Huobi Token - HT

Huobi Token HT price, market cap & charts

Live Huobi Token prices from all markets and HT coin market Capitalization. Stay up to date with the latest Huobi Token price movements and discussion. Check out our snapshot charts and see when there is an opportunity to buy or sell Huobi Token

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

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Huobi Token Key Financial Information

Mkt. CapUSD 117,022,973Volume 24HUSD 67,236,616
Mkt. Share0.07 %Available Supply50,000,200
Change % (1H)-0.47 %Max Supply0
Change % (24H)-1.76 %Total Supply500,000,000
Change % (7D)1.10 %Proof
AlgorithmUpated: 38 minutes ago

Huobi Token Historical Data

DatePriceVolume

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.
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