Encounter probabilities question?

[Ogrork the Mighty]

Let's say there is a 5% chance per hour of an encounter. What is the chance over an 8 hour period that an encounter will occur? I realize that each check is independent of the others, but isn't there a way to figure out an overall % chance for the period?

[it's been a while since I took statistics in school ]

 

[Jeff Wilder]

Let's say there is a 5% chance per hour of an encounter. What is the chance over an 8 hour period that an encounter will occur? I realize that each check is independent of the others, but isn't there a way to figure out an overall % chance for the period?

I'm lousy at probability, but this is one of the few I do know. I'm unlikely to be first to respond, because I'll try to explain as I go.

First, understand that the probability of two events both occurring is (chance of one event) times (chance of second event). The chance of rolling 12 on 2d6 is 1 in 36, which is (1 in 6) times (1 in 6).

Second, understand that something not occurring is also an event. The chance of not rolling a 6 on 1d6 is 5 in 6.

Finally, understand that the probability for all outcomes, summed, has to equal exactly 1.

Given these three things, you can say that for each of 8 hours, the probability of not having an encounter is 19 in 20. So, for all 8 hours, the probability of not having an encounter is (19 in 20) times (19 in 20) ... and so on, or (19/20)^8. That comes out to 0.6634 or so (66%). Given that, and since you know that the (chance of not having an encounter) plus (the chance of having an encounter) has to equal 1, you end up with a probability of having an encounter of 0.3366 (34%) or so. Or close enough to 1 in 3 not to sweat the difference.

 

[sfedi]

you end up with a probability of having an encounter of 0.3366 (34%) or so.

That's the probability of having AT LEAST one encounter.

In that chance, there's the chance of having just one encounter, and of having 8 encounters all sumed up.
BTW, the chance of having 8 encounters is 0.05^8 = 0,0000000000390625 = 0,00000000390625 %

 

[Jeff Wilder]

That's the probability of having AT LEAST one encounter.

True, and a good point. I do think it's what was asked for, though.

BTW, can you explain, step-by-step, how to calculate the same probability, but for one (and only one) encounter? (Like I said, I'm lousy at probability.)

 

[the Jester]

I've taken to checking for encounters 1 in 6 each day and each night. If one's indicated, I check again to see if there's another that day. Saves me the tedium of rolling multiple times per day to confirm that nothing wanders up.

 

[Ogrork the Mighty]

I find the hour-by-hour checks to be useless for my campaign. When the PCs are traveling, we don't play through the wilderness hour-by-hour so a daily check is much better for us. I'd really only be looking for a max of one encounter per 8-hour period, so any more than that and I'd probably be converting them into non-encounter encounters (maybe a natural environmental condition, or strange noise, or something non-combat like that).

 

[Felnar]

BTW, can you explain, step-by-step, how to calculate the same probability, but for one (and only one) encounter? (Like I said, I'm lousy at probability.)

someone can correct me if needed, but here goes...
1. there are 8 events, each one is either 'an encounter' of 'no encounter'
2. chance of 'an encounter' is 1 in 20, chance of 'no encounter' is 19 in 20
3. to have exactly one 'an encounter', the other seven events must be 'no encounter'
4. so (1/20)^1 * (19/20)^7 = 3.491%

hopefully the school system works...

 

[Jack Simth]

someone can correct me if needed, but here goes...
1. there are 8 events, each one is either 'an encounter' of 'no encounter'
2. chance of 'an encounter' is 1 in 20, chance of 'no encounter' is 19 in 20
3. to have exactly one 'an encounter', the other seven events must be 'no encounter'
4. so (1/20)^1 * (19/20)^7 = 3.491%

hopefully the school system works...

*8, as you don't care which of the 8 rolls comes up with the "yes" result, as long as exactly one of them does.

To be a little more generic, if you have probability Y for yes, there is an encounter at any given check, N (=1-Y) for no, there is not an encounter at this check, and you want to find the probability for exactly X encounters given Z checks, the formula works out to:
(Y^X)*(N^(Z-X))*(Z CHOOSE X) ... if I am remembering my math correctly.

 

[Felnar]

*8, as you don't care which of the 8 rolls comes up with the "yes" result, as long as exactly one of them does.

To be a little more generic, if you have probability Y for yes, there is an encounter at any given check, N (=1-Y) for no, there is not an encounter at this check, and you want to find the probability for exactly X encounters given Z checks, the formula works out to:
(Y^X)*(N^(Z-X))*(Z CHOOSE X) ... if I am remembering my math correctly.

yes, excellent, 3.491% * 8 = 27.93%
soon after hitting submit i realized i hadnt summed anything but was pretty sure i was supposed to, thanks for the correction

could you explain why it's *(Z CHOOSE X) and not just *Z
and what "CHOOSE" is for that matter

 

[Jack Simth]

CHOOSE is a function wherein I can't quite remember the standard representation at the moment - basically, it's used for exactly what was asked - figuring out what the probability is of a given number of known and equal-probability events occuring on a given number of chances.

You probably remember Pascal's Triangle:

    1
   1,1
  1,2,1
 1,3,3,1
1,4,6,4,1

Where each # in the triangle is the sum of the two above it, once you get past the "seed" 1 at the top (there is a faster way to do Z Choose X, but this is more easily comprehensible and doesn't requre any operations more complex than addition and counting).
Well, the Choose function simply selects a number from there; the 1,1 row is Z = 1, the 1,2,1 row is Z = 2, and so on; X is the number along the row, with the 1 on the far left being position 0; so (3 CHOOSE 2) = 3, (4 CHOOSE 2) = 6. If you have three "decision points", there are: 1 way to come up with no positives (---), 3 ways to come up with one positive(+--, -+-, --+) 3 ways to come up with two positives(++-, +-+, -++) and one way to come up with 3 positives (+++). The CHOOSE function just tells you how many ways there are to come up with exactly the specified number of positives without you having to go through and come up with all of them. It's also a function that is built into the higher-end calculators nowadays. Note that whenever you are looking at "exactly once", it works out to the number of chances for occurance - checking 8 times, there are 8 cases where exactly one encounter happens.

I'm sure that was all clear as mud.....