- #1
toothpaste666
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Homework Statement
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 first. The winner of the game is the first 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)
Homework Equations
[itex] \frac{a}{1-r} [/itex]
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:
[itex] \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}) [/itex]
[itex] \frac{1}{6} + (\frac{5}{6})^2 \frac{1}{6} +(\frac{5}{6})^4 \frac{1}{6} [/itex]
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:
[itex] \frac{\frac{1}{6}}{\frac{36}{36} - \frac{25}{36}} [/itex]
[itex] \frac{1}{6} * \frac{36}{11} [/itex]
[itex] \frac{6}{11} [/itex]
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?