Question

# A slot machine has 3 dials. Each dial has 30 positions, one of which is "Jackpot"....

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. 1/30 = 0.03 or 3%

2. 3/(30+30+30) = 3/90 = 0.33 or 33%

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

4. 1/(30×30×30) = 1/27000 = 0.00003 or 0.003%

We have 30 positions in each dial, with 1 position being "Jackpot".
Prob. of one play resulting in "Jackpot" in the first dial = 1/30.
Prob. of one play resulting in "Jackpot" in the second dial = 1/30.
Prob. of one play resulting in "Jackpot" in the third dial = 1/30.

Hence, odds of one play winning the jackpot = Prob. of one play resulting in "Jackpot" in the first dial x Prob. of one play resulting in "Jackpot" in the second dial x Prob. of one play resulting in "Jackpot" in the third dial = 1/(30x30x30) = 1/27000 = 0.00003 = 0.003%.

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