Brain Teasers
Rob and Bob
Rob and Bob played a challenging game several times, betting one pebble on the outcome every time. Rob won seven pebbles, while Bob won seven times. There were no ties.
How many times did they play?
How many times did they play?
Answer
If Rob is up seven pebbles, that means he has won seven more times that he has lost. He lost seven times. Therefore Rob won 7+7=14 times.games played
= Rob's wins + Bob's wins
= 14 + 7=21
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Comments
I think that was rather easy....
math is fun................
yet very easy
good teaser though
yet very easy
good teaser though
Although the math is easy, there is also a matter of somantics that it doesn't seem was intended. My answer: they played at most 21 games, but as few as 7. If Rob bet "on the outcome everytime" that he would lose the game, he could lose 7 games, winning a pebble each time.
Nov 13, 2005
If Bob wins the first seven and Rob wins the next seven the teaser is resolved in 14 games. Rob will have seven pebbles and Bob will have won seven games.
Good point chilepepper. Other than that I started to look for tricks because it seemed way too easy.
Lose 7 finish 7 ahead must have won 14. Where's the teser?
Lose 7 finish 7 ahead must have won 14. Where's the teser?
This answer is not correct; or at the least cannot be proven. No where in the problem description does it say that the pebbles won can subsequently be lost. For instance, it does not say that they each started out with only one, or at most two pebbles to bet, and there after had to bet their winnings. Without the assumption that they have to bet the winning, the answer would be 14 or could, as pointed out before, be even 7. This teaser does not have enough information to get an accurate answer.
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