Peter had 2/3 as many marbles as David at first. In a game, Peter lost 1/6 of his marbles to David. What was the ratio of the number of Peter's marbles to the number of David's after the game?
Let's say David initially had x marbles.
Peter had 2/3 * x = (2/3)x marbles.
After the game, Peter lost 1/6 * (2/3)x = (1/9)x marbles to David.
The total number of marbles remaining with Peter is (2/3)x - (1/9)x = (4/9)x.
The ratio of the number of Peter's marbles to the number of David's after the game is (4/9)x / x = 4/9. Answer: \boxed{\frac{4}{9}}.
Let's assume that David had x marbles initially.
Given that Peter had 2/3 as many marbles as David at first, Peter had (2/3)x marbles initially.
In the game, Peter lost 1/6 of his marbles to David. So, Peter gave away (1/6)*(2/3)x = (1/9)x marbles to David.
After the game, Peter will have (2/3)x - (1/9)x = (6/9)x - (1/9)x = (5/9)x marbles.
And David will have x + (1/9)x = (9/9)x + (1/9)x = (10/9)x marbles.
The ratio of the number of Peter's marbles to the number of David's after the game is (5/9)x : (10/9)x, which can be simplified as 5:10, or 1:2.
So, the ratio of the number of Peter's marbles to the number of David's after the game is 1:2.