“When you guys are finished arguing about this…lemme know, and I’ll get a REAL math guy (aka BH) to weigh in, if someone would kindly restate the question.”
So we have three Math experts who will be reviewing our musings here?
BH,
BSM’s cousin
and
DCB’s brother
uh oh :o(
ok everyone,
have a good night,
be back tomorrow for more OT (slide rule edition)
I get what you’re saying.
I actually shortened the explanation of the experiment for the sake of brevity.
With the two slits you would get several lines on the phosphorescent screen because of what’s called an “interference” pattern between light waves/particles.
In essence, though the idea is that the photons aimed at both slits produced multiple lines.
Probability would predict that if one were to slow the light beam down and shoot only a single photon through a single slit, you would expect to get a single line because you’ve cancelled out any chance of “interference” and cancelled out any random stray photons shooting through the second slit.
In reality though, you still get the several lines on the phosphorescent panel any way you shoot the photons.
The accepted explanation is that the photons occupy all possibilities however improbable.
Oh for crying out loud, turns out there are about a million websites that explain the birthday problem we’ve been talking about. Basically I’m right in concept although my actual formula last night wasn’t quite right, there is a simpler way to do the formula. My treating each additional person as an “event” and calculating the odds of a non-match, and doing it in sequence, was correct. Here, from the wikipedia entry on the birthday problem, talking about 23 people in a room:
“The 23 independent events correspond to the 23 people, and can be defined in order. Each event can be defined as the corresponding person not sharing their birthday with any of the previously analyzed people. For event 1, there are no previously analyzed people. Therefore, the probability, P(1), that person number 1 does not share their birthday with previously analyzed people is 1, or 100%. Ignoring leap years for this analysis, the probability of 1 can also be written as 365/365, for reasons that will become clear below. For event 2, the only previously analyzed people are person 1. Assuming that birthdays are equally likely to happen on each of the 365 days of the year, the probability, P(2), that person 2 has a different birthday than person 1 is 364/365. This is because, if person 2 was born on any of the other 364 days of the year, people 1 and 2 will not share the same birthday. Similarly, if person 3 is born on any of the 363 days of the year other than the birthdays of people 1 and 2, person 3 will not share their birthday. This makes the probability P(3) = 363/365”
Someone just posted to an email list I’m on that Target has rented the Carson Pirie Scott store at 1 S. State St. in Chicago.
By cobblehiller on February 16, 2011 6:26 PM
“When you guys are finished arguing about this…lemme know, and I’ll get a REAL math guy (aka BH) to weigh in, if someone would kindly restate the question.”
So we have three Math experts who will be reviewing our musings here?
BH,
BSM’s cousin
and
DCB’s brother
uh oh :o(
ok everyone,
have a good night,
be back tomorrow for more OT (slide rule edition)
“By daveinbedstuy on February 16, 2011 6:42 PM
In the world of statistics and probability, the birthday problem is basically Kindergarten work.”
Yeah, OK there Rosie Ruiz. That would carry a lot more weight if you had contributed your knowledge before we got the answer.
In the world of statistics and probability, the birthday problem is basically Kindergarten work.
I just needed to simplify the formula a little bit. Formula wasn’t wrong, it could just be simplified.
bxgrl,
I get what you’re saying.
I actually shortened the explanation of the experiment for the sake of brevity.
With the two slits you would get several lines on the phosphorescent screen because of what’s called an “interference” pattern between light waves/particles.
In essence, though the idea is that the photons aimed at both slits produced multiple lines.
Probability would predict that if one were to slow the light beam down and shoot only a single photon through a single slit, you would expect to get a single line because you’ve cancelled out any chance of “interference” and cancelled out any random stray photons shooting through the second slit.
In reality though, you still get the several lines on the phosphorescent panel any way you shoot the photons.
The accepted explanation is that the photons occupy all possibilities however improbable.
So yeah, I was right about how to do this.
“This analysis continues until person 23 is reached, whose probability of not sharing their birthday with people analyzed before, P(23), is 343/365.
P(A’) is equal to the product of these individual probabilities:
(1) P(A’) = 365/365 × 364/365 × 363/365 × 362/365 × … × 343/365”
Oh for crying out loud, turns out there are about a million websites that explain the birthday problem we’ve been talking about. Basically I’m right in concept although my actual formula last night wasn’t quite right, there is a simpler way to do the formula. My treating each additional person as an “event” and calculating the odds of a non-match, and doing it in sequence, was correct. Here, from the wikipedia entry on the birthday problem, talking about 23 people in a room:
“The 23 independent events correspond to the 23 people, and can be defined in order. Each event can be defined as the corresponding person not sharing their birthday with any of the previously analyzed people. For event 1, there are no previously analyzed people. Therefore, the probability, P(1), that person number 1 does not share their birthday with previously analyzed people is 1, or 100%. Ignoring leap years for this analysis, the probability of 1 can also be written as 365/365, for reasons that will become clear below. For event 2, the only previously analyzed people are person 1. Assuming that birthdays are equally likely to happen on each of the 365 days of the year, the probability, P(2), that person 2 has a different birthday than person 1 is 364/365. This is because, if person 2 was born on any of the other 364 days of the year, people 1 and 2 will not share the same birthday. Similarly, if person 3 is born on any of the 363 days of the year other than the birthdays of people 1 and 2, person 3 will not share their birthday. This makes the probability P(3) = 363/365”