#### Price

^{BTC}0.00040997

USD

4.03

#### Market Cap

USD

201,300,775

#### Change % (1H)

%

^{} 0.01

#### Change % (24H)

%

^{} -0.37

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

## Huobi Token Wiki

beta## Huobi Token Key Financial Information

Mkt. Cap | ^{USD} 201,300,775 | Volume 24H | ^{USD} 101,955,021 |

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

Change % (1H) | 0.01 % | Max Supply | 0 |

Change % (24H) | -0.37 % | Total Supply | 500,000,000 |

Change % (7D) | 7.24 % | Proof | |

Algorithm | Upated: 30 minutes ago |

## Huobi Token Historical Data

Date | Price | Volume |
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## Huobi Token Videos

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

Review nhanh Sò Toshiba C5200 200w,tụ nguồn HT coin

## 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 ....

- I just wrote some VBA code in Excel to simulate the game of coin flipping. My simulation program with say 10,000 tests (which takes only a few seconds to run) results in an average of about 6 tosses to get two consecutive Heads.
- 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