Party time at Peter's and Paul's

[bonanova]

Peter and Paul, who are neighbors, each threw a party last Friday. Bad scheduling, to be sure, but that's life. Even worse, their guest lists were identical: all 100 of their friends were sent invitations to both parties. When guests arrived, the happy sounds of those already present could be heard through the two open doors, and the old phrase "the more the merrier" figured in their choice of which party to attend: If at any point there were a people present at Peter's party and b people present at Paul's party, the next guest would join Peter with probability a/(a+b) and join Paul with probability b/(a+b).

To illustrate: When the first guest arrived only the two hosts were present. (a = b =1.) So that choice was a tossup, and let's say that the first guest chose Peter's party. (a = 2; b =1.) Now the second guest would follow suit, with probability 2/3, or choose Paul's party, with probability 1/3. And so on, until all 100 guests arrived.

What is the expected number of guests at the less-attended party?

Edited January 18, 2018 by bonanova
Give probability examples

[Izzy]

  On 1/22/2018 at 1:58 PM, bonanova said:

I think it boils down to that, but how to justify doing it?

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Maybe it helps to note that "The less-attended party" doesn't a priori attach to either Peter or Paul. If the question were to ask: "If Peter's were the less-attended party, what would its expected attendance be?" would the equations change? Or, once we see that the distribution is flat, (which really is the <surprising> crux of the puzzle) and we know the less-attended party would have fewer than 50 guests, it's simply asking for the expected value of a flat distribution of numbers less than 50.

 

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Ah, you're right. I was assuming that the lesser amount of guests were all at the same party, but never considered whose party. Since there are actually two different outcomes to consider, the expected total value should be (expected value of Peter's party, given that his is less-attended) + (expected value of Paul's party, given that his is less-attended) = ~25 guests. 

Edited January 22, 2018 by Izzy
Wording.

[Izzy]

 

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It is not so hard to see that the probably that n guests attend the lesser party P(guests = n), for the n guests arriving in any order, is ((101-(n+1))!)(n!)/(101!). There are (100 choose n) possible rearrangements of these guests arriving, giving that P(guests = n) = ((101-(n+1))!)(n!)/(101!) * (100!)\[(100-n!)(n!)] = 1/101 (surprisingly). 

Thus, the expected value is 1/101 * (1 + 2 + ... + 50) = 12.63 guests at the party with less guests.

 

[bonanova]

  On 1/22/2018 at 7:20 PM, Izzy said:

 

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Hey

@bonanova, I tried to PM you, but the forum says you can't receive messages. 

It only took like ten years and some prodding, but I finally solved one of your riddles.  

I've wondered recently, what is your mathematics background like? When I was a kid, I just assumed it was normal for adults to be really good at math, but now I suspect you've done a degree in math or at least CS? I actually just started doing a math PhD, and honestly part of the reason was probably because I always had so much fun doing your riddles. So, thanks for that! 

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I have a PhD in Electrical Engineering with a minor in math. I'm retired now and I got my EE degree way before CS was even a discipline. I  worked 40 years at a (fairly well known) computer company doing EE and physics research until in 1980 they formed a CS department and (there being then no CS grads) populated it with EE types and I volunteered. Ten years after that I was the editor of their Journal of Research and Development for eight years, then webmaster of their internal and executive communications department. Four careers all inside what was then a very special company. It was quite a ride.

I'm a little in awe of your math plans; math at the PhD level is hard. Good luck with that. I'm also humbled if I've in any way been an encouragement.  Thanks for sharing.

 

 

[bonanova]

  On 1/21/2018 at 8:54 AM, Izzy said:

 

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It is not so hard to see that the probably that n guests attend the lesser party P(guests = n), for the n guests arriving in any order, is ((101-(n+1))!)(n!)/(101!). There are (100 choose n) possible rearrangements of these guests arriving, giving that P(guests = n) = ((101-(n+1))!)(n!)/(101!) * (100!)\[(100-n!)(n!)] = 1/101 (surprisingly). 

Thus, the expected value is 1/101 * (1 + 2 + ... + 50) = 12.63 guests at the party with less guests.

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@Izzy  As you say, the distribution is surprising.
To be certain of this expected attendance at the smaller party, you might want to ...

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Calculate the expected attendance at the larger party.

 

[Izzy]

  On 1/22/2018 at 3:54 AM, bonanova said:

@Izzy  As you say, the distribution is surprising.
To be certain of this expected attendance at the smaller party, you might want to ...

  Reveal hidden contents

Calculate the expected attendance at the larger party.

 

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Ah, shoot. That gives 37.37. I've lost 50 guests somehow! 

[bonanova]

@Izzy Yes.
 

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And that number is a great clue to how to fix both answers.
You've basically solved the puzzle -- just lost a bit of thinking about it.

How many parties are there?

 

[Izzy]

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@bonanova Intuitively, I just want to multiply my answer by 2, but I'm trying to figure out where I under counted. I am positive that to get n guests to attend your party (without considering all the possible ways to get n guests to attend, i.e. this counts how to get the first n people to all come to your party, or the last n people, etc.) is (100-n)!(n!)/(101!). The (100 choose n) would give me all the possible rearrangements of the n people coming to the party. Multiplying these gives the surprising 1/101, which *should* be the chances of getting exactly n people at your party, but this must be wrong? 

Is it because this calculates the chance to get exactly n people at some party, but there are actually two parties, so we must multiply by 2? 

[bonanova]

  On 1/22/2018 at 4:48 AM, Izzy said:

 

  Reveal hidden contents

 

@bonanova Intuitively, I just want to multiply my answer by 2, but I'm trying to figure out where I under counted. I am positive that to get n guests to attend your party (without considering all the possible ways to get n guests to attend, i.e. this counts how to get the first n people to all come to your party, or the last n people, etc.) is (100-n)!(n!)/(101!). The (100 choose n) would give me all the possible rearrangements of the n people coming to the party. Multiplying these gives the surprising 1/101, which *should* be the chances of getting exactly n people at your party, but this must be wrong? 

Is it because this calculates the chance to get exactly n people at some party, but there are actually two parties, so we must multiply by 2? 

Expand  

I think it boils down to that, but how to justify doing it?

  Reveal hidden contents

Maybe it helps to note that "The less-attended party" doesn't a priori attach to either Peter or Paul. If the question were to ask: "If Peter's were the less-attended party, what would its expected attendance be?" would the equations change? Or, once we see that the distribution is flat, (which really is the <surprising> crux of the puzzle) and we know the less-attended party would have fewer than 50 guests, it's simply asking for the expected value of a flat distribution of numbers less than 50.

 

[Izzy]

  Reveal hidden contents

Hey

@bonanova, I tried to PM you, but the forum says you can't receive messages. 

It only took like ten years and some prodding, but I finally solved one of your riddles.  

I've wondered recently, what is your mathematics background like? When I was a kid, I just assumed it was normal for adults to be really good at math, but now I suspect you've done a degree in math or at least CS? I actually just started doing a math PhD, and honestly part of the reason was probably because I always had so much fun doing your riddles. So, thanks for that!