*Posted by Seungln in : potd , trackback
*
After getting sick of Sreenath’s jokes about spiders on his head, Seungln challenges Sreenath to play a series of Starcraft games against him. The rules are as follows:

- Condition A: If at any time Seungln is ahead of Sreenath by 2 wins, then Seungln wins and Sreenath has to kill all the spiders in the world.
- Condition B: If at any time Sreenath is ahead of SeungIn by 2 wins, then Sreenath wins and Seungln has to be locked up in a room full of spiders.
- Condition C: If neither of A or B happens, then more matches are played until either Condition A or Condition B occurs.

If Seungln has a 60% chance of beating Sreenath, and there is no draw in Starcraft, then the probability that Sreenath will have to kill all the spiders in the world can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime. Find $n-m$.

show

Let X be the event of SeungIn beating Sreenath, and let Y be the event of Sreenath beating SeungIn. Condition A is satisfied if the number of Xs exceeds the number of Ys by 2. (Notice that the number can increase or decrease by 1 each time, so as soon as the first 2 occurs, it’s over.) Condition B is satisfied if the number of Ys exceed the number of Xs by 2. We also note that, since the number must differ by exactly 2 in order for us to determine the winner, the number of matches have to be even.

The first two matches could be XX, XY, YX or YY. The two middle cases are the only ones that need two extra matches, since it is a 1-1 tie. Then the results for the first four rounds become XYXX, XYXY, XYYX, XYYY, YXXX, YXXY, YXYX, YXYY. Cases 2, 3, 6 and 7 need more matches. Continuing this pattern, we noticed that the probability that Condition A is satisfied becomes $P(X)^{2} + 2P(X)^{3}P(Y) + 4P(X)^{4}P(Y)^{2} + \cdots$, where $P(X)$ is the probability that X will be true in one round. Thus the entire thing becomes

$$\begin{eqnarray} P(X)^{2} &\cdot& (1 + 2P(X)P(Y) + 4P(X)^{2}P(Y)^{2} + \cdots) \\ &=& P(X)^{2} \cdot \frac{1}{1 – 2P(X)P(Y)} \\&=& (\frac{3}{5})^{2} \cdot \frac{1}{1 – 2 \cdot \frac{3}{5} \cdot \frac{2}{5}} \\&=& \frac{9}{25} \cdot \frac{1}{\frac{13}{25}}\\ &=& \frac{9}{13} \end{eqnarray}$$

Thus, the answer is $13 – 9 = \boxed{004}$

## Comments»

no comments yet - be the first?