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Subject: How to figure Dice Probabilities? rss

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I am having some difficulty trying to figure out a formula to tell me the probabilities of some rolls of dice. I am the first to admit I am not the sharpest knife in the drawer when it comes to math, so hopefully there is a genius out there that can help me out.

Ok, here is what I am wondering. The chances of rolling a 6 on 1D6 are one in six. How do you figure the chances of rolling two sixes by rolling 1D6 over and over again until you get your second six? Technically you could roll two sixes in a row, but that isn't probable. How do you determine what is the average for that happening?

To make matters a bit worse, I have to figure out the same thing but rolling a 5+ and a 5+ twice... How many times on average do you need to roll a D6 before you roll a 5 or 6 twice.

All the way down from 4+, 3+ and finally 2+... How do you figure this out?

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Jason Leidich
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I believe the chances of rolling a 6 one time is the same as rolling it a second time (16.67%) if each roll is done individually - each time you roll you have a 1 in 6 chance of getting a specific number. Think of it like flipping a coin. Each flip you have a 50/50 shot of getting tails.

However, I think there's a deeper calculation involved when you are looking to roll the same number twice in a row and it would equally apply to any number on your die. Therefore the probability of rolling two 6s in a row is exactly the same as the probability of rolling two 4s in a row.

Check this tool out: http://anydice.com/

It will help you figure out dice probabilities but maybe not as in depth as you appear to be looking for with multiple rolls. Either way though I hope you find it useful to some degree.

Full disclosure: I'm not a statistician.
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Jason Leidich
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well... that is of course if there's a difference - mathematically - between rolling 2d6 and rolling 1d6 twice. In which case to answer your first question = 2.78% if there is no difference.

And one more addendum, after re-reading your post and seeing all the + signs. Are you saying you need to calculate the odds of rolling say a 4 on a d6 and then rolling a 4, 5 or 6 next (4+)? If so, that's crazy talk and we need some kind of math scientist.
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Franz Kafka
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It sounds like you're asking "how many rolls of a d6 would it take to get two sixes?"

The long long answer: http://en.wikipedia.org/wiki/Negative_binomial_distribution

Filling out some of those formulas for your first case:
--"k" is the number of non-sixes you will have rolled by the time you've rolled "r"=2 sixes.
--"p" is the probability of rolling a non-six on a given roll, which is 5/6
--The average (mean) number of non-sixes rolled is (5/6)*2/(1-(5/6)) = 10, and the average number of total rolls is 10 + 2 = 12
--If you want to get really detailed, the odds (pmf) of rolling "k" non-sixes in this process is C(k+2-1,k)*[(1-(5/6))^2]*[(5/6)^k]

Also, here are the average total rolls for other numbers, by replacing the 5/6 with 4/6, etc.:
6+: 12 rolls
5+: 6 rolls
4+: 4 rolls
3+: 3 rolls
2+: 2.4 rolls
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James Champagne
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Or buy Catan . It lists the dice probabilities right in the rulebook, and then again on the components you stare at while playing!

Basically, though, all you need to know for most games is that the most likely number to come up on two 6-sided dice is 7. The farther you get from 7, the less likely the number. 2 and 12 are the least likely numbers to show up.
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Ian Taylor
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I think you need to clarify what exactly you need to know. I assume you realise that theoretically you could roll a dice forever and never manage to get a two 6s, although it is increasingly unlikely the more attempts you make.

Franz's answer is definitely along the right lines, but might not be exactly what you're asking. Why don't you give us some context and let us know what you are trying to calculate and we'll go from there.
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James Champagne
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Sorry, I misunderstood your question. Here's a simple answer as I understand it.

The odds of rolling a 6 are 1/6. The odds of rolling two 6s in a row are 1/6 * 1/6, or 1/36.

The odds of rolling a 5 or 6 are 1/3. The odds of doing this twice are 1/3 * 1/3, or 1/9.

And the pattern repeats....
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Larry Levy
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Franz' formulas are correct, but there's a simpler and more intuitive answer (although the math to prove it is a little complex). The average number of rolls it takes to achieve a result that has a probability of X is 1/X. So, for example, the odds of rolling a 6 on a single D6 roll is 1/6, so the average number of rolls it would take before a 6 appears is the inverse of that, or 6 rolls. Intuitively, this is exactly what we would expect. To figure out how many rolls on average it takes to roll a result happening twice, just double the number it takes to roll it once. This also makes sense: to get two 6's, for example, first you have to roll one six (which takes an average of 6 rolls to get), then you have to roll your second six (which will take an average of 6 rolls after you roll your first one). So the answer is 6 + 6 = 12 rolls.

The general formula for getting X results twice is 12/X. So if you're trying to roll a 4 or more, then there are 3 results that represent success and the formula tells us that it will take 12/3, or 4 rolls on average to achieve this. This is exactly what Franz' formula shows.
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Herb
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JosefK wrote:
It sounds like you're asking "how many rolls of a d6 would it take to get two sixes?"

The long long answer: http://en.wikipedia.org/wiki/Negative_binomial_distribution

Filling out some of those formulas for your first case:
--"k" is the number of non-sixes you will have rolled by the time you've rolled "r"=2 sixes.
--"p" is the probability of rolling a non-six on a given roll, which is 5/6
--The average (mean) number of non-sixes rolled is (5/6)*2/(1-(5/6)) = 10, and the average number of total rolls is 10 + 2 = 12
--If you want to get really detailed, the odds (pmf) of rolling "k" non-sixes in this process is C(k+2-1,k)*[(1-(5/6))^2]*[(5/6)^k]

Also, here are the average total rolls for other numbers, by replacing the 5/6 with 4/6, etc.:
6+: 12 rolls
5+: 6 rolls
4+: 4 rolls
3+: 3 rolls
2+: 2.4 rolls



The negative binomial distribution is correct for two sixes in n tries.

The problem statement is vague.

Do you need the probability distribution for rolling two sixes, and then two fives, and then two fours, and so on down to the ones?

Or are each independent? So the average number of rolls for two sixes starting from 0 rolls. Then the average number of rolls for two fives starting from 0 rolls? And so on...
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Ian Taylor
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Larry Levy wrote:
Franz' formulas are correct, but there's a simpler and more intuitive answer (although the math to prove it is a little complex). The average number of rolls it takes to achieve a result that has a probability of X is 1/X. So, for example, the odds of rolling a 6 on a single D6 roll is 1/6, so the average number of rolls it would take before a 6 appears is the inverse of that, or 6 rolls.


This is technically correct, but gives the mean not the median, which is skewed by some freak occasions when it might take, say 20+ rolls to get a six. It will take less than 6 attempts far more often than it will take more.
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( ͡° ͜ʖ ͡°)
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Hi All, thank you very much for the replies. I think I am on the right track, but have a lot to read and try to take in. I will post another message on here in the next day or so to further explain where I am at and determine if I am thinking of it correctly!

 
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Herb
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In probability it is all about the wording of the problem. Think about the difference in the following questions.

* What is the probability of getting at least two sixes in five throws of a die?

* What is the problem of getting exactly two sixes in five throws of a die?

* What is the probability of getting a second six on the fifth throw of a die?

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Brandon
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The negative binomial distribution is correct for determining how many rolls it will take to have rolled some number so-many times. As for the probability of combinations of numbers, the trick is an easy language-to-math conversion: when you say "and", you mean "*" and when you say "or", you mean "+".

So, in one roll, what is the probability of rolling a 3 or a 6? The probability of rolling a 5 is 1/6, as is the probability of rolling a 6. Therefore, the probability of rolling a 5 or a 6 is 1/6 + 1/6 = 2/6 = 1/3 = 0.333

Given two dice, what is the probability of rolling a 6 and a 6? Again, the individual probabilities are 1/6. Thus the probability of rolling both with two dice is 1/6 * 1/6 = 1/36 = 0.0278

What about a 3 and a 6? Then you can roll either (3 and 6) or (6 and 3), so the probability is (1/6 * 1/6) + (1/6 * 1/6) = 2/36 = 1/18 = 0.0556

You can get more complicated of course. Given two dice, what is the probability of rolling a total of 4? Any of the following permutations will work: 1-3, 3-1, or 2-2. So, the probability of rolling the permutation (1 and 3) or (3 and 1) or (2 and 2) is (1/6*1/6) + (1/6*1/6) + (1/6*1/6) = 3/36 = 1/12 = 0.0833

As for sequential rolls, each roll is independent, so the probabilities can be calculated without considering past rolls. The chance of rolling a 6 after having just rolled a 6 is still 1/6. In fact, you can think of sequential rolls in the same way as rolling a bunch of dice at the same time; in the above examples, the dice did not interfere with each other so the probability of rolling any given number on that die was always 1/6.

If you want to know the probability of rolling a specific sequence, it's just like knowing the probability of rolling a specific permutation with multiple dice, so you can use the same trick. What's the probability of rolling four 6's in a row (a 6 and then a 6 and then a 6 and then a 6)? 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 = 0.000772. What's the probability of rolling a 6 and then a 5 or a 6? 1/6 * (1/6 + 1/6) = 2/36 = 1/18 = 0.0556.

With two dice, what's the probability of rolling first a sum of 4 and then a sum of 6 (1-5, 5-1, 2-4, 4-2, or 3-3)? ((1/6*1/6)+(1/6*1/6)+(1/6*1/6)) * ((1/6*1/6)+(1/6*1/6)+(1/6*1/6)+(1/6*1/6)+(1/6*1/6)) = (3/36)*(5/36) = 15/1296 = 5/432 = 0.0116.
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Ian Taylor
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jakobcreutzfeldt wrote:

Given two dice, what is the probability of rolling a 3 and a 6? Again, the individual probabilities are 1/6. Thus the probability of rolling both with two dice is 1/6 * 1/6 = 1/36 = 0.0278


No, it's 1/18, seeing as they can be rolled in either order.
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Brandon
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piemasteruk wrote:
jakobcreutzfeldt wrote:

Given two dice, what is the probability of rolling a 3 and a 6? Again, the individual probabilities are 1/6. Thus the probability of rolling both with two dice is 1/6 * 1/6 = 1/36 = 0.0278


No, it's 1/18, seeing as they can be rolled in either order.


Oops, you're right (order is accounted for in the other examples; I just missed it there). Thank's for catching it! I fixed it.
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Trent Boardgamer
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Yer, permutations not combinations.

I'm still not quite sure what the OP is trying to work out though???

Me rolling a 6 this roll has absolutely no impact on what I will roll next.

Is he wanting probabilities of rolls (including sequential or multiple dice together) or wanting to know wether he should bother trying to roll another six, since he just rolled one???
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Ian Taylor
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Bearhug78 wrote:
Yer, permutations not combinations.

I'm still not quite sure what the OP is trying to work out though???


I don't think any of us are. We're kind of hoping he comes back and lets us know at some point.
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Brandon
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Bearhug78 wrote:
Yer, permutations not combinations.


Urgh. You're right. Fixed.
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Ok, I think I am over complicating things by trying to figure it all out at once. Let me start off with the first step and work from there.

Let's say we have a war game and you have two units that are virtually identical, except one unit can hit with his gun on a 4+ roll on 1d6 and the other can hit with his gun on a 3+ roll.

Obviously the one who can hit with the 3+ is better, thus he would cost more points if trying to figure out a point value for these two units.

If the first unit (hitting on a 4+) costs 10 points in the game. How many points would the second unit (hitting on a 3+) need to be in order to be equal so far as the probability of hitting each other.

This is the first part of the problem I am trying to figure out and I am just not grasping the math behind it.

Would 10 guys hitting at a 4+ be an equal fight between 7 guys hitting at a 3+, etc.

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Scott Hill
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I might be wrong on this, but a 3+ chance of missing is 33.333% or 1:3, whereas, at 4+ the chance of missing is 50% or 1:2.

Therefor, I think the points costs for the two units should be in the ratio 3:2.

I.e. if the 3+ unit costs 12pts, then 4+ unit should cost 8pts.
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Scorpion0x17 wrote:
I might be wrong on this, but a 3+ chance of missing is 33.333% or 1:3, whereas, at 4+ the chance of missing is 50% or 1:2.

Therefor, I think the points costs for the two units should be in the ratio 3:2.

I.e. if the 3+ unit costs 12pts, then 4+ unit should cost 8pts.


Hi Scott, Thanks for the reply. That is kind of about where I got and started thinking I was wrong.

Let's use those points you just mentioned. Then two of the 12 point units should be equal to three of the 8 point units.

But that is where I was thinking "What are the odds of rolling a 4+ three times or a 3+ twice?"

For instance, if those two 12 point units were in combat with the three 8 point units, is there is a equal chance that each side would win the combat?
 
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Scott Hill
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Metal Slayer wrote:
Scorpion0x17 wrote:
I might be wrong on this, but a 3+ chance of missing is 33.333% or 1:3, whereas, at 4+ the chance of missing is 50% or 1:2.

Therefor, I think the points costs for the two units should be in the ratio 3:2.

I.e. if the 3+ unit costs 12pts, then 4+ unit should cost 8pts.


Hi Scott, Thanks for the reply. That is kind of about where I got and started thinking I was wrong.

Let's use those points you just mentioned. Then two of the 12 point units should be equal to three of the 8 point units.

But that is where I was thinking "What are the odds of rolling a 4+ three times or a 3+ twice?"

For instance, if those two 12 point units were in combat with the three 8 point units, is there is a equal chance that each side would win the combat?


What do you mean by "win the combat"?
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( ͡° ͜ʖ ͡°)
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Player A has 3 units that kill the enemy on a d6 roll of 4+
These units are 8 points each for a total of 24 points.

Player B has 2 units that kill the enemy on a d6 roll of 3+
These units are 12 points each for a total of 24 points.

Example: These units enter combat together, each player rolls their dice and scores of 4+ for player A or 3+ for player B kill the enemy units.

But are these two units really equal, or does an advantage lie with the player having more dice rolls? If you up the stakes to a 240 "point" game, where Player A now has 30 units rolling 4+ for a kill and player B now was 20 units rolling 3+ for a kill.

Do the odds lie in favor of Player A since he is rolling 10 more dice?

Using the above mentioned formula, these two units would be 'equal' in the probability of the dice rolls, but it seems to me that the person with the more units with a less chance of hitting has an advantage.

How do you figure out the actual math behind this to determine?
 
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Scott Hill
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So, each roll of 4+ that Player A makes kills one of Player B's units?
And, each roll of 3+ that Player B makes kills one of Player A's units?

Is that correct?

(it may seem like I'm asking obvious questions, but how you work out combat affects the math)
 
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Justin R
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Metal Slayer wrote:
Player A has 3 units that kill the enemy on a d6 roll of 4+
These units are 8 points each for a total of 24 points.

Player B has 2 units that kill the enemy on a d6 roll of 3+
These units are 12 points each for a total of 24 points.

Example: These units enter combat together, each player rolls their dice and scores of 4+ for player A or 3+ for player B kill the enemy units.

But are these two units really equal, or does an advantage lie with the player having more dice rolls? If you up the stakes to a 240 "point" game, where Player A now has 30 units rolling 4+ for a kill and player B now was 20 units rolling 3+ for a kill.

Do the odds lie in favor of Player A since he is rolling 10 more dice?

Using the above mentioned formula, these two units would be 'equal' in the probability of the dice rolls, but it seems to me that the person with the more units with a less chance of hitting has an advantage.

How do you figure out the actual math behind this to determine?


Still totally unclear what your question is. Does every unit have to roll the 4+ (or 3+, as applicable)? Or does merely 1 unit have to roll it? Are all the units given a single roll, or some subset? Is the number of units merely the number of rolls you have (say, serially...which yields no different odds than a simultaneous group roll), and some number strictly less than the number of rolls must yield the "hit" criterion to win the combat? If you can explain exactly how this combat works, I can answer your question (including whether the data point you're looking for is probability or expectation value). You can feel free to PM me if you like so that I can ask follow-ups as needed.
 
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