Tell me more ×
Cross Validated is a question and answer site for statisticians, data analysts, data miners and data visualization experts. It's 100% free, no registration required.
up vote 1 down vote favorite

Suppose I have a die and I throw it a large number of times (a billion times) and have proven that it lands on each side an equal number of times so it is proven to be fair.

Next suppose I started a second round of throwing of the same die and it lands on 1 through 5 say a 10,000 times each and the count for 6 is 100.

Suppose someone ask you to bet on the next billion throws.

Would you bet on 6?

share|improve this question
3  
I'd bet on anything but... the die has already proven itself unfair. – John Apr 29 '12 at 2:24
1  
I think both the title and the actual question of interest should be updated to be more clear. – cardinal Apr 29 '12 at 3:00
1  
How can "lands on each side an equal number of times" be a true statement since you rolled the die a billion times and $10^9$ is not divisible by $6$? If you amend your statement to say (for example) that when you said "a large number of times" you really meant six billion times, then how does seeing each face turn up one billion times in six billion throws prove that the die is fair? It does nothing of the sort! – Dilip Sarwate Apr 29 '12 at 19:31

2 Answers

up vote 6 down vote accepted

If the die rolls are independent, then you'd no sooner bet on $6$ now than you would've on the first throw - the probability is $1/6$ now just like it is on every throw.

The scenario you've described seems to be related to the common probabilistic fallacy known as the law of averages, an example of which is the belief that events that are "due" to happen are somehow more likely.

There certainly could be a dependence structure in a time series that makes events that have occurred less frequently than expected somehow more likely in the future (an extreme example of this is a Self Avoiding Random Walk), but when the rolls of the die are independent, this reasoning doesn't make sense.

Note that the law of large numbers implies that the observed relative frequencies of each outcome converge to the true probabilities as the sample size goes to $\infty$, so it does "even out" in the long run. But, this does not imply that the subsequent throws have an increased probability of being '6's - if exactly $1/6$ of the throws for the rest of eternity were $6$s, then the limit implied by the law of large numbers still applies. In your example, only $100$ of the first $50100$ throws were $6$s, but

$$ \lim_{n \rightarrow \infty} \frac{100 + n/6}{50100 + n} = 1/6; $$

so, you can see that although $6$s are wildly underrepresented early on in the sequence, no $6$s need to be 'made up' for things to even out in the long run.

share|improve this answer
but doesn't law of averages refer to small samples? what if it were a big sample such as a billion throws? – mfc Apr 29 '12 at 2:02
3  
any finite sample is 'small' - the law of averages refers to expecting that asymptotic (i.e. infinite sample) results about random numbers should apply in finite samples. – Macro Apr 29 '12 at 2:08
up vote 0 down vote

This is a very crazy scenario. Why would the probability of heads change so much? If it did and you believe the future is best represented by the recent past then if it is fair to assume that the rolls are independent then betting on 6 would be foolish.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.