Harder 3Unit (1 Viewer)

[OLDMAN]

Anyone for harder 3Unit?

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?

 

[Constip8edSkunk]

I'm not sure if this is right:

Considering case by case
In the case of 9 birds excaping, ie 1 shot, there are 10ways
In the case of 8 birds excaping, ie 2 shot, there are 10C2*2^8*2!=23040 ways
In the case of 7 birds excaping, ie 3 shot, there are 10C3*3^7*3!=1574640ways
In the case of 6 birds excaping, ie 4 shot, there are 10C4*4^6*4!=20643840ways
In the case of 5 birds excaping, ie 5 shot, there are 10C5*5^5*5!=94500000ways
In the case of 4 birds excaping, ie 6 shot, there are 10C6*6^4*6!=195955200ways
In the case of 3 birds excaping, ie 7 shot, there are 10C7*7^3*7!=207446400ways
In the case of 2 birds excaping, ie 8 shot, there are 10C8*8^2*8!=116121600ways
In the case of 1 birds excaping, ie 9 shot, there are 10C9*9*9!=32659200ways
In the case of 0 birds excaping, ie 10 shot, there are 10!=3628800ways

therefor the average for all cases is approx. 3.339, thus the average number of birds expected to escape is 3

 

[turtle_2468]

I would disagree... well this is short so it might be wrong, but anyway, here goes...

Consider a random bird (say, the one with a cheesy grin on its face) out of the 10. Its probability of surviving is (0.9)^10 (as 0.9 prob that one of the other birds gets chosen, and all shots go off at same time so doesn't matter whether that bird has been shot before) = 0.3487
But this was just a bird we randomly chose right? So the expected no. of birds killed would be 3.487 which rounds off to 3... but dodgily, 'cause 3.5 is much more accurate.

 

[OLDMAN]

Turtle(congrats!) stole my punchline.
Anyway here is the true story.
Consider a particular duck out of the ten, call him Donald. The probability
of Donald not turning into a swan in afterlife is the probability of all ten
hunters ignoring him, that is 0.9^10. Therefore the probability Donald
surviving is 0.349. Now since there are 10 ducks, the expected survival
is 10*0.349=3.49. Seems that Donald, Huey, Dewey and half of Louie might still go on quacking.

 

[turtle_2468]

lol... haven't done 4U for too long, my mind is rusty but well done for trying to make it interesting for ppl...

 

[spice girl]

Umm...for n shooters tryin their luck on n ducks
as n -> infinity
you have survival chance of lim{n -> infinity}(1 - 1/n)^n = 1/e

i think.....

 

[OLDMAN]

Good one, Spice.

Interesting to see how this question morphs out to another question. A duck-friendly examiner who is into violent video games might come up with the following.

I) Natural Born Killers (NBK) are all perfect shots. Ten NBK's are on a rampage. All 10 pick another NBK at random to shoot at, and all 10 NBK's fire at the same time. How many NBK's could be expected to survive.
II) If instead of 10, we have n NBK's. What is the expected proportion of NBK's that will continue wreaking havoc in this world? Hence prove that Hell is 1/e^2.

 

[bobo123]

oldman do you play tfc? NBK was this clan from long time ago

 

[OLDMAN]

No, too old for video games. But I do keep an eye on what my kids are playing.

Hint : Consider a particular NBK, call him Cheap Shot.

 

[OLDMAN]

Correction:

Might have put some off track. Prove Hell is 1/e not 1/e^2.

 

[OLDMAN]

Just to provide a closure, before this thread drifts downwards to oblivion.
I) Consider a particular NBK, Cheap Shot. Now consider another particular NBK, Ear Shot. The probability that Cheap Shot doesn't get shot by Ear Shot is 8/9 or 1-1/9. Since there are 9 NBK's like Ear Shot that might possibly get a shot at Cheap Shot, the probability that Cheap Shot survives is (1-1/9)^9. And, since there are 10 NBK's, the number of NBK's expected to survive is 10(1-1/9)^9=3.46. Quite close to the duck result, showing how quickly (1-1/n)^n approaches 1/e.
II) The general result for the proportion of NBK's expected to survive with n NBK's, give (1-1/(n-1))^(n-1). To prove hell is 1/e, prove that (1-1/(n-1))^(n-1) --->1/e as (n-1)--->00 (infinity).
This however is not 4U material as we have to invoke l'Hospital's rule, which states: If f(a)=g(a)=0 and if the limit f'(t)/g'(t) exists as t--->a, then lim(t--->a) f(t)/g(t)= lim(t--->a) f'(t)/g'(t).
Sufficient to prove the generaL case (1-1/x)^x=1/e as x--->00.
Let y=(1-1/x)^x. Take natural logs both sides,
ln(y)=x*ln(1-1/x), as this has form 00*0, we must rearrange to
ln(y)=ln(1-1/x)/(1/x). Now this has the form of the indeterminate 0/0, and after differentiating numerator and denominator separately as prescribed by Dr. l'Hospital, we get
lim ln(y)= lim[-(1-1/x)(1/x^2)/(1/x^2)] x--->00
Thus lim ln(y)=-1 as x--->00. Which gives y--->e^(-1)=1/e.