Expected number of coin tosses before landing heads

  • Thread starter Thread starter Unit
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around determining the expected number of coin tosses required to land heads when flipping a fair coin. Participants are exploring the concept of expected value in probability theory.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the expected value using probabilities associated with sequences of coin tosses. They question the probabilities for different outcomes and whether the expected value can be derived from an infinite series.

Discussion Status

Some participants provide insights into the expected value and clarify the distinction between expected value and most likely outcomes. There is an ongoing exploration of the implications of these concepts in practical scenarios.

Contextual Notes

Participants are discussing the definitions and interpretations of expected value and mean, as well as the implications of running a large number of trials in a real-life context.

Unit
Messages
181
Reaction score
0

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

Thanks!
 
Physics news on Phys.org
not quite if you average things you will get an expected value of 2 (mean)
E(x) = \sum_{n=1}^{\infty} n p_n = \sum_{n=1}^{\infty} n \frac{1}{2^n}

"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.
 
Thanks lanedance! So what is the real-life explanation of "expected value" in this case?
 
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
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 23 ·
Replies
23
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K