Calculating Probability of 4 Heads in a Fair Coin Flip

[Somefantastik]

Given 4 flips of a fair coin, what is the probability of {H,H,H,H}?

I thought since flips of a fair coin are independent, then P(H&H&H&H) = P(H)P(H)P(H)P(H) = 1/(2^4) = 1/16. Am I close?

 

[EnumaElish]

Since these are 4 independent trials, yes.

You can verify this & similar problems by figuring out how many ways 4 coins can "land."

 

[Somefantastik]

thanks; you're right. I used the tree-notation and got N(ways HHHH)/N(all ways) = 1/16. Thanks again. I'm not entirely sure why that works.

Is it possible to compute the probability that one pattern comes before the other one? Like THHH will happen before HHHH?

 

[chroot]

Each of the 16 possible total outcomes (HHHH through TTTT) is equally possible, so there is no way to predict which will occur before another.

- Warren

 

[Somefantastik]

Thanks. I should have known that.

 

[CRGreathouse]

Remember to distinguish the probability that "THHH" comes before "HHHH" in a series of discrete tuples of four flips from the probability that the subsequence "THHH" comes before "HHHH" in a continuous series of individual flips.