#### Price

^{BTC}0.00042882

USD

2.34

#### Market Cap

USD

117,022,973

#### Change % (1H)

%

^{} -0.47

#### Change % (24H)

%

^{} -1.76

## Huobi Token (HT) Historical Price & Volume Charts

## Huobi Token Wiki

beta## Huobi Token Key Financial Information

Mkt. Cap | ^{USD} 117,022,973 | Volume 24H | ^{USD} 67,236,616 |

Mkt. Share | 0.07 % | Available Supply | 50,000,200 |

Change % (1H) | -0.47 % | Max Supply | 0 |

Change % (24H) | -1.76 % | Total Supply | 500,000,000 |

Change % (7D) | 1.10 % | Proof | |

Algorithm | Upated: 38 minutes ago |

## Huobi Token Historical Data

Date | Price | Volume |
---|

## Huobi Token Videos

Counter-Intuitive Probability. Coin Flips To HH Versus HT Are Not The Same!

Huobi Token HT - Why Huobi Token is raising Green Flags

## Huobi Token (HT) Reviews & Critics

What is the expected number of coin flips until you get two heads in a row (HH)? What is the expected number of coin flips until you get a heads followed by a ....

- What is the difference between E(HH) and E(HH|T)?
- This is very clear. Thanks for making this
- I solved it a little differently:For HH:There is a 1/4 probability of HH on the first try (E=1); a 1/2 probability of the second spin being T, which "wastes" 2 rolls (E=2+E); and a 1/4 probability of second roll being H, further subdivided based on the third roll, with a 1/8 probability of success at E=2 (3 rolls total), and 1/8 probability of getting a T, "wasting" a total of 3 rolls (E=3+E), summed up thusly and solved for E, to get 5. This corresponds with the 6 from the video since the video is counting the final roll. I only count the first of the successful roll sequence to keep the equality P = 1/E.E = 1 (.25) + 2 ( .25 * .5 ) + (3+E)(.25*.5) + (2+E)*.5 = .25 + .25 + .375 + .125E + 1 + .5E = .875 + .125E + 1 + .5E = 1.875 + .625E = E.375E = 1.875E = 1.875 / .375 = 15 / 3 = 5For HT:Similarly, there is a 1/4 probability of HT on the first try (E=1); a 1/4 probability of the second spin otherwise being a T, "wasting" 2 spins (E=2+E); and a 1/2 probability of second roll being an H, further subdivided infinitely based on number of rolls to reach the T: 1/4 chance of E=2, 1/8 chance of E=3, 1/16 chance of E=4, et cetera, summed up thusly:E = 1(.25) + 2 (.5 * .5) + 3(.5*.5*.5) + 4(.5^4)+...+ (2+E)*.25x = 2(.5^2)+3(.5^3)+...+n(.5^n)2x = 2(.5)+3(.5^2)+...+n(.5^(n-1))2x - x = 2(.5) + .5^2 + .5^3 + .5^4 + ....5^nx = 1 + .5 = 1.5E = 1(.25) + 1.5 + (2+E)*.25 = 1.75 + .5 + .25E.75E = 2.25E = 2.25 / .75 = 9/3 = 3
- I ran a simulation with Matlab. Imagine 10,000 people each given a fair coin. They keeping flipping coins until each of them have at least one HH and one HT sequence. They then average their results on how many flips it first took to get the first HH and HT. One result was 5.9843 for HH and 3.9926 for HT. I then had these 10,000 people do their flip experiment 999 more times. From these 1000 results I made distribution histograms. You can see the histograms here: https://drive.google.com/file/d/1sWWWLYsLq8oXTcqwZwBwLXcQbzsYmD5Y . Keep in mind that Talwalkar's results are equivalent to an infinite number of flip experiments.