Question 1117249: This is the old Monopoly question but I think the question is ambiguous.
In the game of Monopoly, a player who is in jail can get out of jail by throwing a double on the two dice on any of his or her next 3 turns. Both dice show 1, 2, 3, 4, 5, and 6. What is the probability that a player will get out of jail by throwing a double on the next 3 turns?
I interpreted "by throwing a double on the next 3 turns" as being a double on any of the next 3 turns as you only need to throw one double to get out of jail. But, my friend interpreted as being throwing a double on all three turns.
So, for me, I figured it would be 1/6 * 1/6 * 1/6 = 0.0046
My friend though is arguing it should be 1/6 + 1/6 + 1/6 = 0.5
Right now, we are both chasing our alegrabraic tales and just confusing each other. Thanks so much.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The problem is worded poorly, making different interpretations possible. However, since the question is
"What is the probability that a player WILL GET OUT OF JAIL by throwing a double on the next 3 turns?"
I would agree with you that it is only asking the probability that the player will get out of jail by throwing a double on ONE of the next three turns.
However, neither calculation you show is correct for the respective interpretations.
For your friend's interpretation, that the player gets doubles on all three of the next turns, the probability is the calculation you show for YOUR interpretation: (1/6)*(1/6)*(1/6) = 1/216.
For your interpretation, the probability has to interpreted as
(1) get a double on turn 1, OR
(2) not get a double on turn 1 AND get a double on turn 2, OR
(3) not get a double on turn 1 AND not get a double on turn 2 AND get a double on turn 3
Replacing the ORs with addition and the ANDs with multiplication, the calculation for that probability is
Notice the probability can also be calculated as 1 minus the probability that he does NOT get a double on any of the next three turns:
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