# 3 players, first heads wins, 2 methods

• Oct 28th 2012, 03:29 PM
tcRom
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.
• Oct 28th 2012, 04:35 PM
Plato
Re: 3 players, first heads wins, 2 methods
Quote:

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 $\displaystyle S=\{1,~0001,~0000001,~0000000001,\cdots\}$
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 $\displaystyle (.5)+(.5)^4+(.5)^7+(.5)^{10}+\cdots$
Can you find $\displaystyle \sum\limits_{k = 1}^\infty {\left( {0.5} \right)^{3k - 2} }$
• Oct 28th 2012, 05:14 PM
tcRom
Re: 3 players, first heads wins, 2 methods
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 $\displaystyle \sum\limits_{k = 0}^\infty {\frac{1}{2^{3k + 1} }$.

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.
• Oct 28th 2012, 06:47 PM
Plato
Re: 3 players, first heads wins, 2 methods
Quote:

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 $\displaystyle \sum\limits_{k = 0}^\infty {\frac{1}{2^{3k + 1} }$.
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?
• Oct 28th 2012, 07:01 PM
tcRom
Re: 3 players, first heads wins, 2 methods
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.
• Oct 29th 2012, 03:47 AM
Plato
Re: 3 players, first heads wins, 2 methods
Quote:

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 $\displaystyle 0.5$ probability of winning on the first turn but $\displaystyle (0.5)^4$ on A's second flip.