Adrian, author of such novels as, ?'My Dingo has just eaten a Joey' and ?'A Joey has just been eaten by my Dingo' returns from the wilderness. Great to see you mate, you don't say much but when you do, not only is it cryptic but also correct.
"Infinites tend to spawn other infinites."
You can create an infinite overhang.
Let us reverse the problem: how far can brick 1 be from brick 0?
Let us assume that the brick is of length 1.
To determine the place of the center of mass a(n):
a(1)=1/2
a(n)=1/n[(n-1)*a(n-1)+[a(n-1)+1/2]]=a(n-1)+1/(2n)
Thus
n 1 n 1
a(n)=Sum -- = 1/2 Sum - = 1/2 H(n)
m=1 2m m=1 m
Needless to say the limit for n->oo of half the Harmonic series is oo.
Knowing your involvement with the auto industry, try this;
What is the fastest way to make a 90 degree turn on a slippery road
Mark enlightens us with his reply and asks a further question which I hope I have made clear.
HATS
"176214841/479001600 = 0.367879441
which is awfully close to 1/e, which is where the probability converges as the number of hats/men approaches infinity."
?'12 men leave their hats with the hat check. If the hats are randomly returned, what is the probability that nobody gets the correct hat?'
Suppose we are handing out hats to n people. First we start with all the possible outcomes. Then we subtract off those that assign the right hat to a given person, for each of the n people. However, this double-counts each outcome that assigned 2 hats correctly, so we have to add those outcomes back in.
But now we've counted one net copy of each outcome with 3 correct hats in our set, so we have to subtract those off again. But now we've taken away each 4-correct-hat outcome once too often, and have to put it back in, and so forth ... until we add or subtract the outcome that involves all n people getting the correct hats.
Putting it all in probabilities, the measure of the original set is 1. For a given subset of k people, the probability that they all get their correct hats is (n-k)!/n!, while there are (n choose k) such subsets of k people altogether. But then
(n choose k)*(n-k)!/n! = (n!/((n-k)!*k!))*(n-k)!/n! = 1/k!
is the total probability measure we get by counting each subset of k people once each. So we end up generating the finite series
1 - 1/1! + 1/2! - 1/3! +- ... +/- 1/n!
which is of course just the first n+1 terms of the Taylor series expansion for f(x) = e^x centered at 0 and evaluated at -1, which converges to 1/e quite fast. You can compute the exact probability for any n >= 1 simply by rounding n!/e to the nearest whole number, then dividing again by n!.
Moreover I think you will find you are always rounding down for odd n and rounding up for even n.
For example,
12! / e = 176214840.95798...
which is within 0.05 (absolute error, not relative) of the correct intermediate result, 176214841.
Can you help?
Harry has lost his car keys! He's not using a very efficient search; in fact, he's doing a random walk. He starts at 0, and moves 1 unit to the left or right, with equal probability. On the next step, he moves 2 units to the left or right, again with equal probability.
For subsequent turns he follows the pattern 1, 2, 1, etc. His keys, in truth, were right under his nose at point 0. Assuming that he'll spot them the next time he sees them, what is the probability that poor Harry will eventually return to 0
