(1 – (5/6)^n) * 100

Where n is the number of dice purchased.

ex. 3 dice = (1 - (5/6)3) * 100 = (1 - 125/216) * 100 = 42% (rounded down)

I have written a proof for this, but the scanned image is too large to attach per forum standards (I think the max is around 256Kb, and the file is 1.16Mb). If anyone is interested I can type the proof and post it later, but if not thats ok b/c I am lazy by nature.

]]>Recently I came up with an equation for determining the chance of success when rolling for tech in A&A Revised. It is as follows:

(1 – (5/6)^n) * 100

Where n is the number of dice purchased.ex. 3 dice = (1 - (5/6)3) * 100 = (1 - 125/216) * 100 = 42% (rounded down)

I have written a proof for this, but the scanned image is too large to attach per forum standards (I think the max is around 256Kb, and the file is 1.16Mb). If anyone is interested I can type the proof and post it later, but if not thats ok b/c I am lazy by nature.

even some experts around here like <cmdrjennifer>have incorrect statistical proofs in some of their posts. here’s the proof for this one (nice job <syntaxerror111>). P() is the probability function:

**P( success ) = 1 - P( all fail )**

and

**P( all fail ) = P( trial#1 fails AND trial#2 fails AND … )**

assuming rolls of a die are independent and identically distributed, then:

**P( all fail ) = [ P( single fail ) ]number_of_trials**

assuming a fair die:

**P( single fail ) = 5/6**

let n=number_of_trials, and

**P( success ) = 1 - (5/6)n**

if you prefer probabilities in %, multiply by 100%. i personally do not.

here’s a table:

trials P( all fail ) P( success )

0 1.00 0.00

1 0.83 0.17

2 0.69 0.31

3 0.58 0.42

4 0.48 0.52

5 0.40 0.60

6 0.33 0.67

7 0.28 0.72

8 0.23 0.77

9 0.19 0.81

10 0.16 0.84</syntaxerror111></cmdrjennifer>

That would be correct, if multiple successes when rolling for a single tech gave you additional techs, like in original A&A. Let me give you a demonstration.

Let’s say that in Revised you wanted to research Heavy Bombers, and you bought 2 dice. This means that you a six on at least one of those dice to succeed- we can create a table to represent this.

1 2 3 4 5 6

1 - - - - - x

2 - - - - - x

3 - - - - - x

4 - - - - - x

5 - - - - - x

6 x x x x x x

This table represents all possible combinations between the two dice, where the x’s represent a successful attempt. Now divide the number of successes by the total number of possibilities, and you get 11/36, not the 12/36 as you were expecting. This is because a result of 2 sixes does not grant two technologies, only one. Therefore each dice we purchase has diminishing returns on our chances of success. You can negate the diminishing returns by only purchasing one dice each round. That way if you are successful two rounds in a row, you receive two technologies and not one.

Another way to look at it is say the chance of success was indeed 1/6 * #dice. If you bought 6 dice, would that guarantee success every time? If you have ever done this in a game and failed (the whole reason I worked on this in the first place) then you know the answer is no.

Does this make things clearer?

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