"Maths help.with D6." Topic

My brain is tired and even when working properly is not great with maths problems.

I need some stats for D6 rolling please.

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Assuming that on a D6 a result of 5 or 6 is needed.

I want to calculate the average number of hits when rolling a certain amount of D6 at once.

That is 4D6, 6D6 and 8D6.

I am guessing that with 6D6 the average number of hits will be 2, seems obvious but could be way off?

Does this average change the more times it is rolled?
Or is that means and medians, I never can recall the difference.

Your help on what is no doubt a very simple matter for many is appreciated.

Michael.

If you need a 5 or 6, each die will succeed 1/3 of the time.

The expected number of hits will be 1/3 the number of dice you roll. Yes, 6D6 will give you 2 hits on average.

The particular shape of the curve will change, though. Minimum will always be 0 (chance of that will go down with more dice), and the maximum will be the number of dice rolled. Those are both pretty obvious. But the chance of getting the maximum value goes down as well, because every die has to hit.

andy

Your average chance chance of rolling 5 or 6 on 1 X D6 is 2/6 = 1/3.

Your average chance chance of rolling 5 or 6 on 4 X D6 is 4 X 2/6 = 8/6 = 1 1/3.

Your average chance chance of rolling 5 or 6 on 6 X D6 is 6 X 2/6 = 12/6 = 2.

Yes, the more D6 you roll in one throw, the higher your average chance becomes of rolling at least one 5 or 6. The odds increase in a straight line, addinf 1/3 for each additional dice.

Throwing dice multiple times doesn't change the odds for each throw.

Thanks.

Quick question though:

If say the average hits from XD6 is 1.5, does that equate to rolling XD6 on 2 occasions mean that I will have likely made 3 hits?

100/6 = 16.6% per face.

Assuming independent events X/6 * D where X = number of sides that produce the event and D = the number of dice thrown will get you the results you want.

Dependent events are more complex to calculate. For example, exploding dice have no theoretical maximum. So lets say you have 1D6ex where the result is the face value of the die, unless a 6 is rolled, in which case it is 6 + 1D6ex. If that is the case the only value that has no chance of coming up is exactly 6. You can figure out the chance for "at least 6", "at least 12", "at least some multiple of 6" easily enough. So the probability of at least 18 (3*6) is 1/6 * 1/6 * 1/6 = 1/216.

It is great to ask these questions because if you are trying to figure out the choice between different strategies or doing game design it is important to know which are the better choices.

Yes: "If say the average hits from XD6 is 1.5, does that equate to rolling XD6 on 2 occasions mean that I will have likely made 3 hits?"

In your 6d6 example, 2 is your expected number of hits on average. Not your chance to hit. That is usually more useful for game design because you can gauge how many turns it would take to kill a unit /man or whatever.

Aye, I am not after chance to hit but average number of hits.

4D6 will be 1.3 hits on average
6D6 will be 2 hits
8D6 will be 2.6 hits

Fractions of hits are no good to me, so if we roll 4D6 10 times we get 13 hits
6D6 20 hits
8D6 26 hits

Now we have whole numbers which makes more sense in terms of how many times something is hit in a wargame.

Right I think I can work with that.

Cheers.
Oh to have paid more attention at school and less time mucking about.

Actually fractions are good for you. Since you can get more or less hits in a round it is useful to calculate the fractions kn as well.

Google 'anydice'. Best website for calculating dice.odds of.any kind.

Don't confuse expected value with actual number thrown.

E.g. when rolling a single D6, the average outcome (expected value) is 3.5. But you will never roll that number with a single roll, of course. The 3.5 just says: "When rolling my D6 an infinite number of times, the average will be 3.5"

Does that mean that when rolling 100D6, and take the average, you will get 3.5? No, of course not. However, the more dice you roll, the closer you will get to the average. How fast how close you get to the average with more dice rolls, is expressed by the variance – another stochastic measure.

In your problem, the chance of rolling 5 or 6 on a single D6 is 1/3. If you say you score one hit in that case, and zero hits when rolling 1,2,3,4; the average number of hits with a single die roll is 1/3. Of course, you will always score either 0 or 1 hits. But when rolling an infinite amount of D6, the average will be 1/3 per roll.

When rolling x dice, whether sequentially, in groups, or all in one batch, the average number of hits will be x/3.

Will you always roll that number? No, of course not. You might score fewer hits (you can even score 0 hits with 1000 dice), or you might score more hits. But on average, you will score x/3. The probability with which you will deviate from this average number, again, is the variance.

There's nothing mysterious about having fractions in average values.

But when rolling an infinite amount of D6, the average will be 1/3 per roll.

No, when you roll an infinite number of dice you will score an infinite number of hits, provided you have a finite chance to hit.

Angelbarracks, work with whole numbers if that is good for you, but the fractions can matter – the average number of hits on a roll of 4D6 will be 1.3333….

TMP link

Noodle around on the web under Probabilities of dice. There's a site which gives probabilities of 1 to I think 6 d6.

[Results Rounded]

Expected Sucesses

4 Dice:
0: 20%
1: 40%
2: 30%
3: 10%
4: 1%

6 Dice:
0: 9%
1: 26%
2: 33%
3: 22%
4: 8%
5: 2%
6: 0%

8 Dice:
0: 4%
1: 16%
2: 27%
3: 27%
4: 17%
5: 7%
6: 2%
7: 0%
8: 0%

Site: anydice.com

Formula: output d{0, 0, 0, 0, 1, 1} + d{0, 0, 0, 0, 1, 1}

Many thanks for the anydice link. It's just what I needed.

It's a powerful tool. BTW, sharkbait's math is also thorough and correct.

@religon – thanks for checking my (Excel's) math! For those that didn't check the link, here's the chart for d6 probabilities using Excel:

[URL=http://s715.photobucket.com/user/ddka_warren/media/D6_Probabilities_zps1897c3ed.jpg.html]

[/URL]

advocate wrote:

But when rolling an infinite amount of D6, the average will be 1/3 per roll.

No, when you roll an infinite number of dice you will score an infinite number of hits, provided you have a finite chance to hit.

There may well be an infinite number of hits, but there will still be twice as many misses if the chance of a hit is 1/3.

All the best,

John.

Both chances will be infinity. You will throw an infinite number of hits AND an infinite number of misses. Infinity is not just a big number it is an endless and hence fictional number. That said I know what you meant John, for any large but not infinite number of dice rolled the split will be 2:1 between misses and hits and the nearer to infinity the closer to the exact split the result becomes. Need to stop now all this thinking about infinity is making my brain hurt.

Elenderil wrote:

Infinity is not just a big number it is an endless and hence fictional number.

Your head might hurt less if you thought that infinity isn't a number, it's a limit.

All the best,

John.

Infinity is not just a big number it is an endless and hence fictional number.

I beg to differ.

Ever since Cantor (a.o.) defined countability of sets, infinity is something you use in computations. All infinities are just not equal ;-)

See also NaN in floating point representations ;-)