View Full Version : Help me calculate the averages, please

The Troubadour

2011-12-31, 06:10 PM

Hey there, guys! I was wondering if anyone mathematically inclined would mind helping me calculate the averages? :-)

Here's how the system would work: the player would roll a number of d10s and keep only a single one, the highest. If any d10 comes up with a 10, that die is re-rolled and the new value is added to the 10.

As an example: let's say I roll 4d10 and get 2, 3, 7 and 10. I re-roll the 10 and get a 4. So the four results I got on a 4d10 roll were 2, 3, 7 and 14, and thus, my total result for the 4d10 check was 14.

So, with all that in mind, what would be the averages for the following checks:

1d10

2d10

3d10

4d10

5d10

6d10

7d10

A many thanks for any help provided! :-D

Ravens_cry

2011-12-31, 06:20 PM

Add the smallest possible number you can roll with a particular set of dice to the largest and divide by two, averaging them.

That's my system anyway.

DonDuckie

2011-12-31, 06:25 PM

I will do this :D I am not going to be doing the math, because it would involve integrals with the posibility of rolling infinite 10's... which is no fun - I just failed two statistics courses :(

however, i will write a simulation of this and run it many times, and it will give a good estimate.

to be continued...

It seems like a cool system. Similar in a number of ways to WoD. May be a bit slow, but if most checks and combat checks only need one such roll it shouldn't work too slowly. As far as averages, I could help if we didn't factor in the exploding die mechanic. That makes averages hard.

Ravens_cry

2011-12-31, 06:35 PM

Ooh, didn't see the exploding die part. You're right, that does make averages more difficult.

AlterForm

2011-12-31, 06:38 PM

Hmmm. Each face of the d10 contributes 1/10 its value to the expected value of the die roll in total. The 10 face has an effective face value of 10 + the expected value of the roll. Symbolically:

.1 * (1 + 2 + ... + (10 + X) ) = X

Solving:

5.5 + .1X = X (Distribute)

5.5 = .9X (Subtract)

X = 6.111 (Divide)

Any multiple of these dice rolls is simply a multiple of this value.

[EDIT]: Eugh, roll and keep the highest? That's gonna take longer than I expected ... the Monte Carlo Simulation suggested upthread is probably the best way to handle that unless someone stumbles on a generalized EV for "roll and take the highest." I expect the impact of additional dice to quickly become negligible, however.

DonDuckie

2011-12-31, 06:44 PM

1d10s avg: 6.10286

2d10s avg: 12.2348

3d10s avg: 18.3018

4d10s avg: 24.4446

5d10s avg: 30.5327

6d10s avg: 36.6289

7d10s avg: 42.7469

8d10s avg: 48.8544

9d10s avg: 54.9971

10d10s avg: 61.1585

with 100,000 rolls for each(it's enough)

forget that, missed the "keep the highest" as well... brb

Vitruviansquid

2011-12-31, 06:45 PM

Are you accounting for the fact that you're not adding the dice, but choosing the highest one out of each pool?

DonDuckie

2011-12-31, 06:50 PM

1d10s avg: 6.11176

2d10s avg: 8.33768

3d10s avg: 9.69978

4d10s avg: 10.6871

5d10s avg: 11.4927

6d10s avg: 12.1573

7d10s avg: 12.7436

8d10s avg: 13.2767

9d10s avg: 13.7596

10d10s avg: 14.1765

with 1,000,000 simulations each :D happy new year

Actually, if you don't want exact numbers but want a good estimate I have a WoD die roller which automatically rolls an extra die for every ten result on my computer. I could send it to you and you could run each one a couple hundred times. That should work.

Nevermind- already done for you.

Howler Dagger

2011-12-31, 11:20 PM

Any Dice (http://anydice.com/) may help you on this quest.

Perhaps this will help (http://anydice.com/program/d1c).

(Give it a bit of time to calculate. Also, the graph is easier to read than the default.)

DonDuckie

2012-01-01, 06:25 AM

Perhaps this will help (http://anydice.com/program/d1c).

(Give it a bit of time to calculate. Also, the graph is easier to read than the default.)

That's pretty cool, it took a while for me to understand the output, and it doesn't explode more than twice (which is a bit weird)... but still cool.

EDIT: the point

but I am a computer scientist - I re-invent the spoon before i have breakfast :D

The Troubadour

2012-01-01, 09:46 AM

First of all, thanks a lot for all the help, guys! :-) Second, a Happy New Year to all! Hope you had a great night. :-D

1d10s avg: 6.11176

2d10s avg: 8.33768

3d10s avg: 9.69978

4d10s avg: 10.6871

5d10s avg: 11.4927

6d10s avg: 12.1573

7d10s avg: 12.7436

Huh. Seems like, after a certain point, adding dice doesn't really do all that much on a mathematical level, eh?

mucco

2012-01-01, 11:25 AM

Yeah, because no matter how many dice you add only one can be then rerolled. If the OP had asked for a reroll of all max dice, things would have gone up slightly faster.

Chess435

2012-01-01, 02:03 PM

"The GitP friendly advice dispensers: Giving helpful advice on character builds, optimization, class balance, advanced probability and statistics, quantum physics, and Cantorian Set Theory since 2004." :smallwink:

The Troubadour

2012-01-01, 02:33 PM

Yeah, because no matter how many dice you add only one can be then rerolled. If the OP had asked for a reroll of all max dice, things would have gone up slightly faster.

...But all dice that read "10" can be re-rolled, not just a single one. :-) You just have to keep the highest of them.

For instance, if I rolled two 10's, and then re-rolled a 5 on the first one and a 3 on the second one, I'd have two results - 15 and 13 -, and would keep the 15.

I imagine what actually limits so much the benefits of additional dice after a certain point is that you can only keep one of them.

Again, thanks a lot for the help, guys! :-D

Lapak

2012-01-03, 12:10 PM

Each die can only explode once though, correct? Because if your maximum possible value is theoretically infinite it complicates things a bit. :smallyuk:

Binks

2012-01-03, 02:16 PM

Each die can only explode once though, correct? Because if your maximum possible value is theoretically infinite it complicates things a bit. :smallyuk:

Amusingly enough, no. For the result to be usable it has to be finite, as otherwise you just keep re-rolling the die forever and never get a result. So you don't have to consider infinity as a possible value for calculating the average, as it will never be the result of any finite series of rolls, and we don't care about the result of an infinite series of rolls for this scenario.

Allowing multiple explosions of dice won't do much to the end result as the probability increases by a factor of 1/10 for each +10 to the result, ie the odds of getting, say, a 45 are 10 times lower than the odds of getting a 35, so there's far lower impact on the average for the higher numbers. Not saying it wouldn't skew things, but it's not going to send the average that much higher than a single explosion per die. Infinite series can have finite results after all.

I ran the math for my own amusement a while back (was going to post it, but it was for rolling X exploding d10's and summing the result, not taking the best) and if you allow an arbitrary d10 to explode any number of times it only averages 6.1, a whole .6 over a normal d10. It's really not that big of an impact on the average (though the standard deviation and min/max go WAY out of whack).

meschlum

2012-01-06, 03:03 AM

Time to bring out the math!

When rolling n dice, each a k-sider, and keeping the highest, the expected result is:

k - (Sum (i = 1, k-1) (i^n)) / k^n

Examples:

1d9: 9 - (1+2+3+4+5+6+7+8)/9 = 9 - 4 = 5. Correct!

2d2: 2 - (1)/4 = 3.75 Correct! (1/4 roll two ones. Otherwise, at least one two)

2d9: 9 - (1+4+9+16+25+36+49+64)/81 = 9 - 68/27 = 6.48148148... (175/27)

Call this number M(n,k).

So now you know the expected value when taking the highest among n dice. But you're also rerolling!

If you don't roll any tens, you've got the maximum among dice that go from 1 to 9, so n = 9 and done.

If you roll any tens, the first time through, count how many dice you're rerolling, add ten, and you're set! Repeat as needed.

With one die:

9/10: M(1,9) (since the die roll is 9 or less)

1/10: reroll, so 9/100 of 10 + M(1,9), 1/100 of another reroll. Repeat

You get Value(1) = 9/10 M(1,9) + 1/10 (10 + Value(1)), so

Value(1) = 9/10 M(1,9) + 1 + 1/10 Value(1)

9 * Value(1) = 9 M(1,9) + 10

Value(1) = M(1,9) + 10 / 9 = 5 + 1.111111 = 6.1111 Success!

With two dice:

81 / 100: M(2,9) (both dice are 9 or less)

18 / 100: 10 + Value(1) (one die is 10, add the expected value for 1 die on top)

1 / 100: 10 + Value(2) (both dice are 10, so repeat)

Value(2) = 81 M(2,9) / 100 + 18 * 16.11111 / 100 + (10 + Value(2)) / 100

99 * Value(2) = 81 M(2,9) + 300

Value(2) = 9 M(2,9) / 11 + 300 / 99 = 8.3333

With three dice:

729 / 1000: M(3,9) (all three dice at 9 or less)

243 / 1000: 10 + Value(1) (one die is 10)

27 / 1000: 10 + Value(2) (two dice are 10)

1 / 1000: 10 + Value(3) (all dice are 10)

Value(3) = 9.694694...

And so on. This is probably useless, but you now have the tools to determine the exact average for any number of dice.

DonDuckie

2012-01-06, 06:02 AM

Time to bring out the math!

When rolling n dice, each a k-sider, and keeping the highest, the expected result is:

k - (Sum (1, k-1) (k^n)) / k^n

I just need to make sure i understand this function.

The underlined k in the sum is not a constant but the variable you iterate over from 1 to k-1

meschlum

2012-01-06, 10:17 PM

Yes, it is. My bad. Off to edit.

There probably are equations for M(x,y) out there besides the obvious, but I'm somewhat lazy at this point. Still...

M(1,y) = y - (y-1) / 2 = (y+1) / 2

Test: M(1,9) = 10/2 = 5. Correct!

M(2,y) = y - (y-1)(2y-1) / 6y = (4y^2 + 3y - 1) / 6y = (y+1)(4y-1) / 6y

Test: M(2,9) = (324 + 27 - 1) / 54 = 175 / 27. Correct!

M(3,y) = y - (y-1)^2 / 4y = (y+1)(3y-1)/4y

Hmmm...

As a hypothesis, it looks like it could be worth testing

M(x,y) = (y+1)(2xy + 1 - x)/2(x+1)y

If it's valid, you've got an exact (and relatively easy to assess) equation for any Value you're after.

The things I do to relax.

The Troubadour

2012-01-07, 08:59 AM

...

My mind! It is broken!

DonDuckie

2012-01-07, 09:54 AM

As a hypothesis, it looks like it could be worth testing

M(x,y) = (y+1)(2xy + 1 - x)/2(x+1)y

If it's valid, you've got an exact (and relatively easy to assess) equation for any Value you're after.

The things I do to relax.

It's not exact. I did some testing of the two functions. it has some problems with small dice. It's done with 200,000 simulations each. Everything using double floating point precision.

I spoillered the output:

M1: M(n,k) = k - (Sum (i = 1 to k-1) (i^n)) / k^n

M2: M(n,k) = (k+1)(2nk + 1 - n)/2(n+1)k

----------------------------------------

# || Sim | M1: | M2

1d2| 1.49965| 1.5| 1.5

2d2| 1.75042| 1.75| 1.75

3d2| 1.87466| 1.875| 1.875

4d2| 1.9375| 1.9375| 1.95

5d2| 1.96871| 1.96875| 2

6d2| 1.98443| 1.98438| 2.03571

7d2| 1.99236| 1.99219| 2.0625

8d2| 1.99645| 1.99609| 2.08333

9d2| 1.99808| 1.99805| 2.1

10d2| 1.99904| 1.99902| 2.11364

----------------------------------------

# || Sim | M1: | M2

1d3| 1.99943| 2| 2

2d3| 2.44724| 2.44444| 2.44444

3d3| 2.66885| 2.66667| 2.66667

4d3| 2.79047| 2.79012| 2.8

5d3| 2.86455| 2.8642| 2.88889

6d3| 2.91127| 2.91084| 2.95238

7d3| 2.94183| 2.94102| 3

8d3| 2.96028| 2.96083| 3.03704

9d3| 2.9735| 2.97394| 3.06667

10d3| 2.98252| 2.98264| 3.09091

----------------------------------------

# || Sim | M1: | M2

1d4| 2.49831| 2.5| 2.5

2d4| 3.12536| 3.125| 3.125

3d4| 3.43741| 3.4375| 3.4375

4d4| 3.61693| 3.61719| 3.625

5d4| 3.73029| 3.73047| 3.75

6d4| 3.80633| 3.80615| 3.83929

7d4| 3.85887| 3.85864| 3.90625

8d4| 3.8959| 3.89597| 3.95833

9d4| 3.92328| 3.92296| 4

10d4| 3.94289| 3.94271| 4.03409

----------------------------------------

# || Sim | M1: | M2

1d6| 3.50341| 3.5| 3.5

2d6| 4.46948| 4.47222| 4.47222

3d6| 4.95638| 4.95833| 4.95833

4d6| 5.24592| 5.2446| 5.25

5d6| 5.43166| 5.43094| 5.44444

6d6| 5.56154| 5.56029| 5.58333

7d6| 5.65495| 5.65412| 5.6875

8d6| 5.72326| 5.72435| 5.76852

9d6| 5.77961| 5.77818| 5.83333

10d6| 5.82047| 5.82016| 5.88636

----------------------------------------

# || Sim | M1: | M2

1d8| 4.49934| 4.5| 4.5

2d8| 5.80682| 5.8125| 5.8125

3d8| 6.47154| 6.46875| 6.46875

4d8| 6.85651| 6.8584| 6.8625

5d8| 7.11517| 7.11475| 7.125

6d8| 7.295| 7.29497| 7.3125

7d8| 7.42614| 7.42765| 7.45312

8d8| 7.5283| 7.52868| 7.5625

9d8| 7.60698| 7.6076| 7.65

10d8| 7.67141| 7.67048| 7.72159

----------------------------------------

# || Sim | M1: | M2

1d10| 5.50302| 5.5| 5.5

2d10| 7.1573| 7.15| 7.15

3d10| 7.97727| 7.975| 7.975

4d10| 8.46575| 8.4667| 8.47

5d10| 8.78603| 8.79175| 8.8

6d10| 9.02407| 9.0216| 9.03571

7d10| 9.1885| 9.19196| 9.2125

8d10| 9.32325| 9.32269| 9.35

9d10| 9.42508| 9.4257| 9.46

10d10| 9.50927| 9.50857| 9.55

----------------------------------------

# || Sim | M1: | M2

1d12| 6.50219| 6.5| 6.5

2d12| 8.48101| 8.48611| 8.48611

3d12| 9.47781| 9.47917| 9.47917

4d12| 10.077| 10.0722| 10.075

5d12| 10.4626| 10.4653| 10.4722

6d12| 10.7434| 10.7441| 10.756

7d12| 10.9552| 10.9516| 10.9688

8d12| 11.1129| 11.1114| 11.1343

9d12| 11.2418| 11.2379| 11.2667

10d12| 11.339| 11.3402| 11.375

----------------------------------------

# || Sim | M1: | M2

1d20| 10.5169| 10.5| 10.5

2d20| 13.8292| 13.825| 13.825

3d20| 15.4948| 15.4875| 15.4875

4d20| 16.4755| 16.4833| 16.485

5d20| 17.1461| 17.1458| 17.15

6d20| 17.6104| 17.6179| 17.625

7d20| 17.9629| 17.9709| 17.9812

8d20| 18.2425| 18.2445| 18.2583

9d20| 18.4603| 18.4626| 18.48

10d20| 18.6388| 18.6403| 18.6614

----------------------------------------

Razanir

2012-01-07, 11:50 AM

i'm still working on the "best of n" part, but I can easily help with the exploding part:

Ignoring the exploding, we have E=.1*1+.1*2+.1*3 ... +.1*10=5.5

Because it explodes, we roll another of these dice for 10, so we add on .1*(.1*1+.1*2+.1*3 ...)

This becomes an infinite geometric series, a_1=5.5, a_n = .1*a_(n-1) or in other terms, a = 5.5, r = .1

The sum is a/(1-r) = 5.5/(1-.1) = 5.5*10/9 = 6.1111...

Generalizing with n sides instead of 10, we get n(n+1)/2(n-1)

So long story short, I can't help that much with the "best of" part, but for an exploding die with n sides, the expected value is n(n+1)/2(n-1)

Rorrik

2012-01-07, 07:30 PM

Ha, and I was going to do the probability by hand because I don't have a compiler on this computer. Good work, men!

Razanir

2012-01-07, 11:13 PM

Ha, and I was going to do the probability by hand because I don't have a compiler on this computer. Good work, men!

Compiler... that gives me in idea!

One java program later...

I have 95% confidence intervals for the average of 1d10 - 5d10, based of 50 data points each and a Monte Carlo approach with 1,000,000 trials each :smallbiggrin:

1d10- Min: 6.102775 Mean: 6.111526 Max: 6.120277

For comparison, the true value was previously determined to be 6.1111...

The program was right

2d10- Min: 8.323373 Mean: 8.331783 Max: 8.340192

3d10- Min: 9.686582 Mean: 9.693847 Max: 9.701111

4d10- Min: 10.680596 Mean: 10.689970 Max: 10.699343

5d10- Min: 11.478561 Mean: 11.488551 Max: 11.498542

...But all dice that read "10" can be re-rolled, not just a single one. :-) You just have to keep the highest of them.

For instance, if I rolled two 10's, and then re-rolled a 5 on the first one and a 3 on the second one, I'd have two results - 15 and 13 -, and would keep the 15.

I imagine what actually limits so much the benefits of additional dice after a certain point is that you can only keep one of them.

Again, thanks a lot for the help, guys! :-D

Did you mention what happens when someone rerolls a 10?

Do they get 20?

or 10+10+next roll?

Is there a limit?

What you've got more or less is

the sum of 1-9 plus the average score if you roll a 10 divided by 10 per dice.

I have no idea how to calculate that as say an 87 requires 8 10's to be rolled which is rare so each consecutive 10 will reduce the amount by which that roll effects your average, except that higher values increase your average also.

Excession

2012-01-08, 10:03 PM

Going further than averages I wrote some Python (http://python.org/) code:

from __future__ import division

import random

def one_die():

score = random.randint(1, 10)

if score == 10:

score += random.randint(1, 10)

return score

def n_dice(n):

return max(one_die() for i in xrange(n))

def simulate(n, iterations=100000):

counts = [0] * 20

for i in xrange(iterations):

score = n_dice(n)

counts[score - 1] += 1

chances = [count / iterations for count in counts]

return chances

def main():

for n in xrange(1, 10):

print '%dd10' % n

chances = simulate(n)

for i, chance in enumerate(chances):

score = i + 1

print ' %2d %6.3f%%' % (score, chance * 100)

main()

Unless I screwed something up, this uses simulation to determine the chance of each score between 1 and 20 for a number of d10s ranging from 1 to 10. It turns out 9s are pretty popular. These numbers are over 100,000 iterations:

1d10

1 10.056%

2 9.889%

3 10.058%

4 10.002%

5 10.074%

6 10.049%

7 9.968%

8 10.026%

9 10.044%

10 0.000%

11 0.957%

12 0.955%

13 1.001%

14 1.009%

15 0.954%

16 0.973%

17 0.977%

18 1.015%

19 0.985%

20 1.008%

2d10

1 0.983%

2 3.047%

3 5.009%

4 7.117%

5 8.970%

6 10.901%

7 13.123%

8 14.988%

9 16.896%

10 0.000%

11 1.894%

12 1.856%

13 1.814%

14 1.907%

15 1.917%

16 1.806%

17 1.944%

18 1.967%

19 1.988%

20 1.873%

3d10

1 0.108%

2 0.678%

3 1.984%

4 3.633%

5 6.086%

6 9.258%

7 12.654%

8 16.876%

9 21.544%

10 0.000%

11 2.490%

12 2.511%

13 2.540%

14 2.639%

15 2.637%

16 2.794%

17 2.821%

18 2.868%

19 2.905%

20 2.974%

4d10

1 0.008%

2 0.142%

3 0.669%

4 1.789%

5 3.704%

6 6.782%

7 10.953%

8 16.934%

9 24.695%

10 0.000%

11 2.972%

12 3.024%

13 3.239%

14 3.247%

15 3.417%

16 3.503%

17 3.562%

18 3.740%

19 3.772%

20 3.848%

5d10

1 0.000%

2 0.022%

3 0.220%

4 0.815%

5 2.075%

6 4.661%

7 9.058%

8 16.058%

9 26.165%

10 0.000%

11 3.289%

12 3.528%

13 3.729%

14 3.832%

15 3.994%

16 4.228%

17 4.326%

18 4.410%

19 4.667%

20 4.923%

6d10

1 0.000%

2 0.005%

3 0.068%

4 0.336%

5 1.155%

6 3.121%

7 7.038%

8 14.445%

9 27.066%

10 0.000%

11 3.776%

12 3.844%

13 3.933%

14 4.389%

15 4.574%

16 4.718%

17 5.032%

18 5.216%

19 5.584%

20 5.700%

7d10

1 0.000%

2 0.001%

3 0.024%

4 0.151%

5 0.594%

6 2.022%

7 5.612%

8 12.665%

9 26.488%

10 0.000%

11 3.867%

12 4.076%

13 4.396%

14 4.761%

15 5.026%

16 5.326%

17 5.626%

18 6.138%

19 6.382%

20 6.845%

8d10

1 0.000%

2 0.000%

3 0.005%

4 0.062%

5 0.297%

6 1.243%

7 4.150%

8 11.129%

9 26.142%

10 0.000%

11 3.990%

12 4.355%

13 4.668%

14 4.961%

15 5.412%

16 5.809%

17 6.273%

18 6.569%

19 7.254%

20 7.681%

9d10

1 0.000%

2 0.000%

3 0.003%

4 0.028%

5 0.190%

6 0.825%

7 3.086%

8 9.318%

9 25.260%

10 0.000%

11 4.119%

12 4.388%

13 4.871%

14 5.182%

15 5.770%

16 6.123%

17 6.762%

18 7.445%

19 7.880%

20 8.750%

The Troubadour

2012-01-08, 11:44 PM

Did you mention what happens when someone rerolls a 10?

Terribly sorry, it seems I haven't!

I'm actually of two minds about it, but I'm leaning towards letting the player keep on "exploding" the die. Or, as you put it:

Do they get 20?

or 10+10+next roll?

Is there a limit?

The second option, and no, there's not limit.

Gralamin

2012-01-09, 12:39 AM

Assuming infinite dice exploding, and you take the highest of each die after it explodes...

So the case for a single die can easily be done with a recursive function:

The average roll, on a single die, is:

E(x) = sum(1 to 9)/10 + 1/10*(10+E(x))

That is, on a roll of 1 to 9, you just take the result, but on a roll of 10, it will be 10 + Whatever you expect due to having a free roll.

The sum from 1 to n is (n)(n+1)/2, so we can easily change this to:

E(x) = 9*10/(2*10) + 1/10(10+E(x)) = 9/2 + 1 + 1/10E(x)

9/10 E(x) = 11/2

E(x) = 11/2*10/9 = 6.111...

Now that we have the roll per single die, when need to figure out the function to equate having m possible rolls. Since you keep the highest single roll, the result is not independent of the number of dice, and so we cannot use:

mE(x) = E(mx)

Finding statistically dependent averages is hard by hand. This is when a computer comes in. In this case, I made a simple python script:

import random

MAX_DEPTH = 50 # Change me to get more accurate results through deeper recursion

NUM_ITERATIONS = 1000000 # Change me to get more accurate results through more iterations.

MAX_NUM_DICE = 10 # Change me to get more or less results

def rollAD10(depth=0):

if depth >= MAX_DEPTH:

return 0

num = random.randint(1,10)

if num == 10:

return num + rollAD10(depth=depth+1)

return num

def rollND10(n):

results = [rollAD10() for index in range(0, n)]

return max(results)

def getAverage(n):

results = [rollND10(n) for index in range(0, NUM_ITERATIONS)]

return float(sum(results)) / float(len(results))

for index in range(0, MAX_NUM_DICE):

numDice = index + 1

print "%dd10: %f" % (numDice, getAverage(numDice))

raw_input("Press Enter to continue...")

This gives us averages of:

{table="head"]Dice|Average

1|6.11

2|8.33

3|9.69

4|10.68

5|11.48

6|12.17

7|12.75

8|13.28

9|13.75

10|14.19

[/table]

If we do a recursion on this, we find that we can expect our values to match pretty closely to:

3.5152 * ln(n) + 5.9335

Dr.Epic

2012-01-09, 12:43 AM

Hey there, guys! I was wondering if anyone mathematically inclined would mind helping me calculate the averages? :-)

Here's how the system would work: the player would roll a number of d10s and keep only a single one, the highest. If any d10 comes up with a 10, that die is re-rolled and the new value is added to the 10.

As an example: let's say I roll 4d10 and get 2, 3, 7 and 10. I re-roll the 10 and get a 4. So the four results I got on a 4d10 roll were 2, 3, 7 and 14, and thus, my total result for the 4d10 check was 14.

So, with all that in mind, what would be the averages for the following checks:

1d10

2d10

3d10

4d10

5d10

6d10

7d10

A many thanks for any help provided! :-D

I took a class in introductory statistics. And the average for a d10 is 5.5. Just multiple this by the number of dice and there you go.

Gralamin

2012-01-09, 01:47 AM

I took a class in introductory statistics. And the average for a d10 is 5.5. Just multiple this by the number of dice and there you go.

Won't work because:

A) Exploding

and B) Dependant variables

However, now that I think about it, this should resemble the idea of a "Record" (as in a snowfall record).... Which means the overall function should be writeable in the terms of a harmonic series.

Hmm; Suppose that the result of the die roll is the amount of snowfall we get in a year. The first die roll must be a record. The second die roll is a record if the roll is greater then the first year (Symmetry suggests this should be 1/2 probability). The third will be a probability of 1/3rd. This trend continues. So we know the overall record is likely to fall into any of the dice rolled (if m rolls, then 1/m). Now if I can just figure out how this relates to the probability of values, then I could express an exact answer...

Edit: I think the correct answer is going to be very close to:

6.11 * sum(i=1 to n) of 1/(i+(i-1)*k)

Where k is a constant that I do not remember how to calculate, but a quick estimate suggests it should be ~0.650295 to ~0.75. Maybe I will remember tomorrow...

DonDuckie

2012-01-09, 02:03 AM

Just to clarify my results.

I assumed dice keep exploding on rerolled 10s.

and to continue the above:

and C) it's the maximum

The Troubadour

2012-01-09, 06:58 AM

Stop breaking my mind!

Seriously though, thanks for all the help, guys. You're awesome. :-D

Oh, and if anyone's curious what's all this for, it's a little idea I had about playing the old SAGA System without using the cards.

Rorrik

2012-01-09, 09:48 AM

I just noticed that there is no way to roll a ten in this system. Your result will always be more or less than 10, never actually 10.

Tyndmyr

2012-01-09, 12:07 PM

Hey there, guys! I was wondering if anyone mathematically inclined would mind helping me calculate the averages? :-)

Here's how the system would work: the player would roll a number of d10s and keep only a single one, the highest. If any d10 comes up with a 10, that die is re-rolled and the new value is added to the 10.

As an example: let's say I roll 4d10 and get 2, 3, 7 and 10. I re-roll the 10 and get a 4. So the four results I got on a 4d10 roll were 2, 3, 7 and 14, and thus, my total result for the 4d10 check was 14.

So, with all that in mind, what would be the averages for the following checks:

1d10

2d10

3d10

4d10

5d10

6d10

7d10

A many thanks for any help provided! :-D

This sounds like the 7th Sea system(albeit with sometimes less kept dice). There's a huge probability table in the back of the players handbook.

The answer to your first odds question is slightly above six. I'm too lazy to do the math on the others atm.

Gralamin

2012-01-09, 09:58 PM

Stop breaking my mind!

Seriously though, thanks for all the help, guys. You're awesome. :-D

Oh, and if anyone's curious what's all this for, it's a little idea I had about playing the old SAGA System without using the cards.

Neat idea.

And now I remember the correct answer: Order statistics (http://en.wikipedia.org/wiki/Order_statistic) Which are ugly so just use the scripts provided.

I just noticed that there is no way to roll a ten in this system. Your result will always be more or less than 10, never actually 10.

or any multiple of 10.

Rorrik

2012-01-10, 10:06 AM

or any multiple of 10.

Very astute!

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