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DarkBASIC Professional Discussion / How to code Probability?

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Sakuya
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Posted: 10th Apr 2010 17:32
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Ive been searching now for at least 1 hour through Google and can't really find something usefull.

What I try to achieve is a Probability Function like for RPG Games where the Character moves around and suddenly enters a Battle.

I set the Probability for an Encounter in a Script like this:

The problem's that I just can't imagine how to code the Probability System like:

I have 37% chance to encounter an Enemy / enter a Battle.
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dark coder
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Posted: 10th Apr 2010 17:34
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By using a random number generator! DBPro comes with one that can be used via the 'rnd()' command.

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Sakuya
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Posted: 10th Apr 2010 17:42
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well yes Im using the rnd but how exactly am I to use it?
something like this?

that would give me for the 37% example a random number between 1 and 37 if I didnt write something wrong

but how do I do the checking?
like, what do I do with that random number?

I think Ive got a wrong idea here...
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dark coder
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Posted: 10th Apr 2010 17:56
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More like: encounter = rnd(99) < percentageChance

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Green Gandalf
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Posted: 10th Apr 2010 17:58
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If you want a 37% chance of doing something then you can do something like



This works because rnd(99) returns random values from 0 to 99, i.e. 100 different values. The condition is true when the value is any number from 0 to 36, i.e. 37 different values or 37%.

If you need fractional values just increase the number of digits, e.g. to get 37.3% use "rnd(999)<373" instead.

I'm afraid I didn't understand how your code was supposed to work.
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Sakuya
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Posted: 10th Apr 2010 18:19
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thanks I really was on the wrong way...
sometimes I can't think straight :/

@Green Gandalf

yeah that was my Problem as I had no idea what to do so I tried some random stuff (I usually just try random stuff if I completely have no idea)
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TheComet
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Posted: 10th Apr 2010 18:23 Edited at: 10th Apr 2010 18:24
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But this solution gives you a 37% chance of encountering an enemy per loop. Which means in a maximum of about 63 loops the enemy will come after you, and that is about 1 second.

If you want it to wait longer, you can check every 100 loops or so like this:



TheComet

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Sakuya
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Posted: 10th Apr 2010 18:26
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well Im checking not the Encounter based on loops but on Movement.
If I move 1 Tile it checks if there will be an Enemy Encounter
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TheComet
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Posted: 10th Apr 2010 18:29
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Oh, ok. Sounds good

TheComet

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dark coder
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Posted: 10th Apr 2010 19:15
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Quote: "Which means in a maximum of about 63 loops the enemy will come after you"


Where did you learn probability?

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Green Gandalf
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Posted: 11th Apr 2010 01:11
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It's just as well we don't refer to "permega" instead of "percent" - if we did then he'd have to wait a maximum of about 630000 loops.
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NickH
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Posted: 11th Apr 2010 02:53 Edited at: 11th Apr 2010 03:40
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(Removed...doesn't matter...sleep posting again )
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TheComet
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Posted: 11th Apr 2010 13:10
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Quote: "Where did you learn probability?"


From this guy:



TheComet

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Mnemonix
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Posted: 11th Apr 2010 18:52
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I calculate a 82% probability that you didn't learn it from him.

Your signature has been erased by a mod because it's larger than 600x120
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TheComet
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Posted: 11th Apr 2010 20:10
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OK, I screwed up. I meant 1 in 3 loops will make the enemy come after you. Is that correct?

TheComet

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Green Gandalf
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Posted: 11th Apr 2010 23:50
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I think 37 in 100 would be more correct - and that assumes you check the condition every loop.
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Dia
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Posted: 12th Apr 2010 14:31
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unless you have the means in place to stop rolling multiple encounters, if the probability is checked every loop, then the probability of getting AT LEAST 1 encounter is given by

E = 1 - (1-p)^n

1 loop, 37% chance of encounter
2 loops, 60.3% chance of encounter
....
5 loops, 95% chance of encounter
10 loops, 99% chance of encounter

etc etc etc

computer crashed once when I was spawing computer opponents to play each other once.... because I ended up trying to spawn a character about 20 times per second. didnt take long for me to run outa memory

This is not the Sig you are looking for....
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veltro
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Posted: 12th Apr 2010 15:22
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Quote: "1 loop, 37% chance of encounter
2 loops, 60.3% chance of encounter
....
5 loops, 95% chance of encounter
10 loops, 99% chance of encounter"


I don't think so

Each chanche of encounter is indipendent from previous chanches: so for each step the probability is always 37%
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Green Gandalf
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Posted: 12th Apr 2010 16:41 Edited at: 12th Apr 2010 16:41
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Quote: "Each chanche of encounter is indipendent from previous chanches: so for each step the probability is always 37% "


True - and not the issue.

Dia has correctly given the chance of an encounter by the end of the nth loop - assuming independence between loops. Obviously an encounter will eventually occur - just as you will eventually throw a "head" when tossing a fair coin.
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Hawkblood
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Posted: 12th Apr 2010 18:31
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You may want to change the random seed each time they move (use time to do that). That way it seems more random.

The fastest code is the code never written.
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veltro
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Posted: 12th Apr 2010 18:59
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Quote: "
Dia has correctly given the chance of an encounter by the end of the nth loop - assuming independence between loops. Obviously an encounter will eventually occur - just as you will eventually throw a "head" when tossing a fair coin. "


This is correct. Sorry I was misunderstanding Dia post.
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