[DL] Heavy Mathematical Answer to Dirk's Question - Long, too.

[J Tolle <jwtolle@altavista.com>]

On Tue, 25 July 2000, Dirk wrote:
> J Tolle wrote:
> 
> > and it uses limits and series to calculate 
> > probabilities.  Just ask me sometime!)
> > Jeff "Thanks for indulging me" Tolle
> 
> Ok... I'll ask.
> 
> --Dirk
------------------------------------------------

Short Answer:

average roll on 1d4 (open ended) =

the limit (as n goes to inifinty) of the sum (for all n from 1 to infinity) of [6(2n-1)(1/4^n)] =

well if i told you, would you read on?

Long Answer:

Just remember - you asked.  Also mathematicians, please forgive the sloppy notation.  I know, I know, I should be shot, but I don't know how to write sums notationally in HTML - Sorry.

Ok, remember the old paradox about the frog trying to cross the room.  Maybe not, but it goes like this: Mr Frogger had to travel half the distance to the wall with each jump.  The question is, will he ever get there.  The answer is YES.  Easy right.  Well easy to intuit - VERY difficult to prove.  Proving the answer involves Newton, but it highlights some interesting points on open ended dice rolls.

The question above, if it were put into mathematical terms, would read:

does 1/2 (Mr Frogger's first jump) + 1/4 (jump # 2) + 1/8 + ...... + 1/2^n ever approach a number (ever get to the other wall), or does it continue to grow and grow with each sum?  This is know in mathematics as a series.

or a more mathematical (and shorter) way of saying this is:

does the Sum all n from 1 to infinity of 1/2^n approach a number?

Just like the frog jumping across the room the answer is YES, the frog can reach the wall.  The sum does approach a number.  Intuitively, this is obvious, but mathematically it is not easily proved.  Mainly because who has the time too add up all the numbers untill infinity.  More importantly, we may know there is a number out there, but what is it, and how do we calculate it?

Well Newton had the answer.  With a little twist called limits he was able to mathematically codify the intuition.  It was now possible to tell if a series did approach an end result, and better still, what that end result may be.

So the problem above can be reduced to:

the limit (as n goes to inifinty) of the sum (for all n from 1 to infinity) of 1/2^n = 1

It really does.  I will NOT got into the limit proofs of calculus here.  Run out the first 10 numbers on your calculator just to see tho.

"So what" You may cry?  Well in Deadlands you have an open ended roll.  It turns an ordinary die roll into a series, not unlike the one shown above.  Allow me to demonstarte:

If I wanted to see the average 'result' of rolling a 1d4 (not open ended) the procedure is as follows.  Take the probability of each outcome, and multiply it by the 'value' (ie the number shown on the die) of each outcome.  In this example, each value has an equal chance of happening so the Expected result E:

E = 1(1/4) + 2(1/4) + 3(1/4) + 4(1/4)
  = 1/4 + 2/4 + 3/4 + 4/4
  = 10/4 or 2.5

As result many gamers are intimately familiar with.

But if we add the open ended twist, things change just a little.  Following the same formula we get:

E = [1(1/4) + 2(1/4) + 3(1/4)] + 
    [(4+1)(1/4^2) + (4+2)(1/4^2) + (4+3)(1/4^2)]
    + ...... + 
    [(4(n-1)+1)(1/4^n) + (4(n-1)+2)(1/4^n) + (4(n-1)+3)(1/4^n)]

  = [6(1/4)] + [18(1/16)] + .. + [6(2n-1)(1/4^n)]
  = 6/4 + 18/16 + .. + [6(2n-1)(1/4^n)]


Go back and look at it again.  I divided each die roll off with a [ ] for clarity.  Also note that there never is a result that gives you a multiple of '4' as a result.  Thats cause if you roll a '4' you immediately pick up the die to roll again.  I mean, you just got an ace didn't ya?

It should look familiar in it's final form.  Lemme write it this way for ya:

Sum (for all n from 1 to infinity) of [6(2n-1)(1/4^n)]

See it now!  Right - it's a limit problem.  Put it like this:

average roll on 1d4 (open ended) =

the limit (as n goes to inifinty) of the sum (for all n from 1 to infinity) of [6(2n-1)(1/4^n)]

So open ended rolls involve limits, and series to calculate probabilities.

Oh yeah the 1d4 sequence does converge (come to a final number), and the result is 3 1/3.

Surprised?

I was when I saw how much an effect making a die open ended had on the average roll.  (from 2 1/2 to 3 1/3)

(BTW, all this is assuming a result of bust is numberically expressed as 1)

Jeff "Do NOT ask me about average rolls on pick the best of 4d12 with busts figgured in.  I'm just not there yet.  If you really want to know, I can send you the discertation when I get it done, but it is *thorny* at best." Tolle


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