Even easier probability problem(s)...
This is a discussion on Even easier probability problem(s)... within the Off Topic & Humor Discussion forums, part of the The Back Porch category; the probabilities for the first 3 are the same if the number is truely random....

April 24th, 2010 11:05 PM
#16
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the probabilities for the first 3 are the same if the number is truely random.

April 25th, 2010 05:14 PM
#17
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Originally Posted by
OPFOR
Mr. Gray is correct. Jon, you're close... :)
4: Here, you are making TWO "bets"  you must pick the winning number (in the example, the number 1) TWICE and be right both times before the winner is chosen for the first week. The chances are 1:1000x1:1000 or 1:1,000,000.
The OP question # 4 was:
4. What are the odds of the winning number being 1, two weeks in a row?
You didn't say anything about picking the actual two weeks.
So, any two consecutive weeks would meet the condition as stated in the OP question # 4.
Therefore the answer to # 5 also applies to # 4.
Μολὼν λαβέ
I'm just one root in a grassroots organization. No one should assume that I speak for the VCDL.
I am neither an attorneyatlaw nor I do play one on television or on the internet. No one should assumes my opinion is legal advice.
Veni, Vidi, Velcro

April 25th, 2010 05:28 PM
#18
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No. In example 4, the first number MUST be 1, and the second number MUST be 1. In example 5, the first number can be any number, and the second number must just be whatever number happened to come up last week. In number 4, you are requiring both numbers to be a PARTICULAR number. In number 5, you are requiring them to be the same number, but not limiting it to any particular number.
Put another way, again: In number 4, you must first have week number one come up 1. This is a 1:1000 chance, right? Now, you must then have week number 2 come up 1. This is another 1:1000 chance. These multiply because we are requiring both statements to be true (week 1 = 1, week 2 = 1).
In number 5, we are not requiring week one to be ANYTHING. We have a 1:1 chance (100%) of getting SOME number. Now, we have to have week number 2 be equal to that number. Whatever number it was/is, there is a 1:1000 chance of that number coming up in week number 2. Multiply again (1:1x 1:1000), and you get 1:1000.
Again, I didn't invent these, and I wasn't the first to solve them. They are demonstrable mathematical facts used to show probability. They aren't wrong.
A man fires a rifle for many years, and he goes to war. And afterward he turns the rifle in at the armory, and he believes he's finished with the rifle. But no matter what else he might do with his hands  love a woman, build a house, change his son's diaper  his hands remember the rifle.

April 25th, 2010 05:36 PM
#19
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In problem #5 we don't care what the first week’s number is. Our probability of hitting a winning number (any number) is 100%. They are all winners for us in week one. It's only in week two that we need to win with a matcher. That probability is 1 in 1000.
It's only in week two that week ones number becomes relevant, hence your odds are only 1 in 1000.
Last edited by atctimmy; April 25th, 2010 at 05:38 PM.
Reason: ETA oops OPFOR was too fast for me
I American and I Ameriwill!

April 25th, 2010 06:12 PM
#20
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Originally Posted by
OPFOR
Again, I didn't invent these, and I wasn't the first to solve them. They are demonstrable mathematical facts used to show probability. They aren't wrong.
Doesn't matter who wrote them.
As the OP is written:
4. What are the odds of the winning number being 1, two weeks in a row?
The Op says nothing about when the two week in a row occur or any particular two week.
As stated, those "two weeks in a row" could be any "two weeks in a row." There is no requirement that it be any particular "two weeks in a row."
Once the number "1" comes up (whenever it comes up), the odds of it coming up the following week are the same as it would be for any other number coming up "two week in a row."
Therefore, we have 1:1 chance (100%) of getting ANY NUMBER INCLUDING THE NUMBER"1". Then once it comes up, we have to have week number 2 be equal to that number. Whatever number it was/is, there is a 1:1000 chance of that number coming up in week number 2. Multiply again (1:1x 1:1000), and you get 1:1000.
Had the question been what are the odds of picking a number (say "1") and having that number come up the next two week would require both statements to be true (week 1 = 1, week 2 = 1).
The issue is a language problem not a math problem.
Μολὼν λαβέ
I'm just one root in a grassroots organization. No one should assume that I speak for the VCDL.
I am neither an attorneyatlaw nor I do play one on television or on the internet. No one should assumes my opinion is legal advice.
Veni, Vidi, Velcro

April 25th, 2010 06:40 PM
#21
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OK, point taken; it would be better written "What are the odds that the winning number will be 1 the next two weeks in a row."
Even if the wording is slightly ambiguous, the reasonable assumption would be we (we being the author) meant THIS two weeks in a row (whenever "this" happens to be).
A man fires a rifle for many years, and he goes to war. And afterward he turns the rifle in at the armory, and he believes he's finished with the rifle. But no matter what else he might do with his hands  love a woman, build a house, change his son's diaper  his hands remember the rifle.

April 25th, 2010 07:00 PM
#22
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Originally Posted by
OPFOR
OK, point taken; it would be better written "What are the odds that the winning number will be 1 the next two weeks in a row."
Even if the wording is slightly ambiguous, the reasonable assumption would be we (we being the author) meant THIS two weeks in a row (whenever "this" happens to be).
Ahhh, I feel slightly vindicated, since I said #'s 13 were 1:1000 and 45 were 1:1,000,000!
There can be no ambiguity in mathematics!
"Democracy is two wolves and a lamb voting on what to have for lunch; Liberty is a wellarmed lamb contesting the vote."
 Benjamin Franklin

April 25th, 2010 07:16 PM
#23
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Jon, number 5 has always been 1:1000  no argument there. Daves argument is that 4 could ALSO be 1:1000 if we assume that it's over any given two week period, not the NEXT (or any specific) two weeks. His argument only makes you more wrong...
A man fires a rifle for many years, and he goes to war. And afterward he turns the rifle in at the armory, and he believes he's finished with the rifle. But no matter what else he might do with his hands  love a woman, build a house, change his son's diaper  his hands remember the rifle.

April 25th, 2010 07:42 PM
#24
VIP Member
Array
Originally Posted by
OPFOR
Jon, number 5 has always been 1:1000  no argument there. Daves argument is that 4 could ALSO be 1:1000 if we assume that it's over any given two week period, not the NEXT (or any specific) two weeks. His argument only makes you more wrong...
Can you be "more" wrong? I didn't know "wrong" had degrees.
OK, I take it back!
"Democracy is two wolves and a lamb voting on what to have for lunch; Liberty is a wellarmed lamb contesting the vote."
 Benjamin Franklin
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