What Are The Odds???

Maybe it's a set of triplets that didn't fall far from the tree (i.e., the fourth consecutive house number).

None of us are related.

Lots of good info. Filling in some blanks: there are 20 houses on the street, the houses in question are in the middle, average of four people per house.

The houses in question are just the ones that happen to have it happen.

The point of middle or end has to do with how many sets of three a particular house can be involved in. While and end house can only make three with the next two, a house in the middle can make three different sets of three (as long as the street has at least five houses on it).

To really do this right, you wouldn't assume no one in a house shared a birthday. You would find the probability no one shared and work that portion including the possibility of more than one match with an adjacent house and so on, then find the probability of one shared then more shared birthdays in the house and work those.. and then combine by summing. It wouldn't be too far off from our rough estimate.

on reconsideration, I want to retract the above comment. It is not correct.

Please look to Rixters comment below.

What are the odds of celebrating that coincidence with a BLOCK PARTY.

Not sure about the problem as postulated, but for people in a group, it's known as the birthday paradox. The odds of two people having the same birthday is 50/50 in a group of 23 people, and goes up to 99.9% as the group size goes up to 75.

You use the same trick for the birthday paradox, i.e., calculate the odds of nobody having the same birthday and then subtract this from 1. For each new person added, you have one day less out of the year to not have matching birthdays.

P = (364/365) x (363/365) x (362/365) x (361/365) x ...

Each factor is close to one, but when you get about 25 people, the product goes down to about 0.5, the probability of all birthdays being different. The probability that there will be a matching birthday is 1-P, which is also about 0.5.

A lot of people don't understand this and expect that in a group of 25 people that someone will have their birthday.

In other words, you would be better off with Powerball?

Yes yes.. In other words please? 1/1,000? 1/1,000,000,000,000,000,000,000?

I know I'm not a math magician, but I don't understand the answer?

What are the odds?

How can you share these numbers without sounding like a a total poindexter.

You have 8 chances to pick the same number 3 times out of 365. My guess would be...

8/(365)^3.

or 1 out of 6,078,390.

When I was in High school math class the teacher posed this answer to something similar.

He said that out of 30 students in each class the odds that there will be 2 students with the same birth date were near (as I recall) 90%. He had 5 classes that He taught and all but one class had at least 2 people with the same birth date. Just saying. food for thought.

oilcan13

When it happens, it really happens.

First, the current ages of all eight people need to be provided in order for any two ages to be found to possibly coincide...

Second, how many hospitals were up-and-running, within the city limits of your birth-city, on the date of your birth?...

The OP specifies "month and day", year is not relevant.

Oh, you want the easy goal:

For any one of the 3 in the first house, the odds for any one date are 1/(365.25)

For the 3 in the second house, the odds of one matching birthdate are 3/(365.25)

For the 2 in the third house, the odds of one more matching birthdate are 2/(365.25)

The odds of 1 matching birthdate in all three houses is (1x3x2)/(365.25^3)

which is 6/(948,727,112.2031) = 1.23135... E-7 = 0.000 000 1.23135

which corresponds to 1-in-8,121,185.36717, which is less than 1 in 8 million...

Similarly, there is a 1-in-365.25 chance of some one of the same age as you, in any one of your H. S. classes, of having the same birthdate as yours...

Why don't the off-topic inquisitors just explain why, how, or how far, off the subject comment is ''off-topic''?... Doesn't CR$ deserve some kind of an explanation?... Or isn't CR4 worthy enough for such (inquisitors) ''enlightened explanations'' ? ...

One thing is the incessant typographic goofiness in your posts, such as that one.

Phonetically speaking, you say ''tomayto'' and I say'' toemautoe''...

And that one.

A passing footnote to this thread: Two of us were born in the same year, one in Connecticut, the other in Italy. Born on the same date, approximately 4,000 miles apart, now living in consecutive street address. The third person was born two years earlier.

Here is something to consider: the frequency (and thus odds) for that exact particular type of occurrence would be far more remarkable had the criteria been set prior to the discovery.

Because the criteria are formulated upon the discovery it isn't really a very rare occurrence. There are simply far too many possibilities for seemingly similar attributes to end up in close proximity in time/space. If it wasn't consecutive street numbers, perhaps it would be the same street number on consecutive streets,

or consecutive streets numbers on streets in alphabetical order,

or perhaps you each work at building that have consecutive street numbers,

or perhaps you all drive Ford mustangs you bought before knowing each other that are of consecutive years.

Perhaps the three of you have the same or consecutive numbers when you sum your social security number and your zip code.

Perhaps you have the same number of windows that face the street and have each been married the same number of times and have the same number of remaining kidneys.

It may seem like a rare coincidence, but the odds of finding the similarity you have already found is pretty close to 100%

What are the odds? Here is how I would do it.

I'd start with the probability that three houses on the same street have consecutive numbers. That sounds extremely unlikely to me. Most house numbers jump a few numbers in between. Usually in the US you have odd numbers on one side and even on the other, so houses on opposite sides of the street may have two consecutive numbers, but houses on the same side usually jump by 4 or 6 or more, unless they are tightly squeezed together. So what would be the probability that they are that tightly squeezed together? I have no idea.

But, if you mean something other than "consecutive numbers" when you say "consecutive numbers", I would proceed this way.

I'll assume you mean adjacent houses or adjacent house numbers on the street.

Choose the house with 2 people inside. Pick one of those two people. You have a given single birthday.

For each person in the next house, the probability that the next house has a DIFFERENT birthday is 364/365 (approximately). If the next house has 3 people, the probability that NONE of them have that birthday is:

(364/365) **3 = 0.99180331964927

Therefore probability that at least one of them has that birthday is 1.0 minus that probability, or

1.0 - 0.99180331964927 = 0.00819668035073

The probability that at least one of the 3 people in the 3rd house has that same birthday is the same. This is a joint probability, so you multiply them together, and you get:

0.00819668035073 * 0.00819668035073 = 0.000067185568772

Then you have to account for the fact that the first house has another person in it. The probability that the first person's birthday does NOT have matching birthdays in the other two houses is

1.0 - 0.000067185568772 = 0.999932814431228

The probability that the second person also does not have a match is the same. Multiply them together to get the probability that both do not have a match.

0.999932814431228 * 0.999932814431228 = 0.999865633376357

Then calculate the probability that NONE of the sets of 3 consecutive houses on the street share a birthday. To do that, raise the last result by the number of sets of 3 on the street. If you have 20 houses, then you can have 18 sets of 3 adjacent houses.

0.999865633376357 ** 18 = 0.997584161117484

1.0 minus that is the probability that at least one set of 3 adjacent houses on a street with 20 houses shares a birthday in all 3 houses.

1.0 - 0.997584161117484 = -0.002415838882516

I'd say the probability that if you pick 3 adjacent house with 2, 3 and 3 people in them respectively, that they have at least one person with matching birthdays is about 1 in 413.934889134001232.

Give or take a bit.

Much better than the lottery.

I didn't include the probability that 3 adjacent houses have exactly 2, 3, and 3 people in them. I leave that as an exercise for the reader.

The probabilities go way up if you have more people in the houses.

Would somebody please check my math?

Many great responses but some of the questions and solutions leave me questioning the process, or more importantly the data used to arrrive at the solutions.

While there are 3 people in the first house, 2 in the second and 3 in the third the birthdates of 1 in the first, 2 in the second and 2 in the third are known. None of the birthdates of the remaining 3 people, along with the rest of the people on the street are known so why should they be included in arriving at the answer?

(It was also mentioned in a latter post by myself that two of us were born on the same date in the same year, 4000 miles apart, only to find we are across the street from each other but while this is "unusual" it is not really part of the original question).

With all of this considered shouldn't the final answer involve only the 365 days of the year, 100% of the people in the first house and 50% in each of the latter two houses?

It doesn't matter if you don't know the birthdays of the other 3 people or anyone else on the street, because you asked about probabilities, not the actual number. However, if you want to know the odds that those people whose birthdays you know fit this pattern, the odds are different from what I calculated. Yes, you would leave out the unknown persons.

Also, if you only know the birthdays of the people in these 3 houses, then it doesn't matter if they have consecutive numbers or not. The same odds apply to any set of 3 houses found on Earth (or any planet with 365 days in the year).

I could redo my answer to reflect that, but I will leave that as an exercise for the reader. I will say that the odds are much lower.