Rolling exactly 3 6's with 7 die

[ztyL]

Hi there

I have the following problem.

I have 7 dies and if i roll exactly three 6's i win.

I've tried to calculate the possibility but i get different results depending on the method.

First method:
Look at each dice and calculate the following:

(1/6)^3 * (5/6)^4 = 0,00223

Second method:
Look at the number of permutations with 7 dies (with repetition):

6^7 = 279936

Look at how many permutations you can make when choosing 3 out of 7 (with repetition):

7^3 = 343

343/279936 = 0,00122

am i way off or am i missing something?


/ztyL

 

[HallsofIvy]

"Dice", not "dies", is the plural of "die". And later you have "Look at each dice" when it should be "Look at each die".

Yes, the probability of rolling a "6" on a single die is 1/6 and the probability of rolling "not a 6" is 5/6. The probability of rolling "3 6's and 4 not-sixes" in that order is \(\displaystyle \left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^4= 0.00223\), approximately.

And it is easy to show that the probability of rolling "3 sixes and 4 not-sixes", in a specific order, is exactly the same. That is what your "first method" calculates, not the probability of 3 sixes and 4 non-sixes in any order.

We need to multiply by the number of different orders of "SSSNNNN" ("S" is "a six" and "N" is "not a six"). That, the number of ways of ordering seven things, three of them the same, the other four the same, is the "binomial coefficient", \(\displaystyle \frac{7!}{3!4!}= \frac{5040}{(6)(24)}= \frac{5040}{144}= 35\).

35(0.00223)= 0.07805

I really do not understand your "second method".

 

[ztyL]

Nice.. Thank you, then i learned both some english and some more math

My "second" method was an attempt to calculate the ways of ordering three things out of seven.

But i'll stick to my first method and then remember to multiply with the amout of ways ordering them.

Thanks again

/ztyL