1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ The binomial distribution is your friend. $\endgroup$ – Masacroso May 31 '16 at 8:25
  • 2
    $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig May 31 '16 at 8:57
2
$\begingroup$

$$\binom{40}{2}\cdot\left(\frac{1}{20}\right)^{2}\cdot\left(1-\frac{1}{20}\right)^{40-2}\approx27.76\%$$

$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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