## probability of a probability of a probability et cetera

- adamshattuck1985
**Posts:**121**Joined:**Fri Jun 08, 2007 8:54 am

### probability of a probability of a probability et cetera

i found this tech chart by stonefish (i think on this site) but what i need to know is what formula or principle do i need to figure the probability of climbing the tech tree. e.g. Zuul getting Pulsed Graviton Beam through the beam tree versus through the heavy beam tree. i want to know this so can get a general idea for non ballistic weapon research for the zuul and other races. i know some see this as cheating or something else, but i am a math nerd and i have free time.

thanks

AWS

thanks

AWS

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### Re: probability of a probability of a probability et cetera

Well, you need to roll a given tech to have a shot at the next tech in line, barring salvage. So if you have an 80% chance of rolling the link from particle beam to neutron beam and a 70% chance at rolling the link from neutron to positron beam, then at the start of the game there's a 56% chance positron beam will be in your tree. That what you are looking for?

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Fractious Allies -- Hiver vs. Hiver, with allies

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- adamshattuck1985
**Posts:**121**Joined:**Fri Jun 08, 2007 8:54 am

### Re: probability of a probability of a probability et cetera

yeah but as it applies to the chart i posted, i use this to figure which path is more "productive" mathmatically for the zuul to get the pulsed graviton beam.

is it bayes theorem? if not or even if it is can you post the formula (in layman if you please) or show me a link to find it?

is it bayes theorem? if not or even if it is can you post the formula (in layman if you please) or show me a link to find it?

And i say to thee, ACM Mod is great! now go forth an propagate!

### Re: probability of a probability of a probability et cetera

Just calculate the odds of getting it from each possible path (as described previously) and see which has the greatest odds, if that's what you're interested in. If you want to know how much better one path (p1) is than the other (p2), then take a ratio (p1/p2) to determine how much more likely one is than the other.

If you're interested in the odds of getting the tech overall, it's easiest to determine by calculating the odds of not getting it by any path, and then subtracting that from 1.00. So if there are 2 paths to get tech X, and the odds of succeeding to get the tech on each of those to paths is p1 and p2, then the odds of getting it in total are (1.00 - ((1-p1)*(1-p2))).

If you're interested in the odds of getting the tech overall, it's easiest to determine by calculating the odds of not getting it by any path, and then subtracting that from 1.00. So if there are 2 paths to get tech X, and the odds of succeeding to get the tech on each of those to paths is p1 and p2, then the odds of getting it in total are (1.00 - ((1-p1)*(1-p2))).

Zed's TARs (sample):

Fractious Allies -- Hiver vs. Hiver, with allies

Who Let The Bugs Out -- Hiver vs. Tarka and Zuul

Tarka Ascendant -- Tarka vs. Hiver and Zuul

Strategy & Tactics Forum Archive -- More posts on strategy, tactics, and TARs

Fractious Allies -- Hiver vs. Hiver, with allies

Who Let The Bugs Out -- Hiver vs. Tarka and Zuul

Tarka Ascendant -- Tarka vs. Hiver and Zuul

Strategy & Tactics Forum Archive -- More posts on strategy, tactics, and TARs

- adamshattuck1985
**Posts:**121**Joined:**Fri Jun 08, 2007 8:54 am

### Re: probability of a probability of a probability et cetera

maybe i am not understanding but you mean find the average for all techs in between the starting tech and ending tech?

maybe i should clarify what i mean to erase confusion on my part

From:http://library.thinkquest.org/11506/prules.html

_____________________________________________________________________________________________________________________________

When events are dependent, each possible outcome is related to the other. Given two events A and B, the probability of obtaining both A and B is the product of the probability of obtaining one of the events times the conditional probability of obtaining the other event, given the first event has occurred.

P(A and B) = P(A) . P(B|A)

This rule says that for both of the two events to occur, the first one must occur ( (P(A) ) and then, given the first event has occurred, the second event occurs ( (P(B|A) ).

Example:

What is the probability of drawing two aces from a deck of playing cards? Since there are 4 aces in a 52 deck of cards, the probability of drawing one ace is 4/52. Having removed one ace and not replacing it reduces the probabilities of drawing another ace on the second draw. The 51 cards remaining contain 3 aces and therefore the probability of drawing an ace on the second draw is 3/51. Therefore, we can multiply these probabilities and determine the probability of drawing two aces:

4/52 . 3/51 = 1/221

The multiplication rule for dependent events can be extended to several dependent events:

P(A and B and C) = P(A) . P(B|A) . P(C|A and B)

_____________________________________________________________________________________________________________________________

if applied to A=100, B=80, C=60

then (A+B+C)/3=80

BUT

P(A)=100, P(B|A)=90, P(C|B|A)=75

with 5 differences it don't change much but after 5 or so events the percentage could be off by 10-20.

maybe i should clarify what i mean to erase confusion on my part

From:http://library.thinkquest.org/11506/prules.html

_____________________________________________________________________________________________________________________________

When events are dependent, each possible outcome is related to the other. Given two events A and B, the probability of obtaining both A and B is the product of the probability of obtaining one of the events times the conditional probability of obtaining the other event, given the first event has occurred.

P(A and B) = P(A) . P(B|A)

This rule says that for both of the two events to occur, the first one must occur ( (P(A) ) and then, given the first event has occurred, the second event occurs ( (P(B|A) ).

Example:

What is the probability of drawing two aces from a deck of playing cards? Since there are 4 aces in a 52 deck of cards, the probability of drawing one ace is 4/52. Having removed one ace and not replacing it reduces the probabilities of drawing another ace on the second draw. The 51 cards remaining contain 3 aces and therefore the probability of drawing an ace on the second draw is 3/51. Therefore, we can multiply these probabilities and determine the probability of drawing two aces:

4/52 . 3/51 = 1/221

The multiplication rule for dependent events can be extended to several dependent events:

P(A and B and C) = P(A) . P(B|A) . P(C|A and B)

_____________________________________________________________________________________________________________________________

if applied to A=100, B=80, C=60

then (A+B+C)/3=80

BUT

P(A)=100, P(B|A)=90, P(C|B|A)=75

with 5 differences it don't change much but after 5 or so events the percentage could be off by 10-20.

And i say to thee, ACM Mod is great! now go forth an propagate!

### Re: probability of a probability of a probability et cetera

adamshattuck1985 wrote:When events are dependent, each possible outcome is related to the other. Given two events A and B, the probability of obtaining both A and B is the product of the probability of obtaining one of the events times the conditional probability of obtaining the other event, given the first event has occurred.

P(A and B) = P(A) . P(B|A)

This is exactly what I said in the first place.

ZedF wrote:So if you have an 80% chance of rolling the link from particle beam to neutron beam and a 70% chance at rolling the link from neutron to positron beam, then at the start of the game there's a 56% chance positron beam will be in your tree.

So to use your formula with my example:

P(A and B) = P(A) . P(B|A)

P(A) is 80%

P(B|A) is 70%

Thus, P(A and B) is 56%.

adamshattuck1985 wrote:P(A and B and C) = P(A) . P(B|A) . P(C|A and B)

_____________________________________________________________________________________________________________________________

if applied to A=100, B=80, C=60

then (A+B+C)/3=80

BUT

P(A)=100, P(B|A)=90, P(C|B|A)=75

with 5 differences it don't change much but after 5 or so events the percentage could be off by 10-20.

You're using the formula incorrectly.

If P(A) = 1.00, P(B|A) = .80, and P(C|A and B) = .60,

then P(A+B+C) = 1.00*.80*.60 = .48 or 48%.

Each value in the tech chart you linked already has the assumption that you got the prerequisite link(s) built into it. Thus, you just need to multiply them together.

Last edited by ZedF on Thu Nov 11, 2010 6:07 am, edited 1 time in total.

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Fractious Allies -- Hiver vs. Hiver, with allies

Who Let The Bugs Out -- Hiver vs. Tarka and Zuul

Tarka Ascendant -- Tarka vs. Hiver and Zuul

Strategy & Tactics Forum Archive -- More posts on strategy, tactics, and TARs

Fractious Allies -- Hiver vs. Hiver, with allies

Who Let The Bugs Out -- Hiver vs. Tarka and Zuul

Tarka Ascendant -- Tarka vs. Hiver and Zuul

Strategy & Tactics Forum Archive -- More posts on strategy, tactics, and TARs

- adamshattuck1985
**Posts:**121**Joined:**Fri Jun 08, 2007 8:54 am

### Re: probability of a probability of a probability et cetera

thank the great masters for ZedF, you really have helped

thanks!

thanks!

And i say to thee, ACM Mod is great! now go forth an propagate!

### Re: probability of a probability of a probability et cetera

Glad to hear you got the gist of it.

Fractious Allies -- Hiver vs. Hiver, with allies

Who Let The Bugs Out -- Hiver vs. Tarka and Zuul

Tarka Ascendant -- Tarka vs. Hiver and Zuul

Strategy & Tactics Forum Archive -- More posts on strategy, tactics, and TARs

### Re: probability of a probability of a probability et cetera

And for those of you who want an example of the Zuul's total chance of getting Pulsed Graviton Beams by going up the Beam tree, the answer is:

(1.0*.4*.3*.1*.7*.7)*100 = 0.588%

(Red lasers to Particle beams [100% chance] * Particle beams to Neutron beams [40% chance] * Neutron beams to Positron beams [30% chance] * Positron beams to Meson beams [10% chance] * Meson beams to Graviton beams [70% chance] * Graviton beams to Pulsed Graviton beams [70% chance])

Note that dropping the Red lasers to Particle beams 100% out of the calculation and just starting at the Particle beams to Neutron beams number does not change the final answer (that whole multiplying by 1 thing ). Also note that this is just the straight up the tree calculation, which works in this case because each intervening technology only has one path to get to it. If there are multiple paths to get to a individual tech, you must first calculate the total chance to get that individual tech before you can calculate the total chance of getting the next tech up the chain. See the discussion in the thread that Stoneface linked to below to see a more complex example.

Oh and remember that there are three kinds of lies: Lies, Damned Lies and Statistics!

(1.0*.4*.3*.1*.7*.7)*100 = 0.588%

(Red lasers to Particle beams [100% chance] * Particle beams to Neutron beams [40% chance] * Neutron beams to Positron beams [30% chance] * Positron beams to Meson beams [10% chance] * Meson beams to Graviton beams [70% chance] * Graviton beams to Pulsed Graviton beams [70% chance])

Note that dropping the Red lasers to Particle beams 100% out of the calculation and just starting at the Particle beams to Neutron beams number does not change the final answer (that whole multiplying by 1 thing ). Also note that this is just the straight up the tree calculation, which works in this case because each intervening technology only has one path to get to it. If there are multiple paths to get to a individual tech, you must first calculate the total chance to get that individual tech before you can calculate the total chance of getting the next tech up the chain. See the discussion in the thread that Stoneface linked to below to see a more complex example.

Oh and remember that there are three kinds of lies: Lies, Damned Lies and Statistics!

Last edited by Nspace on Fri Dec 10, 2010 5:14 am, edited 1 time in total.

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### Re: probability of a probability of a probability et cetera

There is a java app for this thanks to Arkalius. You can find it here:

viewtopic.php?f=10&t=10207

Extremely helpful. Even works with mods.

viewtopic.php?f=10&t=10207

Extremely helpful. Even works with mods.

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