# Probablity problem using series

1. Apr 13, 2014

### toothpaste666

1. The problem statement, all variables and given/known data

Professors Bob and Fred cannot decide who should buy a new kite. In order to
decide, they play a game. They take it is turns to throw a (standard, six-sided) die, with Professor
Bob going fi rst. The winner of the game is the fi rst one to throw a 4. For example, Professor Bob
wins if he throws a 4 immediately or the results are
non-4 for Bob, non-4 for Fred, 4 for Bob
or
non-4 for Bob, non-4 for Fred, non-4 for Bob, non-4 for Fred, 4 for Bob
and so on.
Find the probability that Professor Bob wins. (Hint: the calculation requires a certain type of
series)

2. Relevant equations

$\frac{a}{1-r}$

3. The attempt at a solution

Bobs first turn he has 1/6 chance of rolling a 4. If it goes to his second turn that means that he did not throw a 4 the first turn and also that Fred did not throw a 4 on his turn so his chances of rolling a 4 on the second turn are 5/6 * 5/6 * 1/6. To analyze the chances of Bob rolling a 4 on his first 3 turns:

$\frac{1}{6} + (\frac{5}{6} * \frac{5}{6} * \frac{1}{6}) + (\frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{1}{6})$

$\frac{1}{6} + (\frac{5}{6})^2 \frac{1}{6} +(\frac{5}{6})^4 \frac{1}{6}$

This is a geometric series where the first term is 1/6 and the common ratio is (5/6)^2
plugging this into a/(1-r) we have:

$\frac{\frac{1}{6}}{\frac{36}{36} - \frac{25}{36}}$

$\frac{1}{6} * \frac{36}{11}$

$\frac{6}{11}$

So Bob's chance of winning would be 6/11. This seems to make sense because they would have just about the same chances of winning except Bob's would be slightly more because he goes first which leaves the chance of him winning without Fred even having a turn. Is this the correct answer/approach to this problem?

2. Apr 13, 2014

### Ray Vickson

Your answer and approach are correct. There is another way that is perhaps more intuitive and more "probabilistic". Let b = probability that Bob wins, given that Bob goes first, and f = probability that Bob wins, given that Fred goes first. We have
$$b = \frac{1}{6} + \frac{5}{6} f \; \longleftarrow \text{do you see why?}\\ f = \frac{5}{6} b$$
The second equation above follows from the fact that if Bob is to win when Fred goes first, Fred's first toss must not be a '4', and then after that, everything is the same, but now with Bob going first.

Solving the two equations gives $b = 6/11, f = 5/11$.

3. Apr 13, 2014

### toothpaste666

Can you explain that a little bit more? It seems a lot more efficient than what I did, but I don't fully understand it.

4. Apr 14, 2014

### Ray Vickson

If Bob starts he wins immediately if he gets '4'; the probability of that is 1/6. Otherwise, Bob wins when Fred goes first following Bob's initial non-4, and the probability of that sequence is (5/6)*f. Those are the two ways Bob can win going first, so b = 1/6 + (5/6)*f. For Bob to win when Fred goes first, two things must happen in sequence: first, Fred has to get a non-4 (probability = 5/6) and then Bob wins eventually while going first (for the remainder of the games). The probability of that sequence is (5/6)*b, so f = (5/6)*b.