Probability Question?

The probability of HH was p^2. The probability of HT or TH was 2p(1-p) and of TT was (1-p)^2. The fact that HH occurred makes one believe that p is greater than 1/2, but I do not know how to make an intelligent guess as to its actual value. The probability of another H is, of course, p. It seems that the other responders didn't understand your question. I understand it but don't know how to solve it.

Not sure of my answer:

A = {HH} --- first two are heads

B = {H} --- third is a head

P(B/A) = P( B ∩ A) /P(A) = p^2 /p = p

The probability that the third toss is a head (given that the first two are heads) is p. (If the tosses were independent, this would be p anyway).

Amount probability

$1 ..... ... p

-$1 ... ... .(1-p)

Expected value = 1(p) + (-1) (1-p) = p +p -1 = 2p-1

2p-1 may be positive or negative depending on what p is.

From this, you may be able to figure out the 'fair' value of the ticket.

Does the wording imply that the ticket will win the lottery only if the result is HHH?

If so, P(HHH) = p^3

$1 ... p^3

-$1....1-p^3

EV = 1(p^3) + (-1) (1 - p^3) = p^3 + p^3 - 1 = 2 p^3 -1 = 0 if p^3 = 1/2 or p = $1/8

Too much ambiguity.

ok, the ticket pays $1 if the 3rd is head, 0 otherwise.

the result of the 3rd toss is clearly independent, thus P(win) = p

now average value of p = ∫ p dp, p= 0 to 1 = 1/2

ie a 50:50 chance, so fair price = $1 <--------

your AD

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"pays one dollar", I took it as what is paid in addition to the stake.

if it is total amount paid, the fair price is $0.50

probability of a third head = p

because the result of the third toss does not depend on the previous tosses.

You don't accept email so the only way I can figure out how to communicate with you is to answer one of your other questions. I wanted to let you know that I calculated the probabilities you were looking for in https://answers.yahoo.com/question/index?qid=20141...

propabbly p always as it doent depend on previous tosses