Author Topic: Arcade House Party Conundrum  (Read 14050 times)

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Offline Weehawk

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Arcade House Party Conundrum
« on: August 31, 2014, 03:18:10 pm »


Jason and Anna Cram went to an arcade party where they and four other married couples were the only people present.

Some of the people at the party had met before, some had not. During the course of the evening, each pair of people that had not met previously made it a point to play a game of Donkey Kong together. No other Donkey Kong was played.

At the end of the party, Jason asked everyone in attendance, including Anna, how many people they played Donkey Kong with. To his surprise, Jason got a different answer from each person.

How many people did Anna play Donkey Kong with at the party?

For the first person who can post the correct answer with an explanation why it is correct who does not already have one, I have a t-shirt:



Please recuse yourself if you were already familiar with the problem.

If you already have the shirt, you can PM your answer to me and I will credit you in the thread.
« Last Edit: August 31, 2014, 03:37:43 pm by Weehawk »
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Offline ChrisP

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Re: Arcade House Party Conundrum
« Reply #1 on: August 31, 2014, 04:11:47 pm »
Are you positive that there isn't a piece of information missing here? ;D

"Someâ„¢" had or had not met before...
http://donkeykongblog.blogspot.com

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Offline ChrisP

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Re: Arcade House Party Conundrum
« Reply #2 on: August 31, 2014, 04:16:40 pm »
<mad>
http://donkeykongblog.blogspot.com

4 Quarters :-* - 800K Avg. Per Qtr. :o - No Restarts 8) - No Proof :'(

7/26/2013   Coin 35,946   710,800   18-1
7/28/2013   Coin 35,947   903,700   22-1
8/16/2013   Coin 35,948   694,100   17-6
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3,201,700: the $1 World Record?
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Offline homerwannabee

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Re: Arcade House Party Conundrum
« Reply #3 on: August 31, 2014, 04:46:23 pm »
6?

Edit:  Too stupid to see the explanation thing.  Suck!
« Last Edit: August 31, 2014, 04:48:43 pm by homerwannabee »
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Offline ChrisP

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Re: Arcade House Party Conundrum
« Reply #4 on: August 31, 2014, 04:46:30 pm »
Here's what I have so far and I don't even know if I'm on the right track at all, so somebody can take this and run with it, fttt this I'm gone.

- There are 10 people total
- The maximum number of people each individual could play DK with is 8 (since they would definitely know themselves and their spouse, but could potentially not have met any of the others).
- The question says that *each* person gave Jason a different answer. I interpret that to mean that he got 9 *distinct* answers (that is, nobody's answer was the same as anybody else's answer).
- There are 9 possible answers, zero through 8.
- However, zero AND 8 could not have BOTH been given as answers. One or the other is possible, but not both. If an attendee played zero games, it means they knew everybody at the party, which means that there could not have been an attendee who played 8 games, because you can't simultaneously have a person who has met everybody and a person who has met nobody.
- So Jason only could have gotten 8 distinct answers.

That's where I'm stuck, and I haven't even gotten to how Anna's answer could be narrowed down from the information given.
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4 Quarters :-* - 800K Avg. Per Qtr. :o - No Restarts 8) - No Proof :'(

7/26/2013   Coin 35,946   710,800   18-1
7/28/2013   Coin 35,947   903,700   22-1
8/16/2013   Coin 35,948   694,100   17-6
8/17/2013   Coin 35,949   893,100   22-1

3,201,700: the $1 World Record?
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Offline homerwannabee

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Re: Arcade House Party Conundrum
« Reply #5 on: August 31, 2014, 04:50:34 pm »
Here's what I have so far and I don't even know if I'm on the right track at all, so somebody can take this and run with it, fttt this I'm gone.

- There are 10 people total
- The maximum number of people each individual could play DK with is 8 (since they would definitely know themselves and their spouse, but could potentially not have met any of the others).
- The question says that *each* person gave Jason a different answer. I interpret that to mean that he got 9 *distinct* answers (that is, nobody's answer was the same as anybody else's answer).
- There are 9 possible answers, zero through 8.
- However, zero AND 8 could not have BOTH been given as answers. One or the other is possible, but not both. If an attendee played zero games, it means they knew everybody at the party, which means that there could not have been an attendee who played 8 games, because you can't simultaneously have a person who has met everybody and a person who has met nobody.
- So Jason only could have gotten 8 distinct answers.

That's where I'm stuck, and I haven't even gotten to how Anna's answer could be narrowed down from the information given.

What you said, and my answer is 7

since 0 and 8 can't be given as answers.   ;D
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Offline VON

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Re: Arcade House Party Conundrum
« Reply #6 on: August 31, 2014, 04:51:48 pm »
Chris, we seriously have to stop meeting like this FailFish  I'll type faster next time...

If I'm reading this correctly...

Jason asked everyone in attendance how many people they played Donkey Kong with, and got a different answer from every person.

There were 10 people in total at the event, or 5 couples, which means the most number of different people any one person could have played with is 8 (because you can't play with yourself or your partner).

Thus, for Jason to have received 9 different answers, those answers must have been 8,7,6,5,4,3,2,1, and 0.

However, if someone played with 0 other people, then suddenly the maximum number of different people any one player could have played with becomes 7, and the problem no longer makes sense.

So, what am I misinterpreting about the Arcade House Party Conundrum?

Offline Weehawk

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Re: Arcade House Party Conundrum
« Reply #7 on: August 31, 2014, 04:59:46 pm »
Here's what I have so far and I don't even know if I'm on the right track at all, so somebody can take this and run with it, fttt this I'm gone.

- There are 10 people total
- The maximum number of people each individual could play DK with is 8 (since they would definitely know themselves and their spouse, but could potentially not have met any of the others).
- The question says that *each* person gave Jason a different answer. I interpret that to mean that he got 9 *distinct* answers (that is, nobody's answer was the same as anybody else's answer).
- There are 9 possible answers, zero through 8.
- However, zero AND 8 could not have BOTH been given as answers. One or the other is possible, but not both. If an attendee played zero games, it means they knew everybody at the party, which means that there could not have been an attendee who played 8 games, because you can't simultaneously have a person who has met everybody and a person who has met nobody.
- So Jason only could have gotten 8 distinct answers.

That's where I'm stuck, and I haven't even gotten to how Anna's answer could be narrowed down from the information given.

What you said, and my answer is 7

since 0 and 8 can't be given as answers.   ;D

The answer is incorrect, and the reasoning you borrowed is fatally flawed.
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Offline The Pro

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Re: Arcade House Party Conundrum
« Reply #8 on: August 31, 2014, 05:03:42 pm »
It's a classic arcade party so obviously there's a few liars and cheats in there, it won't add up.
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Offline ChrisP

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Re: Arcade House Party Conundrum
« Reply #9 on: August 31, 2014, 05:06:41 pm »
ROFL
http://donkeykongblog.blogspot.com

4 Quarters :-* - 800K Avg. Per Qtr. :o - No Restarts 8) - No Proof :'(

7/26/2013   Coin 35,946   710,800   18-1
7/28/2013   Coin 35,947   903,700   22-1
8/16/2013   Coin 35,948   694,100   17-6
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3,201,700: the $1 World Record?
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Offline Weehawk

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Re: Arcade House Party Conundrum
« Reply #10 on: August 31, 2014, 05:09:38 pm »
It's a classic arcade party so obviously there's a few liars and cheats in there, it won't add up.

I naively overlooked this. In my fictional scenario everyone told the truth.
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Offline VON

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Re: Arcade House Party Conundrum
« Reply #11 on: August 31, 2014, 05:21:34 pm »
I've got it!

One of the party goers was a schizophrenic having an episode who either played with themselves or their partner.

T-shirt me baby!!

Or  <Pigger> was on the scene, flying solo.

Boom!  T-shirt me again Kappa

Offline Weehawk

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Re: Arcade House Party Conundrum
« Reply #12 on: August 31, 2014, 05:27:58 pm »
Or  <Pigger> was on the scene, flying solo.

 FailFish
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Offline Barra

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Re: Arcade House Party Conundrum
« Reply #13 on: August 31, 2014, 05:29:24 pm »
You say:

"At the end of the party, Jason asked everyone in attendance"

Can we make any assumption that not everyone was still at the party when Jason asked?
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Offline Weehawk

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Re: Arcade House Party Conundrum
« Reply #14 on: August 31, 2014, 05:34:40 pm »
You say:

"At the end of the party, Jason asked everyone in attendance"

Can we make any assumption that not everyone was still at the party when Jason asked?

No. He asked everyone who attended. Nobody left before he asked.
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