Montana Duck HuntersDate: 11/27/2001 at 21:49:34 From: Nanette Mattos Subject: Probability Montana duck hunters area all perfect shots. Ten Montana hunters are in a duck blind when 10 ducks fly over. All 10 hunters pick a duck at random to shoot at, and all 10 hunters fire at the same time. How many ducks could be expected to escape, on average, if this experiment were repeated a large number of times? Text answer: 3.49 (theoretical probability) I have presented this problem to many high school mathematics teachers, and we are stumped. The closest solution I can come up with is 4.99, using many sample spaces, Pascal's triangle, etc. I hope you can help me understand what I'm doing wrong. Thank you, Nanette Date: 11/28/2001 at 00:29:33 From: Doctor Schwa Subject: Re: Probability Hi Nanette, This is certainly a classic problem! I find several references to it online, such as Virtual Laboratories - Kyle Siegrist Expected Value > Definitions and Properties http://www.math.uah.edu/stat/expect/expect1.html Finite Sampling Models > The Number of Distinct Sample Values http://www.math.uah.edu/stat/urn/urn8.html Here are my various attempts at solving this problem. It starts out pretty long and messy, gets an answer using a computer to help, but then I find a pattern in that answer, and use the pattern to figure it out. The first hunter kills one duck. The second hunter kills 9/10 of a duck on average (he has a 1/10 chance of shooting at the same duck). The third hunter, well, now that's tricky. If there's one duck dead so far, that's a (1/10) chance, and then the third hunter kills 9/10 of a duck. If there are two ducks dead, then the third hunter kills 8/10 of a duck. This looks pretty messy. Is there any way to streamline it? Sure! Let f(n,d) = the probability that d ducks are dead after n hunters have shot. Then f(1,1) = 1. f(2,1) = 1/10 and f(2,2) = 9/10. For the third hunter to end up with 1 dead you must have had two hunters with 1 dead and then hit the same one, so f(3,1) = 1/10 * f(2,1). To end up with 2 dead, you can either have 1 dead after two hunters and then hit any other duck (9/10 chance), or have 2 dead and hit one of them again (2/10 chance). f(3,2) = 9/10 * f(2,1) + 2/10 * f(2,2). Similarly f(3,3) = 8/10 * f(2,2). I'll do one more and hope for a pattern: f(4,1) = 1/10 * f(3,1) [I inserted that space for a reason.] f(4,2) = 9/10 * f(3,1) + 2/10 * f(3,2) f(4,3) = 8/10 * f(3,2) + 3/10 * f(3,3) f(4,4) = 7/10 * f(3,3) I see enough of a pattern now that I can put this into an Excel spreadsheet easily enough, and get (out of 10^10 possibilities) 10 ways that 1 duck dies 45990 ways that 2 ducks die 6717600 ways that 3 ducks die 171889200 ways that 4 ducks die 1285956000 ways that 5 ducks die 3451442400 ways that 6 ducks die 3556224000 ways that 7 ducks die 1360800000 ways that 8 ducks die 163296000 ways that 9 ducks die 3628800 ways that 10 ducks die. A little expected value computation means that on average 6.513215599 ducks die, so 3.486784401 ducks live. The spreadsheet is at http://mathforum.org/dr.math/gifs/mattos11.27.01.xls (Download the file, then open it with Excel.) That's not a very elegant method, though. I wonder if there's an easier way. With numbers that ugly, I'm not so sure. Ah, now I see something (and I added the calculation to the spreadsheet): with 1 hunter, the expected number dead is 1. With 2 hunters, the expected number dead (out of 10) is 1 + 9/10. With 3 hunters, the expected number dead is 1 + (9/10) + (9/10)^2 and so on. That is a great clue to a nice proof! You might want to stop here and try to figure it out for yourself. Okay, the first hunter certainly kills a duck. The second hunter has a 9/10 chance of killing a duck. What's the chance that the third hunter kills a duck? Well, suppose the third hunter shoots at duck number k. What's the chance that neither of the first two hunters shot that same duck? (9/10)^2, of course. Hence the expected number of ducks killed by n hunters is 1 + (9/10) + (9/10)^2 + ... + (9/10)^(n-1). That formula checks out in the small cases, and as n goes to infinity its limit is indeed 10 ducks killed, so I think this answer makes sense. Now to try to figure out the standard deviation... - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ Date: 11/28/2001 at 15:37:45 From: Doctor Anthony Subject: Re: Probability Suppose there are p people and N ducks. From the point of view of a particular duck the probability that someone will shoot at that duck is 1/N. We have p independent trials, so this is binomial probability. The expected number shooting at any duck is p/N. The probability that no one will shoot at that particular duck is P(0) = (1 - 1/N)^p = [(N-1)/N]^p The expected nunber of ducks with no one shooting at them = N.[(N-1)/N]^p (N-1)^p = --------- N^(p-1) If p = 10 and N = 10 this expression gives 9^10 ------ = 3.48678 10^9 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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