[Frankiez]

The problem is following. There are pictures of 15 women and 15 men. Those women and men are really 15 married couples. Your task is to form pairs from those pictures. If you form pairs at random from those pictures how many pairs of pictures should be from married couples on average? Obviously you form pairs of one man and one woman from the pictures but other than that it is randomly selected.

[tickle]

Do you know that we are all volunteers here, with limited time to answer questions, we don't get paid to sit here and figure out your 'expert problem'.

[Frankiez]

I just joined this group and I didn't even know here are "experts"... So what I meant was that to be able to solve this problem you got to be an expert in probability theory and not in this site... So sorry to bother you. Well maybe this is wrong site to try to get an answer to this complex problem but I was just trying my luck.

[joypulv]

Frankiez, there are math experts here. Just wait. We don't usually solve homework problems for anyone, however, so you are expected to try to solve it.

[ebaines]

This is a binomial distribution problem, where for each attempted match either there is a successful match or there isn't. The probability of a success for each attempt is p=1/15 (see note below). In your text they should have provided a couple of useful formulas about binomial distribitions, one of which is the expected value:



where n is the number of trials and p is the probabiliy of success for any one trial. Can you apply this formula to your problem?

Note - at first glance it may seem this is not a true binomial distribution prioblem since each trial is not strictly independent, which is a requrement for using the binomial distribution. For example for the 15th match if that person's spouse has not been picked already the probability of success is 100%, and if the spouse has already been picked it's 0%, so it would seem to be a dependent probability. However, consider that the probability of the spouse having already been picked after 14 trials is 14/15, and the probability of not having already been picked is 1/15, so the probability of the 15th matching is 1/15. Using this logic one can show that the probability of a match on any of the 15 trials is 1/15. Hence p is indeed constant for each of the 15 trials.

[Frankiez]

This is not my homework, I was just trying to solve this for fun. :)

Oh yeah I didn't realized that binomial distribution could be used for this problem, thanks! That note made this problem so much simpler. So the E(X)=15*(1/15)=1.

I also solved this problem using programming in sage, only approx of course. First I made a program that makes a list that contains numbers 1,2,. 15 in random order:

def randomize():
c=[]
c.append(floor(15*random()+1))
while len(c)<15:
a=floor(15*random()+1)
h=False
k=0
while k<len(c):
if a==c[k]:
h=True
k=k+1
if h==False:
c.append(a)
return c

After that I made a program that shows how many of the random list elements are same and in same place than in a list [1,2,. 15]:

def testi(q):
s=0
luku=0
a=[1,2,. 15]
while s<q:
b=randomize()
k=0
while k<len(b):
if a[k]==b[k]:
luku=luku+1
k=k+1
s=s+1
return n(luku/q)

I run the program though 10000 times and it gave me results of 0.99030, 1.0049, 0.9910, 0.9852 and in fifth time 1.00030. So it seems that E(X) really is 1.

[ebaines]

So the E(X)=15*(1/15)=1.

Yes. Note that it doesn't matter how many couples are in the experiment, the expected value of matches is always 1. For example if N=1 (just one couple) there will always be precisely one match. If N=2 then half the time neither couple matches and half the time both couples match, so again the expected value is 1. For three couples there are 6 possible outcomes, one of which involves all three matching, 3 involve just one match, and 2 involve no matches, so the expected value is (1/6)x3 + (3/6)x1 + (2/6)x0 = 1. You can see the pattern here - it's always 1!

[joypulv]

Interesting, both of you, thanks!