Probability of Birthday's landing back to back in a small group

[mostdopeey]

I am in a small group of 12 for my job. It turns out that three of us have birthdays that land back to back - 12/30, 12/31, 01/01

I haven't been able to get rid of my curiosity on what the probability of these combined events. I am stumped on how to go about calculating this. Like is it (1/365)*(1/365)*(1/365) - is the group of 12 irrelevant or is it a factor I should consider?

Can anyone please help?

 

[BigBeachBanana]

I am in a small group of 12 for my job. It turns out that three of us have birthdays that land back to back - 12/30, 12/31, 01/01

I haven't been able to get rid of my curiosity on what the probability of these combined events. I am stumped on how to go about calculating this. Like is it (1/365)*(1/365)*(1/365) - is the group of 12 irrelevant or is it a factor I should consider?

Can anyone please help?

For simplicity, ignore some of the factors such as leap year, twins, and variations in birth rates. Assuming that there are 365 possible birthdays and that each person's birthday is equally likely to be any of these days, independent of the other people in the group.

There are [imath]365^3[/imath] total number of ways 3 people can have birthdays.
There are 365 sets of consecutive birthdays in a year.
There are 3! ways 3 people can have distinct birthdays
Therefore the probability of 3 random people having a consecutive birthday is [imath]\frac{3! \times 365}{365^3}\approx \frac{1}{22,000} \approx 0.0045\%[/imath]

If you have a group of 12 co-workers, there are [imath]{12 \choose 3} = 220 [/imath] ways to choose 3 of them. So on average, you should expect [imath]\frac{220}{22000} = 0.01[/imath] triple consecutive birthdays among your co-workers; which is practically 0. As the group size increases, the expected number of triple consecutive birthdays would also increase.

(Someone can double-check me).

 

[blamocur]

For simplicity, ignore some of the factors such as leap year, twins, and variations in birth rates. Assuming that there are 365 possible birthdays and that each person's birthday is equally likely to be any of these days, independent of the other people in the group.

There are [imath]365^3[/imath] total number of ways 3 people can have birthdays.
There are 365 sets of consecutive birthdays in a year.
There are 3! ways 3 people can have distinct birthdays
Therefore the probability of 3 random people having a consecutive birthday is [imath]\frac{3! \times 365}{365^3}\approx \frac{1}{22,000} \approx 0.0045\%[/imath]

If you have a group of 12 co-workers, there are [imath]{12 \choose 3} = 220 [/imath] ways to choose 3 of them. So on average, you should expect [imath]\frac{220}{22000} = 0.01[/imath] triple consecutive birthdays among your co-workers; which is practically 0. As the group size increases, the expected number of triple consecutive birthdays would also increase.

(Someone can double-check me).

I am too slow at this time to figure out my own answer, but a brute force script with [imath]10^7[/imath] samples yielded 19090 cases, i.e. about 1/520 frequency.

 

[blamocur]

I am too slow at this time to figure out my own answer, but a brute force script with [imath]10^7[/imath] samples yielded 19090 cases, i.e. about 1/520 frequency.

Turns out I am also too slow to write scripts -- I was counting the number of triplets with the same birth days.
Will revise and post again.

 

[blamocur]

Turns out I am also too slow to write scripts -- I was counting the number of triplets with the same birth days.
Will revise and post again.

My revised (and seemingly correct) script yields an approximate probability of 1/25.

 

[blamocur]

Turns out that I am even slower than I thought, so my apologies for spamming this thread with wrong results.

I could not find any errors in @BigBeachBanana 's post, so I've revised my script again, and this time I did get the probability of about 1/109 for 12 people and about 1/22000 for 3 people. My script did not count consecutive days spanning the end of the year (like today and tomorrow), but those should not change the result by much more than 1%.

 

[Steven G]

is the group of 12 irrelevant or is it a factor I should consider?

Imagine there are only 3 people in the group. The probability that they have consecutive birthdays is very small.
On the other hand, if the group size was 20 million, then the probability that three people in the group had consecutive birthdays would be very likely.