Coin Toss Probability: A Vs. B Analysis

Hey guys! Ever wondered about the odds when two people are flipping coins back and forth? Let's dive into a super interesting probability problem where person A and person B are tossing a coin, not just once or twice, but a whopping 100 times each! And here’s the twist: the probability of landing heads is 1/3, making it a bit trickier than your standard 50/50 coin toss. In this article, we’re going to explore the probabilities involved, focusing on a couple of key questions. First, we’ll figure out the probability that A and B get the same number of heads, specifically i heads. Then, we’ll peek into what happens when B keeps tossing even after the initial 100 tosses. So, grab your thinking caps, and let's get started!

Let's break down the problem. We have two individuals, A and B, who are independently tossing a biased coin 100 times each. By biased coin, we mean that the probability of getting heads isn't the usual 1/2, but rather 1/3. This adds an extra layer of complexity and makes the calculations a bit more intriguing. The core of our problem revolves around two main questions:

a) Find P{A = B = i}

This part asks for the probability that both A and B get the exact same number of heads, which we're calling i. It's like asking, "What's the chance they both get, say, 20 heads each?" To solve this, we need to consider all the possible values of i, from 0 heads all the way up to 100 heads. Each value of i represents a specific scenario, and we need to figure out the likelihood of that scenario occurring. This involves using the binomial distribution, which is perfect for modeling the number of successes (in this case, heads) in a fixed number of independent trials (the coin tosses).

b) After 100 tosses B continues...

This second part opens up a whole new realm of possibilities. After their initial 100 tosses, B decides to keep going. This means we're now dealing with an infinite number of tosses for B, which changes the dynamics significantly. We'll need to think about how B's continued tosses affect the overall probabilities and potentially introduce new questions to explore. For instance, we might want to know the probability that B eventually catches up to or surpasses A in the number of heads. Or perhaps we're interested in the long-term behavior of the difference in the number of heads between A and B. This extension adds a layer of depth and requires us to consider concepts like convergence and expected values.

Okay, let's tackle the first part of our problem: finding the probability that both A and B get exactly i heads. This is where the binomial distribution comes into play. Remember, the binomial distribution helps us calculate the probability of getting a certain number of successes (like heads) in a fixed number of trials (the coin tosses), given a constant probability of success (1/3 in our case). To find P{A = B = i}, we need to break it down into smaller, manageable pieces and then put them together. Let’s dive in!

Using the Binomial Distribution

The first thing we need to do is calculate the probability that A gets exactly i heads. We'll call this P(A = i). Since A tosses the coin 100 times, and the probability of heads is 1/3, we can use the binomial probability formula:

P(A = i) = (100 choose i) * (1/3)^i * (2/3)^(100-i)

Whoa, that looks a bit intimidating, right? Let's break it down. The term (100 choose i) represents the number of ways to choose i heads out of 100 tosses. It's also written as ¹⁰⁰C_i or !100 / (i! * (100-i)!). The term (1/3)^i is the probability of getting i heads, and (2/3)^(100-i) is the probability of getting the remaining (100 - i) tails. So, this whole formula is just saying: "The probability of A getting exactly i heads is the number of ways to get i heads times the probability of getting i heads and (100 - i) tails."

Now, we need to do the same thing for B. The good news is, since B also tosses the coin 100 times with the same probability of heads, the formula is exactly the same:

P(B = i) = (100 choose i) * (1/3)^i * (2/3)^(100-i)

Combining the Probabilities

We've got P(A = i) and P(B = i). But we want the probability that both A and B get i heads. Since A and B toss the coins independently (meaning A's tosses don't affect B's tosses), we can simply multiply their probabilities:

P(A = B = i) = P(A = i) * P(B = i)

Substituting our binomial probabilities, we get:

P(A = B = i) = [(100 choose i) * (1/3)^i * (2/3)^(100-i)] * [(100 choose i) * (1/3)^i * (2/3)^(100-i)]

We can simplify this a bit by combining the terms:

P(A = B = i) = [(100 choose i)^2] * [(1/3)^(2i)] * [(2/3)^(200-2i)]

So, there you have it! This formula gives us the probability that both A and B get exactly i heads. We can plug in any value of i from 0 to 100 to find the probability for that specific number of heads. For example, if we wanted to know the probability that they both get 30 heads, we'd plug in i = 30. The (100 choose i) part might seem daunting, but calculators and statistical software can handle these calculations easily. Also, remember Stirling's approximation to simplify the factorial calculations.

Understanding the Result

It's important to take a step back and think about what this result means. The formula we derived gives us the probability for a specific value of i. To get a complete picture, we could calculate P(A = B = i) for all values of i from 0 to 100 and then perhaps visualize the distribution. We'd likely see a bell-shaped curve, with the highest probability around the expected number of heads for both A and B. The expected number of heads is simply the number of tosses (100) times the probability of heads (1/3), which is approximately 33.33. So, we'd expect the highest probabilities to be around i = 33. This makes intuitive sense: it's most likely that A and B will both get a number of heads close to the expected value.

Alright, let's switch gears and dive into the second part of our problem, where B keeps tossing the coin after the initial 100 tosses. This adds a fascinating twist, opening up a world of new possibilities and requiring us to think a bit differently about probabilities. What happens when B gets an unlimited number of tosses? How does this affect the chances of B catching up to or surpassing A in the number of heads? Let's explore!

Setting the Stage

So, A has tossed the coin 100 times, and we have a fixed number of heads for A. Let's say A got a heads, where a is some number between 0 and 100. Now, B has also tossed the coin 100 times initially, but then continues tossing indefinitely. This means B has the potential to keep accumulating heads, and we want to analyze how this affects the relationship between the number of heads A and B have. The key question here is: what's the probability that B will eventually have more heads than A?

Defining the Random Variables

To make our analysis more precise, let's define some random variables. Let A₁₀₀ be the number of heads A gets in the first 100 tosses, and let B₁₀₀ be the number of heads B gets in the first 100 tosses. We already know that A₁₀₀ follows a binomial distribution with parameters n = 100 and p = 1/3, and the same goes for B₁₀₀. Now, let's say B continues tossing the coin. Let B_total be the total number of heads B gets after an infinite number of tosses. This is where things get interesting, because B_total can potentially be any number greater than or equal to B₁₀₀.

Framing the Question

Now, we can rephrase our question more formally. We want to find the probability that B_total is greater than A₁₀₀. In mathematical notation:

P(B_total > A_100)

This might seem like a simple question, but it's actually quite complex because B_total can keep increasing indefinitely. We need to think about this in terms of limits and convergence. How do we handle the fact that B could potentially toss the coin an infinite number of times?

Considering the Long Run

To tackle this, let's think about what happens in the long run. As B continues tossing the coin, the number of heads B gets will tend to increase. The expected number of heads for B in each toss is 1/3. So, for every 3 tosses, we expect B to get about 1 head. If B keeps tossing for a very, very long time, the number of heads B gets will become very large. This suggests that, eventually, B will almost certainly have more heads than A. But how can we make this intuition more rigorous?

Introducing the Law of Large Numbers

This is where the Law of Large Numbers comes to our rescue. The Law of Large Numbers (LLN) basically says that as we repeat an experiment (like tossing a coin) a large number of times, the average result will get closer and closer to the expected value. In our case, the LLN tells us that as B tosses the coin more and more times, the proportion of heads B gets will get closer and closer to 1/3. This is a powerful concept, and it helps us understand the long-term behavior of B's tosses. However, applying the LLN directly to find P(B_total > A₁₀₀) is still tricky because we are dealing with the total number of heads, not the proportion. We need a slightly different approach.

A More Rigorous Approach

To find P(B_total > A₁₀₀), we can consider the difference in the number of heads between A and B. Let's define a new random variable, D_n, as the difference in the number of heads after B has tossed the coin n times beyond the initial 100 tosses:

D_n = B_{100+n} - A_{100}

where B_100+n is the total number of heads B has after 100 + n tosses. Our goal is to find the probability that D_n becomes positive for some value of n. If D_n becomes positive, it means that B has more heads than A.

Now, let's rewrite our probability in terms of D_n:

P(B_total > A_100) = P(D_n > 0 for some n)

This is a bit easier to work with. To solve this, we can think about the expected value and variance of D_n. The expected value of D_n is:

E[D_n] = E[B_{100+n}] - E[A_{100}]

We know that E[A₁₀₀] = 100 * (1/3) = 100/3. The expected number of heads B gets in 100 + n tosses is (100 + n) * (1/3). So,

E[D_n] = (100 + n) * (1/3) - 100/3 = n/3

As n increases, E[D_n] also increases. This makes sense: the more B tosses the coin, the more we expect the difference in heads to favor B. In the limit as n approaches infinity, E[D_n] also approaches infinity. This suggests that it's very likely that D_n will become positive for some value of n.

To make this even more precise, we could analyze the probability that D_n stays negative for all n. This is a bit more complex and might involve using techniques from stochastic processes and martingale theory. However, the intuition is that this probability should approach zero as n approaches infinity. This is because the expected value of D_n is increasing linearly with n, and while there will be fluctuations, the overall trend is for B to get more heads than A.

Putting it All Together

So, after considering all of this, we can conclude that the probability that B eventually has more heads than A is very high, approaching 1. This makes intuitive sense: since B keeps tossing the coin, and the expected number of heads B gets increases over time, it's almost certain that B will eventually catch up to and surpass A.

Well, guys, we've taken a pretty deep dive into this coin-tossing problem! We started by figuring out the probability that A and B get the same number of heads in 100 tosses each, using the binomial distribution. We then explored the fascinating scenario where B keeps tossing the coin, and we discovered that, in the long run, it's almost certain that B will have more heads than A. This problem highlights the power of probability theory and the beauty of how mathematical concepts can help us understand the world around us. Whether it's coin tosses or more complex real-world scenarios, understanding probability is a valuable skill. Keep exploring, keep questioning, and keep those mental coins flipping!