Recall that coupon collecting is equivalent to placing m balls in n bins so that no bin is empty. Last time we derived an upper bound for the probability that some bin is empty which is
Pr[some bin empty] < ne-m/n
Then we showed that if you fix this probability and rearrange the equation, the number of balls you need is:
m > n ln(n) +
W (n)So if m is somewhat bigger than n ln(n), we have a very good chance of hitting all the bins. So if you make about n ln n visits to the supermarket, you have a good chance to get all n coupons.
That also implies the result that we were interested in about stable marriages: After about n ln n proposals under the random algorithm, every female will have been proposed to, and a stable marriage will result. That means if you started with random permutations as preferences, the proposal algorithm would run for about n ln n steps.
Now lets turn to an analysis of the expected value of m to hit all of the bins (or collect all coupons). Let X be the number of balls placed when the last bin is hit.
Let X0 be the number of trials til the first bin is hit (=1).
Let X1 be the number of trials after the 1st bin is hit until the 2nd bin is hit.
Let X2 be the number of trials after the 2nd bin is hit until the 3rd bin is hit.
…
Let Xn-1 be the number of trials after the (n-1)st bin is hit until the nth bin is hit.
Then the total number of trials is:
The trials counted in each Xi are called epochs. In epoch i (corresponding to Xi-1), the probability that any trial hits the next bin is
That’s because there are i bins that have already been hit, and n-i that havent.
Xi is called a geometrically distributed random variable. We will see why in a second.
The value of Xi is the first k such that a certain event happens for the first time. The probability of that event is always the same, pi. So the distribution of Xi looks like this:
Which explains why its called geometric. The sequence of probabilities forms a geometric series with ratio (1-pi).
Let’s compute the mean and variance for a geometric r.v. First the expected value:
We notice that the k(1-pi)k-1 term looks like a derivative…
And now we can sum the geometric series…
So the expected value of a geometric r.v. is just 1/pi. For the variance, we use the formula
We just computed E[Xi] so all we need now is E[Xi2].
And we do some rearranging to make it look like a second and a first derivative:
Replacing the terms with the derivative expressions gives:
Then we can swap the derivatives and sums:
And solving for the sums of the geometric series gives:
The variance is the difference between this term and the square of E[Xi]
We have divided the set of choices into epochs, with Xi being the number of placements during the ith epoch. The expected value of X is
And we change variables and use j=n-i, which gives
for the expected number of rounds to hit all of the bins. The variance is
Applying the substitution j = n-i
The first sum approaches p 2/6 as n ® ¥ , so we have
That is, the standard deviation is approximately (ignoring lower order terms)
Now we can use Chebyshev. E.g. suppose we want to ensure that the probability of an empty bin is less than 0.01. Pick t = 10 in the Chebyshev formula
Which requires that X-E[X] ³ ts X or X ³ n ln(n) + 10np /Ö 6
That’s roughly the same kind of bound we obtained earlier using a direct probability analysis. It’s worth noting however, that if we want to make the probability very small, the Chebyshev bounds don’t help very much. The actual probability falls off exponentially with distance from the mean, where as the Chebyshev bound falls off only as 1/t2.