Maddenation
Got any duck food?
I borrowed this puzzle from the Math Forum an online service for students and teachers from Drexel University. I got to this site while researching Paul Nahin, the author of the Dueling Idiots… book I got for Christmas (or Birthday, I forget).
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?
I won’t give the answer because you can get that from the website if you like. The problem is an interesting one because it’s simple to state, and you have an intuitive feel for the answer because you know there’s a good chance some of the hunters will pick the same duck, thereby hitting it with two or more bullets. My initial guess was that certainly more than one duck escapes, on average, but surely not more than 5. The right answer (on average) is in that range, but not really that easy to derive. Like many probability problems, it’s “tricky” and kind of hard to think about.
There are some other neat problems listed on the forum site. The one on the probability of Joe Dimaggio’s hitting streak is especially interesting because it’s based on a real life case everybody knows about. The answer may surprise you.
Dad • Puzzles • 01/05/05 • 6 comments
Comments
Patrick • 01/06/05 • 1:41 PM:I start thinking about it like this: at least one duck will always die, assuming the hunters never miss (and if they all shoot the same duck, maybe there won’t be much meat left for eating). And all ten ducks will die only if each hunter chooses a different duck (which would be rare). Then I get stuck and look at the answer. Holy mackerel! I guess I’ll take their word for it, but I feel uncomfortable by their apparent chronology. I guess it doesn’t matter (they’re saying “hunter 1 shoots and kills a duck, then hunter 2 shoots and has a 9/10 chance of killing a duck”) but their wording is wrong. All the hunters are shooting at the same time, right? I know, it’s just to simplify the problem, but it still feels weird.
Dad • 01/06/05 • 4:23 PM:I think one of the biggest advantages of the human brain is keeping more than one concept in mind in a kind of “stop-motion time warp.” This allows us to solve problems in slow motion by breaking them apart into separate pieces. One of the mistakes this causes is failure to recognize when chronology matters, such as when events interact with one another. In this case, the problem solver assumes that, while the hunters all shoot at the same time, their actions are independent, as long as during the “time warp” subsequent hunters don’t notice that the duck they were aiming at is already shot.
If you think this answer is complicated, take a look at the one on the hitting streak!
David • 01/08/05 • 11:17 PM:I actually guessed four would escape. (and roommate Tim guessed 3). The average is the right answer. What are the chances? Dad - do you understand all that stuff when you see the answers? I don’t. It means almost nothing to me. But I like it and it’s interesting. You’re right, the hitting streak is very complex.
Here’s what I want to know - what are the chances of someone (me) getting a flat tire, two years in a row, on his birthday? I don’t know - but I just got a flat tire yesterday, Jan. 7, almost one year to the day of last year’s birthday flat tire. Weird.
Dad • 01/09/05 • 5:04 PM:Yes, I understand most of that stuff, but not all. As for your getting a flat tire on successive birthdays, that’s a pretty low probability. (Unless you do things on your birthday that greatly increase the odds of getting flats relative to the rest of the year.)
If you assume your probability of getting a flat on a given day is about 1:1000, then the probability of getting 2 in one year (let alone on your birthdays) would be only 4.6%. this result can be gotten using the binomial distribution (BINOMDIST function in EXCEL) for 2 “successes” in 365 tries (days) with a probability of 1/1000. The probability of getting flats on 2 specific days in that year (or 366 days, to include both birthdays) is just like getting flats 2 days in a row (I think). This is about one in a million, given the assumed daily flat probability of 1/1000. If one day before or after your birthday counts, that’s 3 days, so we must multiply the probability by 3. So in your case, it would come to 9 in a million. Still rare. So how would you be betting on Jan. 6, 2006?
Patrick • 01/10/05 • 3:56 PM:Is that 1:1,000 figure correct? That would mean a flat every 2.75 years, and I don’t think I get them that often. Maybe it’s not too far off, though.
Dave, you should just start saying that it was on your birthday both years. Forget the January 7th detail, because it messes up a better story. Pretty soon, you’ll misremember the date and really think it was on your birthday both years, and then it will be true.
Dad • 01/11/05 • 4:31 PM:How do I know if 1:1000 is correct? I don’t think I’ve had had a flat tire in over 10 years, and then it was a rental car. I figured David doesn’t take very good care of his car, so for him I used 1:1000 number.
By the way, Paul Nahin cites an example of someone (from Princeton) in his book for whom it snowed on 17 straight birthdays. He did not calculate the odds, but one can at least infer that she had a winter birthday.
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