Kick-off
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Jimbo is an ace goal kicker. On the average, he will kick a goal on 2 out of 3 occasions. Unco on the other hand, will normally only kick a goal on 1 out of 2 occasions.
Now Jimbo, being quick to spot an opportunity, challenges Unco to a kicking duel. They will each in turn take one kick for goal until a goal is scored. The winner will be the first to score a goal.
"I'll even give you the first kick", says Jimbo generously to Unco.
What is the probability that Jimbo wins the kicking duel?
HintYou only get to take your next kick if the guy before you misses!
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Answer
Unco kicks: probability of a miss = 0.5
Jimbo kicks: probability of a win = 0.5 x 2/3 = 1/3
{Note Unco must have missed in order for Jimbo to get a kick at all}
Unco kicks: probability of a second miss = 0.5x1/3x 0.5 = 1/12
{Note both Unco and Jimbo must have missed for Unco to get a second kick}
Jimbo kicks: probability of goal = 1/12 x 2/3 = 1/18
Probability that Jimbo wins on first or second or third etc is:
P = 1/3 + 1/18 + 1/108 + ......
Which is a geometric series with limiting sum = 2/5
Answer: Probability that Jimbo wins = 2/5.
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Comments
griphook  
Mar 28, 2003
| I don't particularly like probability teasers but this one is real cool. Congrats. |
jimbo 
Mar 29, 2003
| Thanks Grip. It's so nice to see a positive comment. I should keep that in mind! |
Pheonix_down
Apr 10, 2003
| This was really easy, I liked it though, I was about to make a teaser like this one before I read this... now I'm out of ideas!  |
brianz
May 16, 2005
| Great teaser. I liked it. Not so hard since I just submitted one kinda like this. 
I don't know how you summed up the series, but here's how you could have done it.
You just have to sum up 1/2*2/3*(1/6^0+1/6^1+1/6^2...) or 1/3*(1/6^0+1/6^1+1/6^2...) (1/2 that Unco misses his first shot, 2/3 that Jimbo makes one of his shots when he gets one, and the 1/6^0 is if Jimbo gets to kick on his first try, 1/6^1 is if Jimbo gets to kick on his second try, 1/6^2 is if Jimbo gets to kick on his third try...). This can then be simplified to 1/3*6/5 by the rules for summing up geometric series which then reduces to 2/5, your answer.
That was fun. |
tsimkin 
Dec 31, 2008
| Another way to solve this is as follows:
P(Unco) = 1/2 + (1/2)(1/3)(P(Unco))
This is because Unco either wins on his first shot (with p = 1/2), or misses, and then Jimbo misses, and then Unco is back where he started. This equation becomes:
P(U) = 1/2 + 1/6*P(U)
5/6*P(U) = 1/2
P(U) = 3/5
so P(J) = 2/5
Excellent teaser. |
Jimbo
Dec 31, 2008
| Thanks tsimkin. An excellent method that you have suggested!  |
Zag24
Feb 07, 2009
| Good puzzle. I also used tsimkin's technique, but, of course, I made a stupid arithmetic mistake and got the wrong answer. Doh!  |
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