Patrick and Fred are playing a game in which they alternately toss a pair of dice. The objective of the game is to roll a total sum of 7. Find the probability that Patrick, who tosses after Fred , will win the game on his first turn.


There are 6 ways of landing 7 so I did 6/36=0.166667 but apparently the answer is 0.1388.

Please help, thanks

okay I found it nvm. thanks

If both players roll a sum of 7 with two dice, it is considered a tie (neither player has won).
Hence, the only way Patrick wins the game is if Fred does not roll a sum of 7 while Patrick himself does roll a sum of 7.

Method 1
The two rolls are independent and hence the probability (P) that Patrick wins on his first turn

P = (1 - 6/36)·(6/36) = 5/6 · 1/6 = 5/36 (~13.9%)

Method 2
Let f and p denote the sum rolled by Fred and Patrick, respectively, and consider pairs, {f,p}.
Clearly, 2 ≤ f,p ≤ 12 (where f and p are integers).

Define three events:

A: Fred does not roll 7 while Patrick rolls 7 (A = { {f,7} | 2 ≤ f ≤ 12, f ≠ 7 })
B: Fred and Patrick both roll 7 (B = { {7,7} })
C: Fred rolls any sum while Patrick rolls 7 (C = { {f,7} | 2 ≤ f ≤ 12 })

Then A and B have no elements in common, and thus C = A U B is a union of disjoint events.
We get

P(C) = P( A U B ) = P(A) + P(B) =>

P(A) = P(C) - P(B) = (1 · 1/6) - (1/6 · 1/6) = 1/6 - 1/36 = 5/36

Notice how quick the first method is.