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05-04-2009 , 12:39 PM
I have a fleet of 18 ships. Assuming that I lose 1.5 ships on average per year, meaning that each individual ship has an 8.33% of sinking.

What is the probability of losing exactly one ship? Two ships?

I've been trying to wrap my head around this and getting nowhere
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05-04-2009 , 07:36 PM
Irish,

Think about it this way. Give each ship a name, say "A", "B", ..., "whatever the 18th letter is".

What's the probability of losing ship "A" and ONLY ship "A"?

Well, there's an 8.33% chance of losing ship "A" and a 91.67% chance of NOT losing ship "B" and a 91.67% chance of NOT losing ship "C" and ...

So the probability that you lose exactly ship "A" is:
(0.0833)*(0.9167)^17.

Since it's the same probability that we lose exactly ship "B" or ship "C" or ..., we just multiply this by the total number of ships to get the probability of losing exactly one ship.

There are 18C1 = 18 different ships we could lose if we lose exactly one.

Use a similar argument for exactly 2 ships, only there are 18C2 = 17*9 sets of 2 ships we could lose.

And so on...
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05-04-2009 , 08:00 PM
Another possibly valid (if a little less exact) way to think of it is to think of it in terms of a Poisson distribution. If you know about Poisson distributions, feel free to skip the next few paragraphs.

Spoiler:
Imagine that we flip a coin, where if it comes up heads, we lose three ships, and if it's tails, we lose no ships. That would get us the 1.5 average we're looking for, but it implies there's a 50% chance of losing three ships, a 50% chance of losing no ships, and a 0% chance of any other number. Not very realistic.

Instead, let's flip three coins and count each head as a sunken ship. Then we have a 1/8 chance of losing zero or three, and a 3/8 chance of losing one or two. That's better...but it implies that the chance of losing four or more is exactly zero. The probability of losing four or more ships is probably lower than the probability of losing any fewer number, but it's likely to be more than zero.

So, we could flip six coins and count each head as half a sunken ship. Or, flip twelve coins and count each head as a quarter of a sunken ship. If we continue to add flips and weigh them less, then we start to have the possibility of sinking more ships than we have. But, the numbers from zero to, say, six or seven look very realistic.

That's the Poisson distribution, and it's useful any time we want to estimate the probability a rare event occurs k times, if on average it would occur λ times.

The formula is .


Wyman is totally right too; he's basically using the binomial distribution instead of a Poisson distribution. They're the exact same thing, except I'm using infinitely small intervals. His will be a little more accurate; the binomial distribution will be exact if you know the probability of the sinking of each individual ship (which you do). Mine would be better if you knew you had 1.5 sinkings per year, but not the number of ships total.
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05-04-2009 , 08:49 PM
on the calculator, to find the probability of each integer, do binompdf (n,p,r) where n is total number of trials, p is probability, and r is the integer you want to find the probability of
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05-04-2009 , 10:54 PM
Awesome, thanks for the help guys.
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05-04-2009 , 11:02 PM
you cannot calculate the individual odds unless you know it follows a certain distribution. if the you lose 0 ships the odd years and 3 ships even years, then the chance of losing 1 or 2 ships would be 0%
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05-07-2009 , 12:08 PM
Quote:
Originally Posted by Unabridged
you cannot calculate the individual odds unless you know it follows a certain distribution. if the you lose 0 ships the odd years and 3 ships even years, then the chance of losing 1 or 2 ships would be 0%
I agree with this. 1.5 ships lost per year does not imply that the probability of losing a single ship in one year is 1.5/18. If the ships are together we could assume that losing 2 ships is more likely than expected because they are more likely to go into shallow waters (or some other potentially dangerous scenario) together. This could mean that the probability of losing 1 ship exactly is much lower than 1.5/18 because the probability of losing multiple ships together skews the average.

However, this would require a lot more information to actually solve. It may be a decent approximation to assume that each ship can be treated as an independent trial. In this case a binomial distribution would be appropriate and the maximum likely-hood estimator for the probability would in fact be 1.5/18. The other posters have showed how this calculation could be done. Once you had this, you could then compare to the actual data and determine if the approximation fits the data reasonably.
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05-07-2009 , 12:45 PM
Quote:
Originally Posted by Cooker
I agree with this. 1.5 ships lost per year does not imply that the probability of losing a single ship in one year is 1.5/18. If the ships are together we could assume that losing 2 ships is more likely than expected because they are more likely to go into shallow waters (or some other potentially dangerous scenario) together. This could mean that the probability of losing 1 ship exactly is much lower than 1.5/18 because the probability of losing multiple ships together skews the average.

However, this would require a lot more information to actually solve. It may be a decent approximation to assume that each ship can be treated as an independent trial. In this case a binomial distribution would be appropriate and the maximum likely-hood estimator for the probability would in fact be 1.5/18. The other posters have showed how this calculation could be done. Once you had this, you could then compare to the actual data and determine if the approximation fits the data reasonably.
Your right, I was assuming that each ship was an independent event since the information I had stated that they traveled to very different areas.
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