Topic: "How to hop" has moved - page 5. (Read 16256 times)

When I first published the Slush hop function in How to hop 4 I made a mistake in not bothering to take difficulty into account, since D seemed pretty stable at the time. c00w even asked about how Slush hopping would be affected by difficulty and I lazily replied something like 'not a problem'. I wasn't expecting D to drop by a third!

So I rejigged the hop function to take current difficulty into account. I'll publish the derivation on hoppersden.info when I have time. The new function makes a significant difference (try it against the old one and see) so please pass the word.

Anyway here it is, ready to go:


'c' can be calculated (see 'How to hop 8') if you get some score, share and time values from the pool website, 'hashrate' is the average hashrate of the pool  since start of roundin Ghps, and D is current difficulty. Finding an accurate solution that was still fairly simple took a while, so sorry for the delay.

Edit: fixed formula - brackets in wrong place. Thanks for the heads up, eveofwar

Coinotron's score system is different from Slush's, and mine_c on bitHopper is not appropriate. I would use role:mine instead. I'll explain why at some point, but in the meantime please enjoy following chart porn, just for you:

Word on the street is "c value for Coinatron is 6."  However, I expect the c value to be higher and have had good luck mining with a c value of 3000.  Have you determined a c value for Coinatron yet?  Keep up the great work!

I think long time blog readers will get a kick out of this:


....which I explained way back in How to hop 4:


Have to love it when the real world agrees with the math.

I didn't realise it when I posted How to hop 8, but the Lambert W function in R is actually from the GNU scientific library. Anything in GNU-sl can be used on any platform that can use C and C++ since the source is available.

A good simple reference is here.

lambert_w

How is that implemented in R?

New 'How to hop' post on a simple method to calculate 'c' for a 'Slush scored' pool.

the strategic player himself increases his own profits by being in the game
this effect is eliminated at high enough number of players and at sufficiently large number of players the results would be close to the fractional for all players even for the strategic player himself

Understood - multiple players affect expected winnings. But does it change the fractional profit per strategic player, ie winnings/cost? You might make less per round, but I think the fractional profit will remain the same.

Let me think a bit more on it, I'll do a post on expected profit and see if I can get similar results to you.

because a strategic player actually messes with the distribution of the payouts by his own presense, he can get away with "paying less, and receiving more"

he does this by participating in cheap rounds (rounds with many players in them) and not participating in expensive rounds (rounds with few players in them)

who cares about the EV for all players when some players pay less to receive more?

thought experiment:

the game is getting really stupid now, 9 strategic players play it around the clock professionally
one player plays it fairly

he has to pay $1 on the first toss and $10 on the 6th+ toss
but he only takes ONE share per toss away from the strategic players
and they all pay $1 per round as well until they all stop tossing at 5 tosses because they read the organofcorti guide

it's super cheap for them and super expensive for the poor fair player
in fact, he ends up paying nearly half of the cost but gets only twice as many shares each strategic player, getting about ~20% of the reward

my point is:
strategic players and "it's due for a 10" players change the distribution of the payouts BETWEEN the players, so by playing the game you're changing the distribution of the payout between the players since you're a strategic player

so that's why my comment was your post outlines "the correct strategy as the number of players approaches infinity" at which point your strategic play doesn't disturb the payouts since you're an infinitesimal part of the field

this was already pointed out earlier in 3phase's post where he wrote:
"The number of players participating in the game is relatively stable and big enough so that your participation or your non participation does not make a significant difference (say number of players N=100,000)"

I did not make this assumption and wrote the EXACT EV for the strategic player assuming he was the only strategic player in the game

your non-participation changes the number of players in the game in later rounds

this affects the game differently based on how many players there are to begin with, since your shares vs. their shares scale differently

if there's two other players:

you throw first 5, they throw 10 in that time
then they throw for 5 more and get 10 shares in that time
we all have 25 shares

your share is 5/25 = 1/5 of the total prize pool for $20
you paid 1/3 of $10 five times, so $50/3 = $16.(6)

your profit is $3.33, so 1/5 on top of your original investment

if there's three other players:

you throw the first 5, they throw 15 in that time
they throw 5 more and get 15 more in that time
so total shares is 35, you get 1/7 of the prize pool or almost $14.3

you paid 1/4 of $10 five times, or $12.5
so you made about 14.3% profit or 1/7 on top of our original investment

how do you reconcile these differences under your fractional profit model? When there's a different amount of players, even when the game goes the same way, the profits are DISTRIBUTED differently
this is because the more other players there are, the more each player cuts into the strategic player's shares

so the strategic player could be compensated less than 1.0 even when the fractional is above 1.0
and in smaller games I've found this to be the case, tossing 4 is better than 5 because your fifth share is worth less than the cost

and this is for the player himself, the fractional for all players is still above 1.0, but the EV for one player != fractional for all players

Also, the expected fractional profit for the 5th throw is 1.0, or winnings=costs. Any value above 1.000 is profit, any below 1.000 is a loss. At 1.000 you have neither. The expected fractional profit for throw 5 is 1.00012, so it won't have a likelyhood of producing a profit unless you play thousands of games.

Can you show me what the mistake in the derivation is?

the expected fractional profit does not actually tell you the correct strategy for the game with a lower number of players
this is because the fifth throw has a positive expected fractional profit, but it's a lesser EV than not throwing for the player who is playing strategically

the expected fractional profit ONLY shows the correct strategy as the number of players approaches infinity because of the effect of players gaining shares in later rounds that devaluate your winnings

Expected values for dicecoin: Answers part 2

So here I'll derive the strategy mathematically. I name all variables except number of players and cost as per Analysis of Bitcoin pooled mining reward systems to make the derivation easier to follow. A read of How to hop 7: Expected values will also be helpful.


D = 10, the 'difficulty' of throwing a ten on a ten sided die.
probability of a winning throw, p = 1/D
probability of a losing throw, (1-p) = 1-1/D
I = throw previous to the one being assessed
Y = number of players
N = number of throws
B = total reward per game = 100
C = cost per throw = 10
winnings = 100/(N*Y)
costs = 10/Y per round

Expected profit of the first throw = probability * value

Probabilities:


if last throw = I
probability Nth after Ith throw wins
1/D*(1-1/D)^(N-I-1)

Values:

So winning value Nth throw = B/(N*(Y+1))

Costs:


So the cost value Nth throw = C/(Y+1)

Profit $ value per throw (winnings minus cost)

B/(N*(Y+1)) - C/(Y+1)
= 1/(Y+1)*(B/N-C)


Profit can also be expressed as total income/total outgoings,
eg if it costs $9 to earn $10, the percentage profit is 10/9 = 111.11%, or an 11.11% increase. Expressed as a fraction instead, 10/9 = 1.1111


This allows us to make the number of players irrelevant when determining expected fractional profit of a throw.

Expected fractional profit first throw:

1/D*(1-1/D)^(1-1)*B/C/1 + 1/D*(1-1/D)^(2-1)*B/C/2 + …..

Expected fractional profit second throw:

= 1/D*(1-1/D)^(1-1)*B/C/2 + 1/D*(1-1/D)^(2-1)*B/C/3 + …..

Expected fractional profit Nth throw:



Renaming I (the previous throw) to J for clarity and using wolfram alpha to calculate the first 5 throws:

2.55843, 1.73159, 1.36843, 1.15011, 1.00012

The first twenty throws in terms of fractional profit:




Since the fifth throw is approximately 1.000, a strategy of leaving the game at 4 or 5 throws is equally effective. If you got free drinks per throw, then you'd be better off staying for 5 throws. If instead of drinks there were lots of other tables playing the same game, then you'd be better off leaving at 4 throws - or earlier if another game starts sooner.


I'll post an answer about the expected profits of the game later on, unless someone beats me to it or points out a mistake in my reasoning

You're welcome - glad you enjoyed it. And I don't think you were trying to be clever - you were just interpreting what you that the results told you. A good attempt anyway.

I admitted I was wrong in making that statement, and edited my initial post (post #9) just after iopq's comments. See "EDIT" notes in post #13.

So I missed the prize for trying to play clever  .

Anyway, I did enjoy thinking about it and reading the posts here. Thanks, organofcorti !!!

Expected values for dicecoin - competitions answers

Question 1: is there a winning strategy for this gambling game? What is it?

Both competition entrants iopq and 3phase put a lot more work into this than I expected anyone to make! Good job both of you and I'll make some comments about your results after I give the answer.

I was actually assuming that entrants would simply try the game and see what happens, as I mentioned in one of the hints. If you had tried that you would have obtained something like the following 100 games, which took quite a while using the online dice generator I linked to:


If you weren't going to use probability theory to solve this, the first intuitive leap you had to make was that assessing one throw per game before the costs and prizes were distributed amongst the players seems much easier to assess than trying to work out a per player result (which would depend on how many players were present). If you accept this for now, I'll derive the result mathematically later.

To work out the value of the first throw, we have to find all of the games which contain one or more throws - i.e. all of them. The profit for the first throw of a game before costs and winnings are split amongst players is:
Winnings for the throw minus cost of the throw = $100/(number of throws in the game)-$10


So our average profit (when we define profit as winnings minus cost) for playing only the first throw in a game for this series of games is $15.24


To estimate the value of the second throw, we have to find all of the games which contain two or more throws -

Calculating the average profit as above results in $9.62 per game for the second throw of each game.

Keep in mind that for a constant player, the expected profit is $0 per game.

If we continue calculating the expected profits for a throw as above, we can generate the following results:



When the expected or average value of the throw is below $0, the throw is more likely to reduce your overall winnings than increase them. So for this dataset the strategy would be to stay on for the first 5 throws. I would have accepted this as an answer. It turns out (interestingly) that when the expected values are more accurately calculated the strategy would still be a winner if the 4th throw or the 5th throw is used as the final throw.

Both iopq and 3phase came up with similar results. Although 3phase was first with he later added that any consecutive 5 throws would be a winning strategy (although I can't find that quote in the post anymore, only in the discussion between iopq & 3phase - can you confirm that for me 3phase?)

So, contingent on 3phase confirming his "any consecutive 5 throws" statement, I'm awarding iopq the huge 1btc prize! Congratulations iopq, and well done to both iopq and 3phase.

So I think that's enough for the moment. A little later I'll post a proof for the above, then a discussion of both sets of results from iopq and 3phase, and finally I explain how to calculate the expected profit from the game if you use this strategy.

iopq, please PM me with a bitcoin address to send your prize to.