# A fair coin is tossed four times. What is the probability that the sequence of tosses is HHHT? ?

Relevance
• P(HHHT)=[(1/2)^3](1/2)=(1/2)^4=1/16.

• None of the answers so far, allow for unconventional results. Such as settling on edge or not landing at all. Oh - and they ignore schrodinger principle, where the coin is captured in flight and exists in an indeterminate state. But......this is why I got into trouble in high school math, for being a smartass. I wasn't - these are conditions a good software designer accounts for. Hey - I never even mentioned Heiseberg !!!

• 1 in 16.  there are 16 possible orders that can arise.  that is one of the 16.  2x2x2x2.

• p(HHHT) =

1/2^4 =

1/16

• There are 16 possible sequences, each with the same probability. So the probability of any particular sequence is 1 / 16.

• There are 2 possibilities that can occur at each toss.

so, 2 x 2 x 2 x 2 = 16 ways in total

H, H, H and T is a specific order and one unique combination

i.e. 1/16

:)>

• 1/2x1/2x1/2x1/2=1/16.