What is the probability of winning a game based on coin tosses?

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Discussion Overview

The discussion revolves around calculating the probability of winning a game where two players take turns tossing a fair coin, with the winner being the first to toss a head. Participants explore various methods to derive the probability that the first player wins, considering both mathematical reasoning and intuitive approaches.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant presents a solution using an infinite series to calculate the probability, suggesting that the answer is 2/3.
  • Another participant agrees with the 2/3 probability, explaining the reasoning based on the outcomes of the first player's toss and subsequent turns.
  • A different approach is introduced, where the game is analyzed by considering the scenario if the first player flips tails, leading to a recursive relationship that also results in a probability of 2/3 for the first player.
  • Some participants express uncertainty about the correctness of their calculations and invite corrections or clarifications from others.

Areas of Agreement / Disagreement

Participants generally agree that the probability of the first player winning is 2/3, but there is some uncertainty expressed regarding the clarity and correctness of the presented solutions.

Contextual Notes

Some participants assume familiarity with concepts of probability and infinite series, which may limit the accessibility of the discussion for those less experienced in these areas.

Who May Find This Useful

Individuals preparing for math tests, those interested in probability theory, and participants looking for different methods of problem-solving in games of chance may find this discussion useful.

lhuyvn
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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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HI

Here's my solution...

H = head
T = tails

(AH denotes "A got a head")

Suppose A starts first. Then the different possibilities are tabulated thus:

AH (A gets a head, game stops)
AT,BT,AH (A gets tails, B gets tails, A gets heads, game stops)
AT,BT,AT,BT,AH (A gets tails, B gets tails, A gets tails, B gets tails, A gets heads, game stops)

and so on...

So the probability is given by the sum,

\displaystyle{\frac{1}{2}} + \displaystyle{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}} + \displaystyle{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2} + ...}

the kth term is

(\frac{1}{2})^p

where p = (2k+1) for k = 0, 1, 2, ... note that there are (2k+1) continued products in the kth term)

The game goes on as long as A and B get tails and stops as soon as A gets a head, since A was the one who started the game first.

This is an infinite sum, the value of which is given by

SUM = \displaystyle{\frac{1/2}{1-(1/4)}} = \frac{2}{3}

I think 2/3 should be the answer, but I could be wrong (as usual) ;-)

Someone please correct me if I'm wrong. If any part of the solution is wrong/not clear, please let me know. (I have assumed that you are familar with addition and multiplication in probability and also with geometric progressions, esp containing an infinite number of terms--the kinds that appear in such problems.)

Cheers
Vivek
 
Last edited:
The answer is 2/3

The first player has a probability of 1/2 that both he will take a first toss AND that he will win on that toss.

The second player only has a probability of 1/4 that he will both take his first toss and win on that toss.

The first player then has a probability of 1/8 that he will both require his second toss and win on that toss.

Continuing on like this the first player has a probability of 1/2 + 1/8 + 1/32 + ... and the second player has a probability of 1/4 + 1/16 + 1/64 + ... of winning.
 
For those who like clever answers, you can skip the infinite series. :smile:

Suppose the first player's first flip is a tails. Now, if you look at how the game proceeds, it is identical to the original game, except the first and second player are reversed.

So if p is the probability that the first player in the game wins, then once the first player flips a tails, the second player has a probability p of winning. (and probability 0 of winning otherwise)

Since there's a 1/2 chance the first player will flip tails, the second player has a probability p/2 of winning, and the first player probability p.

Thus, p = 2/3.
 
Thank All for very nice answers !
 

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