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"Math Problem" Topic


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JasonAfrika Inactive Member31 Mar 2016 8:21 a.m. PST

If you have roughly a 17% chance of rolling a 1 on a six-sided die. What are the chances of rolling a single 1 when rolling 2 six-sided dice, or 3. I forget the formula.

Personal logo Doms Decals Sponsoring Member of TMP31 Mar 2016 8:28 a.m. PST

On 2 dice, 1/36 chance of 2 1s, 10/36 chance of a single 1.

On 3 dice, 1/216 of 3, 15/216 chance of 2, 75/216 chance of 1.

photocrinch31 Mar 2016 8:31 a.m. PST

Each 6 die increases the number of possibilities exponentially, so for one 6-sided die – 6 to the 1st power, for 2 die – 6 to the 2nd power, for 3 die, 6 to the third. Since there is only one combination that will end in all die being a 1, the chances are 1/6 (17%), 1/36 (3%), 1/216 (roughly .5%)

Hope that helps.

JasonAfrika Inactive Member31 Mar 2016 8:47 a.m. PST

Dom, could you please translate that into percentages. If you have a 17% chance of rolling a 6 on a six-sided die and a 28% chance of rolling a single 6 using two six-sided dice…what would the percentages be for 3 dice? 31%? How about 4 dice? etc. How much do your chances go up when you keep adding another die? Sorry I'm bad at this sort of thing.
Love your decals btw, use them all the time on my TD and Pico planes!

bbriarcliffe Inactive Member31 Mar 2016 9:28 a.m. PST

DELETED

GildasFacit Sponsoring Member of TMP31 Mar 2016 10:28 a.m. PST

Jason

There isn't a simple rule that what you want can be calculated from. The 'progression' of probabilities by successively adding an extra die comes from calculating the individual probabilities of the events you want and combining them.

for 3 dice :

P(2D6 has single 1) * P(1D6 isn't a 1) = 10/36*5/6 = 50/216
P(2D6 has no 1) * P(1D6 is a 1) = 25/36*1/6 = 25/216

so 75/216 = 35%

… and so on.

Dynaman878931 Mar 2016 12:06 p.m. PST

Are you asking for the chance of rolling a single ones on X dice (never rolling 2 ones for example?) or rolling all ones on all the dice rolled?

Personal logo Doms Decals Sponsoring Member of TMP31 Mar 2016 12:16 p.m. PST

As Tony says, there's no simple shorthand, you have to crunch the numbers every time. If you're specifically looking at exactly one 1 it's simple enough, but also doing 2, 3, 4 etc. makes it a fairly tedious exercise.

Mako11 Inactive Member31 Mar 2016 1:17 p.m. PST

At the last club game, without too many dice rolls, a guy managed to meet the 1/216th odds and roll triples on three dice. Can't recall if it was triple sixes or ones, but it was a bad roll.

We had a good laugh over that, since his luck had abandoned him on a number of die rolls.

CeruLucifus31 Mar 2016 7:43 p.m. PST

You have to know the chance of a given occurrence and also the chance of it *not* happening.

For additive probabilities, the additional factor only matters if it *doesn't* happen in the first instance, so you deflate it by multiplying times the chance of the first not happening, then add them up.

Carry the process forward as you add more factors.

Use fractions, it is most accurate, and only at the end convert to decimal, and multiply that times 100 if you must have percentage. Both of these steps round off.

1 on 1D6: 1/6.
~= .1667 ~= 17%
not 1 on 1D6: 5/6

1 on 2D6: 1/6 + (not 1 on 1D6)x(1 on 1D6)
= 1/6 + (5/6)x(1/6)
= (6 + 5) / 36
= 11/36
~= .3056 ~= 31%
not 1 on 2D6: 25/36

1 on 3D6: (1 on 2D6) + (not 1 on 2D6)x(1 on 1D6)
= 11/36 + (25/36)x(1/6)
= (66 + 25)/216
= 91/216
~= .4213 ~= 42%

But … this way is easier:

What's the chance of getting a result? One minus the chance of *not* getting the result. So take the *not* chance for each factor, multiply out, then subtract from one.

1 in 1D6 = 1 – (not 1 in 1D6)
= 1 – 5/6
= 1/6.

1 in 2D6 = 1 – (not 1 in 2D6)
= 1 – (5/6)x(5/6) = 1 – 25/36
= 11/36.

1 in 3D6 = 1 – (not 1 in 3D6)
= 1 – (5/6)x(5/6)x(5/6) = 1 – 125/216
= 91/216.

etc.

(Phil Dutre) Inactive Member31 Mar 2016 10:22 p.m. PST

If you want to know the odss of exactly a single 1 occuring on multiple dice, you also need to know the combinatorial factors.

E.g. exactly scoring one "1" on 2D6:

(1/6) (scoring a 1 on one die)
* (5/6) (not scoring a 1 on the other die)
* COmbination (1,2) (all possible combinations in which 1 die is selected from 2) = 2
==> 10/36

This is different from scoring at least a 1 using 2 dice, which you can indeed do using the complementary probality = 1-chance of scoring no 1's at all = 1-(5/6)*(5/6) = 11/36

The additional 1/36 chance stems from rolling a double 1.

These are the sort of exercises typically incluided in an intro course on probability and combinatorics.

===

Example for rolling a 1 on 4D6:

Scoring at least a 1:

1-(5/6)*(5/6)*(5/6)*(5/6) = 1-625/1296 = 671/1296

Scoring exactly a single 1:

(1/6)*(5/6)*(5/6)*(5/6)*C(1,4) = 125/1296 * 4 = 500/1296

JasonAfrika Inactive Member01 Apr 2016 5:20 a.m. PST

Wow, sorry I asked, lol. CeruLucifus- can you PLEASE just give me the rest of the results in percentage form. I have NO IDEA what you guys are talking about! You started with 1d6=17%, 2d6=31%, 3d6=42%. Can you keep going up to 8 or 10 d6? just give me the percentage? Thanks!

Puster Supporting Member of TMP01 Apr 2016 6:25 a.m. PST

Its actually pretty easy, and Phil Dute is correct.

General form: n is the number of dices

(1/6) to hit one 1
(5/6)^(n-1) to hit NO 1 with all other dices

Now you just have to consider that ANY of the n dices can be the one that shows the 1, so there are n different configurations.

(1/6) + (5/6)^(n-1) * n

The interesting thing is that once the numbers go up, the actual probability drops again. At the start you need to hit one 1, later on you need to make sure you just hit one 1.
--
for i in range(1,21):
-- print (i, (1/6)*(5/6)**(i-1)*i)

# actually changed for more readable output:
--
1 : 16.67 %
2 : 27.78 %
3 : 34.72 %
4 : 38.58 %
5 : 40.19 %
6 : 40.19 %
7 : 39.07 %
8 : 37.21 %
9 : 34.89 %
10 : 32.3 %
11 : 29.61 %
12 : 26.92 %
13 : 24.3 %
14 : 21.81 %
15 : 19.47 %
16 : 17.31 %
17 : 15.32 %
18 : 13.52 %
19 : 11.89 %
20 : 10.43 %

Puster Supporting Member of TMP01 Apr 2016 10:53 p.m. PST

(1/6) * (5/6)^(n-1) * n
not + …

DesertScrb02 Apr 2016 6:47 a.m. PST

Go here and play around: anydice.com

Puster Supporting Member of TMP02 Apr 2016 10:40 a.m. PST

Anydice – nice tool, but it does not answer this particular problem.

CeruLucifus03 Apr 2016 1:40 p.m. PST

My apologies, as Phil Dutre points out, my numbers work for getting at least one 1 in your roll. His are correct for getting only one 1.

JasonAtAfrika, surely you have a calculator that can divide 2 numbers then shift the decimal two places to get a percentage? Most computer operating systems have one or perhaps you have a physical calculator? So you can derive percentage from a fraction: 11/36 ~= 0.30556 ~= 30.6%

steve1865 Supporting Member of TMP08 Apr 2016 4:17 p.m. PST

A pal of mine says dice over 10 sided are deformed. Use 2 10 sided dice.

(Phil Dutre) Inactive Member09 Apr 2016 2:06 a.m. PST

A pal of mine says dice over 10 sided are deformed.

What do you mean by "deformed"?

Sure, there are only 5 platonic solids (D4, D6, D8, D12, D20).
But that doesn't mean a fair dice has to be one of these. There are plenty of dice designs that roll fair without bias w.r.t. a specific number or set off numbers.

You can even make a fair Dn, with n being any integer number.

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