Ticket Sales

The purchase value of tickets depends on the price in UNIT of the cryptocurrencies deposited and the probability of the outcome.
On day 1 of the round, tickets (yUNIT and nUNIT) are priced in UNIT at the rate of 
1  UNIT=(1−p)  yUNIT,1 \;\textrm{UNIT}= (1-p) \;\textrm{yUNIT},1UNIT=(1−p)yUNIT,
1  UNIT=p  nUNIT,1 \;\textrm{UNIT}= p\;\textrm{nUNIT},1UNIT=pnUNIT,
for ppp
 being the probability of "yes".
On subsequent days, ticket prices depend on the updated probability of the outcome on each day. 
In the Ethereum network, for VVV
 the price of 1 ETH in UNIT and ppp
 the updated probability of "yes" we get that 1 ETH will yield the following amount of yUNIT and nUNIT tickets.
1  ETH=V∗(1−p)  yUNIT,1 \;\textrm{ETH}= V * (1-p) \;\textrm{yUNIT},1ETH=V∗(1−p)yUNIT,
1  ETH=V∗p  nUNIT.1 \;\textrm{ETH}= V * p \;\textrm{nUNIT}.1ETH=V∗pnUNIT.
Example of Ticket Sale Calculation (Day 1)
Question: Will coin A enter The UNIT this round?
Calculate the probability ppp
 that coin A will enter The UNIT in this round given equal 0.50.50.5
 probability that it will satisfy the requirements in any given day.
p=1213[(138)+(139)+(1310)+(1311)+(1312)+(1313)]p = \frac{1}{2^{13}}\biggl[\binom{13}{8}+\binom{13}{9}+\binom{13}{10}+\binom{13}{11}+\binom{13}{12}+\binom{13}{13}\biggr]p=2131​[(813​)+(913​)+(1013​)+(1113​)+(1213​)+(1313​)]
p=23808192=0.29052734375.p =\frac{2380}{8192}= 0.29052734375. p=81922380​=0.29052734375.
Then, if 1 ETH is currently worth 1000 UNIT, then
1  ETH=1000∗0.70947265625  yUNIT=709.47265625  yUNIT,1 \;\textrm{ETH}= 1000*  0.70947265625\;\textrm{yUNIT}=709.47265625\;\textrm{yUNIT},1ETH=1000∗0.70947265625yUNIT=709.47265625yUNIT,
1  ETH=1000∗0.29052734375  nUNIT=290.52734375  nUNIT.1 \;\textrm{ETH}= 1000 * 0.29052734375 \;\textrm{nUNIT} = 290.52734375\;\textrm{nUNIT}.1ETH=1000∗0.29052734375nUNIT=290.52734375nUNIT.
Example of Ticket Sale Calculation (Day 2)
Same Question: Will coin A enter The UNIT this round?
Known Data: On Day 1, coin A met the requirements to enter The UNIT.
Again, calculate the probability p1p_1p1​
 that coin A will enter The UNIT in this round given equal 0.50.50.5
 probability that it will satisfy the requirements in any given day.
p1=1212[(127)+(128)+(129)+(1210)+(1211)+(1212)]p_1 = \frac{1}{2^{12}}\biggl[\binom{12}{7}+\binom{12}{8}+\binom{12}{9}+\binom{12}{10}+\binom{12}{11}+\binom{12}{12}\biggr]p1​=2121​[(712​)+(812​)+(912​)+(1012​)+(1112​)+(1212​)]
p1=15864096=0.38720703125.p_1 = \frac{1586}{4096} =0.38720703125.p1​=40961586​=0.38720703125.
Then, if 1 ETH is currently worth 1000 UNIT, then
1  ETH=1000∗0.61279296875  yUNIT=612.79296875  yUNIT,1 \;\textrm{ETH}= 1000*  0.61279296875\;\textrm{yUNIT}=612.79296875\;\textrm{yUNIT},1ETH=1000∗0.61279296875yUNIT=612.79296875yUNIT,
1  ETH=1000∗0.38720703125  nUNIT=387.20703125  nUNIT.1 \;\textrm{ETH}= 1000 * 0.38720703125 \;\textrm{nUNIT} = 387.20703125\;\textrm{nUNIT}.1ETH=1000∗0.38720703125nUNIT=387.20703125nUNIT.
Last modified 3mo ago