Calculating Probability of Multiple Rolls on a D6

[RDLowrider]

Hi, I play a dice game using a D6 to work out results. Sometimes you need to make multiple rolls and I'd like to be able to work out probability of getting certain results.

1st roll of a D6 needing 3+
If the first roll fails you get a second try again needing 3+
If either roll gets 3+ you roll again this time needing 4+
If succesfull you roll once more needing 4+

Any help with how to work this out would be greatly appreciated.

Cheers,
RD

 

[HallsofIvy]

The probability of getting "3+" (3 or more?) is 4/6= 2/3. The probability of not getting a "3+" is 1- 2/3= 1/3 and if that happens you roll again, the probability of getting "3+" on the second roll is again 2/3. The probability of getting a "3+" on the first roll or on the second roll is (2/3)+ (1/3)(2/3)= 6/9+ 2/9= 8/9.

Assuming that happens, your probability of getting "4+" (4 or more) on the next roll is 3/6= 1/2. If you are successful in that, the probability of getting "4+" on your last roll is also 1/2. The probability of all of those things happening is (8/9)(1/2)(1/2)= (8/9)(1/4)= 2/9.

 

[RDLowrider]

Many thanks.