# Solved – unfair coin flip probability calculation

bernoulli-distributionconditional probabilityprobability

Suppose I have an unfair coin, and the probability of flip a head (H) is p, probability of flip a tail (T) is (1-p).

If I flip the coin 6 times, wondering if the probability of HTT???, and the probability of THT???, and the probability of TTH??? are the same? Suppose each flip is independent. ? means do not care if head or tail. Thanks.

I calculated they are the same, ask here to get advice from expert if my calculation is correct.

Yes, all the three events are independent. The cumulative law gives you:

t * h *t = h * t * t = t* t* h

If you enter number you will get:

0.3 * 0.7 * 0.3 = 0.7 * 0.3 * 0.3 = 0.3 * 0.3 * 0.7 = 0.063

Edit: It does not matter what the first three times are because of the law of conditional probability.

P(A $\bigcap$ B) = P(A) * P(B) $\Leftrightarrow$ P(A) = $\frac{P(A)P(B)} {P(B)}$ = $\frac{P(A)\bigcap P(B)} {P(B)}$ = P(A|B)

You can also take an example. All equations have the same product.

0.3 * 0.7 * 0.3 *0.3 ^ 3 = 0.001701

0.7 * 0.3 * 0.3 *0.3^3= 0.001701

0.3 * 0.3 * 0.7 * 0.3^3= 0.001701

The last could be any combination of head and tail. 0.3^3 is just an example, but it works out equally if one or two or three of the variables is 0.7.