Probability of rolling x, y times sequentially with two dice

[hunter.croil]

Two dice are used in Backgammon. With the understanding that the probabilities for each roll of the dice is a discrete event, not connected to or affecting the roll before or after the present roll, what are the possibilities of rolling (or not) x (on one die) y times sequentially?

[hiteshwd]

Do you mean getting a number i.e. X (can be any no. from 1 to 6) in every throw for y throws ?
not familiar with Backgammon, please incude any info that you want me to consider for the solution

[hunter.croil]

One throws two dice in Backgammon and moves the checkers according to the numbers rolled. The numbers on the two dice constitute separate moves. For example, if a player rolls 5 and 3, he may move one checker five spaces to an open point and another checker three spaces to an open point, or he may move the one checker a total of eight spaces to an open point, but only if the intermediate point (either three or five spaces from the starting point) is also open.

When rolling two dice, what's the probability of rolling an 'x' (1 to 6 on either die) 'y' times sequentially, or conversely what's the probability of not rolling an 'x' 'y' times sequentially.

Using a one die example, it seems to me that though the chances of rolling 'x' (1 to 6) on the first roll is 1/6, and the chance of rolling 'x' on the second roll is within itself also 1/6, etc, the probability of rolling the same 'x' becomes progressively more unlikely with each successive roll.

Each event is insular within its own probabilities, but is there an over arching rule of probability affecting obtaining the same result with each successive roll?

[hiteshwd]

Thats well explained and there is no such over arching principle
If a dice gets 6 in first throw, there's an equal probability of 6 to appear in the second throw

To answer the question:

1) One dice: probability of getting any number y times = (1/6)^y

2) Two dices:
Several cases can be worked on. For eg:
a) A on one dice and B on other dice (without swapping of dices) = (1/6)(1/6)^y
b) With swapping: Multiply previous answer by 2
c) getting a fixed sum on each of y throws of two dices: the answer here would vary as per the sum figure. For eg: probability of getting sum 6 in y throws of two dices= (5/6)^y

Similarly other different cases can be worked on. Let me know in case you need any further help

[hunter.croil]

Interesting that there's no overarching principle. Does that mean that the odds of one die rolling a '1' ten times in a row are the same as one die rolling a '1' 1/6 times over the course of ten rolls?

[hiteshwd]

Yup
In the case you mentioned, the odds just keep on getting multiplied by 1/6 for each extra throw. For two throws= (1/6)^2
For five: previous no. * (1/6)^3 and so on

[hunter.croil]

That's great! I'll send more questions like this your way.

Tx!

[jlucero]

Good explanations here hiteshwd.