Charles C. Berry:
Are the numbers 1:30 equiprobable??
If so, you can find the probability by direct enumeration.
Or by a simple formula:
* Probabilities of Consecutive Integers in Lotto
* Author(s): Stanley P. Gudder and James N. Hagler
* Source: Mathematics Magazine, Vol. 74,
Anthony28 wrote:
I need to use R to model a large number of experiments (say, 1000). Each
experiment involves the random selection of 5 numbers (without replacement)
from a pool of numbers ranging between 1 and 30.
What I need to know is what *proportion* of those experiments contains two
or
How about this
result - numeric(10)
for(i in 1:10){
x - sample(1:30, 5, replace = FALSE)
x - sort(x)
result[i] - any(diff(x) == 1)
}
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Anthony28
Sent: Tuesday, April 29, 2008 8:52 AM
To:
This will work:
my.list - c(2, 28, 31, 4, 27)
sort(my.list)
diff(sort(my.list))
any(diff(sort(my.list)) == 1)
the middle two lines are only to illustrate what's going on.
Best wishes!
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
Hey Anthony,
There must be many ways to do this. This is one of them:
#First, define a function to calculate the proportion of consecutive
numbers in a vector.
prop.diff=function(x){
d=diff(sort(x))
prop=(sum(d==1)+1)/length(x)
return(prop)}
#Note that I am counting both numbers in a
On Tue, 29 Apr 2008, Anthony28 wrote:
I need to use R to model a large number of experiments (say, 1000). Each
experiment involves the random selection of 5 numbers (without replacement)
from a pool of numbers ranging between 1 and 30.
What I need to know is what *proportion* of those
Hey Anthony,
My previous function may not work in all cases. Say one of the
experiments yields these numbers:
1,2,3,6,7
Would you say that the proportion of consecutive numbers is 100%? If
so, this will work:
prop.diff=function(x){
d=diff(sort(x))
prop=sum((c(0,d==1)+c(d==1,0))0)
I'd just like to thank all you guys for stepping in so promptly with help. I
haven't yet had a chance to implement any of your code yet, but just by
looking over what you've suggested, I think I have enough to guide me. So
thanks once again!
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