I have a question:

On a fair 20-sided die, the number $20$ comes up once every $20$ rolls. In forty rolls, it's expected that about two rolls of $20$ will happen. What are the actual odds that, after forty rolls, there will be EXACTLY two rolls that come up $20$?

What I did so far: Since there are $40$ rolls, we choose $2$ of them to be $20$'s, which gives us $780$ ways to get $2 \times 20$'s arranged among $40$ rolls ($\binom{40}{2}$). Since there are $40$ rolls, we have a total of $20^{40}$ different combinations of rolls. Hence my derived answer is $780/(20^{40})$ which is wrong.

I'm not sure where I went wrong but I assume it's the way I'm calculating the number of combinations that have exactly $2 \times 20$'s. (Which should be $\binom{40}{2}$ no?)

Please help me understand where I went wrong and many thanks in advance.

  • 1
    $\begingroup$ The binomial distribution is your friend. $\endgroup$ – Masacroso May 31 '16 at 8:25
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    $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig May 31 '16 at 8:57


  • $\begingroup$ It's a self-explanatory answer, so I hope that it helps you to understand where you've gone wrong (but feel free to let me know otherwise). $\endgroup$ – barak manos May 31 '16 at 8:30
  • $\begingroup$ Hi, could you elaborate on how you came up with the equation? $\endgroup$ – Christopher Leong May 31 '16 at 8:33
  • $\begingroup$ I only get the 40 choose 2 part, the rest is vague to me. $\endgroup$ – Christopher Leong May 31 '16 at 8:34
  • $\begingroup$ @ChristopherLeong: en.wikipedia.org/wiki/Binomial_distribution. $\endgroup$ – joriki May 31 '16 at 8:35
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    $\begingroup$ @ChristopherLeong: Step #1: Choose $2$ out of $40$ places for the specific result that you want (in this case, "20"). Step #2: The probability for each of these places to get that result is $\frac{1}{20}$, independently of each other. Since you need it it $2$ places, raise this probability to the power of $2$. Step #3: The remaining places can get any value except for your specific result (in this case, "1,...,19,21,...,40"). So there are $40-2$ remaining places, and the probability for each of them to get a "good value" is $1-\frac{1}{20}$... independently of each other... get it? $\endgroup$ – barak manos May 31 '16 at 8:38

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