# Probability of a roll of the dice

• February 19th 2010, 12:25 PM
AnAmericanInNederlands25
Probability of a roll of the dice
The following question came up on a practice exam and I am having trouble understanding the formulation to solve the problem.

A) Calculate the possibility of throwing six 6’s in six throws with an unbiased
die.
B) Calculate the possibility of throwing a 6 exactly once in six throws with
an unbiased die.

For A), I have considered that each roll of the dice is always 1/6. Therefore is it correct to assume that the formula would be (1/6).(1/6).(1/6).(1/6).(1/6).(1/6). Is this correct? And I am lost when it comes to B).

Thank you...

Matty
• February 19th 2010, 12:45 PM
ebaines
Quote:

Originally Posted by AnAmericanInNederlands25
The following question came up on a practice exam and I am having trouble understanding the formulation to solve the problem.

A) Calculate the possibility of throwing six 6’s in six throws with an unbiased
die.
B) Calculate the possibility of throwing a 6 exactly once in six throws with
an unbiased die.

For A), I have considered that each roll of the dice is always 1/6. Therefore is it correct to assume that the formula would be (1/6).(1/6).(1/6).(1/6).(1/6).(1/6). Is this correct?

Yes - you got it right.

Quote:

Originally Posted by AnAmericanInNederlands25
And I am lost when it comes to B).

You can think this through by considering that on any throw the probability of a six is 1/6, and the probability of anything but six is 1 - 1/6 = 5/6. Hence, the probability of getting a 6 on the first throw and no more sixes for the next 5 throws is: (1/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6). But you can also win this game by rolling the 6 on the second throw instead of te hfirst, or the the third, forth, fifth, of sixth. The probability of any one of these occurring is the same as above. So give that anyone of these 6 outcomes is a winner, the total probability is: 6*(1/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6).

In math terms, for independent events like throwing a dice, if the probability of something occuring is p, then the probability of that thing occurring k times out of n throws is equal to:

$
P(k) = C(n,k)p^k (1-p)^{(n-k)}
$

Where C(n,k) means the combination of n things taken k at a time, which is equal to:

$
C(n,k) = \frac {n!} {(n-k)!}
$

So for your problem you have:
$
P(1) = \frac { 6!} {5!} (\frac 1 6 ) ^1 ( \frac 5 6 ) ^5 = (\frac 5 6) ^5
$
• February 19th 2010, 12:50 PM
AnAmericanInNederlands25
Thank you ebaines. It makes much more sense now. I had a friend try to tell me that you increase your chances every time you roll the dice. But that did not make any sense, since you still have a 1/6 chance of rolling a 6 every time. Again, thanks!
• February 19th 2010, 12:54 PM
vince
Quote:

Originally Posted by AnAmericanInNederlands25
Thank you ebaines. It makes much more sense now. I had a friend try to tell me that you increase your chances every time you roll the dice. But that did not make any sense, since you still have a 1/6 chance of rolling a 6 every time. Again, thanks!

ouch your friend probably is an avid lotto player(Sun) ..hey i play it too when the prize is over a certain amount...LOL...more likley to get hit by lightning many times over