Prob of 6 - 6 when rolling dies one after the next

[jon_1van]

Say I roll a single die....over and over

I stop when I rolled 6 and then another 6 on the very next roll.

what is the expected number of rolls?

[BruceZ]

[ QUOTE ]
Say I roll a single die....over and over

I stop when I rolled 6 and then another 6 on the very next roll.

what is the expected number of rolls?

[/ QUOTE ]

The expected number of "tries" is 36, where a try will be 1 roll 5/6 of the time when we don't get a 6, and 2 rolls 1/6 of the time when we get a 6. The average number of rolls per try is 5/6 * 1 + 1/6 * 2 = 7/6 rolls. Therefore the expected number of rolls is 36 * 7/6 = 42 rolls.

[pzhon]

State 1: You have not just rolled a 6. You start in this state.
State 2: You have just rolled a 6.
State 3: You have just rolled two sixes in a row.

In these terms, BruceZ's argument was that the expected time to go from state 1 to state 3 or to state 1 is 7/6 of a roll, and you start in state 1 an average of 36 times before you get to state 3.

Here is another method:

Let A be the expected number of rolls remaining if you are in state 1 (including the start). Let B be the expected number of rolls remaining if you are in state 2.

A = 1+ 5/6 A + 1/6 B

That is, you always increase the number of rolls by one. 5/6 of the time you are back in state 1, and have A remaining, on average. 1/6 of the time you have moved to state 2, where you have B rolls remaining on average.

B = 1 + 5/6 A + 1/6 (0)

Solving these equations gives

A = 42
B = 36

The expected number of rolls it takes from the start is 42.

[jon_1van]

thanks