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This is tricky because we don't care about the probability that Bob WON but the probability that he has won on his second turn GIVEN the fact that he won at all. | This is tricky because we don't care about the probability that Bob WON but the probability that he has won on his second turn GIVEN the fact that he won at all. | ||
Latest revision as of 01:41, 18 February 2009
This is tricky because we don't care about the probability that Bob WON but the probability that he has won on his second turn GIVEN the fact that he won at all.
Therefore, we need to calculate the conditional probability.
P(Bob wins on his second try), which equals 5^3/6^4 is calculated fairly simply: Sue rolls NOT 6 twice and Bob rolls NOT 6 and 6 each once.
The probability to roll NOT 6 is 5/6 while the probability to roll 6 is 1/6.
Therefore, the probability for Bob to win on his second try is 5^3/6^4 or 125/1296
Now, let's look at P(Bob Wins at all):
There are two ways to go about this...the first is use of an infinite series which is too painful to dream of at this time.
Let's look at another way!
Let x = P(a player (Sue) can win)
Then, by definition, 1-x = P(the other player (Bob) wins)
But, if Sue doesn't throw a six then the probability of Bob winning on a throw is 5/6*x because Sue rolled a NOT 6 and Bob won. So 5/6*x is the probability that the other player won.
Thus we can solve for x!
x+5/6*x = 1
Solve, and you'll get x= 6/11...which is 1-P(Bob Won)...so P(Bob Wins)=5/11
So now we plug these into our conditional probability statement:
(5^3/6^4)/(5/11) which is about 21 percent.
