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dice probability question (Can't Stop)

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Daggaz
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NewbieDesigner wrote:Thanks

NewbieDesigner wrote:
Thanks Daggaz. Another question regarding the percentages in case I am missing something obvious or asking a new question. Based on my movie question above about rolling three dice (locking two) and choosing two theatres based on the results. Are your most recent results saying that I will be able to choose theatre six 14.51% of the time I roll the 3 dice (for comparison sake, I would be able to choose theatre six 14% of the time with just rolling two dice)?

That strikes me as low intuitively based on the 3 dice selection so I'm assuming I'm misunderstanding something (or perhaps it is that low) or asking a different question.

The results say that you will get to choose theater six 14.51% of the time if you roll three dice.

If that seems low to you, remember that you are also adding lots of possibilities to the other results as well, and the final probability is the number of results you want divided by the total number of possibilities.

Look at the entire distribution from 1 to 12 and you can see that it is very skewed. You added a new value 1, and the distribution is no longer centered perfectly on 7.

NewbieDesigner
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Thanks for confirming.

Thanks for confirming.

Daggaz
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No problem. It was a fun

No problem. It was a fun problem and now I will have to try not to solve it explicitly (math) and instead finish my manuscript. Good luck on your game and don't be afraid to ask questions.

X3M
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No need to become aggresive

Just trying to understand the way how you think. That is all. I can learn from that. Either way.

When looking at that last table, I got that impression. It looks a bit vague to me. 1 to 12 by 1 to 6.

I have written out for the numbers that I got, just to be sure. A list of 216 (rolling 6*6*6). Also looking at the 3 arrangements (A,B+C/B,A+C/C,A+B). Which gives a total of 648. By the rules that NewbieDesigner has given, thus using both value's. We get a total of 1296.
And I was simply counting each value, 1 to 12. Really, counting.

When using permutations, you are discarding some numbers. This is what I don't understand. When AND where do you discard, what you have rolled? I already offered to email the excel. You can point out where I went wrong?

You don't have to believe me. Lets agree that we disagree.

Daggaz
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I discard anything that has

I discard anything that has already been counted, because the rules do not give you extra points for rolling your value twice. You can only use it once. In other words, if a given permutation has multiple paths to getting the desired outcome (many of them do), you only count that permutation once.

Regarding 1296, you dont get this number because not all values have a choice. You can ONLY roll a 12 with two dice. You can ONLY roll a 1 with one. So you dont get to double the entire spread.

X3M
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Daggaz wrote:I discard

Daggaz wrote:
I discard anything that has already been counted, because the rules do not give you extra points for rolling your value twice. You can only use it once. In other words, if a given permutation has multiple paths to getting the desired outcome (many of them do), you only count that permutation once.
Where did he say that? I was under the impression that if you rolled for example 6, 3, 3. That you could turn that into:
6 and 6, 9 and 3 and another 9 and 3.

Regarding the 6 and 6, they both can be used, stated by his rules. Right? So that counts twice. But....

Are you saying that this 9 and 3, which can be created twice, is counted only once? Because, then I finally get where you where at.

I'll look into my table once more. Since I got many of those doubles indeed. I am going to (dis)count them, literly. So please be patient on this one :)

FrankM
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Oops

Daggaz wrote:
Hey Frank,

Im not sure without looking at your sheet and understanding your method where it went wrong, but the total percentages have to add up to 100% as a first thing.


There were two outcomes per state, but I only divided by the 216 states. That made everything sum up to 200%.

* hangs head in shame *

It wasn't clear to me what to make of, for example, 4 and 2+2. If the player would be allowed to move that track's pawns twice, double-counting is perfectly appropriate. If the player is prohibited from picking the same track twice, then the second 2~6 should be excluded.

Daggaz
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Im too tired to explain this

Im too tired to explain this clearly, so I will go to your example.

Given
6 3 3

6 6 Both of these would be counted by my method, assuming you want a 6

9 3
9 3 Only one of these would be counted, this is a degenerate result. This is degenerate whether you want a 9, or a 3, because in either situation, you can only take one set of either one single die, or one pair of die. NOTE that you do count the nine for one outcome (I want a nine), and you do count the 3 for another (I want a three), but you do NOT count two nines, or two threes.

The difference is that in the first situation, even though there are two numbers that are equal, you get them by different methods. Whereas in the second example, the outcome is duplicated, and this happens because you pick the same method twice, using the duplicated primary dice. Duplicated outcomes are known in combinatorics as "degenerate" and you dont count them twice unless there is a specific reason to do so (one such reason comes up in the physics of electron energy levels, but thats another topic).

The odds I give are for a single theater showing up in a given cast of three dice by either method (single or sum) allowed. If you want the odds of the second theater, GIVEN that the first theater is chosen, then it gets yet again more complicated and at this point I really dont feel like running that out. But it's also besides the point. In the design of the tracks as OP describes, he will only be concerned with the primary chance of any one theater showing up. The secondary statistic is simply the primary statistic divided amongst the allowable neighbors.

Daggaz
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The track choice doesn't

The track choice doesn't matter (I hope). It is the stipulation that you have to take one sum and leave one dice alone.

For 4, 2, 2, you cannot move track 6 twice. But you could move track 4 twice, and I count that in my method (OP did not say otherwise I believe).

If OP doesnt want to allow the same track moved twice, even by two methods, then you would need to remove on rare occasion a few counts from the statistics I gave, but it wouldnt have a very significant effect as this is an overwhelmingly rare situation.

EDIT:
NOPE. I did throw that out in the case of the third die having value between 2 thru 6, AND matching the value wanted. Went a bit too fast there, sorry.

So for row 2 through 6, if there is a value of 36, it needs to have added to it the following values:

2: 1
3: 2
4: 3
5: 4
6: 5

Its not a huge change to the statistics, the final tally of permutations is now 1062.

X3M
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Also hangs head in shame

Well, I guess I am almost there now. By simply counting. And while most looks the same. A couple of numbers are different. I also looked if I had asymmetry, which should not be the case with this problem. A, for example, 6+6 has been counted twice though. While 9+3 and 3+9 was corrected.

I have spotted the 36, 30 and 25 now. It happens 6 times indeed for the single die.

This is the list that I have currently:

1-91
2-107
3-121
4-137
5-151
6-167
7-90
8-76
9-60
10-46
11-30
12-16

A total of 1092

Obviously, 2 to 6 are different from your value's. I compared the two. And there is a neat little linear difference of 3, 6, 9, 12 and 15. So I was literately pulling my hair out right now. But then I also removed any double, like the example. Thus not moving twice on the same track.

I got the same list and bow down.

NewbieDesigner
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Yes, moving twice on the same

"For 4, 2, 2, you cannot move track 6 twice. But you could move track 4 twice, and I count that in my method (OP did not say otherwise I believe)."

Correct assumption- moving twice on the same track is allowed for what I have in mind so my movie example is flawed since I should have mentioned you can see the same movie twice if one wanted to.

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