
April 24th, 2010 11:36 AM
#1
VIP Member
Array
Even easier probability problem(s)...
A few more, pretty simple probability questions. These get into the idea that past events don't influence future events (the roulette wheel fallacy that has cost a lot of gamblers a lot of money), and should be basically straightforward math.
Consider a very simple weekly lottery: A number between one and one thousand (inclusive) is randomly chosen each week.
1. If you play one number this week, what are your odds of winning?
2. If you won this week, what are the odds of you winning by choosing the same number next week?
3. If you won this week, what are the odds of you winning by choosing a different number next week?
4. What are the odds of the winning number being 1, two weeks in a row?
5. What are the odds of the winning number being the same, two weeks in a row?
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 24th, 2010 12:36 PM
#2
VIP Member
Array
13 answer is 1 to 999. 45 answer is 1 to 999,999, a priori.
Americans understood the right of selfpreservation as permitting a citizen to repel force by force
when the intervention of society... may be too late to prevent an injury.
Blackstone’s Commentaries 145–146, n. 42 (1803) in District of Columbia v. Heller, 554 U.S. 570 (2008)

April 24th, 2010 12:45 PM
#3
Distinguished Member
Array
Questions 13: the answer is 1 in 1,000
Questiosn 45: the answer is 1 in 1,000,000
The odds of me failing a class in probability, very high!

April 24th, 2010 03:30 PM
#4
VIP Member
Array
The odds are the same for each example. 1:1000
It's like choosing the numbers 123456 in a real lottery. Most people would find that crazy, saying it would never happen. But there's exactly as much chance for that group as ANY random 6 numbers.
"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 24th, 2010 04:20 PM
#5
VIP Member
Array
No one has answered all of the questions correctly yet...
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 24th, 2010 04:52 PM
#6
Senior Member
Array
they are all 1 in 1 billion
If I gave a crap about what you think about my guns.....it was early this morning and I already flushed it!

April 24th, 2010 04:58 PM
#7
VIP Member
Array
Pass the goat please.
CCW permit holder for Idaho, Utah, Pennsylvania, Maine and New Hampshire. I can carry in your country but not my own.

April 24th, 2010 05:04 PM
#8
VIP Member
Array
OK, I know the correct answer is either behind Door #1, Door #2, or Door #3. So I will wait to see which door OPFOR opens.
"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 24th, 2010 05:09 PM
#9
Senior Member
Array
OPFOR;
Sorry,I have to agree with Jon in NY; 1 in 1,000 for each situation.
Please explain your logic for a different answer.
" Keep On Packin' On The Bimah"

April 24th, 2010 05:12 PM
#10
Member
Array
Assuming the numbers are chosen from a uniform distribution and are independent,
1. 1 in 1000
2. 1 in 1000
3. 1 in 1000
4. 1 in 1,000,000
5. 1 in 1000

April 24th, 2010 05:14 PM
#11
VIP Member
Array
OK, I'm changing answers 4 and 5 to 1:1,000,000. Here's why:
Suppose we play this with a die. The odds of rolling a 6 the first time, is obviously 1:6. The odds of rolling a 6 the 2nd time, in a row, are not quite random, because you are looking for a given number, not a random number. So, if it were a die, it would be (1/6) * (1/6) or 1:36 chance of rolling consecutive 6's.
So, if the original odds in the lottery are 1:1000, then it stands to reason that to get the same number twice in a row, it would be (1/1000)*(1/1000) or 1:1,000,000.
Yes? No?
"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 24th, 2010 05:26 PM
#12
VIP Member
Array
Mr. Gray is correct. Jon, you're close... :)
1: Straight odds, 1:1000
2 and 3: Again, straight odds. The number for the second week is entirely independent of the first week; it doesn't matter that you won, or with what number  each week in these examples is its own separate event, with 1:1000 odds of you winning.
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.
5: Here again, the first week doesn't matter. We are simply determining the odds that the second week will be the same as whatever number came up the first week. Since the first week was a random number, the chance that that same number will come up again is the same chance as for ANY random number; 1:1000.
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 24th, 2010 05:45 PM
#13
VIP Member
Array
Oh well, close but no cigar.
"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 24th, 2010 08:02 PM
#14
Senior Member
Array
I disagree with #5 answer. It wasn't defined that the first week was already chosen. If you start all answers prior to first week being picked, then the answer is the same as #4.

April 24th, 2010 10:39 PM
#15
VIP Member
Array
You're welcome to disagree. You'll just be wrong.
The first winning number is random, correct? It occurs, and then it's over. Assume for the sake of argument that the first number is 5.
What are the odds that a 5 will come up in week number 2? Exactly the same that a 1, or a 2, or 999 will come up: that is to say, precisely one in one thousand. The actual number in week number one is meaningless  it's just a number. The odds that that number will come up again in week number two have nothing to do with what number it was  they are two entirely distinct events.
Note the difference in problem 4; in that problem, you had to pick BOTH weeks correctly (by picking the number 1, in that example). In problem 5 you get all 1000 numbers the first week (because whatever number it lands on, all you are doing is taking that number and "playing" it the next week.) Put another way, all you are doing in problem 5 is letting someone else (the lottery ball) pick your number for you for week number 2.
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.
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