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You flip a coin. If you get heads you win \$2 if you get tails you lose \$1. What is the expected value if you flip the coin 1000 times? I know that the expected value of flipping the coin once is $\frac{1}{2}(2) - \frac{1}{2}(1) =0.50$ Would the expected value be 500?
Flip a coin repeatedly, keeping track of the last three outcomes. (Save time, if you like, by assuming the first flip was T and proceeding from there.) Stop whenever the last three are THT or TTH. If the last three were THT, select option 1. Otherwise flip the coin one more time, choosing option 2 upon seeing T and option 3 otherwise.
(Thinking another way: there's a 1/2 chance you flip heads the first time, then a 1/2 of 1/2 = 1/4 chance you don't flip heads until the second time, etc.) The expected value of the number of flips is the sum of each possible number multiplied by the probability that number occurs.
To solve this lets start by naming the two heads and a tail in three coin flips. Lets name the heads as H-a and H-b. Lets name the tail as T. Now based on permutation we can find the arrangements of H-a, H-b and T in the three coin flip positions we have by computing 3p3 = 6. The actual permutations are listed below:
we have 2 results for one flip : up or down so flip 4 times, we have 4x2 = 8 results total. Two results for each of four coin flips. When ways to perform tasks in series, we multiply. So that is $2\times 2\times 2\times 2$ results in total. That is $2^4$ or $16$. For the favourable case we need to count the ways to get $2$ heads and $2$ tails.
(a): How many such outcomes are possible, given 8 coin tosses? For each toss, there are two possible outcomes: heads, or tails. Using the rule of the product , we have that after $8$ tosses, the possible outcomes/sequences is equal to $$\underbrace {2\times 2 \times \cdots \times 2}_{\large 8\; \text{tosses}} = (2)^8= \bf 256$$
In each of the five sequences of coin tosses in which exactly one head appears, no two heads are consecutive. In the only sequence of five coin tosses in which no heads appear, no two heads are consecutive. Hence, the number of sequences of five coin tosses in which no two heads are consecutive is $0 + 0 + 1 + 6 + 5 + 1 = 13$, as you found.
if I need to find a random sample of coin flips $\{0,1\}$, I consider the successive digits (after the decimal point) of π modulo 2. That is, if the digit is even, the outcome is $0$, and if it is odd, the outcome is $1$.
If the coin were fair, then the standard deviation for $1000$ flips is ${1\over2}\sqrt{1000}\approx16$, so a result with $600$ heads is roughly $6$ standard deviations from the mean. If you're familiar with Six Sigma , you'll have grounds for suspecting the coin is not fair.
In one of his interviews, Clip Link, Neil DeGrasse Tyson discusses a coin toss experiment. It goes something like this: Line up 1000 people, each given a coin, to be flipped simultaneously; Ask each one to flip if heads the person can continue; If the person gets tails they are out; The game continues until 1* person remains