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Lech and all;
This morning as I was shaving, I figured out the best way to explain the issue discussed late last night (the probability of 2 people out of a party of 20 having the same birthday).
The best way to demonstrate my point is to look at a roll of the dice. Let’s ask the simple question: “If I roll the die 6 times, what are the odds that a single number (let’s say the number 1) will appear one time”
Using the methodology that you and Legion discussed last night, the answer would be (1/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6)=0.063. This would indicate that the odds would be roughly 1/16 and it would be mid-leading. If you were to go ahead and make a series of bets based on these odds, you would be a loser.
What the above calculation does is calculate the probability for ONE SPECIFIC SEQUENCE OF SIX ROLLS OF THE DIE. As common sense tells us, however, the actual odds for the number “1” to appear one time in six rolls is 1/6, and if we were to place a series of bets, these are the odds we would use. We can understand this just from the simple shape of the die.
This simple example points out the difference between calculating the probability of one specific event, and the underlying statistical distribution (the “odds”). The statistical distribution can be deduced from the nature of the phenomena being characterized, OR, from the law of large numbers, characterized from repeated measurements.
In the case we were discussing: what you and Legion were calculating was the probability that two people would have the same birthday at one given, specific party.
To find the statistical distribution, however, one would have to use the Poisson distribution that I discussed last night. This is the distribution that describes an event that occurs randomly, has a certain periodicity, and for which past events have no bearing on what will happen in the future.
In this case, the event would be meeting a person who has the same birthday as me, and the periodicity is 1/365.
Another phenomenon that has a Poisson distribtion is radioctive emissions, and that is probably the best way to understand the odds. Think of it as follows: my encounters with other humans occurs on a random basis, and on average (the law of average numbers) every 365th one I meet will have the same birthday as me (this is the Poisson periodicity). Hence, to calculate the odds for this party situation, I would insert this periodicity into the Poisson equation (1/365) and calculate the odds after 20 units.
Put another way, here is the exact analogy of the party situation: put a Geiger counter next to a radioactive material that emits, on average, one spurt of radiation every 365 seconds, and characterize how many “blips” register on it after 20 seconds.
If you want, I can make this calcualtion, though no time now. I’m super-busy, preparing for a heavy period of travel.
UGH – anyone watch the Westminster Dog Show?
I just want to complain!!!!!!
They NEVER EVER pick the dogs that I love and are adorable – they always pick these boring or ugly or weird looking dogs in each group!
I am sooooo angry – they never pick Chow Chows, or Golden Retrievers or Huskies or Malamutes EVERRRRRRRRRRRRRRRRRRRRRRRRR
I’ve known a lot of gay couples who were faithful & stayed together long times – probably more than hetero couples I’ve known. (None w/ kids though – maybe that makes a diff.)
Lech and all;
This morning as I was shaving, I figured out the best way to explain the issue discussed late last night (the probability of 2 people out of a party of 20 having the same birthday).
The best way to demonstrate my point is to look at a roll of the dice. Let’s ask the simple question: “If I roll the die 6 times, what are the odds that a single number (let’s say the number 1) will appear one time”
Using the methodology that you and Legion discussed last night, the answer would be (1/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6)=0.063. This would indicate that the odds would be roughly 1/16 and it would be mid-leading. If you were to go ahead and make a series of bets based on these odds, you would be a loser.
What the above calculation does is calculate the probability for ONE SPECIFIC SEQUENCE OF SIX ROLLS OF THE DIE. As common sense tells us, however, the actual odds for the number “1” to appear one time in six rolls is 1/6, and if we were to place a series of bets, these are the odds we would use. We can understand this just from the simple shape of the die.
This simple example points out the difference between calculating the probability of one specific event, and the underlying statistical distribution (the “odds”). The statistical distribution can be deduced from the nature of the phenomena being characterized, OR, from the law of large numbers, characterized from repeated measurements.
In the case we were discussing: what you and Legion were calculating was the probability that two people would have the same birthday at one given, specific party.
To find the statistical distribution, however, one would have to use the Poisson distribution that I discussed last night. This is the distribution that describes an event that occurs randomly, has a certain periodicity, and for which past events have no bearing on what will happen in the future.
In this case, the event would be meeting a person who has the same birthday as me, and the periodicity is 1/365.
Another phenomenon that has a Poisson distribtion is radioctive emissions, and that is probably the best way to understand the odds. Think of it as follows: my encounters with other humans occurs on a random basis, and on average (the law of average numbers) every 365th one I meet will have the same birthday as me (this is the Poisson periodicity). Hence, to calculate the odds for this party situation, I would insert this periodicity into the Poisson equation (1/365) and calculate the odds after 20 units.
Put another way, here is the exact analogy of the party situation: put a Geiger counter next to a radioactive material that emits, on average, one spurt of radiation every 365 seconds, and characterize how many “blips” register on it after 20 seconds.
If you want, I can make this calcualtion, though no time now. I’m super-busy, preparing for a heavy period of travel.
UGH – anyone watch the Westminster Dog Show?
I just want to complain!!!!!!
They NEVER EVER pick the dogs that I love and are adorable – they always pick these boring or ugly or weird looking dogs in each group!
I am sooooo angry – they never pick Chow Chows, or Golden Retrievers or Huskies or Malamutes EVERRRRRRRRRRRRRRRRRRRRRRRRR
Happy Anniversary Biff
Congratulations Biff on 10 year wedding Anniversary.
Get the Mrs. the diamond ring considering……
“Can you explain as simply what others were trying to say?”
No I can’t because I don’t think they got it right.
“I attended 2 weddings in California for same sex couple who had been together for more than 10 years.â€
And…..were they monogamous during the 10+ year period?
Jessi, I bill for work in bed, the shower and the like at 50% off
I’ve known a lot of gay couples who were faithful & stayed together long times – probably more than hetero couples I’ve known. (None w/ kids though – maybe that makes a diff.)
Let’s Get Ready to Rumble!
OK I HAVE TO stop procrastinating and mark up this document.
Not paying attention to OT for a while.