Find the Probability of Winning a Coin Toss Game | GRE Math Practice

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In a coin toss game where two players take turns, the first player has a winning probability of 2/3. The sequences that lead to the first player's win include various combinations of heads and tails, which can be summed to find the probability. The other options, 1/4, 1/3, and 1/2, can be ruled out as they do not account for the advantage of going first. An alternative approach to the problem confirms that the probability is less than 3/4. The discussion highlights the importance of understanding probability sequences in game theory.
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Hi members,
I have traveled this forum sometimes, But this is my first question. I hope to get your help so that I can prepare better for my GRE Math test.

Following is my question.

In a game two players take turns tossing a fair coin; the winner is the firt one to toss a head. The probability that the player who makes the first toss wins the game is:
A)1/4
B)1/3
C)1/2
D)2/3
E)3/4

Thanks in advance.
LuuTruongHuy
 
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It has to be an advantage to make the first toss, so you can immediately rule out (A), (B), and (C).

Here are the sequences that give the win to the first player:

H
TTH
TTTTH
TTTTTTH
TTTTTTTTH
...

Just add the probabilities for the sequences above:

Probability of first player winning = (1/2) + (1/2)^3 + (1/2)^5 + ...

The infinite sum works out to 2/3, so (D) is the right answer.
 
Janitor said:
It has to be an advantage to make the first toss, so you can immediately rule out (A), (B), and (C).

Here are the sequences that give the win to the first player:

H
TTH
TTTTH
TTTTTTH
TTTTTTTTH
...

Just add the probabilities for the sequences above:

Probability of first player winning = (1/2) + (1/2)^3 + (1/2)^5 + ...

The infinite sum works out to 2/3, so (D) is the right answer.

Also, this sum is clearly smaller than (1/2) + [(1/2)^3 +(1/2)^4 + (1/2)^5 +...]

and the number inside [ ] is clearly (1/2)^2 or 1/4.

So the answer would have to be less that 3/4.

This variation to the approach doesn't save a whole lot of time - maybe half a minute at most - and it may leave you feeling uncertain about picking a choice without seeing any direct evidence for its correctness.

Anyways, just thought I'd drop it in as a useful elimination trick.
 
Hi

Apparently, the same question was posted in the general math forum...so we replied to it about an hour back. (I've changed the equations a bit though they are essentially the same things).
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

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