In the absence of Tryst:
The answer is 2*min(p,1-p).
Think of this problem instead as one of a random walk along a number line. Let 0 be the starting point, p be the probability of moving to the right, and q be the probability of moving to the left.
Let B=probability of ever moving one to the left from where you are.
Let A=probability of revisiting current square from the right.
B = q + Aq + A^2q + A^3q + ... = q/(1-A).
A=pB --> B=A/p --> B=A/(1-q).
q/(1-A) = A/(1-q) --> A=q,1-q.
However, A must be less than or equal to both p and q, thus the reasonable solution is A=min(q,1-q).
Redo the above only reverse the words left and right and A will still equal min(q,1-q). One ramification of this is that the probability of revisiting 0 is the same from both the left as the right. This stands to reason since any path has a mirror image on the other side of equal probability. So the answer is 2*min(q,1-q). Where I am a little uncomfortable is dismissing the other solution of A. I believe I can do so but cannot put into words why.
Below are the results of computer trials which have verified the answer above. For each probability p 100,000 trials were conducted, in which a trial ended when either the number of heads equalled the number of tails, or the difference was more than 25. It seems reasonable to assume that for p<=.4 if the difference ever equals 25 it is very unlikely that the gap ever be closed.
A 'win' means the number of heads was ever equal to the number of tails. The number of losses would be 100,000 minus the number of wins.
Code:
p wins
----- ------
0.4 80058
0.3 59799
0.2 39957
0.1 19803
Looks like Slippy was on the right track after all!
You are the foreman in charge of license plate production. Your equipment can produce two different gauges of steel, one weighing about 90% of the other.
Normally you use the heavier gauge but one week you leave that responsibility to someone else who sets the gauge randomly every day. After a week you realize that some days the license plates may have been made of the lighter gauge steel.
Every days license plates are kept in separate piles. You may use a digital scale once which will give an exact weight of whatever you decide to weigh.
Assume that the heavy gauge license plates weigh 1 unit each and the light ones 0.9 units each.
You may use the scale only once.
What do you do
