Easy math question.

[Conradas]

I know this is very basic calculations, but my highschool math seems lost by now.
First one: If you throw a die 3 times, what are the chances you throw a 6 at least one time?
Second one: If you were to play russian roulette (very hypothetical [img]/images/graemlins/smile.gif[/img] ) and there is one bullet in the chamber where it could have been six, how do you calculate your chances for survival? (50% seems pretty right here..., but how to calculate by using the numbers 1/6, 1/5, 1/4?)

[BruceZ]

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First one: If you throw a die 3 times, what are the chances you throw a 6 at least one time?

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1 minus the probability that you throw 3 non-sixes.

1 - (5/6)^3 =~ 42.1%.


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Second one: If you were to play russian roulette (very hypothetical [img]/images/graemlins/smile.gif[/img] ) and there is one bullet in the chamber where it could have been six, how do you calculate your chances for survival? (50% seems pretty right here..., but how to calculate by using the numbers 1/6, 1/5, 1/4?)

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Playing 3 times it is the same as the probability of throwing 3 non-sixes or (5/6)^3 =~ 58.9%.

[AaronBrown]

One minor addition to BruceZ's excellent answer. The Russian roulette number assumes you spin after each shot. If you instead spin once then pull the trigger three times, you have only a 50% chance of survival. Spinning is better because one time in six you end up with the same chamber you used the first time, which you know is empty.

I mention it because when these problems are paired, it's usually to make this point.

[BruceZ]

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One minor addition to BruceZ's excellent answer. The Russian roulette number assumes you spin after each shot. If you instead spin once then pull the trigger three times, you have only a 50% chance of survival.

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That's right, if you don't spin each time, then it's 1/2. To answer the OPs question about how to get this, you note that the probability of each chamber having a bullet is 1/6, and since you are effectively choosing 3 chambers, the probability that one of these 3 chambers contains the bullet is 1/6 + 1/6 + 1/6 = 1/2. You can add these probabilities because the events are mutually exclusive, that is, only 1 of the chambers can actually have the bullet, and P(A or B or C) = P(A) + P(B) + P(C) when A, B, and C are mutually exclusive events.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
One minor addition to BruceZ's excellent answer. The Russian roulette number assumes you spin after each shot. If you instead spin once then pull the trigger three times, you have only a 50% chance of survival.

[/ QUOTE ]

That's right, if you don't spin each time, then it's 1/2. To answer the OPs question about how to get this, you note that the probability of each chamber having a bullet is 1/6, and since you are effectively choosing 3 chambers, the probability that one of these 3 chambers contains the bullet is 1/6 + 1/6 + 1/6 = 1/2. You can add these probabilities because the events are mutually exclusive, that is, only 1 of the chambers can actually have the bullet, and P(A or B or C) = P(A) + P(B) + P(C) when A, B, and C are mutually exclusive events.

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Note how this differs when we spin each time. The probabilities for each chamber are still 1/6, but we can't just add 3 of these because now it is possible for the same chamber to come up more than once, that is, the events of the selected chamber containing the round are not mutually exclusive. Of course the chamber with the round won't really come up more than once, because once it does we blow our brains out, and there is no one left to spin again. This is no matter (for the sake of the calculation I mean). We still pretend that we are going to spin 3 times no matter what, and we say that we can't add 1/6 + 1/6 + 1/6 because it is possible that 2 or 3 spins have the round, and this would have the effect of counting the times that 2 spins have it twice, and it counts the times that 3 spins have it 3 times. So we would actually have to subtract the probability that 2 rounds have it, and then subtract twice the probability that 3 rounds have it. The probability that 2 rounds have it is C(3,2)*(1/6)^2*5/6, and the probability that all 3 round have it is (1/6)^3. So the probability that at least 1 round has it is:

1/6 + 1/6 + 1/6 - C(3,2)*(1/6)^2*5/6 - 2*(1/6)^3 =~ 42.1%.

So the probability that we surive is 1 minus this or 58.9%. This agrees with the (5/6)^3 that we computed before more easily.

[1huskerfan]

Second one, with those numbers:

(1 - 1/6)*(1 - 1/5)*(1 - 1/4) = 1/2

[BruceZ]

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Second one, with those numbers:

(1 - 1/6)*(1 - 1/5)*(1 - 1/4) = 1/2

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Right, that's an alternative to 1 - ( 1/6 + 1/6 + 1/6) for the case where we don't spin each time.

Instead of 1 - [P(A) + P(B) + P(C)], you have P(A')*P(B'|A')*P(C'|A'B'), where ' means NOT, P(B'|A') means the probability of B' GIVEN A', and A'B' means A' AND B'.

[1huskerfan]

Yeah. I thought the OP wanted to know how to get 1/2 using 1/6, 1/5, and 1/4.

[BruceZ]

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Yeah. I thought the OP wanted to know how to get 1/2 using 1/6, 1/5, and 1/4.

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Oh, I get it. I thought those were just examples.

[Conradas]

Thanks to you all for the replies. Get how it works now. About the roulette it is right that I assumed you didnt spin between each time.