Tossing Coins

We throw 4 fair coins simultaneously and we repeat the experiment once more. What is the probability that the second throw will result in the same configuration as the first throw when the coins are indistinguishable.

Give your answer to 3 decimal places.


The answer is 0.273.

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2 solutions

Sam Bealing
Apr 27, 2016

We can look at the probability of different outcomes:

No. Heads 0 0 1 1 2 2 3 3 4 4
p p ( 4 0 ) 1 2 4 = 1 16 \binom{4}{0} \dfrac{1}{2^4}=\dfrac{1}{16} ( 4 1 ) 1 2 4 = 4 16 \binom{4}{1} \dfrac{1}{2^4}=\dfrac{4}{16} ( 4 2 ) 1 2 4 = 6 16 \binom{4}{2} \dfrac{1}{2^4}=\dfrac{6}{16} ( 4 3 ) 1 2 4 = 4 16 \binom{4}{3} \dfrac{1}{2^4}=\dfrac{4}{16} ( 4 4 ) 1 2 4 = 1 16 \binom{4}{4} \dfrac{1}{2^4}=\dfrac{1}{16}
p 2 p^2 1 256 \dfrac{1}{256} 16 256 \dfrac{16}{256} 36 256 \dfrac{36}{256} 16 256 \dfrac{16}{256} 1 256 \dfrac{1}{256}

p 2 = 1 + 16 + 36 + 16 + 1 256 = 70 256 = 35 128 0.273 \sum p^2=\dfrac{1+16+36+16+1}{256}=\dfrac{70}{256}=\dfrac{35}{128} \approx 0.273

I think this is simply

n = 0 4 ( ( 4 x ) 16 ) 2 = 35 128 \sum_{n=0}^{4} \left( \dfrac{\displaystyle{{4 \choose x}}}{16} \right)^2 = \boxed{\dfrac{35}{128}}

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