Do you have any other information?
Miranda and Rob play a game by each tossing a fair coin. The game consists of tossing the two coins together, until for the first time either two heads appear when Miranda wins, or two tails appear when Rob wins.
1) Show that the probability that Miranda wins at or before the nth toss is
2) Show that the probability that the game is decided at or before the nth toss is
Sorry guys, made a bit of a mistake in the question. Should have looked at it better
1. A little weird, will have to think about it for a minute.
2. The question doesn't say Miranda or Rob has to throw the winning toss, so on either persons turn there are only four possible outcomes: {HH, TT, TH, HT}. Thus on any one turn, there is a 0.5 chance the game will end in a win for any one of the players, which makes sense as there is a 0.5 chance the game will end on the "first" toss:
This is a sort of "cummulative" (well it is) distribution for "the probability that that game will end in 'n' number of turns"; which is what the question is asking. For example, what is the probability that the game will end at or before the 2nd turn - 0.75.
Yeah the 1st one put me off so that's why I asked
And for the second one, I think you're saying the formula is
where is the turn number (e.g 1st turn, 2nd turn etc)
Is that right? I'm still getting to grips with statistics.
PS. "cumulative" only has one "m".
EDIT: Well it is in England anyway, I don't know where you're from and whether it's spelt the same there (e.g "Colour" in England is "Color" in America)
EDIT 2: Ahh the formula's in the question :P Didn't see it there
I don't see how the formula in part 1 can be correct, since for n = 1, it gives 3/8. The probability that Miranda wins at or before the 1st toss is 1/4, not 3/8.
As for part 2, you can try this explanation on for size.
Like ANDS! said, the probability of a game ending for any round is 1/2. So, let p(n) denote the probability that the game ends by the nth round. Then
p(1) = 1/2
p(2) = 1/2 + (1/2)(1/2)
p(3) = 1/2 + (1/2)(1/2) + (1/2)(1/2)(1/2)
...
where the numbers come from: the probability that the game has ended on a previous round plus the probability that the game has not ended on a previous round times the probability that the game ends on this round given that the game has not ended on a previous round.
is an geometric series. The general formula is
.
Apply this and you will get the expression given in the problem.
What I gave for part 2 above is also in terms of a geometric series, by the way, I just didn't write using the summation symbol.