Again the same old coins

Probability Level 1

What is the expected number of heads in 101 tosses of a fair coin? If your answer comes in the form of p q \dfrac{p}{q} , find p − q p-q .


The answer is 99.

Anish Harsha by Anish Harsha , 20, India

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1 solution

Sravanth C.
Jan 24, 2016

We know that the probability of getting a head if we toss a coin is equal to 1 2 \dfrac 12 . Because there are only two possible outcomes, either a head or a tail.

So, when we toss a coin multiple times repeatedly, the number of times you get a head would be 1 2 × No. of times you toss the coin \dfrac 12\times\text{No. of times you toss the coin} .

Hence if we toss a coin 101 101 times, the number of times you can get a head would be 1 2 × 101 = 101 2 \dfrac 12\times 101=\boxed{\dfrac{101}2} . So p − q = 101 − 2 = 99 p-q=101-2=\boxed{99} , which is the required answer.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

This is clearly wrong and against even simple intuition. As we approach infinity, the expected number of heads is roughly equal to the expected number of tails (~1/2 of total tosses will be heads and ~1/2 tails), so in that case the expect number of heads should be 50.5.

Also, "when we toss a coin multiple times repeatedly, the probability of getting a head would be 1/2 * no. of times you toss a coin", so by this logic if we toss it 2 times then the probability of getting a head is 1/2*2 = 1, which is - needless to say - wrong, as we assume that the tosses are independent and the coin is fair. Even worse, what if we toss it 3 times? Then the P(H) > 1, wow!

Przemek Zientała - 4 years, 9 months ago

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