'craps' Q: prob. of 2 throws w/o landing '7' on either

[jcraigjcraig]

The probability of throwing a seven with two fair dice is 1/6. What is the probability of completing two random throws without a seven on either throw - considering that if you throw one on the first throw, the turn is over?

 

[Deleted member 4993]

Re: craps roll question

The probability of throwing a seven with two fair dice is 1/6. What is the probability of completing two random throws without a seven on either throw - considering that if you throw one on the first throw, the turn is over?

Please show us your work/thoughts indicating exactly where you are stuck.

 

[jcraigjcraig]

Re: craps roll question

Ok-
I need to know the probability of being able to complete exactly two rolls without throwing a seven. Is it First roll (1/6) + second roll (1/6) = 1/3, or 66.7 chance of success? Or, do you look at it like this - 1/6 = .167, so is it First roll (.167) TIMES Second roll (.167) = .0279, so you have a (100-27.9) or a 72.1% chance of success? Or, since the probability of success on any one roll is 5/6 (.833) would you say that it is First roll (.833) chance of success TIMES second roll (.833) chance of success = .694? Or, would you say that since the two rolls are independent events, the probability remains 83.3%? Not sure how to look at it.
Thanks for your help!!

 

[pka]

Re: craps roll question

The probability of not throwing a seven is \(\displaystyle \frac{5}{6}\).
That is independent of any other throw of the dice.
So the answer is: the probability of no seven the first times the probability of no seven the second.