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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.
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
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:
(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.
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 × 2 × 2 × 2 2 × 2 × 2 × 2 results in total. That is 24 2 4 or 16 16. For the favourable case we need to count the ways to get 2 2 heads and 2 2 ...
Since the flip sequence must consist of runs of H H and T T, the sequence ending in HT H T consists of zero or more T T s, one or more H H s and a T T in that order. The expected number of flips needed is then the sum of. Hence the answer is 1 p + 1 1−p 1 p + 1 1 − p, which is 4 4 when p = 1 2 p = 1 2.
We assume that the coin is fair and is flipped fairly. There are $2^5$ equally likely strings of length $5$ made up of the letters H and/or T. There are precisely $5$ strings that have exactly $1$ H and $4$ T.
Probability of getting a head in coin flip is $1/2$. If the coin is flipped two times what is the probability of getting a head in either of those attempts? I think both the coin flips are mutually exclusive events, so the probability would be getting head in attempt $1$ or attempt $2$ which is:
It follows that tn =Fn+1 , the (n + 1)th Fibonacci number. So the number of sequences of n tosses with no consecutive heads is tn +hn = Fn+2 . For n = 5 this gives F7 = 13 as the number of such sequences, and for n = 100 Wolfram alpha gives. F102 = 927372692193078999176. as the number of such sequences.
$\begingroup$ Wow, 6 answers :) And some of them are quite good, I have to say; the phone numbers one would work very well for me, for example (on koaning.io/human-entropy.html, taking the last 4 digits, discarding any that is 8 or 9, and using the the 3 bits of every other, I generated without much effort about 1 bit/second, and the result was undistinguishable from that of a pseudorandom ...