# Math Help - Chance of being murdered

1. ## Chance of being murdered

Hello all,
The news here keeps showing stories of murders in the city. My GF is getting freaked out. I'm trying to calculate the odds of being murdered and compare it to regular events (like win lottery, etc). I’m totally confused. This is what I know

city population this year= 2,116,581
# of murders this year = 117

Is this how I solve?:
117 / 2,116,581 = 0.0000552778
1 - 0.0000552778 = 99.9944722172
So, can I say there is a 1 in 17,000 chance of being murdered?

I have no training in this stuff, so any help is greatly appreciated.

2. 1) You are forgetting time. If this information leads directly to your probability of being murdered, that would make it about 1/18000 PER YEAR. If you live the first year 17999/18000, there is more probability of your demise in the second year. This might lead to the final answer:

1/18000 + (17999/18000)*(1/17000) + [(17999/18000)^2]*(1/17000) + [(17999/18000)^4]*(1/17000) + ...

If you add these up for 101 years, this suggests about a 0.6% chance over a 101 year lifespan.

However, there is some chance you would die of some dread disease or auto accident or something else, so this 0.6% over a lifetime is a gross over-estimate. Still more, this value assumes the probability stays the same over the entire span of 101 years. This is an assumption that is very unlikely to be observed.

2) Now the real answer. Your 1/18000 is a population average. You absolutely MUST NOT apply population averages to individuals. It is irresponsible! Your chance of being a homocide victim is FAR MORE related to your behavior than to your city. Your neighborhood is more important than your city, too, but your neighborhood may influence your behavior. There are some murders that are very "random". but not very many.

3. Would it be proper to find the probability using a binomial distribution?

$pr(X=1)={100\choose 1}(\frac{1}{18,000})(\frac{17,999}{18,000})^{99}\a pprox 0.005525\approx 0.5525\%$

4. Originally Posted by downthesun01
Would it be proper to find the probability using a binomial distribution?

$pr(X=1)={100\choose 1}(\frac{1}{18,000})(\frac{17,999}{18,000})^{99}\a pprox 0.005525\approx 0.5525\%$
No, that will give you the probability of being murdered exactly once in 100 years, assuming that the probability of murder in any one year is 1/18000. But if this was valid then:

$pr(X=2)={100\choose 2}\left(\frac{1}{18,000}\right)^2 \left(\frac{17,999}{18,000}\right)^{98}\ne 0$

would be the probability of being murdered twice in those 100 years etc..

This is clearly nonsense.

CB

5. Originally Posted by TKHunny
1) You are forgetting time. If this information leads directly to your probability of being murdered, that would make it about 1/18000 PER YEAR. If you live the first year 17999/18000, there is more probability of your demise in the second year. This might lead to the final answer:

1/18000 + (17999/18000)*(1/17000) + [(17999/18000)^2]*(1/17000) + [(17999/18000)^4]*(1/17000) + ...

If you add these up for 101 years, this suggests about a 0.6% chance over a 101 year lifespan.

However, there is some chance you would die of some dread disease or auto accident or something else, so this 0.6% over a lifetime is a gross over-estimate. Still more, this value assumes the probability stays the same over the entire span of 101 years. This is an assumption that is very unlikely to be observed.

2) Now the real answer. Your 1/18000 is a population average. You absolutely MUST NOT apply population averages to individuals. It is irresponsible! Your chance of being a homocide victim is FAR MORE related to your behavior than to your city. Your neighborhood is more important than your city, too, but your neighborhood may influence your behavior. There are some murders that are very "random". but not very many.
But however you slice this you are still more llikely to be murdered than win the lottery (especially if you don't buy lottery tickets)

CB

6. Originally Posted by CaptainBlack
But however you slice this you are still more llikely to be murdered than win the lottery (especially if you don't buy lottery tickets)

CB
With that, Sir, I cannot argue.