This definitely sounds like a homework, but I assure you that it's not. You're probably familiar with the Risk game. Now, friend of mine rolled 3 aces in one hand. I reckon that probability of such event is

$$C(n,k) = \frac{(n-1+k)!}{(n-1)!k!}$$

so that's $8!/(5!*3!) = 56$, so the probability is $1/56$. Am I correct?

Problem starts here: he rolled 3 aces in his 2nd attack, so he asked me: now, tell me 'bout the odds, you do statistics! And I must admit that I'm still stunned by his tremendous attacks (he lost 3 tanks both times).

Is the probability of such two consecutive events $1/56^{2}$?

  • $\begingroup$ By 'such event' do you mean exactly 3 aces, or at least 3 aces? $\endgroup$
    – Larry Wang
    Aug 9, 2010 at 0:20
  • $\begingroup$ That's right, exactly 3 aces, because you roll exactly 3 dices. $\endgroup$
    – aL3xa
    Aug 9, 2010 at 1:39
  • $\begingroup$ Taking to heart the excellent points made in @glen_b's answer, note that the chance of rolling $k=3$ aces in a hand of $n=6$ dice equals $\binom{6}{3}(\frac{1}{6})^6=5/11664=1/2332.8$, which is much smaller than $1/56$. Upon observing three aces the first time it might be natural to ask "what is the chance that will happen next time?" This number answers that question. Its tiny size suggests the dice were inadequately shaken between rolls. $\endgroup$
    – whuber
    Sep 26, 2013 at 15:34
  • 1
    $\begingroup$ A subsequent thread looks at odds in Risk-like games in great detail, including formulas and working code. $\endgroup$
    – whuber
    Jul 27, 2015 at 14:06
  • $\begingroup$ @whuber I don't understand the calculation in your comment (Sep 26 '13). First, isn't the event-in-question rolling $k=3$ in a hand of $n=3$ dice (or this twice). Second, (exactly) $k=3$ aces in a hand of $n=6$ dice would be the binomial probability ${6 \choose 3}\,(\frac{1}{6})^3\,(\frac{5}{6})^3$, $5^3$ times higher than the result in your comment? Or perhaps I misunderstand what rolling $k$ aces in a hand of $n$ dice means? $\endgroup$ Aug 17, 2016 at 17:36

1 Answer 1


You should not do a calculation of probability for an event deemed surprising post hoc as if it were an event specified before it was rolled (observed).

It's very difficult to to do a proper calculation of post hoc probability, because what other events would have been deemed at least as surprising depends on what the context is, and also on the person doing the deeming.

Would three ones twice in a row at an earlier or later stage of the game have been as surprising? Would you rolling three ones have been as surprising as him rolling them? Would three sixes be as surprising as three ones? and so on... What is the totality of all the events would have been surprising enough to generate a post like this one?

To take an extreme example, imagine a wheelbarrow-full-of-dice (ten thousand, say), each with a tiny individualized serial number. We tip the barrow out and exclaim "Whoah, what are the chances of getting this?" That is, of seeing say a "3" on die 1, a 6 on die 2, ... etc. If we work it out, $P(d_1=3)\cdot P(d_2=6)\cdot \ldots P(d_{10000}=2)$ is $6^{-10000}$. The chance of that result is incredibly small. If we repeat the experiment, we get another equally unusual event. In fact, every single time we do it, we get an event with probability so astronomically unbelievably small that we could almost power a starship with it. The problem is that the calculation is meaningless, because we specified the event post-hoc.

Even if it were legitimate to do the calculation as if it were a pre-specified event, it looks like you have that calculation incorrect. Specifically, the probability (for an event specified before the roll) of taking three dice and rolling $(1,1,1)$ is $(1/6)^3 = 1/216$, because the three rolls are independent, not $1/56$, and the probability of doing it twice out of a total of two rolls is the square of that - but neither the condition of being pre-specified nor the "out of two rolls" actually hold.


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