Probability and Paul Harvey

[Jimbo]

While listening to Paul harvey at lunch today he recited the following short story. A man flips a coin every Sunday morning to determine if he goes to church or plays golf. This man is quoted as saying "Ocassionally I must flip ten times to go play golf".

My question is how many times in one year would he likely end up flipping the coin ten times (or more) in order to play golf rather than attend church? We must assume he randomizes his guesses, and does not call heads or tails only, until the coin lands on golf today.

Jimbo

[Bozeman]

We must assume he randomizes his guesses, and does not call heads or tails only, until the coin lands on golf today.

Why?

Craig

[Jimbo]

Because.......... Do you prefer I say I request you assume he does this in case it has relevance to the computation? If so this is why! If not this is still the reason why.

Jimbo

[Homer]

P(1 flip) = 1/2
P(2) = 1/4
P(3) = 1/8
...
P(9) = 1/512

P(1) + P(2) + ... + P(9) = P(1-9) = .998046875

P(10 or more) = 1 - P(1-9) = .001953 (or 1/512)

So it happens about .713 times per year, or around twice every three years.

Is this the answer or am I missing something??

-- Homer

[bad beetz]

P(Has to flip ten times)=P(Church comes up nine times)

=(1/2)^9=1/512

1/512*Chances per year = 1/512 * 52 = .101563

Many thanks to Homer and bad beetz.

Jimbo

[Homer]

oops Jimbo...note that my answer is 7 times that of bad beets. I was thinking he did it every day, instead of only on Sundays. Hence, bad beets answer is actually correct....

-- Homer