To the Crunchy Boys out there: Some probability help

[Stalker0]

I'm working on a game mechanic, and I could use some help with the probability if anyone wouldn't mind.

A player rolls two dice. The second dice is open-ended (this means that if he rolls the maximum value on the dice, he rolls it again and adds the numbers, if he rolls maximum again he can continue to roll). Assume for the sake of simplicity that on a d6 the best he could do would be to roll a 6, roll another 6, and finally roll one more 6.

For each scenario I need the following
A) The average result
B) I need to know the integer number that the dice will roll above:
a)50% of the time
b) 75% of the time
c) 25%
d) 33%
e) 10%


The scenarios:
1) A d6, and a d4
2) A d6, and a d6
3) A d6, and a d10
4) Scenarios 1-3, but this time the open-ended value is increased by 1 (meaning that on a d6, a 5 or a 6 result will result in another dice roll, on a d4 a 3 or 4, etc).

Thank you for any assistance.

 

[orsal]

I'm working on a game mechanic, and I could use some help with the probability if anyone wouldn't mind.

A player rolls two dice. The second dice is open-ended (this means that if he rolls the maximum value on the dice, he rolls it again and adds the numbers, if he rolls maximum again he can continue to roll). Assume for the sake of simplicity that on a d6 the best he could do would be to roll a 6, roll another 6, and finally roll one more 6.

For each scenario I need the following
A) The average result
B) I need to know the integer number that the dice will roll above:
a)50% of the time
b) 75% of the time
c) 25%
d) 33%
e) 10%

I'll give you (a) right now, because it's elegant. When I've got a little more time (or more likely when someone else has a little more time, because chances are someone will beat me to it) I'll do the others. None of them are hard, but apart from (a) I'd have to construct the whole probability distribution.

Here's how you can find the average: Just average the results for each die, and add them up. For the die that you may or may not roll, you multiply its average by the probability that it will be used. To answer for your specific scenarios (and to illustrate, so that you can do this yourself in the future):

The scenarios:
1) A d6, and a d4
2) A d6, and a d6
3) A d6, and a d10
4) Scenarios 1-3, but this time the open-ended value is increased by 1 (meaning that on a d6, a 5 or a 6 result will result in another dice roll, on a d4 a 3 or 4, etc).

(1) 3.5 + 2.5 + (1/4)(2.5) = 6.625
(2) 3.5 + 3.5 + (1/6)(3.5) = 7.583
(3) 3.5 + 5.5 + (1/10)(5.5) = 9.55
(4.1) 3.5 + 2.5 + (2/4)(2.5) = 7.25
(4.2) 3.5 + 3.5 + (2/6)(3.5) = 8.167
(4.3) 3.5 + 5.5 + (2/10)(5.5) = 10.1

If you want to add a possible fourth die, to be rolled only if (let's take scenario #2 here) the second and third dice both land on 6, then the multiplier for it is (1/6)^2=(1/36). For a fifth die, (1/6)^3=(1/216), and so forth.

Thank you for any assistance.

You're welcome.

 

[orsal]

B) I need to know the integer number that the dice will roll above:
a)50% of the time
b) 75% of the time
c) 25%
d) 33%
e) 10%

The scenarios:
1) A d6, and a d4
2) A d6, and a d6
3) A d6, and a d10

Here are the probability distributions for scenarios 1-3 (haven't done #4 yet)
The first column is the total; then each scenario gets a column. The top row is what each column is out of. So for example in scenario #2, the probability of getting 8 is 25/216 (25 from the row that begins with 8, 216 from the top of that column)

- 96 216 600
2 4 6 10
3 8 12 20
4 12 18 30
5 12 24 40
6 13 30 50
7 14 30 60
8 11 25 60
9 8 20 60
10 4 15 60
11 4 10 50
12 3 5 41
13 2 6 32
14 1 5 23
15 - 4 14
16 - 3 5
17 - 2 6
18 - 1 6
19 - - 6
20 - - 6
21 - - 6
22 - - 5
23 - - 4
24 - - 3
25 - - 2
26 - - 1

Thus:

Scenario #1
(a) median = 6
(b) 1st quartile = 4.5
(c) 3rd quartile = 8
(d) 67th percentile = 8
(e) 90th percentile = 11

Scenario #2
(a) median = 7
(b) 1st quartile = 5
(c) 3rd quartile = 9
(d) 67th percentile = 8
(e) 90th percentile = 12

Scenario #3
(a) median = 9
(b) 1st quartile = 6.5
(c) 3rd quartile = 12
(d) 67th percentile = 11
(e) 90th percentile = 15

Hope this helps!

 

[tarchon]

If you want to add a possible fourth die, to be rolled only if (let's take scenario #2 here) the second and third dice both land on 6, then the multiplier for it is (1/6)^2=(1/36). For a fifth die, (1/6)^3=(1/216), and so forth.

Several game systems use that bonus roll mechanic, and the geometric series is often useful for computing its affect on a dice sum mean.
IIRC, it turns out that 1dS with added rolls on Ss (or actually rerolling and adding on any other result) has a mean

m=S(S+1)/(2(S-1))

or you can also say it multiplies the old mean by S/(S-1).
It's easy to modify it for rerolling d6 on 5 or 6 too, like

m=3.5*(1/(1-1/3))=3.5*1.5=5.25

It's fairly easy to compute the probabilities as well (iterated convolution works numerically, or there are numerous asymptotic approaches if you prefer symbols), but I'm sure someone else would be far more interested to do so than I am before dinner.

 

[Stalker0]

Could you explain what quartile and percentile mean as far as those numbers go please?

 

[orsal]

Could you explain what quartile and percentile mean as far as those numbers go please?

Well, they're the numbers you asked for (I kept your lower-case Roman numerals). The 90th percentile is the cutoff point with 90% of the distribution below it (and thus 10% above it). The first quartile is the 25th percentile; the third quartile is the 75th percentile.

-- crunchy orsal

 

[Stalker0]

Well, they're the numbers you asked for (I kept your lower-case Roman numerals). The 90th percentile is the cutoff point with 90% of the distribution below it (and thus 10% above it). The first quartile is the 25th percentile; the third quartile is the 75th percentile.

-- crunchy orsal

*puts on the dunce cap*

hehe, damn I feel stupid now, but thank you.