Computer Dice rollers

[Pinotage]

You're missing that it doesn't generate extremely rare events (which even 1 in 1000 is NOT) statistically more frequently than just rolling the same die 10,000 times would.

Ah, I see. I was thinking of getting specific sequences of numbers and not the more general case, that would apply in this instance, of getting any pair or triplet.

And no, I'm not going to do a statistical analysis of that data set. It just struck me as odd that the pair of 7's came up so frequently.

Pinotage

 

[Arkhandus]

One question though, if the seed of a RNG is the time, wouldn't that make the outcome somewhat more predictable, or at least calculatable, as the time is a calculated, constantly evolving source?

Most likely, time isn't the only thing used to determine the 'random' numbers, and anyway, the computer's probably measuring passage of time in nanoseconds or milliseconds, so the time variable used in the calculation would still be changing rapidly.

But now I have a diffent, most likely very less contraversial question. How does rolling 4d6 and taking the top three increase the chances of rolling better scores than simply rolling three? I know it's most likely an obvious question for most, but I just wonder what the equasion would look like.

This is mentioned in the Hero Builder's Guidebook, one of the old 3.0 thin softcover books by Wizards of the Coast (meant for newbies, but I thought it was handy enough for generating names and character ideas, which it is). It explains the difference in probability from the 3rd Edition standard of 4d6-drop-lowest for ability scores, compared to the old standard of just 3d6.

I could do the math myself, but I hate doing probability calculations and making sure I got the equation set up right. So I'll just tell ya what the book says. In the old method of 3d6 for ability scores, there was a roughly 0.5% chance of rolling a score of 18 (based on a supposed set of 200 rolls of 3d6, only 1 could be expected to be an 18 under the laws of probability). Roughly 3% of characters using that method could be expected to have a single 18 among their ability scores. The average value of each ability score would be 10.5 (in other words, an even mix of 10s and 11s among the scores).

In the 3E method of 4d6-drop-lowest, rolling the extra die and discarding the lowest of the four means you're adding another chance for a good roll into the mix, simply through the law of averages. Even though that extra die still has only a 1-in-6 chance of rolling any particular result (like a 6). I'm no expert in the laws of probability and the law of averages, so I won't bother trying to coherently explain the reasoning, but basically, when you look at a larger number of instances of a particular event, the likelyhood that their results will average out increases, because across a larger sampling, the law of averages has a greater probability of showing through in the outcome.

The 4d6-drop-lowest method produces an average ability score of 11.5, because the extra die roll produces a steeper bell curve of results according to probability.....so the 'middle ground', the peak of the bell curve, is broader and you're more likely to achieve an average result out of that sampling. It's the drop-lowest part that produces this slightly more favorable 'average'; if the method was drop-highest, it would be the reverse, and if it was drop-the-roll-closest-to-3-or-4 or somesuch, then it would probably produce much the same 'average' as the 3d6 method.

With the slightly more favorable set-up of 4d6-drop-lowest, your odds of rolling one score of 18 among your set of six abilities is roughly 9%, rather than 3%. Also because of the slightly-favorable nature of the drop-lowest part, your odds of rolling a really low score are diminished.

 

[palleomortis]

Ok, so anyone that has the great ability of probability then, how much better are my odds of getting an 18 ability if I roll say 5 or 6 die and take the three highest, rather than simply 4 and take the three highest?

edit
(sorry, didn't see the above post when I posted this one.)

 

[Doug Sundseth]

I don't have the numbers for 6d6 keep three best, but here are the probabilities that a particular number will be rolled in a single instance for 4d6 and 5d6:

4d6

3 0.077160%
4 0.308642%
5 0.771605%
6 1.620370%
7 2.932099%
8 4.783951%
9 7.021605%
10 9.413580%
11 11.419753%
12 12.885802%
13 13.271605%
14 12.345679%
15 10.108025%
16 7.253086%
17 4.166667%
18 1.620370%


5d6

3 0.01286%
4 0.06430%
5 0.19290%
6 0.52726%
7 1.15741%
8 2.18621%
9 3.80658%
10 6.04424%
11 8.55195%
12 11.32973%
13 13.56739%
14 14.85340%
15 14.28755%
16 12.02418%
17 7.84465%
18 3.54938%

Mean 13.43016975
Median 14
Mode 14
St.Dev. 2.603635405

 

[Asmor]

[sblock=nm, error]Note: C(n, k) means choose Y elements from X possible elements. Below, the elements are dice, and we're choosing which dice are rolling 6s.

C(n, k)=n!/(k!(n-k)!)

Rolling 3 dice:
Choose three dice to roll six: C(3, 3): 1
Total ways to roll 18: 1
Total possible rolls: 6^3=216
Possibility of 18: 1/216=0.46%

Rolling 4 dice:
Choose three dice to roll six: C(4, 3): 4
Possible combinations of remaining die: 6
Total ways to roll 18: 4*6=24
Total possible rolls: 6^4=1296
Possibility of 18: 24/1296=1.85%

Rolling 5 dice:
Choose three dice to roll six: C(5, 3): 10
Possible combinations of remaining 2 dice: 6^2=36
Total ways to roll 18: 10*36=360
Total possible rolls: 6^5=7776
Possibility of 18: 360/7776=4.63%

Hopefully you can extrapolate this to whatever amount of dice you might want to roll. Note that this is the possibility of rolling 18 on one roll... If you're looking for the possibility by character, just multiply the percentages by 6 (for 6 scores)[/sblock]

Hmm, I screwed up somewhere... Just remembered that there's 21 possible ways to roll an 18 on 4d6, and my reasoning above leads to 24. Anyone with a background in statistics see what I did wrong?

 

[Zelc]

Ok, so anyone that has the great ability of probability then, how much better are my odds of getting an 18 ability if I roll say 5 or 6 die and take the three highest, rather than simply 4 and take the three highest?

For 18, the calculation is much easier since the dice must be 6. For other numbers, you have to factor in the different ways they can add up to X.

We can model it with the cumulative binomial distribution with k=2.5 (2.5=3-.5 to correct for discreteness), n= number of dice rolled, and p=1/6. The probability of rolling at least 3x 6's is 1-binomcdf(2.5, n, 1/6). I did this in Excel real quick and got:

3 0.462963%
4 1.620370%
5 3.549383%
6 6.228567%
7 9.577546%
8 13.484689%
9 17.825959%
10 22.477320%
11 27.322488%
12 32.257381%
13 37.192273%
14 42.052395%
15 46.777513%
16 51.320896%
17 55.647928%
18 59.734569%
19 63.565794%
20 67.134093%

Hmm, my results are different from Asmor's. Let me double-check...

 

[Zelc]

Hmm, I screwed up somewhere... Just remembered that there's 21 possible ways to roll an 18 on 4d6, and my reasoning above leads to 24. Anyone with a background in statistics see what I did wrong?

Ahha! You quadruple-counted the 6, 6, 6, 6 possibility. So subtract (n-1) scenarios for success for each of your calculations. Hmm, correcting that is going to be a bit more complicated. Lemme think.

OK, to correct it in general, you have to subtract the union and add in the unions of the union, etc. Just use the cumulative binomial distribution please .

 

[Zelc]

But now I have a diffent, most likely very less contraversial question. How does rolling 4d6 and taking the top three increase the chances of rolling better scores than simply rolling three? I know it's most likely an obvious question for most, but I just wonder what the equasion would look like.

A helpful way to think through things like this is to go for extremes (also good for stuff like the Monty Hall problem).

Suppose you roll 3d6. Now, all of your dice have to come up 6, so the odds are small that this will happen.

Suppose instead you roll (a billion)d6 take top 3. Obviously, the odds of at least three in a billion dice coming up 6 is extremely large. The more dice you roll, the more likely it is that at least some dice will come up with higher numbers. Since all we care about are the three highest, then the more dice we roll, the more likely it is the top three will be good.

 

[palleomortis]

Thank you O' great math peoples, we revel in your perfect circles of greatness, or somthing. But yea, thanx, I got the stats I was wondering about.

 

[resistor]

The probability for 2 in a row is 0.01, for 3 in a row 0.001 and for 4 in a row 0.0001 or 1 in 10000. You'd expect to probably see 1 or 2 in there, but not as many as observed.

This misconception is a variation on the famous "birthday paradox": how many people do you need to have in a room before there's a greater than 50% chance of at least two of them sharing a birthday?

The answer is easiest to find by calculating the probability of them all having different birthdays, and then subtracting it from one. This can be calculated as:

(365 / 365) * (364 / 365) * (363 / 365) * ... ((365 - n+1) / 365)

You derive this by thinking: the first person can have any birthday at all, the second can have any birthday other than what the first person had, the third can have any other than the first two, etc. If you work out this sequence, you find that at n=23, the probability of them all having different birthdays is less than 50%.

So, in a room of 23 people, there is a 50.7% chance that there are at least two people who share a birthday!

The key to understanding why so few people are required is that the problem places no requirement on which birthday is shared, or on who shares it. If I asked how many people were need in order to have a 50% chance of them sharing a birthday with me, the number would be much higher.

The same principle applies to finding "strings" of repetitions in dice: a "string" is more common than most people think, because most people implicitly think of the probability of getting a string of X's rather than the probability of getting any string at all.