# Math Help - Should I read the rest of the book?

1. ## Should I read the rest of the book?

"Deciding just how much of an outcome is due to skill and how much to luck is not a no-brainer. Random events often come like the raisins in a box of cereal—in groups, streaks, and clusters. And although Fortune is fair in potentialities, she is not fair in outcomes. That means that if each of 10 Hollywood executives tosses 10 coins, although each has an equal chance of being the winner or the loser, in the end there will be winners and losers. In this example, the chances are 2 out of 3 that at least 1 of the executives will score 8 or more heads or tails." The Drunkard's Walk.Leonard Mlodinow.

Shouldn't the right answer be 2 out of 5?

2. ## Re: Should I read the rest of the book?

Below is my (probably flawed) reasoning:
p (any given exec. gets ≥ 8 heads) = p (8 H’s) + p (9 H’s) + p (10 H’s) = ("10 choose 8") + ("10 choose 9")) + ("10 choose 10") * 1 / 2^10
= (45 + 10 + 1) * 1 / 2^10 = 56 * 1/1,024 = 0.0546875
p (any given exec gets < 8 H’s) = 1 - 0.0546875 = 0.9453125
p (NO exec gets ≥ 8 heads) = 0.945312510 = 0.569
p (AT LEAST ONE exec gets ≥ 8 heads) = 1 - 0.569 = 0.431

3. ## Re: Should I read the rest of the book?

I get:

$P(X)=\frac{790888173149955727}{1152921504606846976 }\approx0.6859861404178190366284773027416576951509 34159755706787109375...$

4. ## Re: Should I read the rest of the book?

So that confirms that I was wrong, but I'm not sure where...

5. ## Re: Should I read the rest of the book?

Originally Posted by parex
"Deciding just how much of an outcome is due to skill and how much to luck is not a no-brainer. Random events often come like the raisins in a box of cereal—in groups, streaks, and clusters. And although Fortune is fair in potentialities, she is not fair in outcomes. That means that if each of 10 Hollywood executives tosses 10 coins, although each has an equal chance of being the winner or the loser, in the end there will be winners and losers. In this example, the chances are 2 out of 3 that at least 1 of the executives will score 8 or more heads or tails." The Drunkard's Walk.Leonard Mlodinow.

Shouldn't the right answer be 2 out of 5?
Have a look at the calculation on this webpage.

That tells us that there is about $0.11$ chance of any of those executives getting at least eight heads or tails.

Because I have never seen that book, I really have no idea what your question means.

6. ## Re: Should I read the rest of the book?

Originally Posted by parex
So that confirms that I was wrong, but I'm not sure where...
First, I computed the probability that 1 of the executives would not get 8 or more heads or tails:

$\left(\frac{1}{2} \right)^{10}\sum_{k=3}^7{10 \choose k}=\frac{57}{64}$

And so the probability that all ten executives would not get 8 or more heads or tails:

$\left(\frac{57}{64} \right)^{10}$

And then using the complement law:

$P(X)=1-\left(\frac{57}{64} \right)^{10}=\frac{790888173149955727}{11529215046 06846976}$

7. ## Re: Should I read the rest of the book?

Hi,
I agree with both Plato and MarkFL. Here's the way I look at it:
Flipping one coin is a Bernoulli trial; define the r.v. X as X=1 means a head results and X=0 is a tail. Also P(X=0)=P(X=1)=1/2. Then the sum of 10 independent r.v's each distributed as X is r.v. Y which is then a binomial r.v. with parameters p=q=1/2. So the event at least 8 heads or at least 8 tails is the event with probability
p1=P(Y=0)+P(Y=1)+P(Y=2)+P(Y=8)+P(Y=9)+P(Y=10) which is approximately .109375.

Now define another Bernoulli trial with success meaning at least 8 heads or 8 tails in 10 tosses. The r.v. Z which is the sum of 10 of these Bernoulli r.v.'s is again a binomial r.v. with parameters p1 and 1-p1. The event of interest is Z>0. So P(Z>0)=1-P(Z=0)=1-(1-p1)10 or approximately .685986.

So the probability is a little more than 2/3. I would not decide to quit reading on the basis of this passage.

8. ## Re: Should I read the rest of the book?

Originally Posted by MarkFL
First, I computed the probability that 1 of the executives would not get 8 or more heads or tails:

$\left(\frac{1}{2} \right)^{10}\sum_{k=3}^7{10 \choose k}=\frac{57}{64}$
That is not the probability of more than eight heads or trails.
Because those are mutually exclusive events the answer is twice that.

9. ## Re: Should I read the rest of the book?

Originally Posted by Plato
That is not the probability of more than eight heads or trails.
Because those are mutually exclusive event the answer is twice that.
That is the probability of 1 executive getting 3H7T or 4H6T or 5H5T or 6H4T or 7H3T, which is equivalent to not getting 8 or more heads or tails.

10. ## Re: Should I read the rest of the book?

Well, thank you all for your responses. This is a great forum with addictive potential.

I have to admit that I misread the original sentence in the book: because the author was making the point that a monkey could pick randomly the next blockbuster out of luck, explaining why some great scripts (or books) are initially turned down, and why some Hollywood executives pick winners for a while only to later eclipse into oblivion, I wanted to read that the parallel math analogy he was proposing was to get 8 or more heads in a row (not tails) - I guess the bias is to think of heads as winners, and tails as losers.

In any event, it was frustrating not to be able to copy and paste from Word the equations with combination brackets - Is there a way?

Given the difficulties with the word processing (any hints appreciated) I'm attaching my calculations of this "modified reading" of the original statement as a jpeg.

Thank you!