Correct. The best way to achieve that is to use imbricated/nested if-checks since - as it was pointed out above - using several greet progs could lead to situations (rare, but possible) where ALL greet_progs will trigger.
I'd do the following :
1. Determine how often you would like incoming PCs to get a rumour. The more wandering mobs you have, the smallest the percentage chance should be (unless you want PCs to be spammed with rumours). Let's say, for the sake of this example, that you want an mob to spurt out a rumour with only a 15% chance. This is the number we are going to use for the greet_prog: it will be a greet_prog 15.
2. Write down all the rumours, and determine whether they should all have the same probability to be given or not. A system where some rumours (the more obvious ones) are given more often than other rumours is realistic, but it's harder to get it right. A system where all the rumours have the same probability is simpler (though still not trivial). For the sake of the example, I'll assume that you want all rumours to be given with the same probability.
Side comment wrote:If I have 3 rumours, I want each of them to have a 33.33333% chance to appear, so I can simply build my greet_prog like this:
Code: Select all
>greet_prog 15~
if rand(33)
sayto $n Rumour 1!
endif
if rand(33)
sayto $n Rumour 2!
endif
if rand(33)
sayto $n Rumour 3!
endif
~
This is not correct though - for the same reason that using 3 greet_progs will not work very well: if we use 3 distinct random if-checks, it can happen that the three of them will succeed. That is, it can happen that the mob is going to say Rumour 1! then Rumour 2! then Rumour 3!, and we want to avoid that.
That means that, instead of using consecutive if-checks, we need to use nested if-checks.
With 3 rumours, the process should go as follows:
- Will the first rumour be given? Well, it should be given with a 33% chance (33.3333...% to be precise). If the first rumour is given, the program should then end.
-
Otherwise (= else), then either Rumour 2 or Rumour 3 should be given. If we are in this case, both rumours should be equally likely to be given. That means that there would be a 50% chance for Rumour 2 to be given.
- Finally, if Rumour 2 is not given, then we give Rumour 3.
Turning that into code, we get something like this:
Code: Select all
>greet_prog 15~
if rand(33) <- 33% chance of Rumour 1 being given
sayto $n Rumour 1!
else
if rand(50) <- Otherwise, 50% chance of Rumour 2 being given
sayto $n Rumour 2!
else
sayto $n Rumour 3! <- Otherwise, well... 100% chance of Rumour 3 being given
endif
endif
~
And we shouldn't forget the "endif"s!
3. Once you know how many rumours you want, create the structure of the programs, with nested if-rand(something)-checks. To figure out the (something) numbers, you can check the post referred under the heading "Using imbricated rand checks" in
there.
If you want some rumours to be more likely to appear than others, you can modify the numbers in the rand() things, but it's very easy to get them wrong.
Code: Select all
greet prog 100~
if rand (80)
if rand (50)
do something half of the time
else
if rand (10)
do something a little of the time
endif
endif
else
do something most of the time
endif
The program above, for example, will work as follows:
Code: Select all
if rand (80) ____________________________________________________
if rand (50) _____________________________ |
do something half of the time ___________| 50% of 80% = 40% |
else |
if rand (10) ------------------| | 80%
do something a little of the time | 4% | 40% |
endif ------------------| |
endif __________________________________________________________|
else
do something most of the time | 20%
endif
The numbers are to be read from right to left. First, we have a rand(80) check. It has a 80% chance of succeeding. Therefore, "do something most of the time" will actually be done only 1 in 5 times (= 20% chance).
If the rand(80) succeeds, we get another rand check. This time, the value is 50%. The "do something half of the time" part will thus be done half of the time IF the initial rand(80) check succeeds. That means that it is done only 4 out of 10 times (= 40%).
If the rand(80) succeeds but the rand(50) fails ... which will happen in 40% of the executions of this programme, we reach another if-check. This if-check will succeed only in 10% of the cases when we reach it. 10% of 40% is 4%. Thus, "do something a little of the time" will only be done with a 4% probability.
And if the rand(10) fails... nothing is done.
Summing this up, this programme will do this:
- 40% probability to "do something half the time"
- 4% probability to "do something a little of the time"
- 36% probability to do nothing
- 20% probability to "do something most of the time"
That shows how tricky nested rand-if-checks can be
