From my course i understand that i have actually N^n Possibilities = 6^2 = 36.6 = sides of dice and 2 = number of dice rolls
Kcurrently i calculated the possibilities for all also values. P = Probability and x = value:
Sums of all P should be the probcapacity i am looking for:
Is this correct?Is tbelow any kind of way to fix this "easier" or by making use of a formula. Since the next job is for 3 dice rolls :-)
Thanks a lot.
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edited Oct 22 "17 at 12:51
N. F. Taussig
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Yes, there is a less complicated means. It is referred to as principle of deferred decisions. It goes like this. Condition on the parity of the first die throw. It is either 0 or 1. Now, the parity determines uniquely what the parity of the second die throw should be to have actually the resultant amount have parity 0. Therefore,
eginalignP( exteven sum) &= P( extalso amount, first was even + P( exteven amount, initially was odd) \&= P( exteven)P( extfirst was even) + P( exteven)P( extinitially was odd)endalign
Now, note that $P( extfirst was even) = P( extsecond is even) = 1/2$. Similarly, $P( exteven) = P( extsecond is odd) = 1/2$. Hence, we have $P( extalso sum) = 1/2left(P( extfirst was even) + P( extinitially was odd) ight) = 1/2left(1 ight) = 1/2$.
ADDENDUM: The principle of deferred decisions helps deal with a lot harder comparable problems. Consider this problem - you throw a die 10 times. What is the probcapability that the sum is a multiple of 6?
eginalignP( extsum $equiv 0 mod 6$) &= sum_i=0^5 P( extamount $equiv 0 mod 6$| extprevious amount was $equiv i mod 6$) cdot P( extprevious amount was $equiv i mod 6$)endalign
Keep in mind that $P( extsum $equiv 0 mod 6$| extprevious sum was $equiv i mod 6$) = 1/6$ bereason tright here is just one particular die value that will certainly give us $(6-i) mod 6$. Thus, $P( extsum $equiv 0 mod 6$) = 1/6left(sum_i=0^5 P( extprevious amount was $equiv i mod 6$) ight) = 1/6left(1 ight) = 1/6.$