Question

Playing a game of dreidl, one of four outcomes are possible on each spin. All four outcomes are equally likely. One outcome, called Gimel, is most desirable. If a player spins the dreidl 68 times, what's the probability they'll get Gimel 56 or more times?

(Re: http://www.timesunion.com/local/article/A-one-in-trillions-dreidel-game-2427950.php)

Answer 1

is it possible to calculate fast without writing 56 57 58..68?

Answer 2

I'd use the CDF which implicitly sums a range of numbers.

Answer 3

and the binomial distribution

Answer 4

yeah binomial distribution but idk how to sum all these except doing by hand

Answer 5

i use mathematica

Answer 6

well i can do with help of wolframalpha too, but is it possible to do just having a pen and paper? :D

Answer 7

with enough time i suppose :-)

Answer 8

NO, i can't do with wolfram, i don't know how to repeat

Answer 9

http://www.wolframalpha.com/input/?i=Sum%5Bx%2C+%7Bx%2C+1%2C+10%7D%5D

Answer 10

but it would be \[1-\left( \left(\begin{matrix}68 \\ 56\end{matrix}\right)\left( \frac{1}{4} \right)^{56}\left( \frac{3}{4} \right)^{12}+ \left(\begin{matrix}68 \\ 57\end{matrix}\right)\left( \frac{1}{4} \right)^{57}\left( \frac{3}{4} \right)^{11}+... \right)\]

Answer 11

hmm i think i know

Answer 12

doesn't work http://www.wolframalpha.com/input/?i=1-%28sum+from+56+to+68+%28%2868+choose+i%29*%28%281%2F4%29%5Ei%29*%28%283%2F4%29%5E%2868-i%29%29%29%29
:D