Dice probabilities

D

drach

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Feb 17, 2020
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Hello I wanted to confirm my thoughts on a probability and make sure I am not missing something.

Situation: you roll 3 even 6 sided dice and subtract 9 from the total rolled - Twice. What is the probability the resulting two totals when added together are >= 5.

I started with the Binary group where either roll is > 14. There are 35 out of 216 (~16.2%) possible combinations for this to occur. As it works if either set does this I took that by itself. (This is where I am trying to figure out if there is a hole in my logic)

Next I worked down the possibilities of adding something from the first set. I started with rolling exactly a 10 with the first then the second set would have to roll exactly a 13 (as all of the possibilities of rolling a 14 or higher are already accounted for. This is 27/216 * 21/216 or 576/46656.)
Summary:
10: 27/216 | 13: 21/216 = 576/46656
11: 27/216 | 12 or 13: 46/216 = 1242/46656 (This is another area where I am unsure as some of the chances of rolling a 13 may be covered in the previous case but I am unsure because I think this is the probability of rolling a 12 or 13 only after rolling an 11)
12: 25/216 | 11,12,or13: 73/216 = 1825/46656
13: 21/216 | 10,11,12,13: 100/216 = 2100/46656

After adding the probabilities from those I got 5743/46656 = 12.31%

My conclusion was that the total probability was 28.51%

Cheers
 
I'm not following your thinking, perhaps in part because of language issues.

I think where you say "even" you mean "fair".

I think "Binary group" may be intended to be "binomial distribution".

But can you explain why you picked >14, and how you got 35 out of 216? Once I understand that, I'll read further.

As for my own thinking, I'd start by restating the problem. If we roll the set of dice twice (getting totals of A and B, respectively), and each time subtract 9, and then add these results, the result is (A - 9) + (B - 9) = A + B - 18, so the event asked about is equivalent to A + B - 18 >= 5, so we just want the sum of 6 rolls to be at least 23. That simplifies the problem. Or am I misreading the problem?
 
Yes by even I mean fair

I was using the term binary group to relate that any grouping of dice (equal to or) above 14 automatically gives you the resulting >=5. There are no other factors to take into account. This is also why I used 14 (14-9 = 5). As for 35 out of 216 you have 15 possible combinations to add to 14 + 10 combinations to add to 15 + 6 for 16 + 3 for 17 + 1 for 18 = 35 possible combinations out of 6^3.

Your reworking of the problem wouldn't work but does clarify that I have to specifically state that negative numbers don't count. So a roll of 3 on the first set of dice just counts as a 0 not a -6. Thank you for that :)

I also forgot the equal to for the grouping above 14. Man I am bad at this whole asking questions thing.
 
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This is equivalent to rolling 6 dice and subtracting 18 from the total.

This sum being greater or equal to 5 is equivalent to rolling 6 dice and having the total greater than or equal to 23.

mmmm the simplest way to count these I think is to sum up all the coefficients of the powers of [MATH]x [/MATH] that are [MATH]23 [/MATH] and greater in the

polynomial (x+x2+x3+x4+x5+x6)6\displaystyle (x+x^2 + x^3+x^4+x^5+x^6)^6, and then normalize that by dividing by 66=46656\displaystyle 6^6= 46656

This turns out to be p=564715552≈36.31%\displaystyle p=\dfrac{5647}{15552} \approx 36.31\%
 
I think that my clarification from the previous post that the sum of the roll after the subtraction cannot be < 0 would also make this invalid. I am sorry, I didn't put enough of the constraints.

But your answer makes me question my original conclusion. I would think that the percentage could only be higher as yours takes the situation where the first 3 dice roll a 14 and the second 3 dice roll < 9 and doesn't count that. In my original thinking this situation would result in the 5 or greater. So I am thinking that my % is way to low.
 
Romsek said:
This is equivalent to rolling 6 dice and subtracting 18 from the total.
This sum being greater or equal to 5 is equivalent to rolling 6 dice and having the total greater than or equal to 23.
I think is to sum up all the coefficients of the powers of [MATH]x [/MATH] that are [MATH]23 [/MATH] and greater in the
polynomial (x+x2+x3+x4+x5+x6)6\displaystyle (x+x^2 + x^3+x^4+x^5+x^6)^6, and then normalize that by dividing by 66=46656\displaystyle 6^6= 46656
This turns out to be p=564715552≈36.31%\displaystyle p=\dfrac{5647}{15552} \approx 36.31\%
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drach said:
I think that my clarification from the previous post that the sum of the roll after the subtraction cannot be < 0 would also make this invalid. I am sorry, I didn't put enough of the constraints.
But your answer makes me question my original conclusion. I would think that the percentage could only be higher as yours takes the situation where the first 3 dice roll a 14 and the second 3 dice roll < 9 and doesn't count that. In my original thinking this situation would result in the 5 or greater. So I am thinking that my % is way to low.
Click to expand...
I think you are confusing everyone. Romsek is correct.
Look at this expansion In it the exponents are the sums rolled and the coefficients are the number of ways the sum can be gotten out of 66=466566^6=46656
Why is it that you must exclude negative results?
 
pka said:
I think you are confusing everyone. Romsek is correct.
Look at this expansion In it the exponents are the sums rolled and the coefficients are the number of ways the sum can be gotten out of 66=466566^6=46656
Why is it that you must exclude negative results?
Click to expand...
I would like to just edit my original post and put in more clarifications but I cannot figure out how (or just can't).

This is based off of rolling in a game system. If either roll is greater than 5 then it would count as a success as the roll would have succeeded.
 
Ok.. just to get things straight then.
You roll 3 dice and sum them and subtract 9.
The result is the minimum of 0 and the total minus 9. I.e. no negatives.

You do this twice and sum the result.

You want the probability this final sum is greater than or equal to 5.

Is that all correct?
 
Romsek said:
Ok.. just to get things straight then.
You roll 3 dice and sum them and subtract 9.
The result is the minimum of 0 and the total minus 9. I.e. no negatives.

You do this twice and sum the result.

You want the probability this final sum is greater than or equal to 5.

Is that all correct?
Click to expand...
Yes that is correct.

Sorry for the initial confusion.
 
drach said:
I would like to just edit my original post and put in more clarifications but I cannot figure out how (or just can't).
Click to expand...
There is a 30 min. time limit on being able to edit a posting.
 
Here is the distribution of the final result

\(\displaystyle
\left(
\begin{array}{cc}
0 & \frac{9}{64} \\
1 & \frac{3}{32} \\
2 & \frac{7}{64} \\
3 & \frac{17}{144} \\
4 & \frac{203}{1728} \\
5 & \frac{91}{864} \\
6 & \frac{4189}{46656} \\
7 & \frac{281}{3888} \\
8 & \frac{847}{15552} \\
9 & \frac{889}{23328} \\
10 & \frac{43}{1728} \\
11 & \frac{7}{432} \\
12 & \frac{19}{1944} \\
13 & \frac{7}{1296} \\
14 & \frac{7}{2592} \\
15 & \frac{7}{5832} \\
16 & \frac{7}{15552} \\
17 & \frac{1}{7776} \\
18 & \frac{1}{46656} \\
\end{array}
\right)\)

[MATH]P[\text{final result $\geq 5$}] = \dfrac{727}{1728}[/MATH]
Clipboard01.jpg
 
Romsek said:
Here is the distribution of the final result

\(\displaystyle
\left(
\begin{array}{cc}
0 & \frac{9}{64} \\
1 & \frac{3}{32} \\
2 & \frac{7}{64} \\
3 & \frac{17}{144} \\
4 & \frac{203}{1728} \\
5 & \frac{91}{864} \\
6 & \frac{4189}{46656} \\
7 & \frac{281}{3888} \\
8 & \frac{847}{15552} \\
9 & \frac{889}{23328} \\
10 & \frac{43}{1728} \\
11 & \frac{7}{432} \\
12 & \frac{19}{1944} \\
13 & \frac{7}{1296} \\
14 & \frac{7}{2592} \\
15 & \frac{7}{5832} \\
16 & \frac{7}{15552} \\
17 & \frac{1}{7776} \\
18 & \frac{1}{46656} \\
\end{array}
\right)\)

[MATH]P[\text{final result $\geq 5$}] = \dfrac{727}{1728}[/MATH]
View attachment 16704
Click to expand...
Can you please explain the ln 157 block
 
After you've rolled your 6 dice and subtracted 9 there is a distribution of the results. That's p.

Then what we do is combine the two iterations of this. What the Outer function and f[p1, p2] do is for each result/probability
in p, it adds the values and multiplies the probabilities.

So for example if I have (1,1/8) and (6,5/108) I end up with (5, 5/864).
The result of that is a big matrix of possible results and the probability of each. But there will be duplicates.
So we then make a separate list (psum, possible sums) by removing these duplicates, and then for each item on this list,
we count up how many in the big matrix have the same roll value and sum all their probabilities to get a final probability for that roll value.
 
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Comment & Rant: This is no more that a practical problem from some gaming group. I really doubt that it is an assigned probability question from any academic institution. If I am wrong please, please prove that I am wrong. I for one am fed up with people posting applied mathematics problems that are not assigned by an academic institution. Such a question has no place on a free help site. Answering it is to deprive working applied mathematics mathematicians of revenue.
 
pka said:
Comment & Rant: This is no more that a practical problem from some gaming group. I really doubt that it is an assigned probability question from any academic institution. If I am wrong please, please prove that I am wrong. I for one am fed up with people posting applied mathematics problems that are not assigned by an academic institution. Such a question has no place on a free help site. Answering it is to deprive working applied mathematics mathematicians of revenue.
Click to expand...

Really? Do you honestly expect ordinary people that happen upon a math problem they could use help with to go "hire an applied mathematician" ?
Where do you even look for such? Google Maps had none listed in my area.

I took a close look at the posting specs and while it's mentioned that posters are mostly students there's nothing in there about that being a requirement.

Further, when I detect that the poster is just an average person who'd like a bit of help figuring out a math problem that is affecting their life somehow I tend to subject them to less fuss than I would students. Students should be trying to learn. Ordinary people with math problems just want to get on with their lives and get this thing out of the way. So far no one has complained.

You're always free to ignore a question if you feel it's not appropriate.

If I'm wrong about this Mods please chime in so I can modify my behavior in the future.
 
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