Circus of Luck 3

Probability Level 4

Larry enters a newly opened circus which houses 11 11 tents (Tent A , B , C , . . . , K A, B, C, ... , K ) that each features a free-to-play game. The game of Tent A A involves a computer that will flash one from 2 2 symbols. If the contestant guesses what symbol will be flashed, he wins 2 $ 2\$ . The game of Tent B B involves a computer that will flash one from 3 3 symbols. If the contestant guesses what symbol will be flashed, he wins 3 $ 3\$ . The game of Tent C C involves a computer that will flash one from 4 4 symbols. If the contestant guesses what symbol will be flashed, he wins 4 $ 4\$ . And so on. Tent A ( 1 2 chance of winning , 2 $ prize ) , Tent B ( 1 3 chance of winning , 3 $ prize ) , Tent C ( 1 4 chance of winning , 4 $ prize ) , . . . , Tent K ( 1 12 chance of winning , 12 $ prize ) \text{Tent} \space A \space (\dfrac{1}{2} \space \text{chance of winning}, \space 2\$ \space \text{prize}), \space \text{Tent} \space B \space (\dfrac{1}{3} \space \text{chance of winning}, \space 3\$ \space \text{prize}), \space \text{Tent} \space C \space (\dfrac{1}{4} \space \text{chance of winning}, \space 4\$ \space \text{prize}), \space ... \space , \space \text{Tent} \space K \space (\dfrac{1}{12} \space \text{chance of winning}, \space 12\$ \space \text{prize})

Let's say a "scenario" is the experience of Larry playing in the circus (winning all the games, winning all except Tent F F , losing all games, etc.)

Question : If Larry plays all games each once, how many scenarios are there that have more than 1 % 1\% chance of happening?

Try Circus of Luck 1


The answer is 20.

Kaizen Cyrus by Kaizen Cyrus , 18, Philippines

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1 solution

Kaizen Cyrus
May 21, 2019

Let the fraction below be the probability of losing/winning in a game.

n m \dfrac{n}{m}

Since the amount of dollars as a prize is equal to m m ( 1 4 \frac{1}{\textcolor{#20A900}{4}} chance of winning, 4 $ \textcolor{#20A900}{4}\$ prize), n n could either be 1 \textcolor{#3D99F6}{1} if it's the probability of winning or m − 1 \textcolor{#D61F06}{m-1} if it's the probability of losing.

chance of winning chance of losing Tent C 1 4 3 4 \begin{array}{lcc} & \text{chance of winning} & \text{chance of losing} \\ \text{Tent} \space C &\textcolor{#3D99F6}{\dfrac{1}{4}} & \textcolor{#D61F06}{\dfrac{3}{4}} \end{array}

The equation below is how we calculate the probability of a scenario to happen. The individual factor is the chance of winning/losing in a game while the product is the probability of the scenario .

a 2 × b 3 × c 4 × d 5 × e 6 × f 7 × g 8 × h 9 × i 10 × j 11 × k 12 = x 479 , 001 , 600 \scriptsize \dfrac{a}{2}×\dfrac{b}{3}×\dfrac{c}{4}×\dfrac{d}{5}×\dfrac{e}{6}×\dfrac{f}{7}×\dfrac{g}{8}×\dfrac{h}{9}×\dfrac{i}{10}×\dfrac{j}{11}×\dfrac{k}{12} = \dfrac{x}{479,001,600}

Now, x 479 , 001 , 600 \frac{x}{479,001,600} should be larger than 1 % 1\% . The equations below are the possibilities of scenarios that are larger than the mentioned percentage.

1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 479 , 001 , 600 = 1 12 or 8.33 % \scriptsize \dfrac{\textcolor{#69047E}{1}×2×3×4×5×6×7×8×9×10×11}{479,001,600} = \dfrac{1}{12} \space \text{or} \space 8.33\% 1 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 479 , 001 , 600 = 1 24 or 4.17 % \scriptsize \dfrac{\textcolor{#69047E}{1}×3×4×5×6×7×8×9×10×11}{479,001,600} = \dfrac{1}{24} \space \text{or} \space 4.17\% 1 × 2 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 479 , 001 , 600 = 1 36 or 2.78 % \scriptsize \dfrac{\textcolor{#69047E}{1}×2×4×5×6×7×8×9×10×11}{479,001,600} = \dfrac{1}{36} \space \text{or} \space 2.78\% 1 × 2 × 3 × 5 × 6 × 7 × 8 × 9 × 10 × 11 479 , 001 , 600 = 1 48 or 2.08 % \scriptsize \dfrac{\textcolor{#69047E}{1}×2×3×5×6×7×8×9×10×11}{479,001,600} = \dfrac{1}{48} \space \text{or} \space 2.08\% 1 × 2 × 3 × 4 × 6 × 7 × 8 × 9 × 10 × 11 479 , 001 , 600 = 1 60 or 1.67 % \scriptsize \dfrac{\textcolor{#69047E}{1}×2×3×4×6×7×8×9×10×11}{479,001,600} = \dfrac{1}{60} \space \text{or} \space 1.67\% 1 × 2 × 3 × 4 × 5 × 7 × 8 × 9 × 10 × 11 479 , 001 , 600 = 1 72 or 1.39 % \scriptsize \dfrac{\textcolor{#69047E}{1}×2×3×4×5×7×8×9×10×11}{479,001,600} = \dfrac{1}{72} \space \text{or} \space 1.39\% 1 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 479 , 001 , 600 = 1 72 or 1.39 % \scriptsize \dfrac{\textcolor{#69047E}{1}×4×5×6×7×8×9×10×11}{479,001,600} = \dfrac{1}{72} \space \text{or} \space 1.39\% 1 × 2 × 3 × 4 × 5 × 6 × 8 × 9 × 10 × 11 479 , 001 , 600 = 1 84 or 1.19 % \scriptsize \dfrac{\textcolor{#69047E}{1}×2×3×4×5×6×8×9×10×11}{479,001,600} = \dfrac{1}{84} \space \text{or} \space 1.19\% 1 × 2 × 3 × 4 × 5 × 6 × 7 × 9 × 10 × 11 479 , 001 , 600 = 1 96 or 1.04 % \scriptsize \dfrac{\textcolor{#69047E}{1}×2×3×4×5×6×7×9×10×11}{479,001,600} = \dfrac{1}{96} \space \text{or} \space 1.04\% 1 × 3 × 5 × 6 × 7 × 8 × 9 × 10 × 11 479 , 001 , 600 = 1 96 or 1.04 % \scriptsize \dfrac{\textcolor{#69047E}{1}×3×5×6×7×8×9×10×11}{479,001,600} = \dfrac{1}{96} \space \text{or} \space 1.04\%

Since Tent A A has the same chance of winning in it as losing in it ( 1 2 (\textcolor{#3D99F6}{\frac{1}{2}} chance of winning, 1 2 \textcolor{#D61F06}{\frac{1}{2}} chance of losing), we double the numbers of scenarios above. Doing that, we get 20 \boxed{20} numbers of possible scenarios that has a chance more than 1 % 1\% of happening.

2 Helpful 0 Interesting 1 Brilliant 0 Confused

From the chances of losing each game = 1 12 = \frac{1}{12} , If we instead win game N N we divide by N N .

We want this probability to be above 1 100 \frac{1}{100} , so we can divide up to at most 8 8 . This can be done by 2 , 3 , 4 , 5 , 6 , 7 , 8 , 2 ∗ 3 , 2 ∗ 4 2, 3, 4, 5, 6, 7, 8, 2*3, 2*4 , these, as well as the "lose all games" scenario can all be paired with their equal-likely counterparts of "Win game 1".

This totals 20 20 .

Alex Burgess - 2 years ago

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