Question 1191369
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(1) Since the probability of getting a 5 is less than the probability of getting a 7, in the long run the probability of A winning should be less than 0.5 -- so your answer should be off.  Note the formula you are using to get your answer is not correct....<br>
(2) The probability of A winning depends on whether A or B goes first.  Your statement of the problem doesn't specify....<br>
P(5 on any throw) = 4/36 = 1/9
P(not 5 on any throw) = 8/9
P(7 on any throw) = 6/36 = 1/6
P(not 7 on any throw) = 5/6<br>
If A goes first....<br>
A wins if the sequence of throws is
A 5, or
A not 5, B not 7, A 5, or
A not 5, B not 7, A not 5, B not 7, A 5, or
...
The probability of A winning if A goes first is then<br>
(1/9)+(8/9)(5/6)(1/9)+(8/9)(5/6)(8/9)(5/6)(1/9)+...<br>
Using the formula for the infinite sum of a geometric series (first term 1/9; common ratio (8/9)(5/6)=20/27)....<br>
{{{S = (1/9)/(1-(8/9)(5/6)) = (1/9)/(1-20/27) = (1/9)/(7/27) = (1/9)(27/7) = 3/7}}}<br>
Then it should be true that if A goes first the probability of B winning is 4/7.  Let's verify that using the same kind of calculations as above.<br>
If A goes first....<br>
B wins if the sequence of throws is
A not 5, B 7, or
A not 5, B not 7, A not 5, B 7, or
A not 5, B not 7, A not 5, B not 7, A not 5, B 7, or
...
The probability of A winning if A goes first is then<br>
(8/9)(1/6)+(8/9)(5/6)(8/9)(1/6)+(8/9)(5/6)(8/9)(5/6)(8/9)(1/6)+...<br>
Using the formula for the infinite sum of a geometric series....<br>
{{{S = (4/27)/(1-(8/9)(5/6)) = (4/27)/(1-20/27) = (4/27)/(7/27) = (4/27)(27/7) = 4/7}}}<br>
Now here is a somewhat advanced concept regarding this problem.<br>
Since the common ratio in both sequences is the same -- representing each of A and B rolling the dice and NOT getting the number they want -- the ratio of A winning to B winning is the same as the ratio of A winning on his first throw to B winning on his first throw.<br>
P(A wins on first throw) = 1/9
P(B wins on his first throw) = (8/9)(1/6) = 8/54 = 4/27<br>
{{{(1/9)/(4/27) = (1/9)(27/4)=3/4}}}<br>
Then since the ratio of A winning to B winning is 3:4, the probability that A wins is 3/7 and the probability that B wins is 4/7.<br>
If B goes first....<br>
Again, the statement of the problem did not specify which player goes first, so let's look at the problem again if B goes first.<br>
Using the same argument as in the preceding paragraph, the ratio of the probabilities of B or A winning is the ratio of the probabilities that each player wins on his first throw.<br>
P(B goes first and rolls 7 on first throw) = 1/6
P(B goes first, does not roll 7, and A rolls 5 on first throw) = (5/6)(1/9) = 5/54<br>
The ratio of B winning to A winning is<br>
{{{(1/6)/(5/54) = (1/6)(54/5) = 9/5}}}<br>
In this case (B rolling first), the ratio of B winning to A winning is 9:5, so the probability of B winning is 9/14 and the probability of A winning is 5/14.<br>
You should find those probabilities if you analyze the case of B going first using infinite geometric series, as done in the first part of this response for the case of A going first.<br>