Can someone help me with this statistics problem?

Question

A coin is tossed, and you win a dollar if there are more than 60% tails. Which is better: 10 tosses or 100 tosses?
Thank you!

Answer 1

Because of the law of averages, the more tosses you take on a fair coin, the closer the percentage of tails will be to 50%. Since having more than 60% tails means you would want to be further away from 50%, I would go with fewer tosses. 10 is better.

Answer 2

You compare the results of two binomial distribution calculations.

(Call tails sucess):
1. n = 10, calc p(k>=6)
2. n = 100, calc p(k>=60)
(be careful about the > or >= in problem statement)

The larger number of tosses will vary less from the expected value. For a fair coin, E = np
1. E = 5
2. E = 50

For 2, n is large enough to use the normal approx.

Answer 3

As the number of tries approaches infinity, the number will tend towards the expected value. In this case, the expected value is fifty percent. So, you have a better chance of tossing 6 tails and four heads than 60 tails and 40 heads. Just as you have a better chance of tossing two heads in two tries than 100 heads in 100 tries.

Answer 4

Another approach. Bernoulli trials p=.5, n=10, r=6 Prob = .205
n=100, r=60 Prob=?? 100! too big for me
n= 60, r=48 Prob=.03
n= 30 r=18 Prob=..08

Interesting drop off. 10 tosses a lot better
n=100, r=60, Prob=.00098

Answer 5

10 tosses