At a carnival game, a winning player spins a wheel

Question

At a carnival game, a winning player spins a wheel that always lands on either Prize 1 or Prize 2 to determine which of the two prizes he or she wins. The probability that the prize wheel indicates Prize 2 is double the probability that it indicates Prize 1. If a player does not want the prize that the prize wheel first indicates, then he or she may spin the wheel again. In such cases, the player must accept whichever prize the prize wheel indicates on the second spin.

Select for Prize 1 the number nearest to the probability that a winning player who wants Prize 1 will receive Prize 1 after one or two spins of the prize wheel, and select for Prize 2 the number nearest to the probability that a winning player who wants Prize 2 will receive Prize 2 after one or two spins of the prize wheel. Make only two selections, one in each column.

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Official Answer

Prize 1: 0.5 Prize 2: 0.9

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mitshivi2024 · Accepted Answer

My Approach:

let P(p1) = x, P(p2) = 2x => x+2x = 1 => 3x =1
so x= 1/3.

1. For Prize 1:

Two possible scenarios: 
a. Wants p1 - gets in first time - 1/3
b. Gets p2 - but as it wants p1 - spin again and get p1 => 2/3*1/3
Combined => 1/3 + 2/9 = 5/9 => 0.555 = 0.5 (approx.)

2. For Prize 2:

Two possible scenarios: 
a. Wants p2 - gets in first time - 2/3
b. Gets p1- but as it wants p2 - spin again and get p1 => 1/3*2/3
Combined => 2/3 + 2/9 = 8/9 => 0.888.. = 0.8 (approx.)

Hope that helps ! 
Thanks

chetan2u · Answer

As the Source and OA are correct now, the solution would be

The prize wheel can be taken to be consisting of three parts,   out of which prize 1 one occupies one part and prize 2 occupies the remaining two parts.

(1) Prize 1 after one or two spins-  
 First way:  Total P - no prize 1 after two spins =  1 - (1-\frac{1}{3})(1-\frac{1}{3})=1-\frac{4}{9}=\frac{5}{9}  or 55.55% closest to 0.5 in options. 
 Second way:  P in first spin + P in second spin = \frac{ 1}{3} + (\frac{2}{3}*\frac{1}{3}) = \frac{5}{9}

(2) Prize 2 after one or two spins-  
 First way:  Total P - no prize 2 after two spins =  1 - (1-\frac{2}{3})(1-\frac{2}{3})=1-\frac{1}{9}=\frac{8}{9}   or 88.88% closest to 0.9 in options. 
 Second way:  P in first spin + P in second spin =  \frac{2}{3 }+ (\frac{1}{3}*\frac{2}{3}) = \frac{8}{9}

bhattarbharat

Dbrunik · Answer

got this right but why cant they just put the precise answers. pretty frustrating, given you get access to a calculator here.

DanTheGMATMan · Answer

Go through the 2 scenarios for each prize. Odds of prize one are 1/3 and odds of prize 2 are 2/3:

https://www.youtube.com/watch?v=9wH_Ty74wsg

KarishmaB · Answer

Given: At a carnival game, a winning player spins a wheel that always lands on either Prize 1 or Prize 2 to determine which of the two prizes he or she wins. The probability that the prize wheel indicates Prize 2 is double the probability that it indicates Prize 1. means that there are only two possible outcomes - Prize 1 and Prize 2. So total probability of prize 1 + probability of prize 2 = 1 Also probability of prize 2 is twice that of prize 1 so probability of prize 1 = 1/3 and of prize 2 = 2/3 PRIZE 1 A player who wants prize 1 will either get prize 1 on first spin and walk away - Probability = 1/3 or get prize 2 on first spin and will spin a second time and get prize 1 - Probability = (2/3)*(1/3) = 2/9 Total Probability = 1/3 + 2/9 = 5/9 We know that 1/9 = 11.11% (fraction percent equivalents) so closest answer is 0.5. Select 0.5 PRIZE 2 A player who wants prize 2 and gets it will either get prize 2 on first spin and walk away - Probability = 2/3 or get prize 1 on first spin and will spin a second time and get prize 2 - Probability = (1/3)*(2/3) = 2/9 Total Probability = 2/3 + 2/9 = 8/9 We know that 1/9 = 11.11% (fraction percent equivalents) so 8/9 is 88.88%. Closest answer is 0.9 Select 0.9 Fraction percent equivalents discussed here: https://anaprep.com/arithmetic-simple-a ... rcentages/

Goldenfuture · Answer

1 - prob of neither chance results in prize 2. = 1 - 1/3*1/3 = 8/ 9 Answer to second part

nikfield · Answer

We can treat the two spins as independent events, so the key idea here is that the probability of two independent outcomes happening one after another is the product of their individual probabilities.
Since Prize 1 is 1/3 and Prize 2 is 2/3, the only way a player does not get the prize they want is by missing on both spins.

For someone who wants Prize 1, the "wrong result" each time is Prize 2, which has probability 2/3.
Missing twice is (2/3) × (2/3) = 4/9,
so the chance of eventually getting Prize 1 is 1 − 4/9 = 5/9 ≈ 0.56, rounded to 0.5.

For a player who wants Prize 2, the wrong outcome is Prize 1 with probability 1/3.
Missing twice is (1/3) × (1/3) = 1/9,
so the success rate is 1 − 1/9 = 8/9 ≈ 0.89, rounded to 0.9.

This kind of multiplication rule appears all the time in probability problems, and it shows up naturally in gambling scenarios. A classic example is roulette, where each spin is independent just like in this question. On a European wheel the probability of landing on red is 18/37, while on an American wheel it drops to 18/38 because of the extra pocket. The chance of losing two red bets in a row is therefore (19/37)2 ≈ 0.264 in Europe but (20/38)2 ≈ 0.277 in the U.S. I came across comparisons like this on sites such as  onlinecasinovelemenyek.com  and they really help build intuition for situations like this where independent outcomes combine through multiplication.

cspatadia · Answer

I think it is meant to test the limits of focus and patience in a time constrainst pressure situation. Can you still stay focused to do rounding to the best among available options after doing all these calculations?

At a carnival game, a winning player spins a wheel

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