### Donovans mine formula

Posted:

**Sat Mar 31, 2012 5:01 am**I was looking into mine hits on donovans and was wondering about an inconsistency (or mistake on my part).

http://www.donovansvgap.com/info/spacemines.htm

They state:

Seems to me either the formula should be (d-0) or the example should be (30-1).

http://www.donovansvgap.com/info/spacemines.htm

They state:

But then in the example for 30 ly, they state:The Probability of Hitting a Mine Over A Given Distance

So how can we calculate the probability of hitting one mine when travelling a given distance (x)? To do this we use a probability concept known as: the binomial distribution. To put it simply, if you have some basic notion of probabilities, it is obvious that chance of a minehit follow a binomial distribution.

The probability of hitting a number x of mines when travelling d lightyears is B(x,d,p) if p is the minehit probability per light year travelled.

The function B is therefore:

B(x,d,p) = (d/x) * (p^x) * ((1 - p)^(d-1))

Where (d/x) = d!/(x!*(d - x)!)

and n!=1*2*...*n

Note the parts marked in red. Am I missing something, or does this not match?Assume a 30 light year mine field, with normal configuration p=1% (which means 0.01). We want to know what the probability is of crossing it without minehits:

B(0,30,0.01) = (30/0) * (0.01^0) * ((1 - 0.01)^(30 - 0)) =

= (30! / (0!)*(30 - 0)!) * 1 * ( 0.99 ^ 30 ) =

= (30! / 1 * 30!) * 1 * (0.7397) =

= 1 * 1 * 0.7397 = 0.7397

Seems to me either the formula should be (d-0) or the example should be (30-1).