If I roll 10 dice an infinate amount of time, will I ever roll 10 sixes?

Question

Elaborate on your answer this please....

Answer 1

Yes.

The probability of rolling a six on an unbiased dice is p=1/6

Hence using the Binomial Theorem,
the probability of rolling ten sixes on ten unbiased dice is:
C(10,10)*p^10*(1-p)^0
which is just simply p^10 = 1.65E-8

So, several hundred million trials should do it for you...
you might like to outsource it to a crew of monkeys.

Answer 2

Sure. In fact, if you *don't* roll 10 sixes at some point, it means that the dice are crooked. You can expect (on average) to roll 10 6's about every 60.466,176 rolls. But you could have it *not* happen for 120,932,350 rolls and then happen twice in a row ☺

HTH


Doug

Answer 3

Yes. Let's say you roll the 10 dice for one time. The probability of you getting all 10 sixes is (1/6) to the power of 10. The probability is positive. Therefore this applies to whatever number of rolls u try. Hence for infinite number of rolls to get all 10 sixes is [(1/6) to the power of 10] to the power of infinite.

Answer 4

First of all, you can never say "Look! I have just finished rolling the dice an infinite amount of times!"

But I assume you really mean the limit as the number of rolls increases without bound.

So then the answer to your question is a definite "probably."

Answer 5

Well the chance of one dice of landing on six is a 1 in 6 chance.
So, two dice would be 1/6 x 1/6 which is 1 in 36 chance. If you continue this for 10 dice the answer would be one in 60,466,176 according to my calculator. So theoretically it is possible.

Answer 6

Depends if your dice even have sixes, do they?
And, do you want them in a row, or all together, or what?

As for combos, in an infinite number of rolls, any combination is possible, even a million sixes in a row with every dice displaying a six should happen, so long the rolls are truly infinite.

Answer 7

the answer is 1/6^10 = 1/60,466,176 therefore you have one chance in 60,466,176 on each roll.
To estimate when the chances are approximately equal of occurring (in this case, 10 sixes on ten dice) or that not ocurring you multiply the "odds to one" by .693, the co-log of the hyperbolic log of 2. This will give you an estimation of chances or trials needed to make any event an even or 50-50 propostion.
.693 X 60,466,175 =41,903,059.28
Therefore 41,903,059 rolls of 10 dice is your expected value in this type of proposition
hope this helps

Answer 8

the difficulty right here is determining the kind of approaches the cube can sum to 19. that's complete with a producing function: permit f(x) = (x + x^2 + x^3 + x^4 + x^6)^10 The extensive kind we'd like is the coefficient of x^19 whilst this polynomial is extra suitable. There are pencil-and-paper the form to discover this coefficent, or a lazy guy or woman with the suitable math utility can merely use a working laptop or laptop. that's what I did, with the result that there are 46420 approaches. (This has an identical opinion with a number of your previous solutions.) So the possibility of rolling 19 is 46420 / 6^10 = 7.677 * 10^-4 no longer very probable.

Answer 9

OK you cannot perform such an experiment in real life.
It is very well possible that you dont throw 10 sixes for 100000000000000 times or longer.

mathematicly

you can calculate the probability that you will roll 10 sixes for th e first time ath the N_th trial.

next you can take the limit of N-> infinity

then you will notice that the probality is zero. :

P(10sixesonN_th trial) = (1/6)^10*(1 - 1/6^10)^(N-1)

taking the limit for N->< inf yields nothing , zero ( who invented that number btw )

Answer 10

by odds you would roll all sixes within 1,000 rolls of the dice. but it is a game of chance. it is just odds. with 10 dice rolling 1,000 times it would allow you to get every combination of numbers one time.