0
$\begingroup$

I was recently reading a book on probability "Drunkard's Walk: How Randomness Rules Our Lives" and there is a probability problem below that I cannot seem to replicate:

"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."

I tried replicating the 2/3 (~67%) but I keep running into some sort of error (unless the book is wrong).

Essentially, I know the probability of 8 or more heads or tails in 10 flips is 56/1024 (this is 10 choose 8 + 10 choose 9 + 10 choose 10 for coin flips). However, when I try to replicate the "the chances are 2/3 that at least 1 executive will score 8 or more heads" I do not get 67% but rather closer to 43%.

Here are my binomial inputs: p (success)=56/1024, n (trials)=10, k (successes)=0.

If I run the numbers, 10 choose 0 = 1, p^k = 1, and (1-p)^n-k = 0.57.

This means that the chance that NO executives get 8 or more heads/tails is (1 * 1 * 0.57)=57%, which means that the probability of at l least 1 executive getting it should be 1-this, which is 1-0.57=0.43 (or 43%).

Can someone tell me where I went wrong with this problem?? Thanks!

$\endgroup$
  • $\begingroup$ The $\dfrac{56}{1024}$ is only counting heads. Multiply that by 2 and you'll get the correct probability. $\endgroup$ – Kenny Lau May 5 '16 at 14:58
  • $\begingroup$ Thank you! That makes sense. $\endgroup$ – Nicholas Maggiulli May 5 '16 at 15:13
1
$\begingroup$

As Kenny Lau noted in a comment, you're missing a factor $2$ since both heads and tails are allowed. The desired probability is therefore

$$ 1-\left(1-\frac{112}{1024}\right)^{10}=\frac{790888173149955727}{1152921504606846976}\approx69\%\;. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.