Thread: 3 players, first heads wins, 2 methods

First the problem, then the 2 solutions I have. The question: How do I know which method is correct? ie, test for independence and how would I have known to do that?

Players A, B, and C take turns flipping a fair coin. The first to flip heads (1) wins. They flip in the order presented; A, B, C.
-Describe the sample space
-What is P(A wins)?

Sample space:
S = {1, 01, 001, 0001, ... , 000000...} where the game ends if a heads (1) is flipped.

Method 1:
A has to flip heads on the 1 mod 3 flip to win the game. However, if A is flipping, this means nobody has won the game. Every time A flips, we disregard all prior flips. Thus, every 1 mod 3 flip, P(A wins) = 1/2.

Method 2 (infinite bernouli, I think):
We sum the infinite converging series 1/2 + 1/8 + 1/64 + ... + 1/[2^(3k+1)] where k = 0,1,2,... representing each round/iteration, 3k+1 is A's kth flip. Thus, P(A wins) = 4/7. This math can be done for B and C as well, with all of their probabilities adding to 1.

Originally Posted by tcRom

Players A, B, and C take turns flipping a fair coin. The first to flip heads (1) wins. They flip in the order presented; A, B, C.
-Describe the sample space
-What is P(A wins)?
Sample space:
S = {1, 01, 001, 0001, ... , 000000...} where the game ends if a heads (1) is flipped.

No that is not the sample space for A winning.
It is
Player A can win on the first, fourth, seventh, tenth, etc flip.
But those are player A's first, second, third, fourth flip.
So that is
Can you find

Plato, thank you for your time:

I agree the sample space you've written is the outcome for A winning, but not for the game (which is what I've described). I believe I'm to describe the game, but I'm not certain.

That point aside, I'm more concerned with P(A wins). I get 4/7 from your summation, the same as my second method, but I used .

Why is the first method incorrect? Treating each flip for A as if the game had reset and it was A's first flip again.

Originally Posted by tcRom

Plato, thank you for your time:
I agree the sample space you've written is the outcome for A winning, but not for the game (which is what I've described). I believe I'm to describe the game, but I'm not certain.
That point aside, I'm more concerned with P(A wins). I get 4/7 from your summation, the same as my second method, but I used .
Why is the first method incorrect? Treating each flip for A as if the game had reset and it was A's first flip again.

Well what the h did you describe as the game?
AB&C flip a coin in that order. The first to get a head wins.
What is the probability that A wins?
Is that not what you posted?
If not, then what is the question?

The book doesn't make it clear if it's the whole set or just A winning that should be defined, so I'm not worried about that too much. Yes though, the game is A, B, and C flip a coin, in that order. The first to flip heads wins. So, the sample space for that would be S = {1, 01, 001, 0001, ... , 000000...}. For term 1, 2, 3, and 4 the winner is A, B, C, and A. The sample space I'm describing is all the possible outcomes of the game, not just where A wins.

As for method 2, the summation, we're on the same page.

Why method 1 would be wrong is what I don't understand.

Originally Posted by tcRom

Why method 1 would be wrong is what I don't understand.

Because the probability changes on each flip. A has probability of winning on the first turn but on A's second flip.