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zex20913
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True.

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12-13-2003 at 12:59 AM
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NoahT
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To narrow it down even further, are Beth and Jim fish, and the killer a bird, cat, or dog?

-Noah

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12-13-2003 at 01:23 AM
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zex20913
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You got it. I was thinking cat, but the others could work too I suppose. Maybe not the bird...unless it was an ostrich...or another big bird like that.

But it's your turn for a puzzle now.

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12-13-2003 at 02:26 AM
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DiMono
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I was just going to propose it as a murder-suicide, where the bullets happened to shatter some glass. Not as creative, but also viable.

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12-14-2003 at 04:25 AM
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Allright, Noah, you're holding up the game...if you don't post a new puzzle by 6:00 PST tonight I'm going to have to rule that Matt C. gets to post a puzzle, since he was the first to post that he knew it but was waiting for someone that didn't. That's three full days since your last post, which is more than long enough.

Game on,

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12-15-2003 at 07:17 PM
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Mattcrampy
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How does that one go...?

Tick tick tick tick boom!

Matt

Edit: Oh wait, it's puzzle tag, not lyric tag. Right:

A private detective was first at the scene of a shooting, in a 1920's style bar. Wooden three-legged stools, tables and a small cabaret stage. It was a Tuesday night, so there was only the body, his assailant and one other customer. The detective learnt that the man hasn't paid his debts to the manager of the bar, who was filling in for his sick bartender. However, the video surveillance showed that the bartender had been working the bar the entire time, the date was current, and the only bullets the bartender had were in the rusty old shotgun out the back, which wasn't working.

The only other customer said that he had seen the shooting, kind of. He was down on the floor, putting a napkin underneath his stool to stop it rocking, when he heard loud voices. The bartender was around the back at the time. He then heard a gunshot. He looked up and saw a dark-haired man running for the toilets, but he had escaped through the window by the time the customer had got there.

The video showed the bartender go round the back, and the customer had grabbed a napkin and disappeared from sight per his story.

When the police arrived, the detective explained his story, then suggested the customer be arrested and charged.

Why?

(All the information you need is here.)

Matt

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12-16-2003 at 12:50 PM
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The customer is lying. A three-legged stool cannot rock.

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12-16-2003 at 12:56 PM
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Mattcrampy
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Dammit, that was too easy.

Your turn.

Matt

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12-16-2003 at 01:29 PM
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All right, here's a mathematical one.

While out for a walk one day, mathematics professor Jones meets one of his old friends, the much older teacher Anderson. Anderson has been married twice and has three children.
At the end of the conversation, Jones asks Anderson: "Tell me, how old are your children by now?"
"Well," Anderson responds, "I know you like riddles, so here's one. The product of their ages is 2450, and the sum of their ages is the age of our mutual friend, Mr. Stone. Think about it, and see if you can figure it out." And then he leaves and goes home.
Jones attended Mr. Stone's birthday party a few days ago, so he knows this age. And since he's a skilled mathematican, he figures that he should be able to work out the ages easily. But later that night, he calls Anderson and says:
"That really wasn't nice of you. You know perfectly well that you didn't give me enough information to work out the ages of your children."
Anderson responds: "That's right, but I'll only tell you that my oldest child is younger than you are." Then he hangs up.
With this extra piece of information, Jones was able to work out the ages.

How old is professor Jones?

[Edited by Watcher on 12-16-2003 at 01:44 PM GMT]

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12-16-2003 at 01:42 PM
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Mattcrampy
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What, exactly? I can only narrow it down to a range of five years, between 20 and 25.

Matt

[Edited by Mattcrampy on 12-16-2003 at 02:34 PM GMT: Halved my range]

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12-16-2003 at 02:26 PM
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Schik
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That's a good one!

Okay, 2450 is 7*7*5*5*2. Since Jones knows the sum and the product, but it's still not enough information, that means that there must be two ways to make the same sum. The only two factorizations of 2450 into three numbers that have the same sum are:

7, 7, 50
and
5, 10, 49

The only way that the extra information "My oldest is younger than you are" would be important is if Jones is 50 - if he was 51 or older, then either of the above possibilities would still be possible.

So, Jones is 50.


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12-16-2003 at 02:47 PM
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quote:
Schik wrote:
So, Jones is 50.



That's right! Your turn.

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12-16-2003 at 03:05 PM
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Mattcrampy
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Didin't think of that.

Matt

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12-16-2003 at 04:04 PM
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Schik
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Okay, next puzzle:

There is a "free" action figure in my breakfast cereal. There are four different action figures, and I want to collect all of them. Assuming there is an equal chance of getting any one of the figures, what is the expected number of boxes I must purchase to get all four?

Just to clarify - I'm looking for the average number of boxes it would take to get all four action figures. The minimum is 4 and the maximum is infinite - What's the average?

[Edited by Schik on 12-16-2003 at 05:23 PM GMT]

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12-16-2003 at 04:09 PM
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zex20913
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I'm going to say 11, even though the actual one would probably be more.

The first box you buy is guaranteed to have a *new* toy. The next box has a 3/4 chance. The next has a 2/4, and the last has a 1/4.

I think I remember from a probability section, or something, that if you multiply the reciprocals, you get the answer to questions like this. So, 1*4/3*2*4=10 2/3. Rounding up, this is 11.

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12-16-2003 at 04:48 PM
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Schik
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That is incorrect.

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12-16-2003 at 04:49 PM
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zex20913
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Good thing I didn't post a puzzle then :P I think I see what I did wrong, but I'll leave it to someone else.

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12-17-2003 at 01:27 AM
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Schik
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If you think you know it, feel free to answer again - I don't think there's any rule against making multiple guesses.

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12-17-2003 at 02:23 AM
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Four. If you don't get all four, swap on ebay.
For 7 boxes, 8400 out of 16384 are won.
12-17-2003 at 02:26 AM
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Schik
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While 4 is a creative answer, it's not what I'm looking for. :)

Nor is 7.

Zex's reasoning was pretty close, except for a simple error - what he ended up calculating was (the reciprocal of) the chance that you'll get all 4 with your first four boxes.



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12-17-2003 at 02:31 AM
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Schik
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For 7 boxes, 8400 out of 16384 are won.

Looking at your attachment, it bothered me for a while that your answer was different from mine. Then I realized that you're calculating at what point more than 50% of the permutations of buying k boxes gives you all four figures.

That's different from what I asked - you're basically finding the median, I'm asking for the mean.

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12-17-2003 at 04:38 AM
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DiMono
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To collect them in a given number of tries, the first one will always be new and the last has a 1/4 chance of being new. This means we're really dealing with the middle two, to find out how many tries it takes.

We're going to have some number of 1/4 (first one repeated) followed by a 3/4 (second one), then some number of 1/2 (one of the first two repeated) followed by a 1/2 (third one), followed by some number of 3/4 (one of first 3 repeated) followed by the 1/4 at the end. This really means some number of 3/4 and some number of 1/2 and some number of 1/4, all >= 1.

Since it's all >= 1, we start with 6/64 for finding them in the first 4, or .09375

If it's 5, we have 6/256 + 12/256 + 18/256 = 36/256 = .1406
(* 1/4 or * 2/4 or * 3/4)

If it's 6, we have 6/1024 + 12/1024 + 18/1024 + 24/1024 + 36/1024 + 54/1024 = 150/1024 = .1465
(* 1/4 * 1/4 or * 1/4 * 2/4 or * 1/4 * 3/4 or * 2/4 * 2/4 or * 2/4 * 3/4 or * 3/4 * 3/4)

If it's 7, we have 6/4096 + 12/4096 + 18/4096 + 24/4096 + 36/4096 + 54/4096 + 48/4096 + 72/4096 + 108/4096 + 162/4096 = 540/4096 = .1318

This means that if you're looking for the most likely number of box for finding the fourth figure, it's going to be the sixth one, so the answer is:

6

Edit: I know there's a formula for this, but I don't remember what it is, and don't feel like deriving it at 12:30 in the morning

[Edited by DiMono on 12-17-2003 at 05:33 AM GMT]

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12-17-2003 at 05:30 AM
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Oneiromancer
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I'll answer for Schik--you're wrong. :)

Game on,

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12-17-2003 at 06:10 AM
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DiMono
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Well, the reasonable answers are running thin...

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12-17-2003 at 06:34 AM
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Scott
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I don't know anything about probability so I wrote a program to find out. Ran 1000000 tests and the answer was 8.34 meaning 9.
12-17-2003 at 06:36 AM
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DiMono
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I'd love to be reminded of the formula, could you post it up when the correct answer is provided, Schik?

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12-17-2003 at 06:38 AM
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mrimer
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Scott wrote:
Ran 1000000 tests and the answer was 8.34 meaning 9.
The expected number of tries of get the first novel one is 1 (duh). Then the average number of tries to get the second novel one is 4/3 (i.e. you'll get a new one 3 out of 4 times, which requires slightly more than one try, on average, because you won't get it every time). Then...guess how many tries on average it would take to get one of two new ones when there are four in all: 4/2 = 2 tries. Then, to get the last novel one out of four occurs 1 out of 4 times. In other words, four tries are required on average to get it. To sum up:

1 + 4/3 + 2 + 4 = 8 1/3 tries

(Of course, you say How can you have a third of a try? It's not an actual try, it's an average.)

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12-17-2003 at 07:04 AM
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Schik
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Well, I'm in a bit of a pickle here. Technically, Mike is more correct, but I won't be anal about it. Scott was close enough for me, despite not solving it with the method I had in mind. Mike, thanks for the solution description - it's how I solved it.

Scott, you're up!

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12-17-2003 at 07:14 AM
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No I'll pass to Mike.
12-17-2003 at 07:20 AM
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zex20913
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Yup. That was exactly what I did wrong. I was all like "Moron! You add them, not multiply them!" But I let it be.

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12-17-2003 at 07:22 AM
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