Circus of Luck
Larry enters a newly opened circus which houses 1 1 tents (Tent A , B , C , . . . , K ) that each features a free-to-play game. The game of Tent A involves a computer that will flash one from 2 symbols. If the contestant guesses what symbol will be flashed, he wins 2 $ . The game of Tent B involves a computer that will flash one from 3 symbols. If the contestant guesses what symbol will be flashed, he wins 3 $ . The game of Tent C involves a computer that will flash one from 4 symbols. If the contestant guesses what symbol will be flashed, he wins 4 $ . And so on. Tent A ( 2 1 chance of winning , 2 $ prize ) , Tent B ( 3 1 chance of winning , 3 $ prize ) , Tent C ( 4 1 chance of winning , 4 $ prize ) , . . . , Tent K ( 1 2 1 chance of winning , 1 2 $ prize )
Question : If Larry plays all games once each, what is the probability in percentage ( % ) that Larry wins a total of 1 2 $ ?
NOTE : The answer must be rounded to the nearest hundredths.
Try Circus of Luck 2 , 3 , and 4
The answer is 4.55.
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2 solutions
Nicely explained!
Chances of losing every game = 2 1 3 2 . . . 1 2 1 1 = 1 2 1 . If we instead winning a game worth N $ , we divide this by ( N − 1 ) .
We can make 12$ by:
s u m 1 2 1 0 + 2 9 + 3 8 + 4 7 + 5 7 + 3 + 2 6 + 4 + 2 5 + 4 + 3 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ P r o b a b i l i t y 1 2 ∗ 1 1 1 1 2 ∗ 9 ∗ 1 1 1 2 ∗ 8 ∗ 2 1 1 2 ∗ 7 ∗ 3 1 1 2 ∗ 6 ∗ 4 1 1 2 ∗ 6 ∗ 2 ∗ 1 1 1 2 ∗ 5 ∗ 3 ∗ 1 1 1 2 ∗ 4 ∗ 3 ∗ 2 1
Total = 1 2 1 ∗ ( 1 1 1 + 9 1 + 1 6 1 + 2 1 1 + 2 4 1 + 1 2 1 + 1 5 1 + 2 4 1 ) = 6 6 5 2 8 0 3 0 2 4 1 = 0 . 0 4 5 4 5 6
For Larry to win a total of 1 2 $ , he must either:
Meeting up with either of those scenarios, the prize(s) he'd win will add up to 1 2 $ .
Now, let's compute the probability of each scenario.
Scenario 1 : Winning only Tent K 2 1 × 3 2 × 4 3 × 5 4 × 6 5 × 7 6 × 8 7 × 9 8 × 1 0 9 × 1 1 1 0 × 1 2 1 = 0 . 7 5 7 6 %
Scenario 2 : Winning only Tent A and I 2 1 × 3 2 × 4 3 × 5 4 × 6 5 × 7 6 × 8 7 × 9 8 × 1 0 1 × 1 1 1 0 × 1 2 1 1 = 0 . 9 2 5 9 %
Scenario 3 : Winning only Tent B and H 2 1 × 3 1 × 4 3 × 5 4 × 6 5 × 7 6 × 8 7 × 9 1 × 1 0 9 × 1 1 1 0 × 1 2 1 1 = 0 . 5 2 0 8 %
Scenario 4 : Winning only Tent C and G 2 1 × 3 2 × 4 1 × 5 4 × 6 5 × 7 6 × 8 1 × 9 8 × 1 0 9 × 1 1 1 0 × 1 2 1 1 = 0 . 3 9 6 8 %
Scenario 5 : Winning only Tent D and F 2 1 × 3 2 × 4 3 × 5 1 × 6 5 × 7 1 × 8 7 × 9 8 × 1 0 9 × 1 1 1 0 × 1 2 1 1 = 0 . 3 4 7 2 %
Scenario 6 : Winning only Tent A , B , and F 2 1 × 3 1 × 4 3 × 5 4 × 6 5 × 7 1 × 8 7 × 9 8 × 1 0 9 × 1 1 1 0 × 1 2 1 1 = 0 . 6 9 4 4 %
Scenario 7 : Winning only Tent A , C , and E 2 1 × 3 2 × 4 1 × 5 4 × 6 1 × 7 6 × 8 7 × 9 8 × 1 0 9 × 1 1 1 0 × 1 2 1 1 = 0 . 5 5 5 6 %
Scenario 8 : Winning only Tent B , C , and D 2 1 × 3 1 × 4 1 × 5 1 × 6 5 × 7 6 × 8 7 × 9 8 × 1 0 9 × 1 1 1 0 × 1 2 1 1 = 0 . 3 4 7 2 %
NOTE : All of the results above are rounded to the nearest ten-thousandths.
Adding up those possibilities, the chance of Larry winning 1 2 $ , rounded to the nearest hundredths, is 4 . 5 5 % .