I was trying to solve a probability problem as follows:

Suppose you are flipping an unfair coin. It flips 70% heads and 30% tails, and each flip is independent from the previous.

Now you flip this coin three times. What is the probability you get 3 heads given that you flipped at least 1 head?

My first guess is one of the heads fixed, therefore it boils down to what is the probability that two coins are heads:

$0.7 * 0.7 = 0.49$

However I also solved it another way using Bayes Theorem:

Let probability of flipping 3 heads be $P(H3)$

Let the probability of at least 1 head be denoted as $P(A1)$

Using the Bayes theorem $$P(H3|A1) = P(A1|H3) * P(H3) / P(A1) = 1 * 0.7^3 / (1 - 0.3^3) = 1 * 0.343 / 0.973 = 0.353 $$

Can someone explain which way is right and which way is wrong and also the intuition behind it? I'm think the first way is wrong, but I don't see where the fallacy is.


The second method is correct. The first method would be correct if you were given "first flip is Head" instead of "at least $1$ Head". You could verify this using Bayes Theorem but it's clear that the situation would then be the same as simply flipping the coin twice and requiring $2$ Heads, leading to the probability $0.7^2=0.49$.

The difference in the above two conditioning events is that "at least $1$ Head" eliminates only the outcome $TTT$ but "first flip is Head" eliminates outcomes $TTT,\;TTH,\;THT,\;THH$.

  • $\begingroup$ Thanks for the intuition, it helps a lot! $\endgroup$ – CowZow Feb 18 '15 at 17:16

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