The worlds first riddle!

One way to do the problem is to think of the dwarves' portions as a 7-dimensional vector, and of the redistributions as 7x7 matrices. Multiply the matrices together to get the matrix representation of the composition of all 7 transformations and find the eigenvector that corresponds to an eigenvalue of 1.

Another way is to guess.

They all deserve an entire glass of milk. Mark agrees that the distribution of milk among the seven dwarfs (in ounces) is:
12, 10, 8, 6, 4, 2, and 0.

Haha; Kolyo, you sure do have a good imagination!
Thanks for the extra dimensions; I wonder if it would work in this situation…


I don’t wish to appear immodest, but my buddies and I are all perfect shots - you have to be in this neck of the woods in case an atheist Democrat should show up!

So nine buddies and me are holed up when all of a sudden, 10 Canadian geese fly over with a look that said we are better than native Louisiana wildlife!

That ruffled more feathers than a ‘hanging chad’, so…..

We all pick a goose at random to shoot at, and all 10 of us fire at the same time.

How many geese on average could be expected to escape if this scenario were to be repeated a large number of times?



Please note: No Democrats were harmed in the composition of the conundrum.

Have a good weekend guys; Ima heading for New York as I hear the sidewalks are paved with gold!

That's essentially what I did for a 3-dwarf model. Then I realized there is a pattern.

geese:

10 - 6.513215599 = 3.486784401

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Mark is a darn good shot – but has he hit the mark?

The first hunter kills one goose.
The second hunter kills 9/10 of a goose on average (he has a 1/10 chance of shooting at the same goose).

The third hunter, well, now that's tricky. If there's one goose dead so far, that's a (1/10) chance, and then the third hunter kills 9/10 of a goose. If there are two geese dead, then the third hunter kills 8/10 of a goose.

This looks pretty messy. Is there any way to streamline it?

Sure! Let f(n,d) = the probability that d geese 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 spread sheet easily enough, and get (out of 10^10 possibilities)

10 ways that 1 goose dies
45990 ways those 2 geese die
6717600 ways those 3 geese die
171889200 ways those 4 geese die
1285956000 ways those 5 geese die
3451442400 ways those 6 geese die
3556224000 ways those 7 geese die
1360800000 ways those 8 geese die
163296000 ways those 9 geese die
3628800 ways those 10 geese die.

A little expected value computation means that on average 6.513215599 geese die, so 3.486784401 geese live… Good for them!

Oh, and good for Mark!


Talking about ten:

I have been busy writing ten begging letters and corresponding envelopes.

Then after copious shots of home brew I inserted the letters into the envelopes randomly (one letter per envelope). Hic!!!

Before I mail them, I need to know…
What is the probability that exactly nine letters were inserted in the proper envelopes?

There are 3,628,800 ways to randomly put the letters in the envelopes.
10!/(10-9)! = 3,628,800
If nine letters are in the proper envelopes, it would mean that the tenth would in the correct envelope as well. Therefore, there is only one possible way to have all the letters all in their respective envelopes from the 3,628,800 ways.
This means that the probability of this arrangement is 1/3,628,800 = 2.75573192*10^-7.

envelopes:
zero - can't get exactly nine correct
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I think I see where you are getting at. Are you taking the question to mean that if all 9 letters are inserted in proper envelopes then what is the probability that the last 1 should be in the improper envelope? If then, it is a contradiction, and yes the probability will be zero. The question to me was a little ambiguous.

Yes - the key is "exactly nine."

And it's not as though Tryagain wasn't dropping hints left and right...

How to more naturally follow up a question about multitudes of dead geese than with a little ray of hope for the future of that species -- a goose egg.

Clever!

Hey; I just wanna say that you guys are ok, and take heed of what Mark says as he is by now au fait with my devious, tricky, conniving, scheming, Machiavellian ways.

However I’m still not considered dastardly enough to join the GOP, so I sat the entrance test for the Mafia instead.

It is a 26 question test; eight points were credited for each correct answer and five points were deducted for each wrong answer.

It was in the form of tick the box on multi-choice options. I thought I had dun ok and answered all the questions, so when Tony the Terminator told me my score was zero; I was aghast!

Needless to say to avoid my lamentable academic result becoming public knowledge I gave him a Republican kiss - Tony now swims with the fishes; but I crave your indulgence and seek enlightenment as to:

How many questions did I actually answer correctly? (If any).


Thank you and happy fishing!

Answer: 10 correct, 16 incorrect

Thankfully, Try's just warming up for the day.

Hi DD, AKA Merlin, long time no C, I hope you and your family are doing well. I see you have lost none of your joie de vivre or your acumen in the intervening period.

I have no eye deer how DD arrived at his correct answer, but this is one way:

The number of answers in each category is inversely proportional to the value, so there were [5/ (5+8) ] (26) or 10 correct answers!

I was walking along the street when a cop approached me and asked, "Where were you between 6 and 11?" "At Elementary school" I replied, which is where I started with one bacterium.

Every minute, every bacteria turns into 0, 1, 2, or 3 bacteria with a probability of 25% for each case (dies, does nothing, splits into 2, or splits into 3).

What is the probability that the bacteria population eventually dies out?


The same cannot be said about Politicians, who like diapers, need changing regularly………. And for the same reason!

Eventually dies out? 100%

bacteria population:
I'll say sqrt(2)-1 which is approximately 0.41421. This is a root of p^3 + p^2 - 3p + 1 = 0.
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If each bacteria has a 25% chance of dying every generation, then isn't inevitable that they will eventually all go toes up on one of the iterations?

There's a non-zero probability, with a denominator that increases exponentially as the population grows, that they'll all die out on a given iteration, but it's not a certainty. I approached it this way:

Let p be the probability that they eventually die out.
p = 1/4 + p/4 + (p^2)/4 + (p^3)/4

Hmmm, 1 is also a root of the polynomial I used.

Found this:
Truly, I cant see why it cant be 1... Let qn denote the probability that the nth generation is empty, then q0=0, qn+1=f(qn) for the function f one knows, and the probability of extinction is q=limn→∞qn. Obviously q is a fixed point of f, that is, q=f(q). For every fixed point x>=0 of f, q0<=x and f is increasing hence qn<=x for every n hence q<=x. Thus q must be the smallest nonnegative fixed point.

Note that if the problem is changed such that there are three possible outcomes (0, 1, 2) with probability 1/3 each, then both roots of the resulting polynomial are 1.