Just-Dice.com : Invest in 1% House Edge Dice Game - page 109.

dooglus

legendary

Activity: 2940

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Quote from: chriswen on September 28, 2013, 06:29:15 PM

So say there are two investors. Investor A has invested 100 BTC. Investor B has invested 100 BTC also.

Right now the max profit is set to 0.5%.

If someone bets 0.1btc at 49.5%
Then they both are able to gain 0.05 btc. (or lose)

Now you implement your method. Investor B decides to risk 1% instead.

So now if someone bets 0.1 btc,
Investor A can win ±.33 and investor B can win ±.67.

This doesn't seem fair. Investor A is still okay with betting at this risk level. Why does investor B take some of his bet?

Here's another analogy:

--
A and B both invest 100 BTC. Someone wants to bet 0.1 BTC. Both investors stand to gain 0.05 BTC.

Now investor B decides to double his risk by investing another 100 BTC.

So now if someone bets 0.1 btc,
Investor A can win ±.33 and investor B can win ±.67.

This doesn't seem fair. Investor A is still okay with betting at this risk level. Why does investor B take some of his bet?
--

In both cases investor B has doubled his risk, and as a result 'takes' some of investor A's action. In both cases he has diluted the other investor's share by risking more. In your analogy he did it by doubling his Kelly factor, and in mine he did it by doubling his investment. In both cases he's doubling his exposure, and so doubles his reward.

RationalSpeculator

sr. member

Activity: 294

Merit: 250

This bull will try to shake you off. Hold tight!

Quote from: Peter R on September 29, 2013, 07:29:35 PM

Quote from: RationalSpeculator on September 29, 2013, 07:27:05 PM

Quote from: Peter R on September 29, 2013, 07:22:46 PM

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?

dooglus

legendary

Activity: 2940

Merit: 1333

Quote from: Deprived on September 28, 2013, 04:57:07 PM

Quote from: dooglus on September 28, 2013, 04:47:01 PM

Quote from: mechs on September 28, 2013, 04:34:27 PM

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...

Peter R

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Activity: 1162

Merit: 1007

Quote from: RationalSpeculator on September 29, 2013, 07:27:05 PM

Quote from: Peter R on September 29, 2013, 07:22:46 PM

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.

RationalSpeculator

sr. member

Activity: 294

Merit: 250

This bull will try to shake you off. Hold tight!

Quote from: Peter R on September 29, 2013, 07:22:46 PM

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

Activity: 1162

Merit: 1007

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.

wolverine.ks

sr. member

Activity: 375

Merit: 250

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.

bingjiw

newbie

Activity: 37

Merit: 0

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.

Peter R

legendary

Activity: 1162

Merit: 1007

Quote from: chriswen on September 29, 2013, 04:04:35 PM

Okay, let's see what that would be like.

https://bitcointalksearch.org/topic/m.2778468

Assumptions:
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 BTC
Expected BTC return for gamblers = 990 000 BTC
Expected Profit = 10 000 BTC
Expected Standard Deviation of amount returned to gamblers = 7000 BTC

Expected profit is (10000±7000) BTC
If 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:

Quote from: Peter R on September 28, 2013, 12:38:08 PM

Mean and Variance of a Gambler's Profits after a Sequence of Bets on Just-Dice.com

Let

   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:

= (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 -

)^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:

=

(1) + ... +

(n)

V_total = V(1) + ... + V(n)

QED

maqifrnswa

sr. member

Activity: 454

Merit: 252

Quote from: oda.krell on September 29, 2013, 05:21:52 PM

Quote from: chriswen on September 29, 2013, 05:16:12 PM

Quote from: oda.krell on September 29, 2013, 05:13:29 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?

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).

maqifrnswa

sr. member

Activity: 454

Merit: 252

Quote from: oda.krell on September 29, 2013, 05:22:42 PM

Quote from: maqifrnswa on September 29, 2013, 05:21:38 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.

oda.krell

legendary

Activity: 1470

Merit: 1007

Quote from: maqifrnswa on September 29, 2013, 05:21:38 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.

oda.krell

legendary

Activity: 1470

Merit: 1007

Quote from: chriswen on September 29, 2013, 05:16:12 PM

Quote from: oda.krell on September 29, 2013, 05:13:29 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?

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?

maqifrnswa

sr. member

Activity: 454

Merit: 252

Quote from: oda.krell on September 29, 2013, 05:13:29 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?

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

mechs

full member

Activity: 210

Merit: 100

Is it true FF Ownz went from 4 BTC all the way to 2000 BTC??

chriswen

hero member

Activity: 756

Merit: 500

Quote from: oda.krell on September 29, 2013, 05:13:29 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?

There's no max bet. It's max profit.

chriswen

hero member

Activity: 756

Merit: 500

today, 17:58
all time: https://i.imgur.com/6OESfpm.png
2 weeks: https://i.imgur.com/iL4EA7h.png
week: https://i.imgur.com/IDcQyFE.png
day: https://i.imgur.com/yMKuPUd.png

oda.krell

legendary

Activity: 1470

Merit: 1007

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?

maqifrnswa

sr. member

Activity: 454

Merit: 252

Quote from: mechs on September 29, 2013, 04:59:21 PM

Quote from: maqifrnswa on September 29, 2013, 04:55:16 PM

Quote from: chriswen on September 29, 2013, 04:41:59 PM

Quote from: mechs on September 29, 2013, 04:36:54 PM

Quote from: chriswen on September 29, 2013, 04:35:03 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 Wink

mechs

full member

Activity: 210

Merit: 100

Quote from: maqifrnswa on September 29, 2013, 04:55:16 PM

Quote from: chriswen on September 29, 2013, 04:41:59 PM

Quote from: mechs on September 29, 2013, 04:36:54 PM

Quote from: chriswen on September 29, 2013, 04:35:03 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

Topic: Just-Dice.com : Invest in 1% House Edge Dice Game - page 109. (Read 435457 times)