Odds for odd folk

[Matt2matt2002]

Any help me out with calculating odds?

I will be touring with 3 other folk for a 2 month period.

What are the odds that 2 of us will share a birthday?
( If you can explain that to me, go for 3 or 4 of us)

In any one year I think the odds of 2 folk sharing a B/day is 0.3%

But I get stuck on "4" folk and the 2 month window.

Thanks

Matt
Who takes his shoes off to count past 10

[Andre Jute]

This is a twister with lots of invitations to complications tying you in knots. But the answer is ruled by the stars...

Twelve months in the year. Everyone must have a birthday once a year, so the fact that there are these particular four people is a red herring, unless two or more of them are twins, triplets or quadruplets. Assuming that the odds of birth in any month is the same for the population at large, and not skewed by natural or social factors. Then the odds that anyone will have a birthday during any particular month of the year is 1/12.

So: The odds that any person will have a birthday in during two month tour is 1/6. The odds that two particular people will have a birthday during those two particular months is smaller at 1/6 x 1/6 or 1/36. The odds on three people having a birthday on this two month tour is 1/6 x 1/6 x 1/6 or 1/216. The odds that all four of you will have birthdays on tour is one in about 1300.

[Donerol]

I thought that Matt was asking for the odds of sharing the actual birthday? Doesn't that make the odds smaller?  I've never been able to work this stuff out! 

[Matt2matt2002]

Thanks Andre.
When you write it out like that, it looks so easy.
But I had a headache trying to figure it out.

I'll let you know if the odds came up trumps, when I hear from the other 3 folk.

Thanks again.
Matt
Without a headache now!

[Matt2matt2002]

Thanks Donegal

I guess my question was really along the lines of, " when we are away together will one of us be buying a birthday cake?"
And I think Andre answered that.

Odds on any 2 of us having the same birthday and it being within the 2 month tour period.....?

Oh dear. Another headache coming g on.
But please enlighten me if you are able.

Matt

[Slammin Sammy]

I'll take a stab at that one, Matt.

Since there are 365 days in a year, and making the same assumptions as Andre did, I would expect that the odds of any two people sharing the same birth DATE is 1/365. (Note this is not the same as sharing the same birth DAY, ie the exact same age. This would be a far more complex calculation requiring population and actuarial statistics, etc.)

The odds of the shared birth DATE being in the two month window are 1/6, regardless of how many share the date.

It's meaningless to relate the two odds, but if you insist, just multiply them together.

[David Simpson]

Odds on any 2 of us having the same birthday and it being within the 2 month tour period.....?

Yes, Sammy is correct. I was typing up a longer reply while Sammy answered, so here is my longer reply.

The odds of two people having the same birthday is 1/365 (not counting leap years). The way to think about it like this: person A's birthday is on X (some unknown day in the year). What are the odds that person B's birthday is on X? There are 365 possible days for X, and person B's birthday has an equal chance of being on any of those 365 days. Therefore the probability is 1/365.

For 3 people, it is slightly more complicated. When you want to calculate the odds of two events BOTH happening, you need to MULTIPLY the odds of each event happening individually. If you want to calculate the odds of EITHER event happening, you ADD the odds. (This assumes that the events are independent. That is, the fact that one event happens does not affect the odds of the other event happening.)

So for 3 people, we can ask two questions:
1. What are the odds that all 3 people have the same birthday?
2. What are the odds that any 2 of the 3 have the same birthday?

First, what are the odds that all 3 people have the same birthday? Let's call these people A, B, and C. We have already calculated the odds that A and B have the same birthday: 1/365. Now what is the probability that C also has that same birthday? We want to calculate the odds of two events happening at the same time: A and B have the same birthday, and A and C have the same birthday. The odds of A and B having the same birthday is 1/365. Similarly, the odds of A and C having the same birthday is the same. Therefore the odds of all 3 having the same birthday is (1/365) * (1/365) = 1 / (365*365) = 1 / 133,225.

Now, what are the odds that any 2 of the 3 have the same birthday? In this case, we add the individual odds: (1/365) + (1/365) = 2 / 365.

Your final question: what are the odds that the birthday is during your 2-month trip? The odds of a particular day happening during your 2-month  trip is 2 /12, since there are 12 months in total. To calculate the odds of this day happening during your 2-month trip AND one of the other events happening, we multiply the odds. The odds that all 3 of you have the same birthday and that birthday happens during the trip is: (1/133,225) * (2/12) = 2 / 1,598,700 = 1 / 799,350. The odds that any 2 of you have the same birthday and that birthday happens during the trip is: (2/365) * (2/12) = 4 / 4380 = 1 / 1095.

- Dave

[Andre Jute]

When you write it out like that, it looks so easy.

Statistics are real simple if you work calmly through the logic. It's when you let the wrong complications through the door that it becomes frustrating.

I thought that Matt was asking for the odds of sharing the actual birthday? Doesn't that make the odds smaller?  I've never been able to work this stuff out! 

Then the odds are much smaller, but the logic, assuming I'm right about it in the first instance, is precisely the same.

Everyone has a birthday once a year and there are 365 days in the year (unless it is a leap year, which we will asssume it is not). In most two month stretches there are 61 days but we already know two months is 1/6 of a year.

The chances of anyone at all having a birthday on any particular day in the whole year is 1/365. The chances that two people chosen at random will have a birthday on the same day is (1/365)^2 or 1/133225, that three will have birthday on the same day (1/365)^3, or 1/48,627,125  and so on. The likelihood gets very small very quickly.

If we then narrow the timeframe to not just any day in the year but a day in that one-sixth of the year on which you will all be on tour, the likelihood of a simultaneous birthday on the same day while on tour for two months out of the year will need further reduction as (1/365)^n x 1/6 where n is the number of cyclists having a birthday on the same particular day while on tour for two months. For two cyclists to have a simultaneous birthday on tour the calculation is (1/365)^2 x 1/6, which is 1/799350 or about 0.000001251 or .0001251% or  125 likely occurrences in every million random concatenations of 4 cyclists for two months.

Here's another one for you, Matt. The chances of any one of the four cyclists having a birthday on any specified day during your tour is (1/365 + 1/365 + 1/365 + 1/364) x 1/6 or 4/2190, call it 1/548 or about 0.1825%.

Happy headaches!

[David Simpson]

One question that I find interesting is: what is the lowest number of people that you can have such that the odds of two of them having the same birthday is more than 50%? It is actually quite low: 19 people.

The odds of at least 2 people having the same birthday in a group of N people is: (1/365) + (2/365) + (3/365) + ... + (N/365) = (1+2+3+...+N)/365.

What is the N so that this value becomes greater than 1/2 = 177/365? Answer: N=19

So for any group of 19 or more people, most likely 2 of those people have the same birthday.

- Dave

[AndyE]

Celebrating birthdays above 29 is futile! If you are above and beyond the said age, doubtless you will need reminding of the fact because there is always someone who is younger. What are the odds that one of your companions being at the same age on the same day, one of the twins would be older!

Andy, ( 29 & 288 months)

P.S AKA Peter Pan 

[onrbikes]

Touring with 3 other people, for 2 months.
Sorry to go off track but
The question you may want to ask is,

What are the odds, someone isn't going to get along with the other, and crack the poos.

If one of them isn't your wife/partner the odds are already up.

[Matt2matt2002]

What are the odds, someone isn't going to get along with the other, and crack the poos.

If one of them isn't your wife/partner the odds are already up.

Good point. And yes by all means, drift off topic.

2 of the 4 have toured twice together.
None are my wife.
I am the youngest male but not youngest of the group.
I am the only one from the Northern hemisphere.

So what colour are my eyes?

Ha ha. Sorry. I love those kind of riddles and conundrums

Great to have all the input on stats and odds. I love maths but need guidance.

Regarding getting along; the plan is to kick off together and then hang loose depending on a variety of factors.
Strength and desire of milage. Personal goals on various stretches. Visa time. Etc.

We like the idea of group security and assistance during times of sickness.
Its a pretty remote area and we're no spring chickens. ( total age of group = 240+)
I hope our age has brought maturity to our outlook on life. But, as they say, sh1t happens.

Thanks again for the numbers info.

Matt
Brown eyes

[Danneaux]

Hi All!

Follow Dan's Birthday Math and you'll never be sad nor ever over 18, and you can return to ages you knew and loved periodically.

Just add the two digits together to get your "Birthday Age". Saves on candles and the risk of house fires.

At age 97, my father is once again 16 and lovin' it.  I'm reliving Age 10. Not so bad, now I'm mature enough to enjoy it. 

All the best,

Dan.

[JimK]

Hmmm. Math is tricky!

I think the easiest way to think about this sort of birthday puzzle is to look at it backwards. What is the probability that N people all have unique birthdays, that no birthday is shared?

For 2 people, it's 364/365
For 3 people it's (363/365)(364/365)
For 4 people it's (362/365)(363/365)(364/365) = 0.98364

So for 4 people, there is a 0.016 chance that there will be a shared birthday in there somewhere.

Will there be a shared birthday in a two month window? What a mess! Because there could be two shared birthdays!

The probability that all four have the same birthday is (1/365)^3 = 2E-8

The probability that 3 have the same birthday and the other person's is different, is 4*(364/365) *(1/365)^2 = 3E-05

The probability that there are two pairs of birthdays is 3*(364/365)*(1/365)^2 = 2.25E-5

The probability that two people share a birthday while the other two are distinct, is 6*(363/365)*(364/365)*(1/365) = 0.016

The probability that all the birthdays are distinct is (362/365)*(363/365)*(364/365) = 0.98

Now pulling in that two month window, the only tricky case is when there are two shared birthdays. Seems like

(5/6)*(5/6) that both birthdays are out of the window
(1/6)*(1/6) that both are in the window
2*(1/6)*(5/6) that one is in the window and the other out of the window.

Math!

[Templogin]

You have all missed the obvious.  When it is someone's birthday do they buy the cake or have the cake bought for them!?