Probability question
Probability question
The organizer of our fantasy football league has drawn the #1 draft position out of a basket for the second straight year. He asked me "What are the odds?" I told him the odds were obviously 1-1 since he was the guy in charge. But, seriously, since it is a 12-man league, I assume the odds of him drawing the #1 pick last year was 1-12 and this year the odds were also 1-12. So, am I correct that the odds of him drawing #1 both years is 1-144 (1/12 x 1/12)?? I assume that the odds of him drawing #1 next year will also be 1-12, but is that really correct since the odds of him drawing #1 3 years in a row would be 1/12 x 1/12 x 1/12? I know this sounds crazy, but wouldn't this lower his odds from the expected 1-12?
Re: Probability question
The odds do not change, the odds of him drawing the #1 three years in a row is 1/12x1/12x1/12, but the odds of him picking #1 in any given single year is still 1/12. The belief that him picking the #1 for the last two years lowers his chances of picking it again next year is called "the gamblers fallacy". The outcome of past events has no effect on the odds of a future occurance.
Last edited by John3754 on Fri Aug 29, 2014 2:29 pm, edited 1 time in total.
Re: Probability question
Is your friend the only person picking? Is he picking first?
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Re: Probability question
May be thinking it this way will help:
He has drawn 2 #1 picks. This has already happened. Now, odds of him drawing #1 pick again is 1/12.
Also, odds of him being born on his birthday, meeting you, drawing 2 #1 picks and then drawing #1 pick again are still 1/12.
But odds of him drawing #1 picks for next 3 years in a row are 1/12 * 1/12 * 1/12
He has drawn 2 #1 picks. This has already happened. Now, odds of him drawing #1 pick again is 1/12.
Also, odds of him being born on his birthday, meeting you, drawing 2 #1 picks and then drawing #1 pick again are still 1/12.
But odds of him drawing #1 picks for next 3 years in a row are 1/12 * 1/12 * 1/12
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Re: Probability question
The only reason that the previous two years' "1/12" probability event MIGHT affect the third year is IF there is evidence, once you decide to look into it, etc., that the "1/12" probability wasn't accurate in the first place, or for some reason does in fact change.
If he's using heavier paper for the hat draw (or something similar for other "random" choice methods), that could affect the outcome (purposefully or not).
I put the word "random" in quotes, because most "choosing" techniques like this are not truly random, but are more "arbitrary", and the odds might not be truly what they would be if there was a truly random process.
RM
If he's using heavier paper for the hat draw (or something similar for other "random" choice methods), that could affect the outcome (purposefully or not).
I put the word "random" in quotes, because most "choosing" techniques like this are not truly random, but are more "arbitrary", and the odds might not be truly what they would be if there was a truly random process.
RM
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Re: Probability question
Independent events is the term people are describing here. Past draws do not affect future draws anymore than a coin knows it is time to come up heads because the last 3 flips were tails.
Re: Probability question
To remain actionable (and on-topic), remain focused on helping with the math. Sports discussions are off-topic.
Re: Probability question
The probability would be 1/144 or 0.694%. The odds would be 1:143. Both of these from the point of view before either draft was done. Given they are independent events, the probability and odds would be 1/12 or 8.3% and 1:11 respectively that the person having the number one selection in the first draft would get it in the second. The same probability would be in place next year. The chances are not that small for each individual event.
'It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so!' Mark Twain
Re: Probability question
The difference here is a priori (intial) probability versus conditional probability. To compute the conditional probability of B given A, compute the probability of A and B both happening, then divide by the probability of A. For example, the probability that a die rolls a 6, given that it rolls an even number, is (1/6)/(1/2)=1/3.Pacific wrote:The organizer of our fantasy football league has drawn the #1 draft position out of a basket for the second straight year. He asked me "What are the odds?" I told him the odds were obviously 1-1 since he was the guy in charge. But, seriously, since it is a 12-man league, I assume the odds of him drawing the #1 pick last year was 1-12 and this year the odds were also 1-12. So, am I correct that the odds of him drawing #1 both years is 1-144 (1/12 x 1/12)?? I assume that the odds of him drawing #1 next year will also be 1-12, but is that really correct since the odds of him drawing #1 3 years in a row would be 1/12 x 1/12 x 1/12? I know this sounds crazy, but wouldn't this lower his odds from the expected 1-12?
The probability that he will draw the first pick three years in a row, before any draws have been made, is 1/12^3=1/1728. The conditional probability that he will draw the first pick three years in a row, given that he has drawn it twice, is 1/12. He has 1/12 chance of drawing the first pick in the third year regardless of where he drew in the first two years; alternatively, you could compute this as (1/12^3)/(1.12^2)=1/12.
Re: Probability question
The mathematics behind this is the Binomial distribution, which allows one to calculate the probability of getting exactly k specific outcomes from a series of n independent events. The events have to be able to be modeled as binary events, e.g. success and failure, heads and tails, and so on. In this case, the "success" would be drawing the #1 draft position and the "failure" would be drawing any other position.
Here are a few links on the Binomial distribution:
http://mathworld.wolfram.com/BinomialDistribution.html
http://en.wikipedia.org/wiki/Binomial_distribution
The primary formula to calculate the probability of k successes in n events, where p is the probability of success and q is the probability of failure (q = 1-p), is:
(n choose k) * p^k * q^(n-k)
When n equals k, this simplifies to p^k, since q^0 is 1 and (n choose n) is also 1.
So in this example, p = 1/12, n = 2 and k = 2, so the probability is indeed (1/12)^2 (e.g. 1/12 squared or 1/12*1/12).
Edit: The notation p^k is just a way of representing powers in a text based format. I forget how to do superscripts in phpBB, and ^ is the LaTeX character for doing superscripts in mathematical formulas.
Here are a few links on the Binomial distribution:
http://mathworld.wolfram.com/BinomialDistribution.html
http://en.wikipedia.org/wiki/Binomial_distribution
The primary formula to calculate the probability of k successes in n events, where p is the probability of success and q is the probability of failure (q = 1-p), is:
(n choose k) * p^k * q^(n-k)
When n equals k, this simplifies to p^k, since q^0 is 1 and (n choose n) is also 1.
So in this example, p = 1/12, n = 2 and k = 2, so the probability is indeed (1/12)^2 (e.g. 1/12 squared or 1/12*1/12).
Edit: The notation p^k is just a way of representing powers in a text based format. I forget how to do superscripts in phpBB, and ^ is the LaTeX character for doing superscripts in mathematical formulas.
Re: Probability question
What is your best guess for his integrity? Surely there's some chance he's cheating to his own benefit. If you're talking about the 2-in-a-row question, you'll get very different answers averaging, say, a 1/144 chance (he's honest) with a 1% chance he cheats (100%). Those are of similar magnitudes and would double the expectation, and I've got to think people cheat more than 1% of the time if they know they can't be caught.
Re: Probability question
Your question is retrospective. The probability he would get it twice is simply 100%. Had you asked the question before the two years, it would be 1/144. Had you thought to ask it after the first year, it would be 1/12. But you didn't. You are asking the question in retrospect - data-mining if you will.
You wouldn't be posing it here if it hadn't already happened, so the correct answer is the probability of him already having won twice is 100%.
Now, having thought of the question, the chances of him winning three in a row is 1/12, four in a row is 1/144 and so on. Having posed your hypothesis, he needs to win the next two (four in a row) to be at p<0.05, i.e. statistically significant.
You wouldn't be posing it here if it hadn't already happened, so the correct answer is the probability of him already having won twice is 100%.
Now, having thought of the question, the chances of him winning three in a row is 1/12, four in a row is 1/144 and so on. Having posed your hypothesis, he needs to win the next two (four in a row) to be at p<0.05, i.e. statistically significant.
Re: Probability question
The phpBB BBCode is not available. I don't see a need for adding it to the post editor.Mudpuppy wrote:Edit: The notation p^k is just a way of representing powers in a text based format. I forget how to do superscripts in phpBB, and ^ is the LaTeX character for doing superscripts in mathematical formulas.
Notation of the form p^k is used in spreadsheets and several programming languages. I don't think there's any problem of misunderstanding the intent.
Be careful on the binding of the operators, e.g. "10^n/10" is interpreted as "10^n * 1/10" instead of "10^(n/10)". Use of parentheses to clarify an expression can never hurt (as you've already done in your expression).
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Re: Probability question
This section is in case you feel the organizer has a draft advantage
Does your league use a serpentine draft? i.e. Round 1 team 1 picks first, round 2 team 12 picks first, round 3 team 1...
If that's the case you could argue that even if your friend was skewing the odds he won't get an advantage. Picking in the middle spots allows you to use more draft strategery in between each round.
The #1 draft spot had the 9th highest success rate of winning their respective leagues according to Fantasy Sharks website.
Odds
Back to the statistics I agree with others. 1/12 * 1/12 = 1/44 looking forward. 100% looking back.
Does your league use a serpentine draft? i.e. Round 1 team 1 picks first, round 2 team 12 picks first, round 3 team 1...
If that's the case you could argue that even if your friend was skewing the odds he won't get an advantage. Picking in the middle spots allows you to use more draft strategery in between each round.
The #1 draft spot had the 9th highest success rate of winning their respective leagues according to Fantasy Sharks website.
Odds
Back to the statistics I agree with others. 1/12 * 1/12 = 1/44 looking forward. 100% looking back.
Re: Probability question
"Odds" are expressed as a ratio. The odds in this case are 11 to 1 against each year - so I guess 143 to 1 against for two years?
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Re: Probability question
Apologies for the zombie thread post but the abilities of people here on probability are impressive. It's a subject that's always confounded me. Do you guys have any book- or online course recommendations to really get the hang of probability calculations?
If the instruction style is enquiry-driven, that's even better.
If the instruction style is enquiry-driven, that's even better.
Re: Probability question
Try khan Academy first.
Re: Probability question
OK as long as we've got a zombie thread in the middle of the night:
I always had trouble with this. Why doesn't it work if you frame the question differently?
The probability of getting the #1 pick 3 years in a row is (1/12)^3.
But the probability of getting any other pick besides #1 isn't (11/12)^3.
I always had trouble with this. Why doesn't it work if you frame the question differently?
The probability of getting the #1 pick 3 years in a row is (1/12)^3.
But the probability of getting any other pick besides #1 isn't (11/12)^3.
Re: Probability question
Are you talking about the probability of failing, in all 3 years, to draw #1?
Or the probability of failing, at least once during the 3 years, to draw #1?
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Re: Probability question
mega317 wrote: ↑Sat Apr 11, 2020 12:29 am OK as long as we've got a zombie thread in the middle of the night:
I always had trouble with this. Why doesn't it work if you frame the question differently?
The probability of getting the #1 pick 3 years in a row is (1/12)^3.
But the probability of getting any other pick besides #1 isn't (11/12)^3.
- Probability of all three #1 (P3)=(1/12)^3*(11/12)^0
Probability of only two #1 (P2)=3*(1/12)^2*(11/12)^1
Probability of only one #1 (P1)=3*(1/12)^1*(11/12)^2
Probability of zero #1 (P0)=(1/12)^0*(11/12)^3
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Re: Probability question
While the probability that the organizer gets the first pick twice in a row may be 1/144, the probability just that someone gets the first pick twice in a row is only 1/12.
This can be determined without enumerating the outcomes by just noting that the probability someone gets pick #1 the first year is 1, and the probability that the same person gets it again is 1/12.
If you only know the person after the fact, it is only a 1/12 chance that someone gets pick #1 twice in a row. Once you know the person you may calculate the probability for that particular person as 1/144 and think the event was much less probable than it was when the person to possibly achieve the feat was arbitrary and unknown.
This can be determined without enumerating the outcomes by just noting that the probability someone gets pick #1 the first year is 1, and the probability that the same person gets it again is 1/12.
If you only know the person after the fact, it is only a 1/12 chance that someone gets pick #1 twice in a row. Once you know the person you may calculate the probability for that particular person as 1/144 and think the event was much less probable than it was when the person to possibly achieve the feat was arbitrary and unknown.
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Re: Probability question
I tried it but beyond basic probability, they jump into teaching by formula very quickly. Not like their usual offerings. Perhaps I just need more exposure to problems to develop an instinct for it.
If anyone has any other recommendations, let me know.
Re: Probability question
Deleted.
Last edited by m@ver1ck on Sat Apr 11, 2020 9:47 pm, edited 1 time in total.
Re: Probability question
FWIW - this is the text I have -InvestingGeek wrote: ↑Sat Apr 11, 2020 9:23 pmI tried it but beyond basic probability, they jump into teaching by formula very quickly. Not like their usual offerings. Perhaps I just need more exposure to problems to develop an instinct for it.
If anyone has any other recommendations, let me know.
Introduction to Probability, 2nd Edition https://www.amazon.com/dp/188652923X/re ... KEbECWRJZA
But I found the videos more useful.
I find that one needs to work ramp up on combinatorics first too - most of the problems devolve to permutations and combinations.
Still - would love other suggestions too.
Re: Probability question
OK I got 2 more.
In high school our teacher did this birthday thing but didn't actually teach us, just used to to illustrate probability is not always what you'd expect. So in a group of n people, what is the probability that any 2 share a birthday?
And I never thought of this until my 4 year old was torturing me with it today: In a randomly shuffled deck of cards, what is the probability there will be NO repeats in the sequence? This seems like it's got to be close to 0.
In high school our teacher did this birthday thing but didn't actually teach us, just used to to illustrate probability is not always what you'd expect. So in a group of n people, what is the probability that any 2 share a birthday?
And I never thought of this until my 4 year old was torturing me with it today: In a randomly shuffled deck of cards, what is the probability there will be NO repeats in the sequence? This seems like it's got to be close to 0.
Re: Probability question
Given that n-1 people all have different birthdays, the probability that the nth person has a different birthday from the others is (366-n)/365, since there are 366-n different dates available.
Thus the probability that n people all have different birthdays is
Code: Select all
365 364 363 365-n
---*---*---*...*-----
365 365 365 365
If a repeat is defined as two consecutive cards of the same rank, the probability that a pair repeats is 3/51=1/17. Assuming independence (which is not quite correct), the probability of no repeats is (16/17)^51=.045.And I never thought of this until my 4 year old was torturing me with it today: In a randomly shuffled deck of cards, what is the probability there will be NO repeats in the sequence? This seems like it's got to be close to 0.
The actual probability is slightly lower, because repeats are anti-correlated. To a first approximation, if cards N and N-1 are different, the probability that cards N and N+1 are different is 47/50 rather than 48/51 since one of the cards that does not match N was used as card N-1. Thus a better approximation is (16/17)*(47/50)^50=.043.
Re: Probability question
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