Expected Number of Tosses for a Fair Coin: Calculating E(x) for Heads and Tails

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In summary, the expected number of tosses required to obtain a head when tossing a fair coin repeatedly is 2. This means that only one tail is expected to be obtained before a head is obtained. The formula used was E(x) = 1/p = 2, where p is the probability of getting a head (1/2). This can also be calculated by setting up a sum for the expected value, which would be E(x) = \sum_{n=0}^{\infty} n P(X=n).
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uva123
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Homework Statement



Suppose that a fair coin is tossed repeatedly until a head is obtained for the first time.

(a) What is the expected number of tosses that will be required?
(b) What is the expected number of tails that will be obtained before the first head is
obtained?


Homework Equations



E(x)= from -infinity to +infinity (continuous case)
E(x)= for all x (discrete case)

The Attempt at a Solution



p=head=1/2
q=tail=1-p=1/2

Pr(X>1)=1 for n=2,3,...

Pr(X>n)=qn-1

E(x)=1/(1-q)=1/p=2

-if only 2 tosses are expected to obtain a head then only one tail would be obtained before a head

-does this make sense?
 
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  • #2
helps if you show you steps to make it clear what you did... i'd start like this
[tex] \begin{matrix}
P(1) & = P(H) & = 1/2 & &\to P(X \leq 1) = 1/2 &\to P(X>1) = 1/2 \\
P(2) & = P(TH) &= (1/2)(1/2) &= 1/4 & \to P(X \leq 2) = 3/4 & \to P(X>2) = 1/4\\
P(3) & = P(TTH) &= (1/4)(1/2) &= 1/8 & \to P(X \leq 2) = 7/8 & \to P(X>1) = 1/8
\end{matrix} [/tex]

which looks like you're close for P(X>n) but not quite, assuming X is number of tosses for a head - should be able to write out P(X=n) from here and set up a sum for the expected value
[tex] E(x) = \sum_{n=0}^{\infty} n P(X=n) [/tex]
 
  • #3
not too sure what you meant by
Pr(X>1)=1 for n=2,3,...
 

What is the expected number of tosses for a fair coin?

The expected number of tosses for a fair coin is 2. This means that on average, it takes 2 tosses to get either a heads or tails outcome.

How do you calculate the expected number of tosses for a fair coin?

The expected number of tosses can be calculated using the formula E(x) = 1/p, where p is the probability of the desired outcome (in this case, 1/2 for a fair coin).

What does E(x) represent in the context of coin tosses?

E(x) represents the expected number of tosses needed to get a specific outcome. In the case of a fair coin, E(x) represents the expected number of tosses to get either a heads or tails result.

Is the expected number of tosses the same for getting heads and tails?

Yes, the expected number of tosses is the same for getting heads and tails. This is because a fair coin has an equal probability of landing on either heads or tails.

Can the expected number of tosses be used to predict the outcome of a coin toss?

No, the expected number of tosses is a theoretical concept and cannot be used to predict the outcome of a single coin toss. It only represents the average number of tosses needed to get a specific outcome over a large number of trials.

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