maqifrnswa
|
|
September 29, 2013, 09:55:16 PM |
|
So, I have this awesome idea.
Someone should turn just-dice into a stock. So people can trade it on bitfunder or something. Now you ask why wouldn't you just do a regular stock? That's because now people can buy options on the stock. That would be cool.
A stock would be based off revenue from commissions and ads - that is a much better deal than being an investor. No it would be a third party who invested in just-dice. Doog doesn't need capital, so he doesn't need to sell stocks. You're proposing a pass through. Go ahead and set one up, offer an IPO, invest it all in JD, and distribute dividends above some bankroll if it is available a given week. Seems kind of silly to me since you could just directly invest in JD - but you're right, you want to setup a JD derivative and that needs some underlying security.
|
|
|
|
mechs
|
|
September 29, 2013, 09:59:21 PM |
|
So, I have this awesome idea.
Someone should turn just-dice into a stock. So people can trade it on bitfunder or something. Now you ask why wouldn't you just do a regular stock? That's because now people can buy options on the stock. That would be cool.
A stock would be based off revenue from commissions and ads - that is a much better deal than being an investor. No it would be a third party who invested in just-dice. Doog doesn't need capital, so he doesn't need to sell stocks. You're proposing a pass through. Go ahead and set one up, offer an IPO, invest it all in JD, and distribute dividends above some bankroll if it is available a given week. Seems kind of silly to me since you could just directly invest in JD - but you're right, you want to setup a JD derivative and that needs some underlying security. With almost 3.5M wagered and the site at a loss, I think he looking for a way to short J-D
|
|
|
|
maqifrnswa
|
|
September 29, 2013, 10:05:36 PM |
|
So, I have this awesome idea.
Someone should turn just-dice into a stock. So people can trade it on bitfunder or something. Now you ask why wouldn't you just do a regular stock? That's because now people can buy options on the stock. That would be cool.
A stock would be based off revenue from commissions and ads - that is a much better deal than being an investor. No it would be a third party who invested in just-dice. Doog doesn't need capital, so he doesn't need to sell stocks. You're proposing a pass through. Go ahead and set one up, offer an IPO, invest it all in JD, and distribute dividends above some bankroll if it is available a given week. Seems kind of silly to me since you could just directly invest in JD - but you're right, you want to setup a JD derivative and that needs some underlying security. With almost 3.5M wagered and the site at a loss, I think he looking for a way to short J-D there's an easy way to do that, gamble instead of invest
|
|
|
|
oda.krell
Legendary
Offline
Activity: 1470
Merit: 1007
|
|
September 29, 2013, 10:13:29 PM |
|
Question regarding just-dice house edge
I'm trying to get a better understanding of the results of different investment strategies in j-d. Being the lazy person that I am, I chose to simulate first, before I try to solve anything analytically. But first, let me know if I understood the house edge on j-d correctly:
It is not an edge on the probabilities of the "dice throw" itself (like, say, in Roulette), but it is a reduced payout in case the player wins, right?
Example:
Say there is exactly one investor, who put in 100 btc, at 1% maxbet. A player places a maximum size bet of 1 btc.
In case the players loses, the investor's new balance is: old_balance * (1 + 0.01) = 101 btc
In case the player wins, the investors new balance is: old_balance * (1 - (0.01 * 0.99)) = 99.01, where the factor 0.99 represents the house edge of 1%.
Yes? No?
|
Not sure which Bitcoin wallet you should use? Get Electrum!Electrum is an open-source lightweight client: fast, user friendly, and 100% secure. Download the source or executables for Windows/OSX/Linux/Android from, and only from, the official Electrum homepage.
|
|
|
chriswen
|
|
September 29, 2013, 10:14:12 PM |
|
|
|
|
|
chriswen
|
|
September 29, 2013, 10:16:12 PM |
|
Question regarding just-dice house edge
I'm trying to get a better understanding of the results of different investment strategies in j-d. Being the lazy person that I am, I chose to simulate first, before I try to solve anything analytically. But first, let me know if I understood the house edge on j-d correctly:
It is not an edge on the probabilities of the "dice throw" itself (like, say, in Roulette), but it is a reduced payout in case the player wins, right?
Example:
Say there is exactly one investor, who put in 100 btc, at 1% maxbet. A player places a maximum size bet of 1 btc.
In case the players loses, the investor's new balance is: old_balance * (1 + 0.01) = 101 btc
In case the player wins, the investors new balance is: old_balance * (1 - (0.01 * 0.99)) = 99.01, where the factor 0.99 represents the house edge of 1%.
Yes? No?
There's no max bet. It's max profit.
|
|
|
|
mechs
|
|
September 29, 2013, 10:21:12 PM |
|
Is it true FF Ownz went from 4 BTC all the way to 2000 BTC??
|
|
|
|
maqifrnswa
|
|
September 29, 2013, 10:21:38 PM |
|
Question regarding just-dice house edge
I'm trying to get a better understanding of the results of different investment strategies in j-d. Being the lazy person that I am, I chose to simulate first, before I try to solve anything analytically. But first, let me know if I understood the house edge on j-d correctly:
It is not an edge on the probabilities of the "dice throw" itself (like, say, in Roulette), but it is a reduced payout in case the player wins, right?
Example:
Say there is exactly one investor, who put in 100 btc, at 1% maxbet. A player places a maximum size bet of 1 btc.
In case the players loses, the investor's new balance is: old_balance * (1 + 0.01) = 101 btc
In case the player wins, the investors new balance is: old_balance * (1 - (0.01 * 0.99)) = 99.01, where the factor 0.99 represents the house edge of 1%.
Yes? No?
it is an edge on the probabilities of the dice throw. Eg: gambler wins 2x if RNG picks a number > 50.5 or < 49.5 (roll hi or roll low). So if gamblers hits roll high, house wins from 0-50.5, player wins from 50.5-100
|
|
|
|
oda.krell
Legendary
Offline
Activity: 1470
Merit: 1007
|
|
September 29, 2013, 10:21:52 PM |
|
Question regarding just-dice house edge
I'm trying to get a better understanding of the results of different investment strategies in j-d. Being the lazy person that I am, I chose to simulate first, before I try to solve anything analytically. But first, let me know if I understood the house edge on j-d correctly:
It is not an edge on the probabilities of the "dice throw" itself (like, say, in Roulette), but it is a reduced payout in case the player wins, right?
Example:
Say there is exactly one investor, who put in 100 btc, at 1% maxbet. A player places a maximum size bet of 1 btc.
In case the players loses, the investor's new balance is: old_balance * (1 + 0.01) = 101 btc
In case the player wins, the investors new balance is: old_balance * (1 - (0.01 * 0.99)) = 99.01, where the factor 0.99 represents the house edge of 1%.
Yes? No?
There's no max bet. It's max profit. Right. But let's say we take the usual 50% bet, then max bet = max profit (ignoring for a moment house edge) In that case, did I get calculation of house edge right?
|
Not sure which Bitcoin wallet you should use? Get Electrum!Electrum is an open-source lightweight client: fast, user friendly, and 100% secure. Download the source or executables for Windows/OSX/Linux/Android from, and only from, the official Electrum homepage.
|
|
|
oda.krell
Legendary
Offline
Activity: 1470
Merit: 1007
|
|
September 29, 2013, 10:22:42 PM |
|
it is an edge on the probabilities of the dice throw.
Eg: gambler wins 2x if RNG picks a number > 50.5 or < 49.5 (roll hi or roll low). So if gamblers hits roll high, house wins from 0-50.5, player wins from 50.5-100
Where do you get this from? The j-d FAQ seems to imply otherwise.
|
Not sure which Bitcoin wallet you should use? Get Electrum!Electrum is an open-source lightweight client: fast, user friendly, and 100% secure. Download the source or executables for Windows/OSX/Linux/Android from, and only from, the official Electrum homepage.
|
|
|
maqifrnswa
|
|
September 29, 2013, 10:42:04 PM |
|
it is an edge on the probabilities of the dice throw.
Eg: gambler wins 2x if RNG picks a number > 50.5 or < 49.5 (roll hi or roll low). So if gamblers hits roll high, house wins from 0-50.5, player wins from 50.5-100
Where do you get this from? The j-d FAQ seems to imply otherwise. user: berathea (161188) date: 2013-09-29 18:39:30 betid: 144345199 lucky: 91.4117 target: >50.4999 bet: 170.79869440 payout: 2x profit: +170.79869439 Two ways of looking at it: 1) 2x payout means you roll > 50.5 2) rolling > 50 means 1.98x payout both are true, FAQ points out (2), but the bet log has lots of (1), as shown above.
|
|
|
|
maqifrnswa
|
|
September 29, 2013, 10:44:54 PM |
|
Question regarding just-dice house edge
I'm trying to get a better understanding of the results of different investment strategies in j-d. Being the lazy person that I am, I chose to simulate first, before I try to solve anything analytically. But first, let me know if I understood the house edge on j-d correctly:
It is not an edge on the probabilities of the "dice throw" itself (like, say, in Roulette), but it is a reduced payout in case the player wins, right?
Example:
Say there is exactly one investor, who put in 100 btc, at 1% maxbet. A player places a maximum size bet of 1 btc.
In case the players loses, the investor's new balance is: old_balance * (1 + 0.01) = 101 btc
In case the player wins, the investors new balance is: old_balance * (1 - (0.01 * 0.99)) = 99.01, where the factor 0.99 represents the house edge of 1%.
Yes? No?
There's no max bet. It's max profit. Right. But let's say we take the usual 50% bet, then max bet = max profit (ignoring for a moment house edge) In that case, did I get calculation of house edge right? No. If an investors has 100 BTC in an account and we're at full kelly (1%), a single 2x payout bet would yield +/- 1 BTC. The reason why is that a 2x bet is the range > 50.5 or < 49.5. The odds are different in the roll (see the log I posted above for an example).
|
|
|
|
Peter R
Legendary
Offline
Activity: 1162
Merit: 1007
|
|
September 29, 2013, 11:25:58 PM |
|
Okay, let's see what that would be like. https://bitcointalk.org/index.php?topic=242962.msg2778468#msg2778468Assumptions: p = 0.495 b = 200 //I'll be assuming 5 000 bets of 200 BTC at 49.5% chance of winning c = 5 000 Total BTC bet = 1 000 000 BTCExpected BTC return for gamblers = 990 000 BTCExpected Profit = 10 000 BTCExpected Standard Deviation of amount returned to gamblers = 7000 BTCExpected profit is (10000±7000) BTCIf actual profit was 0, than that would be 1.43 times the standard deviation away from the expected value. If the profit followed a normal distribution, the probability of this distribution occurring through sheer chance is approximately 15.3% I don't think this is quite right (and was the same mistake I made at first). I think the variance *for a single roll* is: V = B^2 (1-E)^2 (1-p) / p which over 5000 rolls gives: House Profit = 10,000 +/- 14,000 BTC using Chris's numbers. Here was my proof: Mean and Variance of a Gambler's Profits after a Sequence of Bets on Just-Dice.comLet B = bet size P = gamber's profit p = probability of a win E = house edge JD betting is a Bernoulli process. There are two outcomes: win or lose. | GAMBLER'S PROFIT TABLE | | ____WIN____ | | ____LOSE____ | probability = p | | probability = 1-p | P = B(1-E-p)/p | | P = -B |
Gambler's expected profit per roll: <P> = (profit if he wins)x(probability of winning) + (profit if he loses)x(probability of losing) = B(1-E-p)/p x p + -B x (1-p) = -E B (of course!) Gambler's variance per roll: V = <(P - <P>)^2> = (B(1-E-p)/p - -E B)^2 x p + (-B - -E B)^2 x (1-p) = B^2 (1-E)^2 (1-p) / p Over multiple rolls his expected profit and variances add: <P_total> = <P>(1) + ... + <P>(n) V_total = V(1) + ... + V(n) QED
|
|
|
|
bingjiw
Newbie
Offline
Activity: 37
Merit: 0
|
|
September 29, 2013, 11:50:20 PM |
|
The profit is negative again!!
Please change the max bet to 0.1% and change house edge to 1.5%.
Why SatoshiDice can always profit and Just-dice always loss all the profits and go to negative?
We need to change something.
|
|
|
|
wolverine.ks
|
|
September 30, 2013, 12:05:31 AM |
|
what about partnering with Coinflow.co to offer graphs to investors and gamblers. it would relieve doog of work, add value to the site, probably some other good stuff too.
thoughts.
|
|
|
|
Peter R
Legendary
Offline
Activity: 1162
Merit: 1007
|
|
September 30, 2013, 12:22:46 AM |
|
How many repeat bets does it it take to push a gambler's +1 sigma line negative?
Equations:
E = house edge, B = bet size, n = number of bets, p = chance of win
total expected profit after n bets = -n E B
accumulated variance = n B^2 (1-E)^2 (1-p) / p
So, the question is when does
-n E B + sqrt(n B^2 (1-E)^2 (1-p) / p) = 0 ?
Solving for number of bets n gives
n = (1-E)^2 (p-1) / (E^2 p)
So, with 1% house edge betting on p = 0.495, it takes 9999 bets. With 2% house edge betting on p = 0.495, it only takes 2450 bets
It takes 4 X longer for a gambler's +1 sigma line to go negative at 1% than at 2% house edge.
|
|
|
|
RationalSpeculator
Sr. Member
Offline
Activity: 294
Merit: 250
This bull will try to shake you off. Hold tight!
|
|
September 30, 2013, 12:27:05 AM |
|
How many repeat bets does it it take to push a gambler's +1 sigma line negative?
It takes 4 X longer for a gambler's +1 sigma line to go negative at 1% than at 2% house edge.
What does that mean, a sigma line going negative?
|
|
|
|
Peter R
Legendary
Offline
Activity: 1162
Merit: 1007
|
|
September 30, 2013, 12:29:35 AM |
|
How many repeat bets does it it take to push a gambler's +1 sigma line negative?
It takes 4 X longer for a gambler's +1 sigma line to go negative at 1% than at 2% house edge.
What does that mean, a sigma line going negative? I mean the curve 1 standard deviation above his expected profits. Until this curve goes negative, it is really not that surprising at all for the gambler to keep popping up into profits.
|
|
|
|
dooglus (OP)
Legendary
Offline
Activity: 2940
Merit: 1333
|
|
September 30, 2013, 12:35:55 AM |
|
Why would I invest 500 BTC at 0.25% when I can just invest 125BTC at 1% then and not have the CP risk for the othet 375 BTC?
Because you would get less variance. Suppose a whale bets max bet and wins 20 times. If you invest 500 BTC at 0.25%, you'll still have 95% of your bankroll left: >>> a=500 >>> for i in range(20): a *= (1-0.0025); print "%.2f" % (a*100/500,), 99.75 99.50 99.25 99.00 98.76 98.51 98.26 98.02 97.77 97.53 97.28 97.04 96.80 96.56 96.31 96.07 95.83 95.59 95.36 95.12 But if you invest 125 BTC at 1%, you'll have only 82% of it left: >>> a=125 >>> for i in range(20): a *= (1-0.0100); print "%.2f" % (a*100/125,), 99.00 98.01 97.03 96.06 95.10 94.15 93.21 92.27 91.35 90.44 89.53 88.64 87.75 86.87 86.01 85.15 84.29 83.45 82.62 81.79 Your risk of ruin is much greater if you invest at 1% than if you invest at 0.25%. Not sure that's true in the second case - as their BR is what they have left of the 125 plus the untouched 375 they have in cold storage. Fair enough. Here it is again taking that into account: >>> a=500 >>> for i in range(20): a *= (1-0.0025); print "%.2f" % (a*100/500,), 99.75 99.50 99.25 99.00 98.76 98.51 98.26 98.02 97.77 97.53 97.28 97.04 96.80 96.56 96.31 96.07 95.83 95.59 95.36 95.12, ..., 0 >>> a=125 >>> for i in range(20): a *= (1-0.0100); print "%.2f" % ((a+375)*100/500,), 99.75 99.50 99.26 99.01 98.77 98.54 98.30 98.07 97.84 97.61 97.38 97.16 96.94 96.72 96.50 96.29 96.07 95.86 95.65 95.45, ..., 75 If the whale keeps winning when they invest 500 at 0.25%, they end up losing everything. If they invest 125 at 1%, they end up only losing the 25% they invested...
|
Just-Dice | ██ ██████████ ██████████████████ ██████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████ ██████████████ ██████ | Play or Invest | ██ ██████████ ██████████████████ ██████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████████████ ██████████████████████ ██████████████ ██████ | 1% House Edge |
|
|
|
RationalSpeculator
Sr. Member
Offline
Activity: 294
Merit: 250
This bull will try to shake you off. Hold tight!
|
|
September 30, 2013, 12:38:21 AM |
|
How many repeat bets does it it take to push a gambler's +1 sigma line negative?
It takes 4 X longer for a gambler's +1 sigma line to go negative at 1% than at 2% house edge.
What does that mean, a sigma line going negative? I mean the curve 1 standard deviation above his expected profits. Until this curve goes negative, it is really not that surprising at all for the gambler to keep popping up into profits. If I remember correctly he took around 15,000 bets according to dooglus in the chat. In such case that is above the 9,999 bets you state, so in the negative curve line as you say. How high is the chance for that to happen according to your estimations/calculations?
|
|
|
|
|