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?

  • 3
    $\begingroup$ I'd bet on anything but... the die has already proven itself unfair. $\endgroup$
    – John
    Apr 29, 2012 at 2:24
  • 1
    $\begingroup$ I think both the title and the actual question of interest should be updated to be more clear. $\endgroup$
    – cardinal
    Apr 29, 2012 at 3:00
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    $\begingroup$ 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! $\endgroup$ Apr 29, 2012 at 19:31

3 Answers 3


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.

  • $\begingroup$ but doesn't law of averages refer to small samples? what if it were a big sample such as a billion throws? $\endgroup$
    – mfc
    Apr 29, 2012 at 2:02
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    $\begingroup$ 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. $\endgroup$
    – Macro
    Apr 29, 2012 at 2:08

Betting on six is not justified for three different reasons.

1) Macro correctly identified the fallacy in the offset argument -- betting that since six is under-represented, then six must OFFSET (be over-represented) to achieve the proper balance per the law of averages. Macro showed that OFFSET is not necessary and that DILUTION -- given enough tries-- will accomplish the same thing.

2) Chernick correctly identified the weakness in assuming the underlying probabilities are unchanged. Chernick noted that this is a "crazy scenario". To see how crazy this, calculate the chance of getting 100 6s (or fewer) in the next 50,100 throws of a fair die. Using Excel:

  • BINOM.DIST(8350,50100,1/6,1)= 5.03 E-01. Chance of 8350 or fewer (1/6th of tries).
  • BINOM.DIST(8000,50100,1/6,1)= 1.26 E-05. Chance of 8000 or fewer
  • BINOM.DIST(6000,50100,1/6,1)= 4.52 E-190. Chance of 6000 or fewer
  • BINOM.DIST(5500,50100,1/6,1)= 2.50 E-284. Chance of 5500 or fewer
  • BINOM.DIST(5400,50100,1/6,1)= 8.72 E-306. Chance of 5400 or fewer

If you had observed only 5,400 sixs, this (or fewer) would be EXPECTED if you rolled 8.72 x 10^306 sets of 50,100 die. [If the probability of an outcome is p, then that outcome is expected in n = 1/p trials.] 8.72x 10^306 is an INCREDIBLY large number of trials. Way more than billions or trillions.

Since getting 100 or fewer 6s in 50,100 tries is so unlikely, we can/should reject the null hypothesis that the underlying probability of a 6 is 1 in six, conclude that the new probability of a 6 is 100 in 50,100 and bet on anything but six [if you believe "the future is best represented by the recent past"].

3) My choice would be to not bet -- to walk away -- given the magnitude of the change and the lack of any evidence that the most recent relative-frequencies are constant.

  • $\begingroup$ This is a well-reasoned answer (+1). However, I do not understand why the numbers 8350, 8000, 6000, 5500, 5400 play such a prominent role: this seems only to distract from the argument. How are these numbers related to the question? If you want to compute the chance of obtaining 100 or fewer sixes in 50100 throws, why not just state its value? (It equals $1.25\ldots\times 10^{-3725}$, which is easily found in Excel using its LOG and GAMMALN functions.) BTW, your calculation is not quite appropriate: you need to find the chance of a run of 100 of something somewhere in the sequence. $\endgroup$
    – whuber
    Aug 14, 2014 at 14:28

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.


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