Expected number of coin tosses before landing heads

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In summary, the expected value of the number of coin flips needed to get heads when flipping a fair coin is 2. This means that on average, it will take 2 flips to get heads if the experiment is repeated many times. The expected value is the average or mean.
  • #1
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Homework Statement


Given a fair coin, how many times will you flip it (on average) before you get heads?

Homework Equations


E(x) = x1p1 + x2p2 + x3p3 + ... + xnpn

The Attempt at a Solution


I know this is easy, but I'm stumped with the probability part.

Probability of heads is 1/2 and probability of tails is 1/2. So, after 1 flip, the probability of having landed heads right away is 1/2. So x1p1 = (1)(1/2). Otherwise, flip again: x2 = 2. My question: does p2 = (1/2)2? I got this by arguing that the probability of the sequence TH is P(T) times P(H). Similarly, for p3, the sequence is TTH, so does p3 = (1/2)3?

I'm also guessing that arbitrarily long sequences TTT...TH (of length n) are possible. Assuming I'm right so far, would I take the limit of E(x) as n approaches infinity?

In symbols,
[tex] \lim_{n \to \infty} E(x) := E = \sum_{n=1}^\infty \frac{n}{2^n} = 2 [/tex]
So is the expected value 2 tosses? Will I usually have gotten heads on my second toss?

Thanks!
 
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  • #2
not quite if you average things you will get an expected value of 2 (mean)
[tex] E(x) = \sum_{n=1}^{\infty} n p_n = \sum_{n=1}^{\infty} n \frac{1}{2^n} [/tex]

"usually have gotten" implies most likely value (mode) which in this case will be 1. ie if you repeat this experiment over and aver you will get heads on the first toss more than any otehr number.
 
  • #3
Thanks lanedance! So what is the real-life explanation of "expected value" in this case?
 
  • #4
The expected value is the average or mean.

For example if someone said they would run the experiment 1,000,000 times, and each time pay you a dollar or each tosses, then you would make:
$2*1,000,000 = $2,000,000
 

1. What is the expected number of coin tosses before landing heads?

The expected number of coin tosses before landing heads is equal to two. This means that on average, it will take two coin tosses to get a heads result.

2. Why is the expected number of coin tosses before landing heads two?

This is because a coin toss is a binary event, meaning there are only two possible outcomes: heads or tails. The probability of getting a heads result on a single coin toss is 1/2 or 0.5. Therefore, the expected number of tosses needed to get a heads result is equal to 1 divided by the probability of getting heads, which is 1/0.5 or 2.

3. Is the expected number of coin tosses before landing heads always two?

No, the expected number of coin tosses can vary depending on the probability of getting a heads result. If the coin is biased and more likely to land on heads, the expected number of tosses will be lower. Similarly, if the coin is biased towards tails, the expected number of tosses will be higher.

4. Can the expected number of coin tosses before landing heads be a fraction?

No, the expected number of coin tosses must be a whole number. This is because it represents the average number of tosses needed to get a heads result. You cannot have a fraction of a coin toss, so the expected number must always be a whole number.

5. How can the expected number of coin tosses be used in real-world scenarios?

The concept of expected number of coin tosses can be applied in various scenarios, such as predicting the number of trials needed to achieve a desired outcome in a scientific experiment, or estimating the number of attempts needed to win a game of chance. It can also be used in statistics to calculate probabilities and make predictions based on data.

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