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Birthday Trio: Newsletter Challenge (April 2014)

Posted March 31, 2014 5:01 PM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from GlobalSpec:

A school teacher notices that 3 students in his class of 30 students share the same birthday. What is the percent chance of this occurring?

Assume all days of the year are equally likely as a birthday, ignore leap years, and assume the class represents a random sample of birthdays.

And the answer is:

There is a 2.85% chance that three people among 30 share the same birthday.

This problem is the triplet variation of the The Birthday Problem. The equation to solve this problem is:

where W is the number of triplets present in the class (there could be more than one set of triplets), n is the number of students in the class, and m is the number of days in a year. [A. DasGupta, J. Statist. Plann. Inference 130, 377 (2005)]

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#1

Re: Birthday Trio: Newsletter Challenge (April 2014)

03/31/2014 5:49 PM

Reminds me of an old joke:

It's Saturday afternoon in Chicago and the bar is empty except for two patrons. One of them staggers over to the other and says, "Good day to you, sir. I can't help but saying I heard you talkin' to the bartender and I thought I heard the sound of my old homeland. Do ye' mind me askin' where you're from?"

The other guy says "I'm from God's own little piece of Heaven here on Earth. I'm from Ireland."

The first drunk says "Brilliant! I'm from Ireland too! Let's have a round for Ireland!" They both drink merrily.

Then the first guy says "So where in Ireland are you from?"

"Born and raised on the banks of the River Liffey, right in the heart of Dublin."

"Dublin? Awesome! I'm from Dublin too! Let's have another round for Dublin!" Once again, they both drink merrily.

Then the first guy asks, "So where did you go to school?"

"Blessed Saint Mary's, class of '62" answers the other guy.

"Incredible! I graduated in '62 from St. Mary's, too! Let's have a round for St. Mary's!" Once again, they suck down another round.

Just then, one of the bar regulars walks in and sits at the bar. He asks the bartender, "So what's going on today?"

The bartender answers, "Nothing much... Just the O'Malley twins getting drunk again."

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#2
In reply to #1

Re: Birthday Trio: Newsletter Challenge (April 2014)

03/31/2014 6:04 PM

Three expatriates are drinking in a NY City bar.


"As good as this is," said the Scotsman, "I still prefer the pubs back home. In Glasgow, there's a wee place called McTavish's. The landlord goes out of his way for the locals. When you buy four drinks, he'll buy the fifth.

"Well, Angus," said the Englishman, "at my local pub in London, the Red Lion, the barman will buy you your third drink after you buy the first two."

"Ahhh, dat's nothin'," said the Irishman, "back home in my favorite pub, the moment you set foot in the place, they'll buy you a drink, then another, all the drinks you like, actually. Then, when you've had enough drinks, they'll take you upstairs and see dat you gets laid, all on the house!"

The Englishman and Scotsman were suspicious of the claims.

The Irishman swore every word was true.

"Did this actually happen to you?" they asked.

"No not meself, personally, no," admitted the Irishman,
"but it did happen to me sister quite a few times."

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#66
In reply to #2

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/07/2014 5:33 AM

Back to coincidents and birthdays.

Four Brits meet in a New York bar after a while they start to introduce themselves; the Irishman says:

"I was born on St Patrick's day so I'm Patrick."

"Amazing" says the Welsh guy, "I was born on St David's day, so I'm David."

The Scotsman then declares with even more astonishment "I'm Andrew because I was born on St. Andrews day."

The three of them all turn to the Englishman wondering if all four of them could be named after their patron saints; he looks a bit sheepish and embarrassed. "Well", they ask "what's your name?"

After another embarrassed pause he replies "Errrr... Pancake."

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#3

Re: Birthday Trio: Newsletter Challenge (April 2014)

03/31/2014 10:29 PM

"What is the percent chance of this occurring?"

Not good.

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#4

Re: Birthday Trio: Newsletter Challenge (April 2014)

03/31/2014 11:12 PM

This sounds nutty but, 48.

It is about the same for getting two the same with a class of twenty two~twenty three people and almost a certainty of two the same from a class of sixty people.

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#5

Re: Birthday Trio: Newsletter Challenge (April 2014)

03/31/2014 11:36 PM

I am not going to do all the math as the system will time out, so I am relying on Excel to make things palatable.

The problem needs to be a multiple set of problems where:

1. Solve for the probability that no one has the same birthday (0.294)
2. Solve for the possibility one pair shares a birthday (0.38)
3. Solve for the probability two pairs share a different birthday
4. Solve for the probability that three pairs share different birthdays…

1+N/2
2+N/2

Then you sum those probabilities and subtract from 1 to get the probability that only three individuals in the set (30 members) share a common birthday…

And the answer is, is, is…. 42.

Not really - just fooling. It is about 0.011 or 1 out of 91.

Probably did something wrong, so I will let the engineering accountants sort out the final score.

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#7
In reply to #5

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 9:55 AM

Oh, my calculations assumed a 365 days per year without thought to leap years, which complicates the whole thing too much for me.

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#8
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 11:01 AM

The question uses only 365 day years.

365!/335!/36530= probability that no two will share a birthday.

But if two do share, they become one and the next one must come from the remaining 28.

365!/336!/36529

100*(1-365!/335!/36530)*(1-365!/336!/36529) = probability of three occurring on the same day.

I ran the years out on excel and got the 48% I posted earlier.

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#25
In reply to #8

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 6:49 PM

The probability of any 2 people out of 30 having the same birthday is roughly 71%. (As AH pointed out.) But now you need one more person of the remaining 28 to have the same birthday as those two.

So it seems to me the chance would be roughly (71%) * (28/365) = 5.4%.

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#69
In reply to #25

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/08/2014 9:51 AM

This seems to me to be the correct answer. I believe that the 48% probability calculated by passingtongreen is actually the probability that three individuals in a group of thirty have someone else in that group with the same birthday but not necessarily the same single birthday among these three. So two and three pairs of birthdays are a "true" result.

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#63
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/04/2014 12:13 PM

61,9413907971742%

nearly 2 of 3 groups of 30 persons contain 3 persons wih the same birthday!

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#86
In reply to #8

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/23/2014 11:01 PM

48,0979138803%

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#6

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 9:49 AM

It has been indicated as fact, that the teacher has 3 of 30 students with the same birthday, therefore 100% chance of this occuring.

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#15
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 12:18 PM

Sounds good to me.

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#9

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 11:24 AM

It seems like the odds of 2 people having the same birthday would be 29/365, and a third would be multiplied by 28/365, which results in about 1 out of 164.

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#12
In reply to #9

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 11:41 AM

I would agree. well stated.

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#14
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 12:07 PM

It is not. It is closer to 71% for two people with the same birthday.

The correct formula for a two person statistic is:

P = (365 * 364 * 363 * … * (365 - n+1)) / 365^n

Where n = the numbers of individuals in the set or sample (i.e., 30).

Doing the math yields 70.6% if n = 30.

Extending the problem to three people complicates the calculations as you need to know the probabilities for several other combinations (see my earlier post), then sum all of those probabilities, and subtract that result from 1.

Prob = 1 - (P + P1 + P2 + … Pw)

Where w = the number of other permeations to consider.

Why? Note that the way the puzzle is worded is such that the question asks what the probability for exactly 3 people out of a total of 30 have the same birthday, so this excludes all other combinations where, say, 4 people have the same birthday, or two sets of 3 people have the same birthdays…

That's why you must calculate the probabilities for every possible case when n=30 and why Excel is the better way to do this.

If you run the numbers for only 3 people having the same birthday and all other 27 individuals having unique birthdays the odds are just about 95 to 1.

However, if you remove that constraint from the puzzle the odds are 35 to 1.

Still in line with my original ad-hoc post of "Not good." :-)

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#21
In reply to #14

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 5:12 PM

I agree on 70.6% for two on the same day. I don't read the question as saying "exactly three", I thought of it as "at least three". I'll try it both ways.

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#23
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 6:24 PM

From the challenge question, "A school teacher notices that 3 students in his class of 30 students share the same birthday."

I am thinking that means exactly 3 out of the 30 students.

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#47
In reply to #23

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 3:52 AM

I think 'a school teacher notices' is far more indicative of 'at least three' rather than 'exactly three'.

.

There is no indication implicit or explicit that the teacher had birthday data for each student to review. Perhaps the teacher noticed because they are identical triplets, or because everyone sang happy birthday to all three that morning.

.

I don't think you can preclude the possibility of another group of three of more students sharing a different birthday from the problem as presented. I do agree with your answer of roughly 1/35...a little less than 3%.

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#87
In reply to #14

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/23/2014 11:08 PM

FORTRAN Source Code wanted?

And/or compiled Program? (for DOS or Windows System?)

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#10

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 11:27 AM

"Assume all days of the year are equally likely as a birthday"

Does this mean in reality that there are days of the year you can't be born or likely not to be born on?

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#11
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 11:39 AM

I would bet that if you did a statistical study that you would find that not every day of the year sees a statistically relevant equal distribution.

Holidays, seasons, and other events tend to skew the actual birth rates.

The following chart is for the USA, but there is no reason that the uSA is unique in the world with respect to birthrates over the course of the year.

If the puzzle did not state that assumption it would be reasonable to assume that a number of engineers here would pile on and criticize its absence.

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#17
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 2:40 PM

Just the wording of the statement. All days are equally likely that there will be a birthday. There is by your graph over 1000 births a day in the USA. So is 100% likely there will be a birth on any given day The difference is in the numbers

The graph peaks July thru October call it the out come of cabin fever.

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#18
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 3:28 PM

You did ask, "not likely to be born on." :-)

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#20
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 5:09 PM

And there's a different curve for weekdays and weekends...

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#33
In reply to #20

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 9:10 AM

What's that? Request for time off to give birth? Declined! Tell 'em to wait 'til the weekend - they ain't doing it on Company time!

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#55
In reply to #20

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 1:43 PM

I don't have my glasses on. I am having a hard time deciphering that picture.

Is that a double exposure of a woman lying on her back?

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#37
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 11:17 AM

As I remeber from the draft days of the early 1970's, there was a large furor over the fact that September birthdays were mostly high numbers (low draft chance) and it was stated then as a fact that more people were born in September than other months. So yes, there is a small percentage favoring a date in September over other months, at least at that time in history. As I recall it was assumed to be because of Christmas parties and the what occurred afterwards, after many alcoholic drinks.

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#13

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 11:52 AM

0.030.....

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#16

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 1:35 PM

Calculate how many sets of triples there are in 30 pupils (order doesn't matter).

N=(30!)/((3!)*(27!); N=4060

Probability that a triple has same birthday: (1/(365*365));

Probability that at least one triple has the same birthday = 4060/(365*365) = 3.047 percent.

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#32
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 8:29 AM

This is not exactly right. I believe #31 is correct.

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#70
In reply to #16

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/08/2014 5:31 PM

Check out this site

http://jaxwebster.wordpress.com/2011/05/29/three-birthdays-on-the-same-day/

The answer of around 3 seems right since this guy did it for 35 people and came up with 4.5%. Very official looking proof.

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#89
In reply to #70

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/24/2014 3:38 PM

Number of people

Probability of three or more people having their birthday on the same day (%)

10
20
30.00075
40.003
50.0075
60.015
70.026
80.042
90.062
100.089
110.12
120.16
130.21
140.27
150.33
160.41
170.5
180.59
190.7
200.82
210.96
221.1
231.3
241.4
251.6
261.8
272.1
282.3
292.6
302.9
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#90
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/25/2014 12:53 AM

How did someone find this comment 'off topic'. It is exactly 'on topic'. Moreover, it is also correct.

I have done my part to counter the erroneous 'off topic' rating.

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#91
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/25/2014 1:02 AM

No formula, no rationale. Correctness unknown.

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#92
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/25/2014 1:11 AM

The formula and rationale are in the preceding link.

.

.

'Correctness unknown', invites the obvious question: 'by whom'?

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#93
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/25/2014 1:20 AM

I was tempted to ask, What preceding link? I did find it, but why was it not given proximately, for those who dislike wild-goose chases? That said, the conclusion is acceptable.

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#19

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 4:49 PM

I'd love to see a clarifier added to this challenge. Is the "same birthday" condition just the same numeric day of the month, the same numeric day and month or the same day, month and year?

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#22
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 6:22 PM

This is a derivative of a classic statistics problem and all of those are about day of the year.

For that matter, I have never seen one different in that respect and the challenge did say to ignore leap years, so I vote for day of the year based on prior precedent.

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#24
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 6:43 PM

I believe that you are correct but the newsletter challenge has repeatedly made ambiguous challenges in the past. More often than not what I considered the least likely meaning was the intended meaning. For once I'd like to get a authoritative clarification of an ambiguity.

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#26
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 6:54 PM

"For once I'd like to get a authoritative clarification of an ambiguity."

Lord! That just takes me back through 4 years of high school!

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#27
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 9:12 PM

I'm sure it's same day and month, e.g., April 1. Also I am assuming that the question refers not to exactly 3 but 3 or more (which is very slightly higher).

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#28
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/01/2014 10:59 PM

If that was the case why didn't the teacher discover 4 or 5 instead of 3?

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#31
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 8:28 AM

If that was the case why didn't the teacher discover 4 or 5 instead of 3?

Short answer: The teacher may not have looked at them all.

Long answer:

The classical problem is calculating the probability of two people sharing a birthday in a group of N. The way it is normally calculated is to compute the probability of all birthdays being different and then subtracting from 1. But what you are really calculating is the probability of 2 or more matching birthdays, which is the converse of no birthdays matching. This could include three that match, multiple groups of two, etc. Semantically, if there are more than two matching birthdays then there are two, and if you want to specify exactly one set of matching birthdays, you need to state the problem in that fashion.

So I am assuming that the problem is calculating the probability there is a set of 3 that match and not exactly one set of 3 and no other matches.

There are 4060 sets (combinations) of 3 in 30 students. For each set, the probability that they match is 1/(365)^2. The probability that they don't match is 1-(1/(365)^2);

The probability that all 4060 sets are a non-match is (1-(1/(365)^2))^4060. Finally, the probability of at least a matching triplet is this number subtracted from 1, which works out to 3.0015191748%.

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#29

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 7:01 AM

I remember watching a talk on TED given by a statistician. He had been an expert witness in a case of SIDS/infanticide. The paediatrician had stated that the mother had, in all likelihood killed her 2 children as the chance of one child dying of SIDS is 1 in several 100s of thousands and the chance of having a second child die of SIDS is in the order of 10s of millions. The statistician warned of people not learned in statistics dabbling in something that may lead to the accused being incarcerated in a travesty.

He said that in fact the chance of a 2nd child dying of SIDS is almost 100%. As the cause is not known, the second and subsequent children are likely to be exposed to the same conditions; genes, care, environment, diet and exposure to substances such as wool, cotton or plastic that may be causative. His expert opinion resulted in a review of several cases and then acquittal.

I have always been amazed that in a group of people there will always be 2 who share the same birthday but I didn't know that the likelihood can be calculated. My hat is doffed to AH and Passingtongreen.

Jim

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#30

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 8:15 AM

"...A school teacher notices that 3 students in his class of 30 students share the same birthday. What is the percent chance of this occurring?..."

.

Two different answers could be correct.

.

1. The teacher has noticed 3 students share the same birthday. The chance of that occurring is 100% because it already occurred. (It might also be due to the fact that the Jabrowski tripplets are in his class this year). I would stick with this answer if the verb conjugation was intentional and correct: '....what is the chance ...'

.

2. If the question intended is better expressed as something like, 'what was the chance that it would occur..' Then the answer is slightly less than 3%.

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#41
In reply to #30

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 2:13 PM

Somebody else that would like a little clarification. It is nice to not be alone.

I do not like any of the numeric answers so far because they do not meet the certainty test. In the classic 2 birthday problem a group of just 1 and 366 people have a certainty probability of zero and one respectively. In this problem a certainty of having three people with the same birthday should happen only once 2*364+1=729 people exist. AH touched upon this with his non-numeric answer but a clear method eludes me, for now.

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#34

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 9:58 AM

Just for everyone's amusement…

Today is a birth day for not one, not two, not even three members of my family…

Wait for it…

Four members in my family have a birthday today (April 2nd)!

The following are celebrating a birthday today; my fiancée's son, myself, my fiancée's father, and her mother.

We just completed our lengthly contrafibulations. ;-)

And, yes, it is true, despite the odds.

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#36
In reply to #34

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 10:52 AM

Have a good day, and many more of 'em!

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#42
In reply to #34

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 5:34 PM

How big is your family? We'll use that for next month's puzzle.

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#43
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 6:12 PM

Between myself and my fiancée; I have my father and one brother and she has her son, one sister, and a mother, a step mother, and father.

That's less than 30 members, too. :-)

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#46
In reply to #43

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 2:30 AM

And if we asked what the odds of that are, similar to the question in the topic off this thread.....

.

I'm tempted to say that the chance approaches 100% (because I see very little chance that you would be purposefully deceiving us about this, and because I see very little chance that you might have any problem assessing the situation you have related.)

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#52
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 7:14 AM

I really don't want the legacy of starting something like another bathtub breaking thread. :)

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#59
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 2:48 PM

Let's see, if there are 10 of you, that's 110 combinations of 4 out of 10 (N=10!/(4!*6!))

The probability for 4 people not having the same birthday is p = (1-(1/365)^3)=.999999979435...

The probability of 110 4 people groups without a quadruple birthday P=p^110=.9999977378...

The probability of at least 1 quadruple birthday in a group of 10 is 1-P = 2.2624x10^-4 percent or 1 out of 442065 families.

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#60
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 2:54 PM

I feel special. :-)

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#56
In reply to #34

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 1:52 PM

Congratulations to you and your family.

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#35

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 10:39 AM

54.8

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#38

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 11:59 AM

The chance of any random person having the same birthday as me is 1/365. This is multiplied by 29 when I am in a room of 30 people, so 29/365 chance that someone else in the room shares my birth date.

Within the room of 30, the chance that yet another person has the same birthday as me becomes 28/365.

Now, to make it exactly 3 out of 30, the chance that NO other people in the room has the same birthday is 1-(27/365).

So, I'm coming up with (29/365)*(28/365)*((1-(27/365)), 0.5644% chance of exactly 3 people in a room of 30 having the same birthday.

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#39
In reply to #38

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 12:35 PM

Unless someone can explain why this isn't accurate, it seems like far simpler math than some of the other examples. It also shows that the chances of 2 people having the same birthday is 7.9% and not the 71% that had been suggested.

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#40
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 12:56 PM

Instead of writing a long reply, go here to understand why...

Scroll down to the chart and the calculated value for 2 birthdays within a group of 30 people is 70.6.

All the math is there, but it may take a little work to digest if you do not have a good handle on statistics and factorials.

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#51
In reply to #40

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 7:10 AM

Ah, it finally makes sense now. I was basically selecting one random person who would then have a 29/365 chance of someone having the same birthday. It took awhile to understand how each unique pairing possibility seems to increase the odds beyond the 30 possible birthdays.

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#48
In reply to #38

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 4:09 AM

That is not the same problem.

.

'The chance that at least one out of a group of 29 will have a certain (your) birthday'...

.

....is much smaller than....

'

'The chance that any two people out of a room of thirty will share a birthday.'

.

By rephrasing the problem with the perspective of the first example, the possibility that two people, neither of which happens to be you might share a birthday while no one else has your particular birthday, is completely neglected.

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#44

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/02/2014 9:08 PM

For those who don't see the 70.6 result for two students, this is from an Excel sheet.

StudentR=Days RemainingIndividual R/365Combined
Number
136511
23640.997260.99726
33630.9945210.991796
43620.9917810.983644
53610.9890410.972864
63600.9863010.959538
73590.9835620.943764
83580.9808220.925665
93570.9780820.905376
103560.9753420.883052
113550.9726030.858859
123540.9698630.832975
133530.9671230.80559
143520.9643840.776897
153510.9616440.747099
163500.9589040.716396
173490.9561640.684992
183480.9534250.653089
193470.9506850.620881
203460.9479450.588562
213450.9452050.556312
223440.9424660.524305
233430.9397260.492703
243420.9369860.461656
253410.9342470.4313
263400.9315070.401759
273390.9287670.373141
283380.9260270.345539
293370.9232880.319031
303360.9205480.293684
100*(1-0.293684)70.63162

I wonder about finding a third. Each remaining student has a 364/365 chance of missing the joint date. From above, that equals 0.99726. put that to the power twenty eight and you have 0.926. 100*(1-0.926)= 7.394.

On that basis the ratio for three =70.631*7.394/100= 5.22%

All well and fine, having identified the triple, what are the odds of having another one with the other 27 students, I didn't eliminate that chance. I hope we don't have to run all of the combinations although AH probably has a shortcut.

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#45

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 12:01 AM

1/365x1/365x1/365x100=0.000002056%.

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#49
In reply to #45

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 4:17 AM

What you have provided is closer to the percent chance that the next three people you talk to all share you same birthday.....which isn't remotely as likely as at least three people out of a group of 30 sharing a birthday.

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#50
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 5:38 AM

Sorry... I didn't notice that. You are right. It is for 3 persons sharing and for 30 persons, it will be 10 times more.

1/365x1/365x1/365x100/10=0.00002056%

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#54
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 1:06 PM

The first one is a gimme'.....

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#53

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 11:47 AM

I believe the students birthdays are independent events.

So, the number of events is 30 and the number of successes is 3.

The chance of each independent event is 1/365 = 0.0027.

Using the binomial probabilities formula yields a percent chance of 0.0074%

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#57

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 1:56 PM

50%

Either they do, or they don't.

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#58
In reply to #57

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 2:42 PM

Either you win the $300 million lottery or you don't. With the odds you cite you should buy a ticket. :-)

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#68
In reply to #58

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/08/2014 12:43 AM

or 2

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#61

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/03/2014 7:19 PM

Here is how to attack the problem: Calculate the opposite, i.e., what is the probability that there are not 3 (or more) students with the same birthday. Then subtract this from 1 to get the probability you want.

Now take every combination of students of 3 out of the 30 students, starting with (1,2,3), (1,2,4), ... (1,2,30), (2,3,4), (2,3,5), ...,(28,29,30). There are 4060 combination sets. Calculate the probability for a single set that the birthdays do not match. This is equal to 1 - (1/365)*(1/365).

Now calculate the probability that all of the 4060 sets of 3 students are non-matching birthdays. This is the product of all the probabilities, or the probability we calculated for one set raised to the power 4060.

Subtract this number from 1 to get the probability that there is at least one set of three matching birthdays.

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#62

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/04/2014 1:17 AM

6.587957273511E-4

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#64
In reply to #62

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/04/2014 12:18 PM

the probability that one of 30 persons has on a special day his birthday is 0,082191780821917808219178082191781=8,2%

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#65
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/04/2014 3:30 PM

The probability that, for your statement to be correct, an assumption of a year having just one (to the nearest integer) special day, approaches unity.

.

Are no other days special? I've heard some claims that 'every day is special', but that seems suspiciously likely to be specious, as well.

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#67
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/08/2014 12:37 AM

a triplet?

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#72
In reply to #67

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/10/2014 1:07 AM

there are 7 billion people on earth - this makes 2million every day of a year, every year!

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#73
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/10/2014 1:09 AM

Move the decimal point one place over--you guess which way.

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#74
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/11/2014 12:33 AM

only a single zero (more or less - 20 millions)

all together now!

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#75
In reply to #74

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/11/2014 8:53 AM

Here is the basic curve I referenced earlier from

http://jaxwebster.wordpress.com/2011/05/29/three-birthdays-on-the-same-day/

He came up with 4.5% for 35 people so 3% is roughly right for 30. You need 87 people to have a 50% probability.

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#76
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/11/2014 12:57 PM

somwhere on earth - saharian or shanghai?

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#71

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/08/2014 8:15 PM

I know that class and those kids and they are triplets so the answer is clearly 100%!!!

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#77
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/11/2014 1:01 PM

I don't know this class - so the answer is 0%!

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#78

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/15/2014 10:55 PM

1 kid having a birthday in a year is 1 in 365. 3 kids having a birthday on the same day is 3 * 365 = 1 in 1095.

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#79
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/15/2014 11:06 PM

Every kid has a birthday.

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#82
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/19/2014 5:21 AM

True.

.

But, littlebetter was describing the chance of having a birthday in a year. While the numbers stated include a significant error, there is something to what was said. The chance is not unity.

.

1/(365+365+365+366) = 1/1096 chance a baby goat will be born on leap day (assuming all days have equal chance....and that anyone cares about goatbirthdays).

.

3/4 chance of a randomly chosen year not being a leap year (some assumptions apply).

.

--> 3/7084 chance that a baby goat born on a random day will not have birthday in a randomly chosen year within its life.

.

≡7081/7084 chance that a baby goat born on a random day will have a birthday in a randomly chosen year within its life.

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#80

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/15/2014 11:09 PM

since there are 30 kids, there are 30 chances greater, so divide the answer by 30. 1095 / 30 = 36.5

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#81

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/15/2014 11:13 PM

actually 1 in 36.5. so 1 / 36.5 = 2.7%.

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#83
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/19/2014 5:29 AM

Interesting approach. Certainly not how I'd do it, but I'm not familiar with the premise and working of your approach. Perhaps you would explain a little further.

.

You divided by thirty because there are thirty students in the class...that raises some questions. I you saying that the chance of three people having the same birthday in a group of thirty is only 30 times greater than the chance of three people having the same birthday in a group of only one person. It seems like once you get below three people in the group, its going to be pretty tough from three members of the group to ...well, to pretty much anything reasonable.

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#84

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/19/2014 1:38 PM

OK, this is my final answer. To heck with an analytic solution, it's time to Monte-Carlo it.

Here is a Matlab program that calculates the probability you will find at least one set of 'exactly n' matching birthdays in a set of 30 people. For each iteration, it creates a sample of 30 birthdays, sorts it, and then checks for exactly n equal elements in a row in the sorted set.

START OF MATLAB PROGRAM

%M possible states, group of N, Q have same state, P = probability

clear;

M=365; %Possible states

N=30; %Sample size

NN=100000; %number of iterations

P(1:10)=0; %Cumulative matches (1-10)

for k=1:NN

Q(1:10)=0; %Match per iteration

I=floor(M*rand(1,N)); %create a random birthday sample of N people

II=sort(I); %sort

n=1; %match counter

for j=2:N

if II(j)==II(j-1)

n=n+1;

else

Q(n)=1; % match of n birthdays exists for this iteration

n=1;

end

end

P=P+Q; %cumulative matches

end

P/NN %print out probability for different number of matches

END OF MATLAB PROGRAM
Here is what I get. It looks like 100% for non matching birthdays, 68.3% for 2 matching birthdays, and about 2.67% for 3 matching birthdays, etc.
ans =
Columns 1 through 7:
1.00000 0.68328 0.02670 0.00048 0.00001 0.00000 0.00000
Columns 8 through 10:
0.00000 0.00000 0.00000

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#85

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/22/2014 5:31 PM

From the excell sheet above, the chance that 2 of 30 have the same birthday is (1 - 0.29368) or 70.6%. Now you have only 29 left to match, which is (1-.319) or 68.1%. So the combined probability is .706 x .681 or 48.07%.

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#88
In reply to #85

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/24/2014 7:37 AM

Welcome. Love your work. Way over my head. Mark Twain says "there are lies, damned lies and then there are statistics" I would like to add "and then there are statistics, and then there are statistics, and then there are statistics and th....."

There have been a few erudite answers to this post and it seems that different methods produce similar answers but noticeably not the same. e.g. Rixter has posted #16 3.04% AND #84 2.67%. And Tinac #30 has it as just under 3%. USBPORT has a similar figure of 5.4% BUT this is almost double. AH #14 has odds of 95:1 which again is close but about half.

I know when dealing with the human body figures end up somewhat rubbery but how come statistics are so in exact?

I hope you enjoy this forum and come back for more

Jim

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#94
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Re: Birthday Trio: Newsletter Challenge (April 2014)

04/28/2014 12:16 AM

I don't think that the range of answers means that there is an inherent uncertainty in getting to a correct answer, rather, that it is a complex statistical problem and people have difficulty in recognising flaws in their methodologies. I could well be one of those.

Most people agree that the probability of no one out of 30 students sharing same birthday is approx. 29.4% meaning that chance of two or more out of 30 sharing same birthday is 70.6% (100 - 29.4). Some people then tried to work out probability of having none or two people with the same birthday and then subtracting from 100% to get the probability that 3 or more people share same birthday. To do this properly all combinations of two people sharing the same birthday need to be considered from one pair out of 30 all the way to 15 pairs sharing 15 different birthdays. I don't think any-one has attempted this. I had a go up to 6 pairs and found that this was sufficient to get a pretty good approximation which was 29.4% (no shared birthdays) + 38.0% (one pair with shared birthday) + 21.3% (two different pairs with shared birthdays) + 6.8% (3 different pairs) + 1.4% (4 different pairs) + 0.2% (5 different pairs) + 0.02% (6 different pairs) = 97.12% resulting in an answer of approximately 2.8% for 3 or more sharing same birthday.

IN the case of exactly 3 sharing same birthday it is an easier calculation. There are 365 ways for the 3 students to share the same birthday, the fourth student only has 364 choices for his birthday since he can't choose the one that the three have in common, the fifth has 364 etc. Multiplying out to 30 students gives the number of ways that 3 particular students (and no others) can share the same birthday. To cater for any 3 students we then need to multiply this by the number of ways of choosing 3 students out of 30 = 4060. Finally we divide by the total ways of 30 students choosing a birthday = 36530 which gives approximately 1% probability of exactly 3 students out of 30 (and no others) sharing the same birthday..

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#95
In reply to #94

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/28/2014 5:11 AM

A GA from me. You get the most common result of just under 3% with an elegant explanation. I like elegant!

Jim

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#96

Re: Birthday Trio: Newsletter Challenge (April 2014)

04/28/2014 5:33 AM

Just out of interest, a program to explicitly check all possible combinations of the 30 students, allowing 10 μs per iteration, would take about 2.3 x 1064 years to run. So not a coffeebreak job .

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