“If an “occurrence” is adding one more person to the group, what you’re asking after X occurrences is what the likelihood is that at each individual event there was no “match.””
Lech;
But this is the logical fault, as Legion just said above. What you are describing is a series of twenty “events”. As Legion just said, it is the equivalent of picking 20 people out of a hat, and the last person has the same birthday as the first.
That is not what we want to know. You want a probability DISTRIBUTION of a SINGLE situation: you have twenty random people together, and what are the odds that two of them will have the same birthday.
Benson I think I got it right. And I think legion recapped it correctly. Accumulated probability of non-occurrence is exacttly what’s being measured.
If an “occurrence” is adding one more person to the group, what you’re asking after X occurrences is what the likelihood is that at each individual event there was no “match.”
I think I didn’t explain the Poisson distribution well enough. It describes a continuous process in which the likelihood of an event occuring is constant through time. One of the key properties of this type of event is that the past is not a predictor of when the event will occur.
Phenomena that fall into this type of distribution are casualties in a battle and the length of time on a phone call (In other words, if a person has been on for one minute, the probability distribution regarding the remaining time on the call is THE SAME as a person who has been on for 30 minutes).
I once used this distribution to model defects in a factory that was causing us to cut the fiber optics short.
hi benson,
welcome back to the OT!
now we know how to coax you back here.
ok,
“you must also take into account the accumulating probabilities of the opposite outcome (that two individuals won’t share the same birthday)
and factor that into the equation.”
This is what I mean to say here,
you are adding together progressive calculations
on the chance that every successive individual in a
line of 30 will not share the same birthday as another.
The “accumulating probabilities” are each successive calculation in the 365!(factorial) series.
The “opposite outcome” is the chance that a person won’t share a birthday with another.
as you said,
the formulas were incorrect because my assumption was incorrect.
I was solving for the probability that one person would share the same birthday as the first individual as opposed to finding two persons that share the same birthday out of 30.
“But the fly boys were certain that they were not playing odds the more flights they flew a la likelihood of a run of “heads” when flipping coins.”
They were correct. Assuming that the level of expertise did not increase with each flight, they were correct. The number of previous missions flown had no bearing on the success of the current mission.
One of the most important statistical distributions (the Poisson) was developed by Poisson as he observed casualties during the Napoleonic war. Basically, what the flyboys were describing above was a Poisson distribution.
“you must also take into account the accumulating probabilities of the opposite outcome (that two individuals won’t share the same birthday)
and factor that into the equation.”
There is alot of confusion on this issue of the probability.
The formulas you are using (multiplying the probability of one event by the next event) is incorrect. Multiplying in such a fashion implies independence of the events. Of course, the simplest example is the roll of a die. One roll of a die is completely independent of the next: they are separate events. Hence, the probability of hitting the same number in two rolls is (1/6) * (1/6). Likewise, in the example you are using, if I wanted to know the probability of picking a person with a certain birthday, and then picking the NEXT person to have the same birthday, it would be (1/365)*(1/365).
That is not what you are talking about here. In this case, there is ONE EVENT: you have a population of 20 people (with a random distribution of birthdays), and you want to know the probability that two of them will have the same birthday. This is the realm of sampling theory.
I can’t remember the formula off my head, but I’ll try to look for it tomorrow.
At one time, I did alot of work in probability, but more in the line of “random walk” studies and Monte Carlo simulations. Certain characteristics of of fiber optics are random, and my job was to prevent a random walk such that the light signal would be too blurred, after traversing the ocean length.
Why the fuck am I still awake, drinking Evan Williams and cruising the InterWebs.
Why the fuck are you reading this?
Good night again, you loosers.
I school, I was good at math but sucked at statistics.
I do like these problems, so I’ll play along tomorrow.
Good night.
“If an “occurrence” is adding one more person to the group, what you’re asking after X occurrences is what the likelihood is that at each individual event there was no “match.””
Lech;
But this is the logical fault, as Legion just said above. What you are describing is a series of twenty “events”. As Legion just said, it is the equivalent of picking 20 people out of a hat, and the last person has the same birthday as the first.
That is not what we want to know. You want a probability DISTRIBUTION of a SINGLE situation: you have twenty random people together, and what are the odds that two of them will have the same birthday.
Think about it.
Pick it up later.
Benson I think I got it right. And I think legion recapped it correctly. Accumulated probability of non-occurrence is exacttly what’s being measured.
If an “occurrence” is adding one more person to the group, what you’re asking after X occurrences is what the likelihood is that at each individual event there was no “match.”
Let’s pick this up tomorrow, it’s fun.
I think I didn’t explain the Poisson distribution well enough. It describes a continuous process in which the likelihood of an event occuring is constant through time. One of the key properties of this type of event is that the past is not a predictor of when the event will occur.
Phenomena that fall into this type of distribution are casualties in a battle and the length of time on a phone call (In other words, if a person has been on for one minute, the probability distribution regarding the remaining time on the call is THE SAME as a person who has been on for 30 minutes).
I once used this distribution to model defects in a factory that was causing us to cut the fiber optics short.
I loved my work in probability theory.
hi benson,
welcome back to the OT!
now we know how to coax you back here.
ok,
“you must also take into account the accumulating probabilities of the opposite outcome (that two individuals won’t share the same birthday)
and factor that into the equation.”
This is what I mean to say here,
you are adding together progressive calculations
on the chance that every successive individual in a
line of 30 will not share the same birthday as another.
The “accumulating probabilities” are each successive calculation in the 365!(factorial) series.
The “opposite outcome” is the chance that a person won’t share a birthday with another.
as you said,
the formulas were incorrect because my assumption was incorrect.
I was solving for the probability that one person would share the same birthday as the first individual as opposed to finding two persons that share the same birthday out of 30.
“But the fly boys were certain that they were not playing odds the more flights they flew a la likelihood of a run of “heads” when flipping coins.”
They were correct. Assuming that the level of expertise did not increase with each flight, they were correct. The number of previous missions flown had no bearing on the success of the current mission.
One of the most important statistical distributions (the Poisson) was developed by Poisson as he observed casualties during the Napoleonic war. Basically, what the flyboys were describing above was a Poisson distribution.
“you must also take into account the accumulating probabilities of the opposite outcome (that two individuals won’t share the same birthday)
and factor that into the equation.”
Legion, my friend: this statement makes no sense.
Folks;
There is alot of confusion on this issue of the probability.
The formulas you are using (multiplying the probability of one event by the next event) is incorrect. Multiplying in such a fashion implies independence of the events. Of course, the simplest example is the roll of a die. One roll of a die is completely independent of the next: they are separate events. Hence, the probability of hitting the same number in two rolls is (1/6) * (1/6). Likewise, in the example you are using, if I wanted to know the probability of picking a person with a certain birthday, and then picking the NEXT person to have the same birthday, it would be (1/365)*(1/365).
That is not what you are talking about here. In this case, there is ONE EVENT: you have a population of 20 people (with a random distribution of birthdays), and you want to know the probability that two of them will have the same birthday. This is the realm of sampling theory.
I can’t remember the formula off my head, but I’ll try to look for it tomorrow.
At one time, I did alot of work in probability, but more in the line of “random walk” studies and Monte Carlo simulations. Certain characteristics of of fiber optics are random, and my job was to prevent a random walk such that the light signal would be too blurred, after traversing the ocean length.