View Full Version : math hammers - looking for dice probabilities

21-11-2009, 05:08
As the title states, just looking for the probablities of all the dice rolls.

I guess my only confusion is for example a 5/6 is 83/100 meaning it has a 17% probability, but is a 5+ dice roll 17% probability or do you include the number as well making it more like 33% probability to roll a 5+.

21-11-2009, 05:27
Dice have a 16.66% (100/6) chance to land on any one side. So for a 5+ roll, the dice would have to land on 5 or 6. And since every side has a 16.66% chance to land, we add up the 2, meaning a 33% chance to roll 5+. An easier way is to look at a die as a fraction: in this example, you want a 5 or 6, which are 2 possible outcomes on a 6 sided die, so a 2/6 chance = 1/3 = 33%!

Hope this helps!

21-11-2009, 05:48
okay that cleared it up for me, i knew it was 17% per side, I was just doing 5/6 which is 83-100 which is 17% which didnt seem right. It is easier to do the 17% per side and just add it up that way. thanks

edit: on a side not, how does re-rolls work into the equation? for either to hit or to wound. How does that adjust the probabilities

21-11-2009, 06:09
Depends what you mean, chance to roll 5+ is always 33%, it's not because you just rolled a 1 that your chances of rolling a 5 or 6 have altered!

However, if you want to know the chance of failing a roll, followed by succeeding the roll (in this example rolling 1-4, followed by 5-6) things become a little more complicated. First we determine the chance of failing, then multiply this by the chance of succeeding. In this example: 4/6 * 2/6 = 8/36 (=22% chance). A better question though, would be the chance of failing the 5+ roll twice in a row, which is 4/6 * 4/6 = 16/36 (=44%).

End result! 33% chance to make it from the first roll, 22% chance to fail, followed by success, 44% chance to fail the roll completely. This all adds up to 22%+33%+44%=99%! This mysterious last 1% was eaten by a hungry Tyranid! (In reality, it's just rounding errors; it's actually 22.22[..]2%, 33.33[..]3% and 44.44[..]4%)

Freman Bloodglaive
21-11-2009, 07:04
If you roll multiple dice it becomes even more entertaining.

For example. If you are rolling two dice, hitting on 3+, wounding on 5+ with re-rolls and ignoring saves.

Probability with one dice.

2/3 (to hit) x (1-(2/3 x 2/3)) = 10/27

With two dice you have.

(10/27)^2 + 2 x ((10/27) x (17/27)) + (17/27)^2 where the first set of numbers is the probability of inflicting two wounds, the second set the probability of inflicting just one wound, and the last the probability of doing no wounds at all.

100/729 + 340/729 + 289/729

100/729 or 0.137 of doing two wounds.
340/729 or 0.466 of doing one wound.
289/729 or 0.396 of doing no wounds.

It gets more intricate as you add more dice.

21-11-2009, 07:05
my head hurts:mad: :mad: :mad:

which probably explains why I try to avoid mathhammer.
especially after last nights Friday night magic, when I had a mana echoes (http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=39571) out and then played 2 goblin offensives (http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=205388) and then a stream of life (http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=83284). when we were on izzet steam maze (http://gatherer.wizards.com/Pages/Card/Details.aspx?multiverseid=198065).

21-11-2009, 09:54
Obviously your problem is that you keep rounding things up to do it out of 100 and are thinking about this wrong

5/6 does not equal 83/100

5/6 does not represent a 5+, it represents a 2+

21-11-2009, 15:00
alright so let me see if i get this, if i have 24 shots hitting on 3+ and wounding on 4+ would the probability to wound be this:

This is assuming no re-rolls: 24x0.66=15.84x0.50=7.92 So would I have a probability to inflict 7.92 wounds from 24 dice on 3+ then 4+?

This is assuming re-rolls on FAILED dice to wound: 24x0.66=15.84x0.5=7.92x1.50=11.8 So I would have 11.8 wounds from 24 dice on 3+ then 4+ and re-rolling failed 4+?

21-11-2009, 16:28
Firstly, USE SIMPLE FRACTIONS!, ie 1/6, 2/3
Do not use percentages, or big fractions.

24 shots
hits on 3+, so 4/6, which is equal to 2/3
wounds on a 4+, so 3/6 , which is equal to 1/2

Therefore 24 x 2/3 x1/2 = 8 kills expected.

By keeping it as simple, small as possible fractions, you make it both more accurate and easier.

with a re roll to wound, it looks like.
24 shots, 2/3rds hit, therefore 16 hits.
1/2 of those 16 wound, so 8 wounds straight off
Re-roll the failed 8 wounds, so another 4 wounds
=12 in total.

This is an what is known as the expectation value, it tells you the most likely number of wounds. It does not tell you how likely you are to actually score this number of wounds, rather than significantly more or less than it, but which average out to the expectation value over time...

It is possible to calculate more exact figures, if you're not rolling too many dice*, but the maths rapidly becomes massive, so isn't worth doing unless you have the programming ability to get a computer to do it for you...

* the only normal circumstances where it is worth it is when trying to kill vehicles or MCs, as you are often just using a few shots.

21-11-2009, 18:36
If you're only worried about average results and not the probability of a specific outcome, the numbers are pretty straightforward:

Roll Success Fraction Success Percent
6+ 6/36 = 1/6 16.67%
6+ w/ reroll 11/36 30.55%
5+ 12/36 = 1/3 33.33%
5+ w/reroll 20/36 = 5/9 55.56%
4+ 18/36 = 1/2 50%
4+ w/reroll 27/36 = 3/4 75%
3+ 24/36 = 2/3 66.67%
3+ w/reroll 32/36 = 8/9 88.89%
2+ 30/36 = 5/6 83.33%
2+ w/reroll 35/36 97.22%