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# dice probability question (Can't Stop)

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NewbieDesigner
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Joined: 01/20/2011

Are the overall odds/dynamics roughly the same if:

1. Instead of 4 dice being rolled, 3 dice are rolled instead.
2. Instead of 3 pieces advancing up the tracks, 4 pieces advance.

So you have one more piece advancing but roll only 3 dice. For the choosing values for advancing from the 3 dice it would be A+B and C+A or C+B. Example: You roll a 2, 5 and 6 and you choose to advance up the 11 spot and 8 spot. Or 11 & 7.

Thanks.

NewbieDesigner
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As a follow up question.

As a follow up question. What does the probability graph look like when rolling three dice, adding two together and using the third one by itself?

So 3,4,5 could be 7&5, 8&4, or 9&3.

BHFuturist
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math...

I don't think you will get much help on the raw math here. I am not even sure that the math is the part that will help with this mechanic.

However you could try spending some time testing things with:

http://anydice.com/program/2ef0

From a design perspective and not a math perspective. Rolling 4 dice and grouping them into two pairs (only using each die once) will give the player less meaningful choices than rolling 3 and being able to clone one of the dice to make two pairs.

Your second post gives the player the same level of meaningful choice as rolling three dice but limits the overall range of movement points for the second move.

This is something that can be playteted and iterated very quickly to find the combination that gives you the "working results" you are looking for. (no math degree required)

Knowing the math is great but the intuitive understanding of how it plays is more important in this case (I think).

The thing you really need to know is what are you trying to give the players? More meaningful choices or a quick roll that is not hard to calculate and keeps the pace of movement to a rate that fits the length of the board and length of playtime.

Just my two cents, meant only as food for thought.

@BHFuturist

X3M
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Joined: 10/28/2013
Finally someone else

Finally someone else suggesting any dice. It is a great website once you know how to use the program perfectly.

Regarding the number of dice. Making 2 pairs with 4 dice gives more options. But those options would indeed give lower levels for the second combination.
If the number of options for players is your goal. You can stick with this one.

Having only 3 dice means, that you have very few options.
6, 5 and 2 can give:
11 + 7 or
11 + 8 or
7 + 8

By copying the highest die, you automatically have better results than rolling this die.

3 dice gives less options, but higher. Good luck with the program.

NewbieDesigner
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Thanks and will check out the

I will check out the website and thanks for linking directly to the 3 dice scenario I mentioned!

Daggaz
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Joined: 12/19/2016
Not enough information. And

Not enough information. And ill-formed question.

What do you mean by "roughly" ..? You want to talk about mathematical probabilities but this is not a mathematically defined concept.

Difference between 3 or 4 dice... what size dice? What are you trying to roll? This is impossible to answer specifically with the lack of information given. One can only say:

The odds will always be different.

BHFuturist
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humm

X3M wrote:
... Making 2 pairs with 4 dice gives more options ... 3 dice gives less options...

I stand by what I said but you might be misunderstanding what I said (because I did not explain it very well)

From a design perspective and not a math perspective. Rolling 4 dice and grouping them into two pairs (only using each die once) will give the player less meaningful choices than rolling 3 and being able to clone one of the dice to make two pairs.

When I roll 4 dice I get 3 options (one choice of combinations):

3, 1, 5, 6,

I can make: 4&11, 8&7, 9&6

When I roll 3 dice I get 3 options (one choice of combinations):

3, 1, 6

I can make: 4&9, 4&7, 9&7

When I roll 4 dice (with a double) I get 2 options (one choice of combinations):

3, 3, 1, 6

I can make: 6&7, 4&9

I was not trying to say that the number of options was more or less just that I felt that there was a more meaningful choice in rolling 3 dice and picking the die to clone.

This was not a math thing for me, it is a philosophical point about how the choice feels to me as a player. I feel like my choice matters more to the end result. My choice "seems" to have more sway over the range results rather than a straight selection of raw values. It is not about the math...

You are assuming that a higher value is better and that the choice will be automatic. But if this is a roll & move game where landing pieces on certain spaces are needed for some actions to happen and you have several pieces you can move. Based on where they are at the time of the roll having a wider range of options is more meaningful in my mind.

Mathematically there is not much difference between rolling 4 and rolling 3 where either way you are making two sets of numbers from adding them together. The range of values will many times be similar if you roll doubles in the set of 4 dice. A set of doubles is more likely the more dice you roll anyway...

It is just a gut feeling nothing more... no scientific justification. I hope that makes more sense.

@BHFuturist

X3M
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Yeah, I guess I explained it

Yeah, I guess I explained it not properly either. To short on words there and going ahead to far, sorry.

This is how I think:

BHFuturist wrote:
You are assuming that a higher value is better and that the choice will be automatic. But if this is a roll & move game where landing pieces on certain spaces are needed for some actions to happen and you have several pieces you can move. Based on where they are at the time of the roll having a wider range of options is more meaningful in my mind.
I did not assume this. Somehow I had the feeling that NewbieDesigner wants players to choose a location instead of trying to get as far as possible.
I know that I said that the fewer options gives higher results. But that is something that I noticed. This is because the best die can be copied. Not necessarily what he wants for his players, honestly.

Please take note, that I am only pointing out that the total sum of all choices to be possible higher with the 3 dice. But the same can be said about, trying to be as slow as possible. Then too, 3 dice yields better options.

I see something different when considering the number of options too:

BHFuturist wrote:

When I roll 4 dice I get 3 options (one choice of combinations):

3, 1, 5, 6,

I can make: 4&11, 8&7, 9&6

I see 4, 6, 7, 8, 9 and 11.
6 options for a certain piece.

BHFuturist wrote:

When I roll 3 dice I get 3 options (one choice of combinations):

3, 1, 6

I can make: 4&9, 4&7, 9&7

I see 4, 7 and 9.
Only 3 options for that same piece.

I do assume here that these pieces don't have to start from the same spot. That what ever you choose for one piece. Another piece will be moving with the other value regardless. An unanswered question will be, which piece? Piece number 2, or any piece?

In either case, things can be different:
With 4 dice, one piece has those 6 options. Considering that the other piece is fixed with 1 option linked to the choice.
4, 6, 7, 8, 9 and 11. Will yield only one other option each for the second piece.
With 4, the other piece can only have 11 and vice versa.
With 6, the other piece can only have 9 and vice versa.
With 7, the other piece can only have 8 and vice versa.

With 3 dice, one piece has 3 options.
4, 7 and 9.
But the joke here is that the second piece will have 2 options open due to the copying mechanic.
With 4, the other piece can have 7 or 9.
With 7, the other piece can have 4 or 9.
With 9, the other piece can have 4 or 7.

It depends now on the other workings of the game.
It raises the question. Will the game provide just enough importance for only the movement of one piece, and the other is simply following up? Or will it be important to have both pieces move in a considerate way?

Basics when regarding the total amount of options; 6x1=6 or 3x2=6. And you got yourself an issue of dividing freedom between the 2 pieces.

@NewbieDesigner
The only thing that confuses me is that you said that one way, 3 pieces can be moved. The other way, 4 pieces. Do they move as a group? Or are the options for 1 piece each? Perhaps they do start from different spots? Or what is exactly the intention here?

BHFuturist
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Gulp...

X3M wrote:
With 3 dice, one piece has 3 options.
4, 7 and 9.
But the joke here is that the second piece will have 2 options open due to the copying mechanic.

Thank you, for saving me from myself... I got the tunnel visions on that "joke". It was the only thing I was focused on... I also did not stop to do good maths. What I did was chuck some dice to see how it felt.

That joke is what makes the roll 3 way feel like more of a meaningful choice to me. You are right that the 4 dice gives a wider range of picks for only the first piece that you want to move but the way the second choice is made for you based on that first choice is what makes the roll 4 way seem like it was less of a meaningful choice.

I stand corrected and am sorry for assuming that you were assuming.

@BHFuturist

NewbieDesigner
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Sorry I haven't checked in

Sorry I haven't checked in sooner. I posted a similar question on another forum and realized I wasn't being as clear as I could be (I blame it on sucking at math...lol). So yeah, essentially, I want to be able to move 4 pieces instead of 3 along 12 paths like Can't Stop using 3 dice instead of 4, and want to know how to properly value those paths now. Here's what I recently posted somewhere else:

Bob goes to the local 12 theatre Cineplex every day and chooses two movies to see based on the results of rolling three dice. One dice is chosen to determine his first theatre choice, and the other two dice are added together to determine his 2nd theatre choice.

Example: Bob rolls three dice and can choose to see two movies in these theatres:

6,6,6= Theatre 12 and 6
1,1,1= Theatre 1 and 2
2,4,5= Theatre 2,4,5,6,7,9
3,4,5= Theatre 3,4,5,7,8,9
4,4,6= Theatre 4,6,8,10

Question- How often (in percentages compared to other theatres) will Bob be able to see a movie in theatre 1? Theatre 2? Etc.

Stormyknight1976
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Probability is chance

What are you rolling three dice for?

1: Adding all three numbers to move ?
2: Adding all three numbers to attack ?
3: Adding all three numbers to defend ?
4: Picking two numbers that are higher out of the three probability chance rolls to do a skill or attack ?
5: Probability of rolling all three die is 666 chances of getting a 123,234,456,111,222,333,444,555,666.

or

254,126,356,654,146,226. etc.

Or just divide the numbers of 666 to get what your looking for or divide 333 . With out knowing what your looking for while rolling 3 dices it is difficult to answer your question.

Stormy

X3M
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Joined: 10/28/2013
You are making things to

You are making things to complicated by changing what you want to know.
- Now the die isn't copied?
- What happened to the situation of 4 dice, that needs to be the comparison?

When each die has its own function:
3 Dice gives 216 possibilities.
4 Dice gives 1296 possibilities.

When you start adding rules, you will get set classes. And each set has their own chance ratio. A simple example is that all dice are added up:
3 Dice gives 16 possibilities.
4 Dice gives 21 possibilities.

With the theatre. Your rule asks for 3 dice. And these 3 dice can create 1 to 3 sets. Now you want to know what value's can be created. And what the chances are on these value's.

???

NewbieDesigner
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X3M wrote:You are making

X3M wrote:
You are making things to complicated by changing what you want to know.
- Now the die isn't copied?
- What happened to the situation of 4 dice, that needs to be the comparison?

When each die has its own function:
3 Dice gives 216 possibilities.
4 Dice gives 1296 possibilities.

When you start adding rules, you will get set classes. And each set has their own chance ratio. A simple example is that all dice are added up:
3 Dice gives 16 possibilities.
4 Dice gives 21 possibilities.

With the theatre. Your rule asks for 3 dice. And these 3 dice can create 1 to 3 sets. Now you want to know what value's can be created. And what the chances are on these value's.

???

Apologies with the changes and evolving question.

In Can't Stop, the 2 and 12 tracks have two spaces to advance on before reaching the finish line. The 7 track has twelve spaces. Seventy-two total spaces.

How many spaces per track should exist on a revised 1-12 Can't Stop board with players having the choice to roll three dice with pairing two of them as a requirement (forget pairing them in any way as I originally mentioned). I assume 1-6 will be much more frequent now (and thus more spaces per track) than 7-12 but I'm just not sure how much?

Daggaz
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NewbieDesigner

NewbieDesigner wrote:
Essentially, I want to be able to move 4 pieces instead of 3 along 12 paths like Can't Stop using 3 dice instead of 4, and want to know how to properly value those paths now. Here's what I recently posted somewhere else:

Bob goes to the local 12 theatre Cineplex every day and chooses two movies to see based on the results of rolling three dice. One dice is chosen to determine his first theatre choice, and the other two dice are added together to determine his 2nd theatre choice.

Example: Bob rolls three dice and can choose to see two movies in these theatres:

6,6,6= Theatre 12 and 6
1,1,1= Theatre 1 and 2
2,4,5= Theatre 2,4,5,6,7,9
3,4,5= Theatre 3,4,5,7,8,9
4,4,6= Theatre 4,6,8,10

Question- How often (in percentages compared to other theatres) will Bob be able to see a movie in theatre 1? Theatre 2? Etc.

Ok.

You have three dice A B and C. You roll them all and Dice A you pick as a stand-alone value.

Dice A gives 1/n chance for each state 1 through n.

Dice B and C are added together. You can ignore which of the three dice is actually Dice A, because the moment you pick Dice A to stand alone, you separate the problem. I am assuming here that the three dice all have the same n number of sides of course, and the same range of values.

For the sum, there are n^2 possible outcomes. The outcomes are ranked as (state 1 + state 1), (state 1 + state 2), ....(state 1 + state n), (state 2 + state n), ....(state n + state n)

So basically you start with both dice in the lowest state, and increment one of the dice until it hits the maximum, then increment the other dice until it hits the maximum, in order to hit all the possible outcome values.

The probability that a given outcome will occur is easy to find. If one of the dice is in state 1, then the state number of the second dice is the number of times that this particular sum can be found, and the probability of finding this outcome is = second dice state number / n^2.

Similarly, if one of the dice is in state n then the number of outcomes is equal to (n - second dice state number + 1) and again the probability is that value / n^2.

For the case of one dice in state 1, and one dice in state n, either formula works.

This is easier to see in a matrix, in this case two dice each of state 1 through 6, here the state values match the state numbers.

1­­­ ­ 2­ ­ 3­­­ ­ ­4­­­ ­ 5­ ­ 6

2­­­­­ ­ 3­ ­ 4­ ­ 5 ­ 6 ­ 7 ­ ­ 1
3­­­ ­ 4­ ­ 5­ ­ 6 ­ 7 ­ 8 ­ ­ 2
4 ­ 5­ ­ 6­ ­ 7 ­ 8 ­ 9 ­ ­ 3
5 ­ 6­ ­ 7­ ­ 8 ­ 9 ­ 10­ ­ 4
6 ­ 7­ ­ 8­ ­ 9 ­ 10 11 ­ 5
7­ ­ ­8­ ­ 9­ ­10 ­11 12 ­ 6

so n = 6, and if you want an outcome value of 7, you have state 1 + state n, which gives n out of n^2 possibilities. Similarly if you want an outcome of 11, you have state n + state 5, for (n - 5 + 1) = (6 - 5 + 1) = (2) possibilities, out of the n^2 = 36 total.

I use state number and not value because maybe your dice dont have values that match the state number (a 6 sided dice with values 2, 4, 8, 20, 30, 50), but now as long as there are no duplicate or negative values on a single dice, you can just use a lookup table to get the final sum. The important thing is the probability will be the same no matter what individual value is linked to the state number.

NewbieDesigner
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Thanks Daggas, very

Thanks Daggas, very impressive to someone terrible at math. I feel bad for asking, but can you list how to calculate the odds using a lookout chart? One number should suffice (e.g., the odds of being able to choose a 7 from three dice...locking one pair...is 50%).

Daggaz
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I listed those odds. A

I listed those odds. A lookup chart is just for the values if you want to play around with the numbers on your dice faces.

The odds for getting a given sum X from two dice out of three are a bit more complicated to write down. I wrote it out the other way because a) its a bit simpler that way and b) I assumed that you would select the first dice to stand alone first, and the other two were then left over.

If you do it with six siders, then just take the matrix I wrote down, and recopy that six times, with the value of the third dice next to it.

So
[matrix] + 1
[matrix] + 2
...
[matrix] + 6

Now you look for your sum in the matrix values.

You also take the column + third dice, and the row + third dice, which gives you the same sum. That creates a column/row cross.

You need to now overlap the column/row cross, with the matrix values that match your sum as well. Count every square only once. Do that for each of the six matrixes.

The probability is, remembering to not count any squares more than once, the total number of hits / 6^3

Sorry if that doesnt make more sense, it would be simpler to explain this visually but I dont have time to draw that out now.

X3M
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Joined: 10/28/2013
Yeah, drawing things is hard

Yeah, drawing things is hard here. And it is hard to follow more often than understanding.
With the link given by BHFuturist, you can get far. And manual calculating with table's too. No formulas here. Just adding up and multiplying will do the trick this time.

1d6:
Sum chance = 6
Rolling - Part
1 - 1
2 - 1
3 - 1
4 - 1
5 - 1
6 - 1

2d6:
Sum chance = 6^2 = 36
Rolling - Part
2 - 1
3 - 2
4 - 3
5 - 4
6 - 5
7 - 6
8 - 5
9 - 4
10 - 3
11 - 2
12 - 1

There are fractions!!!
To add the 2 together in a 1 on 1 addition, they need the same dimension. I am referring here to the sum. 1d6 has 6, 2d6 has 36. This means that the parts of 1d6 needs to be multiplied by 6. And once the parts are added. We get a new total sum, of 72.

1d6:
Sum chance = 6*6=36
Rolling - Part
1 - 6
2 - 6
3 - 6
4 - 6
5 - 6
6 - 6

And let us add them to the other list:
2d6 AND 1d6:
Sum chance = 72
Rolling - Part
1 - 6
2 - 7
3 - 8
4 - 9
5 - 10
6 - 11
7 - 6
8 - 5
9 - 4
10 - 3
11 - 2
12 - 1

And you said that you had a total of 72 in tracks? With the numbers here, that can be absolutely right.
Track 1 has 6 fields, Track 12 has only 1.

Is this correct?

NewbieDesigner
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Assuming your math is

Assuming your math is correct, that looks like exactly what I need! :-)

Daggaz
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This is wrong tho. OP wants

This is wrong tho. OP wants to roll 3 dice, hold one dice for the first value, add two remaining dice together for the second value, and look at all possible permutations of choosing between those two values and getting a particular end value limited to the set (1,2,...12)

You claim six ways to roll a 1. There are 91 unique rolls to get at least one 1, which we then hold and claim. ((6*6) + (6*5) + (5*5)) = 36 + 30 + 25 = 91, where each value is progressively setting one dice to 1 and counting the uncounted total possibilities remaining on the other dice.

91 cannot be reduced to 6 so clearly you have a problem that doesn't involve simplification of your fractions. This is obvious as well in that you only have a sum chance of 72, but it should be 6^3 = 216 total possible primary permutations from which you can create an even larger subset of summations, as a single primary permutation of three dice can give 3 sub-permutations involving hold 1/add 2. Now 72 happens to be a third of 216, and you certainly needed to multiply your set by those three sub-permutations. It still bothers me tho, as you have overlap for the values (2,3,4,5,6) regarding which value you choose, and one sub-permutation can be counted for more than one final value, so I want to see a sum chance that is larger than 216. Given that you can have 2 outcomes per sub-permutation, I would want to see at least 432 for the sum chance. Given that there are 91 possible ways each to get the 1 through 6 using only the single held dice, that pushes it further up towards a minimum of 546. And now you have to add in all the sum possibilities..

Either way, I myself would hesitate to calculate this explicitly without first drawing out the matrices as I described in my previous post, and thus create a visual control over the actual range of possibilities. As you can see, its so easy to mistakes with this kind of math otherwise, if you are not working from personally well-understood theory.

NewbieDesigner
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Thanks Daggaz. I would add

Thanks Daggaz. I would add that the size of the grid like Can't Stop is not essential for my purposes, just figuring out the percentage of rolling each 1-12 value based on the pairing scenario I described.

NewbieDesigner
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Btw, is there an easier

Btw, is there an easier method do determine the odds, of say, within +-5%? Factoring in the luck of rolling dice, the precise odds of being able to choose a particular 1-12 number from rolling 3 dice isn't necessary.

Daggaz
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For this problem you will

For this problem you will find drawing a matrix and counting all the possibilities, listing as you go in a table, really won't take very long.

For more advanced problems, you either solve it explicitly -very carefully-, or you make a monte carlo program that rolls dice a few thousand times and tabulates the odds.

FrankM
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Not sure I understand the question

It sounds like the question is roll three dice, and you would like the distribution of outcomes from a single die and the sum of the other two.

There are four variables here: A, B and C are dice (where Nature chooses a value of 1 to 6 for each) and D (the player's decision of which die to stand alone). This gives you 6x6x6x3 = 648 total states.

But if you want the raw probabilities without any situation-specific strategic thinking by the player, as mentioned above you can simplify this by always picking A as the stand-alone die.

Result A B+C Either
1 36 00 36 / 216 = 16.67%
2 36 06 42 / 216 = 19.44%
3 36 12 48 / 216 = 22.22%
4 36 18 54 / 216 = 25.00%
5 36 24 60 / 216 = 27.78%
6 36 30 66 / 216 = 30.56%
7 00 36 36 / 216 = 16.67%
8 00 30 30 / 216 = 13.89%
9 00 24 24 / 216 = 11.11%
10 00 18 18 / 216 = 08.33%
11 00 12 12 / 216 = 05.56%
12 00 06 06 / 216 = 02.78%

If the plan is to make these track length proportional to the probability, you could divide the number of outcomes by 6. But whatever factor you use, the "6" track is always going to be 11 times the length of the "12" track.

X3M
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FrankM is right, regarding

FrankM is right, regarding the chances (a total of 200%). I went with the minimal result. And forgot that fact, that the player is going to use 2 value's per roll.

Either way, the proportions seem to be the same.

@Daggaz
I don't get how you got those permutations? Please elaborate.

NewbieDesigner
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Thanks for the additional

Thanks for the additional info everyone.

Daggaz
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The sub-permutation of 3

The sub-permutation of 3 comes from the fact that you can pick either dice A, B, or C as the single die. Or {(A :(B+C)), (B:(A+C)), (C:(A+B))}

So you have 6*6*6*3 permutations at this point. But its not done there.

You have the final set of values you are trying to achieve, and you make a choice at this point if you want value X (the single die) or value Y (the summed dice) to achieve this final value.

The final set of values is 1 through 12. But as I mentioned, only (2,3,4,5,6) can be made by either a single die or a summed pair of dice. That leaves seven final values which cannot have a choice, and one of them can only be made with one die, while the other six must be made with the sum.

If every value had a legitimate choice between the two, then 6*6*6*3*2 = 1296 would be the maximum limit, but they dont, so the real number will be lower than this. I am tempted to say it is :

6*6*6*3*2*(5/12) + 6*6*6*3*(7/12) = 540 + 378 = 918.

But just looking at that it seems like there could be a mistake because of the complex overlap of the sub-sets. This just looks too easy. So lets actually count these. I drew out the summation matrix for 2d6, then laid transparent plastic strips over the rows and columns that could give me the target value with or without the third die, and then included any additional counts found on the matrix not covered by these strips, which are summations not involving the third die. This was done iterating six times for the possible values of the third die, see the top row.

.....01 02 03 04 05 06 ..... (this is the value of the third die)
01 : 36 11 11 11 11 11 : 091 : 08.69%
02 : 20 36 12 12 12 12 : 104 : 09.93%
03 : 20 20 36 13 13 13 : 115 : 10.98%
04 : 21 22 21 36 14 14 : 128 : 12.23%
05 : 22 22 22 22 36 15 : 139 : 13.28%
06 : 23 23 24 23 23 36 : 152 : 14.51%
07 : 15 15 15 15 15 15 : 090 : 08.60%
08 : 05 14 14 15 14 14 : 076 : 07.26%
09 : 04 04 13 13 13 13 : 060 : 05.73%
10 : 03 03 03 12 13 12 : 046 : 04.39%
11 : 02 02 02 02 11 11 : 030 : 02.87%
12 : 01 01 01 01 01 11 : 016 : 01.53%

value we want : the number of ways to get the wanted value given the third die : total number of ways to get this value: percent chance

Total permutations : 1047
Sum of percentages : exactly 100% (amazing considering I rounded)

PS: If your probabilities sum to over 100%, you know immediately you have a problem. Total probability MUST be equal to 100%. If it is over, you have overcounted somewhere. If it is under, you are missing some outcomes or undervalued something.

I think this puts the nail in the coffin. Both regarding the answer, and the point that you really should work these out the long way rather than trying to calculate the answer, if you do not know exactly what you are doing and why you are doing it (I certainly didnt).

NewbieDesigner
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Thanks Daggaz. Another

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.

X3M
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Daggaz The first 2 dice are

Daggaz
The first 2 dice are added up. They can't count as a 1.
Adding up asks for 36 possibilities. Then you are stuck with 36 rolls on 1 dice. That is how I get my 6 times a 1.

Please keep in mind that both value's are used at the same time. Thus A and B+C or B and A+C or C and A+B.

When everyone doubts:
I got a table in excel with all the numbers. (including the list of FrankM). Can I email it to you, so that you can take a look?

Daggaz
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Dude... you don't honestly

X3M, your response is rather condescending, which is made all the worse by the fact that you clearly don't understand what I did. I explained the method earlier in this thread and you didnt seem to have a problem with it, so you must have lost track. I will explain it explicitely now.

FIRST I take 3 dice, A B C. I put two of them A and B into a summation matrix, and the third C I hold on the outside. PLEASE NOTE: This does NOT mean that I add the first two, the matrix is simply a tool to keep track of the possible summation values. Its a matrix after all, by itself it does not represent a singular quantity.

THEN I iterate the third dice. First 1. Then 2. Then 3...etc. up to 6

THEN for EACH iteration, I look at the wanted values, 1 through 12, and count the possibilities using the matrix. ALL of the possibilities.

EXAMPLE

IF you have a 3 on the third dice, and you need to get a final value of 4, you can either roll a 4 on either of dice A or B (mark the 4 column and the 4 row), OR you can roll a 1 on either dice A or B which you add to dice C (mark the 1 column and 1 row), OR you can roll A = 2 B = 2 for one more hit. Add all of these possibilities together, making sure not to count anything twice. You get 21. Again, if this is not clear, this includes ALL possible combinations of the three dice which result in 4. Now look at the chart I posted. Third Dice = 3, Wanted Value = 4, Returns 21

X3M wrote:
The first 2 dice are added up. They can't count as a 1.

Dude... you don't honestly think I added two dice to get a one do you?

X3M wrote:
Adding up asks for 36 possibilities. Then you are stuck with 36 rolls on 1 dice. That is how I get my 6 times a 1.
Im not sure how 6 * 1 has anything to do with 36, you are going to need to explain this leap of logic a bit more clearly. Especially the part about 36 rolls on 1 dice, because that never happens. Not even equivalently.

X3M wrote:
When everyone doubts:
I got a table in excel with all the numbers. (including the list of FrankM). Can I email it to you, so that you can take a look?

Now you are not only rude, but you are directly contradicting yourself. If you are so sure of your numbers, why have you not posted them (I did), and why are you asking FrankM to check your work privately in the same message?

If I sound pissed off, it's because I spent time in my busy schedule today to solve this problem instead of working, and while I welcome legitimate criticism quite openly, yours is anything but. And to top it off you are simply wrong.

FrankM
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Joined: 01/27/2017
Where my table came from

I had counted everything from a 216-row exhaustive list of outcomes using Excel, though the results make sense this way: for A there are six possibilities and each one spans a full set of 36 outcomes for the other two dice... and for B+C there are 36 permutations (11 distinct sums) and each one spans a full set of 6 outcomes for the other die. That's why everything in the first column is a multiple of 36 and everything in the second column is a multiple of 6.

Daggaz
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Joined: 12/19/2016
Im not sure without looking

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.

Another issue, which is why I went through each matrix individually, is that the overlap changes depending on the value. Some values are easy to hit because there are lots of combinations, and you can choose to either hold or add. 1 is tricky because you simply have to roll it outright (notice that my counting method returns the same number as my calculation : 91), 7 and higher you have to achieve with two dice.

If you count it out by hand, you can see how this overlap evolves as you mark the rows and columns and look for the extra hits.

Again with respect to 1: notice how I have 36, 11 11 11 11 11. These are the base odds for rolling a value outright with one dice. That matches what you did. But as soon as you move to 2 or higher, this changes. These odds underlie the results for 2 through 6, but no longer contribute to 7 and higher (with the caveat that the 36 moves across the row as the desired number grows).

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