My friends and I play a game occasionally, whereby two players take
turns rolling a pair of dice.  A round consists of each player rolling
the dice one time, and the winner of each round is simply the higher
number.  To win the game, though, a player must win 4 rounds in a row,
so that a game sometimes lasts for many rounds.  Our question is: 
Using statistics, what should be the average length of a game, in
rounds?

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Request for Question Clarification by pafalafa-ga on 11 May 2006 06:38 PDT

I'm tempted to say 32 rounds...that is, 32 rolls of the dice, on average.

However, I can't convincingly back it up, so I'll just post it here as
a teaser, and let someone more competent with numbers post an answer
to either confirm or refute.

pafalafa-ga

What happens if there is a tie?  Does that count as a round?  If so,
that could break up a winning streak even if he doesn't lose a round. 
You did say he had to win four consecutive rounds.

Throwing a dice, there are 6 possible outcomes, each with probability
1/6. Since a tie is a possible outcome, the probability of this is 1/6
* 1/6= 1/36, leaving a probability of 35/36 for one of you winning the
round.

The probability for person A to win the round is then 1/2* 35/36 =
35/72, and the probability for A to win four consecutive rounds is
then (35/72)^4 = 0.0558

The expected number of rounds to play can then be found by using the
geometric probability distribution mean, that is: E = 1 / 0.0558 =
17.9084.

This means one will in average use around 18 rounds before anyone wins a game!

Hope this is an satisfying answer:)

Mark, please clarify if each player rolls one die (as Swendbjarne has
understood) or two dice, which allows totals of 2 to 12.  The latter
problem will obviously complicate the solution.

Svendbjarne,

I am inclined to think your answer is about right, after running 100
simulations in Excel, but I'm not so sure about your methodology.  For
a single event, the waiting time (expected number of rounds) is indeed
based on a geometric distribution, where E = 1/p as you mentioned. 
But here we have four consecutive events.  So the answer couldn't be
1, 2, or 3.  It seems to me this makes things a little different. 
Also, since we don't care which person wins, why by your logic
wouldn't the answer be about 8?  The winner of the first round doesn't
matter, but the winner of the next three rounds has to be the same. 
After 100 simulations, I came up with 16.03 (where ties don't count as
a round, they are a do over).  I don't think this discussion thread
has the answer yet.

ansel001, I think I calculated a little to fast:)

Not really sure how to calculate this if a tie will result in a do
over and not count in, but I think I missed a little bit on the
probability for a tie.

Since the ties can be for all outcomes on the dice(1-1, 2-2...), the
probability for a tie will be 6*1/36 = 1/6, leaving a probability of
5/6 for someone to win the game.

Then it will be a prb of 5/12 for person A to win a round.

Since each round can be assumed to be independent of each other, the
prb of A winning four consecutive rounds will be (5/12)^4=0.0301, and
the expected number of rounds, including ties, will be 1/0.0301=
33.1776.

I'm not 100% sure the calculations are correct, but I think so.
ansel001, how many rounds does the simulation give you if include ties?

if one discards "ties" (i.e. the players ignore them and treat it as a
"do over"), then the problem can be reduced to a coin-flip situation,
i think.

then there is a 50/50 chance of either player winning, so one can
calculate the probability of winning four consecutive hands as such:

(1/2)*(1/2)*(1/2)*(1/2) = 1/32 

or, it would take 32 hands for one player to win four hands in a row. 
however, since there are two players (both with an equal probability
of winning), this number must be multiplied by 2 (since a loss for one
player is clearly a win for the other player), giving my answer of 16.

so, i would say that the "average" match under the terms i listed
above (ties are do-overs) would be 16.