unforum
a.r.g.b.b

Welcome!
 New users, PLEASE read these forum guidelines. New posters, SEARCH before posting and read these rules before posting your killer new campaign. New players may also wish to peruse the ARG Player Tutorial.All users must abide by the Terms of Service.

amandel!

Why do we need your help?
Announcements
 The Project MU Archives Now available in stunning full-color print edition.

 The time now is Sun May 19, 2013 11:45 am All times are UTC - 6 View posts in this forum since last visit View unanswered posts in this forum Calendar

Confirmation code
Subject
Subject description
Message icons
 No icon
Message body

 Emoticons View more Emoticons
 Font colour: Default Dark Red Red Orange Brown Yellow Green Olive Cyan Blue Dark Blue Indigo Violet White Black  Font size: Font size Tiny Small Normal Large Huge Close Tags
Options
HTML is ON
BBCode is ON
Smilies are ON
 Disable HTML in this post Disable BBCode in this post Disable Smilies in this post
 Topic review
Page 1 of 1 [1 Post]
Author Message
Cabbage
FairmontKing wrote:
hobyrne wrote:
Follow-up puzzle:

My birthday is 29 Feb. Where should I be in line to maximize my chance of winning?

The card (or the solve page) says to assume all birthdays have the same probability. The answer works out to be the the same if you use 365 or 366 days in a year. Slightly different probabilities, but still the max occurs at the same spot in line.

Yeah but you are less likely to encounter someone whose birthday is also the 29th Feb as it only occurs once every four years, or 1 in 1461 days as opposed to 1 in 365.

hobyrne - I would give up on getting that prize...

Posted: Sun Oct 01, 2006 1:11 pm
FairmontKing
hobyrne wrote:
Follow-up puzzle:

My birthday is 29 Feb. Where should I be in line to maximize my chance of winning?

The card (or the solve page) says to assume all birthdays have the same probability. The answer works out to be the the same if you use 365 or 366 days in a year. Slightly different probabilities, but still the max occurs at the same spot in line.

Posted: Sat Sep 30, 2006 7:05 pm
hobyrne
Follow-up puzzle:

My birthday is 29 Feb. Where should I be in line to maximize my chance of winning?

Posted: Fri Sep 29, 2006 6:57 am
mqpippin
I took statistics over 20 years ago....

I wish I had paid more attention. The only part I remember was the teacher explained something about Odds and Chances.

We started calling out our birthdates from the first row. Class size was about 33-40 students. We only made it to the 2nd or 3rd person in row 2 when we'd made a match of birthdates.

The teacher explained he'd never had a class that didn't match before the end of the 3rd row until the year that he'd had a student who had to retake the class and he organized it so that all the matched birthdays were the last seats in the last row. They had 3 matches in that class.

He'd also had twins in the class right before mine, but said "Normally, for the purposes of this class, you're supposed to ignore them."

Posted: Fri Sep 08, 2006 9:14 pm
Agent Lex
Those numbers are, I assume, if the line actually gets to you. The card is asking about the chance of not only the most likely person to win if the line gets to them (which would obviously be the 367th, taking leap years into account) but the person most likely to win considering the line has to get to them in the first place. You can't win if the person right in front of you did

Posted: Sat Aug 05, 2006 4:46 pm
Platinumflux
Actually it does say on Wikipedia that there is a more than 50% chance if you are number 23 and a 99% chance if number 60. Only at number 367 is there a 100% chance. but none of these answers are accepted as the answer!

Posted: Sat Aug 05, 2006 4:29 pm
sotonrich
just got the solve on this one and noticed after that:

Spoiler (Rollover to View):
there are 19 people from the door to the title box. The 20th person in the line on the card is hidden by the title box, prehaps its Kurt

Posted: Thu Feb 23, 2006 12:39 am
Rob_Riv
fantasticalan wrote:
Does the guy who's first in the queue really get it in the shorts?.. in the, er... non-rude sense, because he's got around a 0.00% chance of winning. No-one's gonna win 'cos no-one will be first in the queue!

But he gets first choice of ice-cream flavour.

Posted: Sat Feb 18, 2006 3:16 pm
fantasticalan
Does the guy who's first in the queue really get it in the shorts?.. in the, er... non-rude sense, because he's got around a 0.00% chance of winning. No-one's gonna win 'cos no-one will be first in the queue!

Posted: Mon Dec 05, 2005 11:26 am
c1023
A person wins if no person before them has won, and their birthday is the same as someone before them.
Spoiler (Rollover to View):
Probability of winning for person n [Pn] is:
( 1 - probability someone earlier won ) x Number of people infront / Days in a year
( 1 - ( [Pn-1] + [Pn-2] + ... + [P2] + [P1] ) ) x (n-1) / 365.254

Using Excel, column A can be the position in the queue, and column B can their probability of winning.

The first person (row 1) is easy (cannot win):
1, 0
The next person will be:
=A1+1, =(1-SUM(B\$1:B1))*(A2-1)/365.254
Copying the two item in row 2 down to the next thirty rows shows that:

Spoiler (Rollover to View):
The 20th person has the greatest chance: 3.23%

It does not make any difference whether you assume 365 days in a year, or take into account leap years and century years, it just changes the probabilities slightly.

Posted: Fri Nov 25, 2005 10:58 am
ReeKorl
I think we need to work out the chance of no collisions in any birthday in a group of size n. It's not just the next guy in the line checking to see if he's got the same as all of the others, it's all of the others checking too.

Think about it this way - take three dots, and connect them all up to each other. How many lines do you have? 3. Now take four dots, how many lines this time? 6. 5 dots, 10 lines, etc.

Every guy added to the line adds an extra amount of checks needed to get any pair in the line of (n-1).

Now, the chance of no date collisions is going to be (364/365) * (363/365) * ... * ((365-n)/365)

Spoiler (Rollover to View):
We'll only ever get a 100% chance of a pair if 365 people are in the line, but it does mean that we'll get a 50% chance of a pair after the 23rd guy is added to the line.

As we need the greatest chance of being the first duplicate, I think we need the highest increase in chance from the previous guy.

After this 23rd guy, the next guy already has more than 50% chance, but I think Curlytek had it right saying the chances tail off after the 50% mark, as if you take your chance and subtract the chance of the guy before you, the highest difference in chance is after this 50% mark, at 23 people.

Posted: Fri Nov 25, 2005 6:48 am
Curlytek
I think there are as many people in the line as required....the number can't be any bigger than 366 as there will be a match by that point. On the bright side that's only 122 days of guessing! (which of course wont be necessary with the bright unforum folks on the case)

Posted: Fri Nov 25, 2005 6:03 am
Hunting4Treasure
Wouldn't it matter how many people are in the line? They don't tell us.

Posted: Fri Nov 25, 2005 5:54 am
Curlytek
I think that the probability of being the first person who's birthday matches the birthday of someone in front of you will increase to a certain number, N, then decrease. The following isn't an answer, so probably doesn't need to be 'spoiled', but what the heck.

Spoiler (Rollover to View):
Clearly the first person in line has no chance, the second a 1/365 (or more accurately 1/365.25) chance, the third a 2/365.25 chance etc., given the random distribution of birthdays. But, for each person in front of you the probability that they will be the first, and not you, must be considered.

There is probably a formula to sum the probability of matching birthdays in a given position vs the probability that someone else has already matched.....but I'm no mathematician/statistician. My guesses would start around the halfway mark (182/365.25), and tend toward smaller values.

Posted: Fri Nov 25, 2005 5:25 am
European Chris
It's clearly related to the
Spoiler (Rollover to View):