1. Math: Odds/Chance Calculation

Ok, I am struggling to find a link to answer my question and it's been 8 or 9 years at least since my last stats/finite math class. I used to be great at this stuff, but now I can't remember the formula/calculation.

I'll give a simplified scenario:

You roll a die once and you have a 1 in 6 chance of rolling a particular number.

If you roll that die twice, my contention is your odds, or chance increases of rolling that exact number. I can't remember if this is the factorial function or simple multiplication or what.

Basically, if you have to roll a "1" and you have one roll your odds are easily determined, but what if you have 10 rolls to roll a "1". Each roll the odds are the same, but the overall odds are increased. How do you quantify this?

2. Oh and if you have a link to a site that shows/proves this that would be great as well.

3. Sorry, probability is 1/6, odds are 1/5, just correcting myself before someone else does.

4. They independant bernouli trials so the probabilty of each event is multiplied to get the total probability. i.e. rolling a 1 with 1 die 10 times in a row is (1/6)^10

5. Actually, that's not what I asked, but I finally found a site that gave me the answer.

It is Probability1 + Probability2 - (Probability1 x Probability2)

So 1/6 + 1/6 - (1/6 x 1/6) Which I believe works out to 30.1 % versus the original 16.67%.

6. I agree with the mailman. The probability of getting any number for each roll remains the same, but when matching with a previous roll, the probability decreases with each roll.

You have three socks - one black and two white. If you close your eyes and pick two socks are you more likely to get a match, mismatch, or are the outcomes equal?

7. Ok I think the way this works out is by calculating the complement of the successful event. What I mean is the formula is Pr(S) = 1 - Pr(Sbar).

So in your example the pr of success is 1/6, the complement or prob of failure is 5/6.
So if you roll the dice 10 times you get Pr(Sbar) = (5/6)^10 = 0.162. So the prob of success of at least 1 roll of 1, is 1 - 0.162 = 0.838.

methinks...

8. Yeah ELP you got it right.

9. Oh, I think I misread. You want to determine the probability of rolling "1" within 10 rolls?

10. ja

11. the probability of rolling a one at least once within 10 rolls is equal to 1-the product of the probabilities of not rolling it across all 10 rolls

OR

1-[(5/6)^10)]

/edit
LOL i should have read all the responses first

yeah, pusher has it right

it gets more complicated if you want to do anything other than find the probability of rolling 1 at least once, but that'll do for that calculation

12. Which question stupidhead? The socks?

The odds of finding a matched pair are as follows.

You cannot get a matched pair by picking a black sock, as there is only one of them.

So to get a pair you must pick white both times.

Your odds of this is 2/3 the first pull, and 1/2 the second pull.

So it would be 2/3 * 1/2 or .667 * .5 = 33%.

I think.

13. So to finish answering your stupid question, the odds are in favour of a mismatch, as less than half the time you will get a match in theory.

14. Pete I think you need to apply the same rationale to the sock question.

you want to avoid the black sock, so your probability of success is equal to 1 - the product of your probabilities of failure across the two draws. You have the numbers right but the rationale wrong.

Sock Draw #1: 1/3 chance of failure
Sock Draw #2: 1/2 chance of failure

probability of drawing a matched set of socks

1-(.3333*.5)
or
83% chance of successfully dressing yourself with matching socks from that setup.

I think

15. Bleh, not sure now that I think of it.

16. One thing though Cal, you don't need to draw the black sock twice, just once in either pull, so that may not be accounted in your calc.

17. yes but that's why the probability changes from pull one to pull two

the black sock is the only thing that can create 'failure' therefore it is the relevant thing to measure

you have a 1/3 chance of failure in pull 1 and a 1/2 chance of failure in pull 2

18. oh wait, yer right... the prob shouldn't be less than the 1/3 at the outset

hm

19. I would fail my SATs if I wrote 'em today

it's been too damned long since I've done stats

20. Originally Posted by ElPietro
Classic.

21. Socks = twice as likely to get a mismatch than a match.

As for the dice, I know I'm rusty, but ... the probability of rolling a certain number is 1/6, so in theory by 6 rolls you should have rolled that number at least once ... so wouldn't the probability in 10 rolls be greater than 100%?

22. Originally Posted by Anthony
... the probability of rolling a certain number is 1/6, so in theory by 6 rolls you should have rolled that number at least once ... so wouldn't the probability in 10 rolls be greater than 100%?
In six rolls you could have rolled a 1 zero times, unless once you roll a number you can't roll that number anymore, but each event is independent.

23. Originally Posted by ElPietro
Ok, I am struggling to find a link to answer my question and it's been 8 or 9 years at least since my last stats/finite math class. I used to be great at this stuff, but now I can't remember the formula/calculation.

I'll give a simplified scenario:

You roll a die once and you have a 1 in 6 chance of rolling a particular number.

If you roll that die twice, my contention is your odds, or chance increases of rolling that exact number. I can't remember if this is the factorial function or simple multiplication or what.

Basically, if you have to roll a "1" and you have one roll your odds are easily determined, but what if you have 10 rolls to roll a "1". Each roll the odds are the same, but the overall odds are increased. How do you quantify this?
The chance you will throw a 1 at least once out of 10 rolls, is

1-(5/6)^10

24. You can't quantify it. In probability you can only give a probable estimate for a given event. Rolling a die ten times is 10 separate events. The probability is always 1/6 no matter what the previous results have been.

By nature basic probabilty can only deal with events involving a Time (T) which is static. Obviously if we were to try and determine the probability of multiple events at multiple points (T) we would resort to using non-linear equations and venture into the world of quantumn probability.

EDIT: It's the same with the roulette wheel you can't determine a red number will come up next simply because 5 black numbers have already turned up in a row. There will always be a 1/2 chance that it will be a black number again, although common sense might tell us otherwise.

25. Sounds nice but you are wrong. You CAN tell exactly how big the chance is to throw a 1 at least once in 10 throws. its

1- (5/6)^10

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