Prime Dime Time!
A dime is flipped 10 times. What is the probability that it lands on heads a prime number of times?
Give your answer to three decimal places.
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The answer is 0.524.
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1 solution
If we let P n be the probability that in n flips a prime number of those flips are heads, then I'm getting P 2 0 ≈ 0 . 3 2 5 and P 3 0 ≈ 0 . 3 2 9 , (WolframAlpha has problems going past this, for some reason). Which of course makes me wonder how P n behaves as n increases and what n → ∞ lim P n might be. I can't find any suitable references to "prime flipping", so you've come up with another novel question. :)
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Hahaha... I should have suspected that you'd have some way of making my sort of silly little problem much more meaningful! :^)
Looks like it approaches something between 1/e and 1/pi perhaps?
On second thought, shouldn't P n approach zero as n → ∞ ? Since prime numbers get sparser and sparser?
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Yeah, as I was thinking about it after I wrote my note I came to the same suspicion. It would likely approach the same n π ( n ) ≈ ln ( n ) 1 curve as per the prime number theorem. I get P 4 0 ≈ 0 . 2 9 6 , compared to 4 0 π ( 4 0 ) = 0 . 3 , and compare P ( 3 0 ) = 0 . 3 2 9 with 3 0 π ( 3 0 ) = 0 . 3 3 3 ; close matches indeed even for relatively small n . I'm not sure if I should be surprised at how close they are. Still an interesting problem to play around with. :)
The prime numbers from 1-10 are 2,3,5, and 7.
Pascal's triangle can help us visualize the possible outcomes of the coin flips...
Consider starting your flips from the top. A heads correspond to moving down to the right and a tails corresponds to moving down to the left.
If you look at the bottom row (10 flips), there are 45 ways 2 heads could have been flipped, 120 ways 3 or 7 could have been flipped, and 252 ways that 5 heads could have been flipped.
4 5 + 1 2 0 + 1 2 0 + 2 5 2 = 5 3 7
And there are 1 0 2 4 total ways the flips could have landed.
1 0 2 4 5 3 7 = 0 . 5 2 4
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