Basic statistics/probability question

[vanizorc]

Moderator note: moved to homework section[/size]

I'm not quite sure how to frame this question/thought of mine, as I'm not a math/physics student, but here goes: say you're in a game of Russian Roulette where the probability in each round of getting killed is 1/6. I need to show that, with more rounds that occur, the probability of getting killed will correspondingly increase (despite the fact that the 1/6 chance of getting killed in each round will remain constant). So for example, the probability of getting killed after 3 rounds is less than the probability of getting killed after 9 rounds.

Does anyone know how to show the formula for such a thought experiment?

 

[Stephen Tashi]

http://en.wikipedia.org/wiki/Geometric_distribution

 

[HallsofIvy]

I think your question is ambiguous. If you mean "The chance of being killed in three rounds" (i.e on the first, second, or third rounds) is less than the chance of being killed in nine rounds" then you can calculate- the chance of being killed in the first round is 1/6. If you are not killed on the first round (probability 5/6), your chance of being killed on the second round is also 1/6. If you are not killed on the first or second round (probability (5/6)^2= 25/36), your chance of being killed on the third round is also 1/6. Your chance of being killed on one of the first three rounds is 1/6+ (5/6)(1/6)+ (5/6)(5/6)(1/6), the "Geometric distribution" Stephen Tashi references. You chance of being killed on one of the first nine rounds is that plus the probability of being killed on the fourth, fifth, sixth, seventh, eighth, or ninth rounds so, of course, it is larger.

But if you mean "given that you were not killed on the first two rounds, the chance of being killed on the third round" and "given that you were not killed on the first eight rounds, the chance of being killed on the ninth round", one is NOT larger than the other, they are both exactly 1/6.

 

[vanizorc]

http://en.wikipedia.org/wiki/Geometric_distribution

Thanks.

 

[vanizorc]

I think your question is ambiguous. If you mean "The chance of being killed in three rounds" (i.e on the first, second, or third rounds) is less than the chance of being killed in nine rounds" then you can calculate- the chance of being killed in the first round is 1/6. If you are not killed on the first round (probability 5/6), your chance of being killed on the second round is also 1/6. If you are not killed on the first or second round (probability (5/6)^2= 25/36), your chance of being killed on the third round is also 1/6. Your chance of being killed on one of the first three rounds is 1/6+ (5/6)(1/6)+ (5/6)(5/6)(1/6), the "Geometric distribution" Stephen Tashi references. You chance of being killed on one of the first nine rounds is that plus the probability of being killed on the fourth, fifth, sixth, seventh, eighth, or ninth rounds so, of course, it is larger.

But if you mean "given that you were not killed on the first two rounds, the chance of being killed on the third round" and "given that you were not killed on the first eight rounds, the chance of being killed on the ninth round", one is NOT larger than the other, they are both exactly 1/6.

Your former interpretation of my question is what I had meant. Basically, that even though the probability of getting killed in each round stays stagnant at 1/6, you nevertheless have an increasing chance of getting killed with a greater number of rounds you're put through. This question of mine came about because, during a discussion with a friend, my friend claimed that the probability of being killed in Russian Roulette stays the same no matter how many rounds you go through, and I rebutted that this is incorrect because with every additional round, another 1/6 chance of getting killed gets introduced into the picture.

 

[mfb]

Your friend probably refers to the probability of getting killed in a specific round (like "the next one"), if you survive all previous rounds: this is indeed constant (it is 1/6).
The probability to get killed in some round within the first n increases with increasing n, of course.

 

[Ray Vickson]

Your former interpretation of my question is what I had meant. Basically, that even though the probability of getting killed in each round stays stagnant at 1/6, you nevertheless have an increasing chance of getting killed with a greater number of rounds you're put through. This question of mine came about because, during a discussion with a friend, my friend claimed that the probability of being killed in Russian Roulette stays the same no matter how many rounds you go through, and I rebutted that this is incorrect because with every additional round, another 1/6 chance of getting killed gets introduced into the picture.

Both you and your friend are right, because you are (likely) talking about different things, but using the same words.