CODEWITHSUNDEEP
rprobability-density
Bob
Bob

Reputation: 160

In R, how to find the x for which f(x)=0.5

I'm coding the birthday problem. My code:

> k = 100
> for (n in 1:100) {
+ prob =  1 - (0:(n-1))/365
+ k[n] = 1 - prod(prob) }
> plot(k)

I need to find the n for which its ~50% likely that two people share the same birthday. Any help? I tried googling but I can only find information about popular distributions.

Upvotes: 1

Views: 203

Answers (3)

colemand77
colemand77

Reputation: 551

Try the following from the stats package...

 approx(k,1:100, xout = .5)

The above uses linear interpolation between the closest two provided points, which I may get jumped on because it is inaccurate...

But if we're doing the birthday problem I'm guessing we're learning about coding and not launching a rocket, so maybe its close enough?

Upvotes: 1

IRTFM
IRTFM

Reputation: 263342

The "Birthday problem" already has two R functions designed to solve various forms of it:

birthday {stats}    R Documentation
Probability of coincidences

Description

Computes answers to a generalised birthday paradox problem. pbirthday computes the 
probability of a coincidence and qbirthday computes the smallest number of observations 
needed to have at least a specified probability of coincidence.

Usage

qbirthday(prob = 0.5, classes = 365, coincident = 2)
pbirthday(n, classes = 365, coincident = 2)

So the defaults for `qbirthday:: 

 > qbirthday()
 [1] 23

Upvotes: 4

Rick
Rick

Reputation: 898

You've already solved the problem. You can find it "by eye", just type "k" and start counting. Or you can have R find it for you. The first day where the probability is greater than or equal to .5 is day number:

#TRUE is the maximum, pick the first of the ties:
which.max(k >= .5)

We can find the day with the minimum absolute difference from probability .5:

which.min(abs(k - .5))

Upvotes: 1

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