MHB Expected values: three fair coins

  • Thread starter Thread starter oyth94
  • Start date Start date
AI Thread Summary
The discussion focuses on calculating the expected value E(X) when flipping three fair coins, where X is defined as the cube of the number of heads. The initial approach incorrectly assumed probabilities for heads, but it was clarified that X should take values of 1, 8, and 27 for 1, 2, and 3 heads, respectively. The correct probabilities for getting k heads in three flips are derived using binomial coefficients, resulting in P(k=0) = 1/8, P(k=1) = 3/8, P(k=2) = 3/8, and P(k=3) = 1/8. The final expected value can be computed by summing the products of these probabilities and their corresponding X values.
oyth94
Messages
32
Reaction score
0
Suppose you roll three fair coins, and let X be the cube of the number of heads
showing. Compute E(X)

I started this question with X=xi where i = 1,2,3 since there are 3 fair coins and there are either 1 2 or 3 heads that will show up. then it says that X will be the cube of the number of heads so I put
P(getting heads) = (1/2)^(i^3)
So E(X) = $$xiP(X=xi)$$ = $$i(1/2)^(i^3)$$ summation from i=1 to 3..
am i on the right track??
 
Mathematics news on Phys.org
Re: expected values: three fair coins

oyth94 said:
Suppose you roll three fair coins, and let X be the cube of the number of heads
showing. Compute E(X)

I started this question with X=xi where i = 1,2,3 since there are 3 fair coins and there are either 1 2 or 3 heads that will show up. then it says that X will be the cube of the number of heads so I put
P(getting heads) = (1/2)^(i^3)
So E(X) = $$xiP(X=xi)$$ = $$i(1/2)^(i^3)$$ summation from i=1 to 3..
am i on the right track??

Not quite (and you mean flip 3 fair coins...not roll... XD); since $X$ represents the cube of the number of heads, you instead want $x_1=1^3=1$, $x_2=2^3=8$ and $x_3=3^3=27$. Then $\mathbb{P}(X=x_i)$ would represent the probability of getting 1, 2, or 3 heads; in this case, we would have $\mathbb{P}(X=x_i)=\dfrac{1}{2^i}$, $i=1,2,3$. Thus,
\[E[X]=\sum_{i=1}^3 x_i\mathbb{P}(X=x_i)=\ldots\]

I hope this makes sense!
 
Re: expected values: three fair coins

Chris L T521 said:
Not quite (and you mean flip 3 fair coins...not roll... XD); since $X$ represents the cube of the number of heads, you instead want $x_1=1^3=1$, $x_2=2^3=8$ and $x_3=3^3=27$. Then $\mathbb{P}(X=x_i)$ would represent the probability of getting 1, 2, or 3 heads; in this case, we would have $\mathbb{P}(X=x_i)=\dfrac{1}{2^i}$, $i=1,2,3$. Thus,
\[E[X]=\sum_{i=1}^3 x_i\mathbb{P}(X=x_i)=\ldots\]

I hope this makes sense!

Oh I see what you did! Thank you so much!
 
Re: expected values: three fair coins

Chris L T521 said:
$\mathbb{P}(X=x_i)=\dfrac{1}{2^i}$, $i=1,2,3$. Thus,
\[E[X]=\sum_{i=1}^3 x_i\mathbb{P}(X=x_i)=\ldots\]
i have a doubt(probably elementary),
you have assumed probability of getting one head is $$\frac{1}{2}$$.But i think it is the probability to get at least one head ,doesn't that include cases of $$2$$ or $$3$$ heads and why isn't the probability $$\frac{1}{2^3}$$ for all cases
(i mean why aren't we taking exact probability here)
 
Re: expected values: three fair coins

mathworker said:
i have a doubt(probably elementary),
you have assumed probability of getting one head is $$\frac{1}{2}$$.But i think it is the probability to get at least one head ,doesn't that include cases of $$2$$ or $$3$$ heads and why isn't the probability $$\frac{1}{2^3}$$ for all cases
(i mean why aren't we taking exact probability here)

The probability to have k heads in 3 trials is $\displaystyle \frac{\binom{3}{k}}{8}$, so that...

$P\{k=0\} = \frac{1}{8}$

$P\{k=1\} = \frac{3}{8}$

$P\{k=2\} = \frac{3}{8}$

$P\{k=3\} = \frac{1}{8}$

Kind regards

$\chi$ $\sigma$
 
got it,changing the positions we could get 3 possibilities of eight , thank you
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top