Two people are taking turns tossing a pair of coins; the first person to toss two alike wins. What are the probabilities of winning for the first player and for the second player? Hint: Although there are an infinite number of possibilities here (win on first turn, second turn, third turn, etc.), the sum of the probabilities is a geometric series which can be summed; see Chapter 1 if necessary.

Short Answer

Expert verified
The probability that Player 1 wins is \( \frac{2}{3} \), and the probability that Player 2 wins is \( \frac{1}{3} \).

Step by step solution

01

Define the probabilities for each turn

When a person tosses two coins, the possible outcomes are HH, HT, TH, and TT. The probabilities for getting two alike (HH or TT) are \( \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \). Similarly, the probabilities for getting two different (HT or TH) are \( \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \).
02

Calculate the probability that Player 1 wins on the first turn

Player 1 wins on the first turn if two alike coins appear. The probability of this event is \( P_1 = \frac{1}{2} \).
03

Calculate the probability that Player 1 wins on subsequent turns

If Player 1 does not win on the first turn, then Player 2 gets a turn. The probability that neither wins on their first turn and the game continues to Player 1's next turn is \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \). Thus, the probability that Player 1 wins exactly on their second turn is \( \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \). For Player 1's third turn: \( \frac{1}{4}^2 \times \frac{1}{2} = \frac{1}{32} \). Generally, the probability that Player 1 wins on their \( n \)th turn is \( \frac{1}{2} \times \frac{1}{4}^{n-1} \).
04

Sum the geometric series for Player 1's winning probability

The total probability that Player 1 wins can be found by summing the geometric series \( \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \text{...} \). This series can be represented as \( P = \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{1/2}{3/4} = \frac{2}{3} \). Hence, the probability that Player 1 wins is \( \frac{2}{3} \).
05

Calculate Player 2's winning probability

The probability that Player 2 wins can be derived from the fact that the total probabilities must sum to 1. Thus, Player 2's winning probability is \( 1 - \frac{2}{3} = \frac{1}{3} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
A geometric series is a series of terms where each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio. In this exercise, the probabilities form a geometric series because each probability is a power of \(\frac{1}{4}\).
For Player 1, we consider the series \(\frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \text{...}\). Here:
  • Each term represents the probability of winning on a specific turn (first, second, third, etc.).
  • The common ratio between consecutive terms is \(\frac{1}{4}\).
To sum this infinite geometric series, we use the formula for the sum of an infinite geometric series:
\[ S = \frac{a}{1 - r} \] where \(a\) is the first term and \(r\) is the common ratio. Plugging in \(a = \frac{1}{2}\) and \(r = \frac{1}{4}\), we get:
\[ S = \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3} \] This tells us that the probability that Player 1 wins is \(\frac{2}{3}\).
Coin Toss Probability
Probability theory helps us understand the likelihood of different outcomes when tossing coins. A coin toss is a simple random experiment with two possible outcomes: Heads (H) or Tails (T). In this exercise, we toss two coins, and the possible outcomes for each toss are:
  • HH (both heads)
  • HT (one head, one tail)
  • TH (one tail, one head)
  • TT (both tails)
Each of these outcomes has an equal probability of \( \frac{1}{4} \). When we want two alike outcomes (HH or TT), the combined probabilities are:
\[ P(\text{HH or TT}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \] Likewise, the probability of getting two different outcomes (HT or TH) is also:
\[ P(\text{HT or TH}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \] Understanding these basic probabilities is crucial for solving problems that involve repeated trials, like in this exercise.
Independent Events
Independent events are those where the outcome of one event does not affect the outcome of another. In this exercise, each coin toss is an independent event. This means:
  • The result of Player 1's turn does not affect the probability of outcomes for Player 2's turn.
  • Each toss is its own event with a probability of \( \frac{1}{2} \) of getting two alike coins.
Because of this independence, we can calculate combined probabilities by multiplying the probabilities of individual events. For example, the probability that neither Player 1 wins on the first turn nor Player 2 wins on their first turn can be calculated as:
\[ P(\text{No one wins after first turn}) = P(\text{Player 1 loses}) \times P(\text{Player 2 loses}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \] This concept is key to understanding why we multiply probabilities when calculating the odds for Player 1 winning on subsequent turns.
This independence also supports creating the geometric series we summed in the exercise. Each term in that series includes the product of independent event probabilities, ensuring accurate calculation of overall winning probabilities.

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Most popular questions from this chapter

(a) There are 3 red and 5 black balls in one box and 6 red and 4 white balls in another. If you pick a box at random, and then pick a ball from it at random, what is the probability that it is red? Black? White? That it is either red or white? (b) Suppose the first ball selected is red and is not replaced before a second ball is drawn. What is the probability that the second ball is red also? (c) If both balls are red, what is the probability that they both came from the same box?

(a) Note that (3.4) assumes \(P(A) \neq 0\) since \(P_{A}(B)\) is meaningless if \(P(A)=0\) Assuming both \(P(A) \neq 0\) and \(P(B) \neq 0,\) show that if (3.4) is true, then \(P(A)=P_{B}(A) ;\) that is if \(B\) is independent of \(A,\) then \(A\) is independent of \(B\) If either \(P(A)\) or \(P(B)\) is zero, then we use (3.5) to define independence. (b) When is an event \(E\) independent of itself? When is \(E\) independent of "not \(E " ?\)

Given a family of two children (assume boys and girls equally likely, that is, probability 1/2 for each), what is the probability that both are boys? That at least one is a girl? Given that at least one is a girl, what is the probability that both are girls? Given that the first two are girls, what is the probability that an expected third child will be a boy?

Show that adding a constant \(K\) to a random variable increases the average by \(K\) but does not change the variance. Show that multiplying a random variable by \(K\) multiplies both the average and the standard deviation by \(K\).

If you select a three-digit number at random, what is the probability that the units digit is \(7 ?\) What is the probability that the hundreds digit is \(7 ?\)

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