup. Again.

Yeah, the 5:12 vs 7:12 result is correct.

Have to admit, though, I'm doing the calculations differently than you guys.

Assuming 'A' is the die that has to beat die 'B'...

When A is 1, B wins 6 times
When A is 2, B wins 5 times
...
When A is 6, B wins 1 time

So B's total wins are 6 + 5 + 4 + 3 + 2 + 1 = 6 * 7 / 2 = 21 out of 36.

That leaves 15 / 36 for A.

As for the asymmetry, yeah, it's because A must *beat* B, so the dice aren't really equal.

Yamahako
Offline
Joined: 12/01/2010
Here's something I do when

Here's something I do when I'm trying to figure out the probability of die rolls. It takes a bit of work, but once its set-up its easy to tweak to figure things out:

The first tab shows how competing 2d6 rolls happen and what their results are.

I added a tab for these results:

the D12 will win 62.5% of the time
the D8 will win 29.17% of the time
and they will tie 8.33% of the time

The Chaz Offline
Joined: 11/20/2012
stats

First, I will confirm the results that

Quote:
the D12 will win 62.5% of the time
the D8 will win 29.17% of the time
and they will tie 8.33% of the time

Second, a brief statistics primer.

a) List, graph, etc the possible outcomes. This is called the SAMPLE SPACE. The number of possible outcomes is used as the DENOMINATOR when calculating possibilities.

(In this case, we need an 8x12 grid. Since 8*12 = 96, the sample space has size 96)

b) Determine what event you'd like to measure, and calculate how many ways there are for this to happen. This can get complicated and involve combinatorics, but it's simple in our case.
e.g. We want to know in how many ways the 1d8 can equal the 1d12.
11
22
33
44
55
66
77
88
So there are eight ways. The size of the event (X = Y) is 8.

c) Divide. The probability of an event is (size of event) / (size of sample space)
In our example, 8/96

-----------------------
But say you've got an 8x12 grid with all the "equals" spaces marked off. You want to know the remaining probabilities (like when X > Y or Y > X).

Well first you can trim off a 4x8 area to leave an 8x8 grid with a diagonal removed. You have something like a 7x7 triangle below and one above.
The number of outcomes in this triangle is actually the seventh TRIANGULAR NUMBER = 28.
Let's not forget the 4x8 = 32 outcomes that we trimmed off.

So
(# of ways 1d8 > 1d12) = 28, and
(# of ways 1d8 = 1d12) = 8, and
(# of ways ad8 < 1d12) = 28 + 32 (from the 4x8) = 60
----------------------
28 + 8 + 60 = 96, the size of our sample space.

Izraphael Offline
Joined: 01/29/2010
online

Anyone knows if there are some online dice odds calculators?

pelle Offline
Joined: 08/11/2008
anydice.com. I prefer to

anydice.com. I prefer to solve this kind of problem in R or Python, but that online solution seems popular.

Izraphael Offline
Joined: 01/29/2010
anydice.com

pelle wrote:
anydice.com. I prefer to solve this kind of problem in R or Python, but that online solution seems popular.

Expecially for who knows nothing about programming :)

pelle Offline
Joined: 08/11/2008
Well, R is for statistics

Well, R is for statistics people, and was even created by statistics people. It is a weird language if you are used to programming normal programming languages. This is how the problem in this thread can be solved using R plus the optional prob library:
> prob(probspace(expand.grid(1:8, 1:12)), Var2>Var1)
 0.625

There is probably (eh...) a better way to express that, but I am no R expert. For many other problems there is a very easy to use R dice library, but I don't think it can handle comparisons between different dice.

For an intellectual challenge I do enjoy solving probability problems manually, or (almost as manually) in a spreadsheet, but to get something done I can really recommend to try some scripting. It is useful for LOTS of other things when making boardgames anyway, even without doing much actual programming.

Here is a solution using python and a library for solving dice problems I wrote a few years ago:
>>> dice(1, 12).p_gt_other(dice(1, 8))
0.625

pelle Offline
Joined: 08/11/2008
clojure

Could not resist trying to solve it in Clojure as well (probably far from the best solution... currently learning):

user=> (let [c (cartesian-product (range 12) (range 8))] (/ (count (filter (partial apply >) c)) (count c)))
5/8

Somewhat cryptic, yes. It is nice that it gives the exact answer. :)

svenne Offline
Joined: 01/06/2013
Color dice odds

I am cutting in with a new question here because it seems to be nice odds thread.

I have 1d6 with two sides colored with red, the others are gray. Red is for hit and gray for misses.

If I just roll one of those dices the chance of a hit would be 33,33%, but if I am rolling two of them? Three of them and so on?

The thing is that I want a really easy fight system that smaller kids can handle.

gabrielcohn
Offline
Joined: 11/25/2010
thirds

on one die you would get a hit 1/3 of the time.
on 2 dice you would get ONE OR MORE hits 1/3 + 2/3*1/3 of the time (that is 5/9 total).
on 3 dice you would get ONE OR MORE hits 1/3 + 2/3*1/3 + 4/9*1/3 of the time (19/27 total)
on 4 dice you would get ONE OR MORE hits 1/3 + 2/3*1/3 + 4/9*1/3 + 8/27*1/3 of the time (65/81 total)
...and so forth

An easier way to put this, is your CHANCE OF FAILING TO GET ONE OR MORE HITS = (2/3)^x (where x is the number of dice rolled)
1 die = 2/3
2 dice = 4/9
3 dice = 8/27
4 dice = 16/81
...and so forth.

svenne Offline
Joined: 01/06/2013
Thanks gabrielcohn. You

Thanks gabrielcohn. You helped me a lot.
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