OT: Birthday odds trivia

[JR4]

trivia:
The probability that two players on the same team have the same birth day
( 5/17 = 5/17 ..etc .... year doesn't matter)
is over 98% !!

So which players have the same birthday?

The odds that a player has your birth day is not too good?
Does any player have your birth day?
Vince Wilfork is mine

BTW, Matt Cassel is only 26. After this year he will hit Free Agency
at 27 years of age and maybe proved himself.

He has the potential of going from ~500K a year to 6M+ a year! wow!

 

[Patjew]

366 possible birthdays.

53 players.

trivia:
The probability that two players on the same team have the same birth day
( 5/17 = 5/17 ..etc .... year doesn't matter)
is over 98% !!

How?

Edit: I just did some reading. http://www.innovations-report.com/html/reports/statistics/report-112162.html Interesting stuff.

 

[JR4]

366 possible birthdays.

53 players.



How?

Edit: I just did some reading. http://www.innovations-report.com/html/reports/statistics/report-112162.html Interesting stuff.

Here is the math behind it:
http://leaver.org/noca//birthday/other.html

 

[ctpatsfan77]

Here is the math behind it:
http://leaver.org/noca//birthday/other.html

Granted this is even further OT, but worth discussing to mention how "odd" probabilities can be:

If a lottery has one million possible combinations, and you buy one million tickets with random combinations (whether you buy them one per week or ten thousand per week is, believe it or not, irrelevant), what is the probability of you winning the jackpot?

 

[JR4]

I'll bite ... what is it?

Just seems it would have to do with the odds of getting duplicates.
If of the 1 million random tickets there are only 1000 different numbers
then the odd of winning are not very good.

 

[xmarkd400x]

Granted this is even further OT, but worth discussing to mention how "odd" probabilities can be:

If a lottery has one million possible combinations, and you buy one million tickets with random combinations (whether you buy them one per week or ten thousand per week is, believe it or not, irrelevant), what is the probability of you winning the jackpot?

To make the problem a little easier: Roll a 6 sided die 6 times. What are the odds that you will get a six at least once?

On each of the individual tries, your odds are 1/6. I don't really know where to go from there, but you can think through it and know it's less than 100%

 

[ALP]

Granted this is even further OT, but worth discussing to mention how "odd" probabilities can be:

If a lottery has one million possible combinations, and you buy one million tickets with random combinations (whether you buy them one per week or ten thousand per week is, believe it or not, irrelevant), what is the probability of you winning the jackpot?

i hope anyone who would spend that much money would at least make sure no tickets were duplicates, LOL

 

[JoePats]

If a lottery has one million possible combinations, and you buy one million tickets with random combinations (whether you buy them one per week or ten thousand per week is, believe it or not, irrelevant), what is the probability of you winning the jackpot?

Who cares? You'll only get half your money back anyway over time.

 

[xmarkd400x]

To make the problem a little easier: Roll a 6 sided die 6 times. What are the odds that you will get a six at least once?

On each of the individual tries, your odds are 1/6. I don't really know where to go from there, but you can think through it and know it's less than 100%

You have to think of it backwards (just thought of this).

Instead of a 1/6 chance of winning, you have a 5/6 chance of NOT winning.

Then you multiply them together. If you had two rolls, 5/6 * 5/6 = 25/36 = 69% chance of not rolling a 6.

If you got 6 rolls, (5/6)^6 [Five-sixths to the sixth power] = 33% chance of not rolling a 6, meaning a 67% chance of rolling a 6.

(I could be wrong, but I think this is right).

 

[JR4]

Anyway back to original trivia:

I have found three pairs of Players that have the same birthday:

1. Wilhite and Mayo Feb 23rd
2. Jordan and J.Sanders Nov 11th
3. Connolly and Rudd Sept 2nd



Maybe others?

 

[xmarkd400x]

I'll take a stab at the original question.

Pick person P1 with a given birthday of B1. The odds of him having the same birthday as random person P2 with birthday B2 is 1/365.25 (.25 for the leap years) The odds of NOT having the same birthday is 364.25/365.25 (99.7%). 99.7% to the 52nd power (for 52 other people), and your player P1 has an 86.7% chance of NOT having a shared birthday with any of the other 52 people on the team.

For the next player, you compare their birthday with 51 other people (because the first two have been compared).

So, you get the formula for the probablility of players NOT sharing a birthday:
(364.25/365.25)^52 * (364.25/365.25)^51 * (364.25/365.25)^50 * ... * (364.25/365.25)^1

Or, better written (364.25/365.25)^(52!) [three-sixty-four and one-quarter over three-sixty-five and one quarter to the power of fifty-two factorial]

My calculator can't do that math, unfortunatley. But I'm pretty sure that's how you solve that problem.

 

[JR4]

I'll take a stab at the original question.

Pick person P1 with a given birthday of B1. The odds of him having the same birthday as random person P2 with birthday B2 is 1/365.25 (.25 for the leap years) The odds of NOT having the same birthday is 364.25/365.25 (99.7%). 99.7% to the 52nd power (for 52 other people), and your player P1 has an 86.7% chance of NOT having a shared birthday with any of the other 52 people on the team.

For the next player, you compare their birthday with 51 other people (because the first two have been compared).

So, you get the formula for the probablility of players NOT sharing a birthday:
(364.25/365.25)^52 * (364.25/365.25)^51 * (364.25/365.25)^50 * ... * (364.25/365.25)^1

Or, better written (364.25/365.25)^(52!) [three-sixty-four and one-quarter over three-sixty-five and one quarter to the power of fifty-two factorial]

My calculator can't do that math, unfortunatley. But I'm pretty sure that's how you solve that problem.

Actually I posted the math for the 98% above. Here's that link:

http://leaver.org/noca//birthday/other.html

 

[WhoaDirty]

You have to think of it backwards (just thought of this).

Instead of a 1/6 chance of winning, you have a 5/6 chance of NOT winning.

Then you multiply them together. If you had two rolls, 5/6 * 5/6 = 25/36 = 69% chance of not rolling a 6.

If you got 6 rolls, (5/6)^6 [Five-sixths to the sixth power] = 33% chance of not rolling a 6, meaning a 67% chance of rolling a 6.

(I could be wrong, but I think this is right).

Long time since I have taken a math class but I believe it is:

1/6 chance on the first roll PLUS (5/6 of not rolling it the previous time * 1/6 chance, for each of the sucessive rolls. So:

1/6 + (5/6*1/6) + (5/6*1/6) + (5/6*1/6) + (5/6*1/6) + (5/6*1/6) = 31/36 chance of rolling a 6 in 6 rolls.

 

[ctpatsfan77]

Long time since I have taken a math class but I believe it is:

1/6 chance on the first roll PLUS (5/6 of not rolling it the previous time * 1/6 chance, for each of the sucessive rolls. So:

1/6 + (5/6*1/6) + (5/6*1/6) + (5/6*1/6) + (5/6*1/6) + (5/6*1/6) = 31/36 chance of rolling a 6 in 6 rolls.

That's not the right way of looking at it. On each roll, there is a 5/6 chance of rolling anything but a 1. So, the odds of not rolling a 1 on any of the six rolls is (5/6) to the 6th power [(5/6)^6 = 33.5%]. Thus, the odds of rolling at least one 1 is 1 minus that number, or 66.5%.

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To answer my lottery question: with one million randomly-selected tickets for a lottery with one million possible combinations, the probability of winning is 63.2%. Remarkably, if you buy 2 million tickets, the probability is still only 86.4%.