Question

A slot machine has 3 dials. Each dial has 30 positions, one of which is "Jackpot". To win the jackpot, all three dials must be in the "Jackpot" position. Assuming each play spins the dials and stops each independently and randomly, what are the odds of one play winning the jackpot?

1/30 = 0.03 or 3%


3/(30+30+30) = 3/90 = 0.33 or 33%


3/(30×30×30) = 3/27000 = 0.0001 or 0.01%


1/(30×30×30) = 1/27000 = 0.00003 or 0.003%

Answer 1

(1/30)^3, or 3.7·10^(-5)

Explanation:

The probability of each dial reaching the jackpot position is 1/30, as each dial has 30 positions. The behaviors of the three dials are independent, and so the probability of reaching the jackpot on each is (1/30)^3; we merely multiply the three probabilities together. Winning the jackpot consists of all three dials being at 'jackpot' position.