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Falesh



Joined: Aug 02, 2003

Post   Posted: Feb 13, 2004 - 05:37 Reply with quote Back to top

Hello, I've been trying to make a spreadsheet that works out the odds of making multiple dice rolls. This is easy enough but I'm struggling when it comes to adding re-rolls to the mix.

The way I have tried to work out multiple dice rolls with a re-roll is this:

First work out the number of possibilities, so for 2 rolls needed with a re-roll it would be 6*6*6*6. This is the bit that could well be wrong, but bare with me.

Then I work out the number of dice rolls in that that will make me fail. So say x is the first number I need then y is the second I do this formula:


((x-1)*(x-1)*6*6)+((x-1)*(7-x)*(y-1)*6)+((7-x)*6*(y-1)*(y-1))

The first bit "((x-1)*(x-1)*6*6)" works out the number of possible fails if I fail on the first roll and re-roll, so say I needed a 4 or more on the first dice then I would have 4-1 chances to fail * 4-1 on the re-roll then no matter what else I roll on the other dice it is still a failier so I add 6*6.

The next bit "((x-1)*(7-x)*(y-1)*6)" works out the number of possible fails if I fail the first roll, succeed on the re-roll then fail on the second roll. Since I have already used the re-roll, when I fail the second roll I just *6.

The last bit "((7-x)*6*(y-1)*(y-1))" works out the number of possible fails if I succeed the first roll, then fail the second one and the re-roll on that.

Then I work out the % chance of success by first working out the % worth of each possible outcome, 6*6*6*6 = 1296, 100/1296 = about 0.077, then times the total of "((x-1)*(x-1)*6*6)+((x-1)*(7-x)*(y-1)*6)+((7-x)*6*(y-1)*(y-1))" by the result of 100/1296 which gives the chance of failier then I just change that to give the chance of success by doing 100 - chance of failier.


The finished thing looks like this:

=SUM(100-(((C8-1)*(C8-1)*6*6)+((C8-1)*(7-C8)*(D8-1)*6)+((7-C8)*6*(D8-1)*(D8-1)))*(100/(6*6*6*6)))

Where C8=x and D8=y


Where I feel I may well be wrong is that I am adding up the number of possible permutations as if I was rolling 4 dice, i.e. 6*6*6*6, but that would never happen since when I use the re-roll the dice isn't rolled again.

The reason why I did this, and also feel it could be the right way too, is because I think that not doing so would give more % weight then others. I am having trouble describing this, which is probably why I need help with this.

If anyone can help me with this, or give me a better way of working out odds of multiple dice rolls with a re-roll I would much appreciate it!

John
neverborn



Joined: Aug 02, 2003

Post   Posted: Feb 13, 2004 - 06:14 Reply with quote Back to top

Firstly, to help the people imagine what you are trying to do - write a program that you could project which course of action has the best odds

a pass with your orc thrower or 2 gfi's and a hand off, whack em in and see which odds are better...

NOW, your stuff confuses me, if something has a reroll, multiply it by itself then add it to all the other probabilities in your equation and then press equals for your answer...

Only problem there i suppose is using team rerolls, because it can only apply to one of the not automatic things that happens, but not all... But it would be a good approximation to make the hardest non-auto dice roll the team reroll because thats where you are most likely to use it

so where the p* value is the chance of failure (ie, gfi = 1/6)

p1(3) = passing
p2(3) = catch
p3(6) = gfi
p4(6) = gfi

so (((1/p1) * (1/p1))+((1/p2)+(1/p2))+(1/p3)+(1/p4)) = your answer
Falesh



Joined: Aug 02, 2003

Post   Posted: Feb 13, 2004 - 06:35 Reply with quote Back to top

Thanks for the post.

Quote:
Only problem there i suppose is using team rerolls, because it can only apply to one of the not automatic things that happens, but not all


I forgot to say, this is using a team re-roll, hence the trickiness. I was planning to make this go to more then two dice, which is easy enough with the formula that I use in the opening post since it's just a matter of counting more odds that fail.

I'm not really after an approximate. I started this after a game when I was so tired my brain had melted and I couldn't guess at simple odds, but now that I have started it has got me interested and I would like to find the true answer.

Here is the spread sheet which I forgot to put in the first post:

OpenOffice.org format: http://homepage.ntlworld.com/andrew.baldock/Odds.sxc

MS Excel format: http://homepage.ntlworld.com/andrew.baldock/Odds.xls
neverborn



Joined: Aug 02, 2003

Post   Posted: Feb 13, 2004 - 07:02 Reply with quote Back to top

open office format

you beast!!

i'll have a look and let you know
Rimmer



Joined: Aug 19, 2003

Post   Posted: Feb 13, 2004 - 09:08 Reply with quote Back to top

It looks to as if you are doing it the hard way...

Pass (AG 3)+ Catch (AG3) +GFI +GFI:

Without any rerolls the chance of succes = 1/2*1/2*5/6*5/6 (if the throw is accurate)
And 1/3*1*1/4*1/8*1/3*5/6*5/6 (if the throw is inaccurate(scatter roll included)

With a reroll you double the chance of succes on a single roll.

_________________
As always one is always 100% sure about the truth until one learns that it isn´t the truth. Then one is 100% sure that it isn´t the truth.
BloodRunners



Joined: Aug 02, 2003

Post   Posted: Feb 13, 2004 - 09:48 Reply with quote Back to top

doubling the chance of success is only an approximation. That is fine for quick in game odd figuring. But his curiosity lies in finding the exact answer. Not so much for the impact on his game, but for the sake of conquering a problem
Wol



Joined: Jan 08, 2004

Post   Posted: Feb 13, 2004 - 10:31 Reply with quote Back to top

Let's take a single roll, for instance a GFI :
chance of success of the roll : 5/6
chance of failure : 1/6

Three cases here :
- you succeed in the first roll, great : 5/6
- you miss the first roll but succeeed in the next one : 1/6 to fail the first roll, then 5/6 to

succeed in the next one : 1/6*5/6 = 5/36
- you miss both rolls : 1/6 then 1/6 again : 1/6*1/*6 = 1/36

You want to know the odds of succeeding your GFI with a possible reroll : 5/6 + 5/36 = 35/36 (same

result as 1 minus the odds of failing : 1-1/36 = 35/36).
Doubling the chance of success is a crude approximation. It will more or less work if odds are lower than 50%.


Now, with multiple rolls, you have much more possibilities.
And the team reroll makes it a bit more complicated as you can reroll only one of those rolls.

First, you need to write down the odds of succeeding all the rolls on the first try.
Let's say a pass, a catch and 2 GFIs (all AG3, short pass -> no mod, no skill).

Pass : 4+ => 1/2
Catch (pass successful) : 3+ => 2/3
GFI : 5/6

Failed pass (but no fumble) : 2 or 3 -> 2/6 = 1/3
Odds of having the ball return to the square it was aimed to if the pass has failed : about 5% : 1/20
Catch (failed pass) : 4+ => 1/2

First possibility (everyting succeeds) :
1/2*2/3*5/6*5/6 = 50/216

Then, let's say you fail the pass but reroll it successfully (then you can't fail anything else):
(1/2*1/2)*2/3*5/6*5/6 = 50/432

You fail the pass but it lands well :
(1/3*1/20)*1/2*5/6*5/6= 25/4320

And so on, you have to figure out all the possibilities and find the odd of this one succeeding.
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