Anna and bob play a game in which Anna begins by rolling a fair dice, after which bob tosses a fair coin. They take turns until one of them wins. Anna wins when she rolls a 6. Bob wins when the coin lands on heads. What is the probability that Anna will win the game?
Well there are a number of ways for Anna to win like 1st it is 1/6 the second way can be 5/6*1/2*5/6*1/2*1/6 etc. Where do i go from here to find an exact number for the probability?
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To find the exact probability of Anna winning the game, we can use the concept of geometric series.
A geometric series is the sum of the terms in a geometric sequence. In this case, each term represents the probability of Anna winning at a particular turn.
Since Anna wins when she rolls a 6, the probability of her winning on the first roll is 1/6.
For the subsequent turns, Anna must have a failure (not rolling a 6) followed by Bob's failure (not getting heads on the coin toss). So, the probability of Anna winning on the second turn is (5/6) * (1/2) * (5/6) * (1/2) * (1/6).
This pattern continues for each subsequent turn, with the probability of Anna winning on the nth turn being represented as (5/6) * (1/2) * (5/6) * (1/2) * (1/6) repeated n-1 times.
Now we have an infinite geometric series with a common ratio of (5/6) * (1/2) * (5/6) * (1/2) * (1/6). The sum of this series will give us the exact probability of Anna winning the game.
The formula to find the sum of an infinite geometric series is:
S = a / (1 - r)
where S is the sum, a is the first term, and r is the common ratio.
In our case, a = 1/6 (probability of Anna winning on the first turn) and r = (5/6) * (1/2) * (5/6) * (1/2) * (1/6).
Plugging these values into the formula, we can find the exact probability of Anna winning the game.