“Flipping the Coin” is a quest given by Cinder-Tail in Grahtwood‘s Redfur Trading Post. Flipping the Coin has a chess-like puzzle requiring you to move your Thief Statue before the Guard Statues can get to her. On approaching Redfur Trading Post, I came upon Cinder-Tail, a Khajiit in trouble with the local authorities.

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.

2017年1月12日 · 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 $1$.

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?

2021年8月3日 · 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.

2021年12月20日 · For more illustration , lets assume that we indicate both heads and tails with stars such as $"*"$, and we determine whether they are heads or tails according to their position in terms of a bar such as $"|"$.Lets say that the stars which are on the left side of bar are representing the tails and the stars which are on the right side of bar are representing the heads.

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.

2013年10月10日 · The second term is if we get a H (heads) on our first flip (with probability $\frac{1}{2}$), which means that we are effectively starting over with 1 extra flip. Now let Y be the random variable which counts the number of flips till we get a HT sequence.

2021年10月27日 · HHT and HTH appear just as often, but half of the time HTH appears just one flip after HHT. As a suggestion to help your intuition, let's suppose no one wins in the first three coin flips (this remove 1/4 of the tries, half of them wins and the other half losses). Now consider the first HTH of the sequence and ask yourself what was the previous ...

You flip a coin until you get a total of n heads. What is the expected number of flips this will take? In terms of Expectation, E[x] and Variance Var(x). Edit: The heads don't have to be consecutive.