Topic: "How to hop" has moved - page 7. (Read 16281 times)

Great game - much more faithful to bitcoin, and a lot simpler. Wish I'd thought of it.

"Dicecoin" has a few things added in to make it look more difficult than it really is, so while it makes a good exercise in calculating expected values, it's not too easy to follow for someone who is not familiar with bitcoin already. But without having checked your game out properly, it's much more faithful to teukon's original idea for a game, teaching people about how bitcoin mining works.

Just a note on your question phrasing, I think it's better to ask if there *is* a strategy and then ask what it is, rather than to explain what the strategy is and to ask what the variable should be. You're giving away the fact there is one.

For example, if the payout was paid proportionally only to the people making the last ten throws - including the ones in the last game, if it's a short game - is there a winning strategy and what would it be? (no prizes for that answer!)

Nice layout iopq, you explained your thinking so I could follow what you were doing. I noticed that you are assuming the correct strategy is to leave the game at a particular throw. Wouldn't it be better to start from the point of view of not knowing what the strategy is and instead calculate the expected value of each throw?

While I'm at it, here's the third hint for the competition: There's a way of finding the expected profit for a particular throw that makes  the number of players irrelevant.

I am, that's why I divide everything by ten (1/10 probability of success) and multiply each round by 0.9 (which 9/10 probability of fail on previous rounds)

so my summation goes like this

1/10*P(1) + 9/10*1/10*P(2) + (9/10)^2*1/10*P(3) etc.

A more faithful game design would be:

Each person pays $1 to throw two ten-sided dice. When the two dice come up snake eyes (1 and 1 on both dice, which happens 1/100 of the time) the $100 prize is shared by everyone who threw dice in this game proportional to the amount of throws they've made. After how many throws since the game started should you stop throwing, given that the game will keep on going without you?

four player game one toss strategy

toss 1 profit: 100*1/(1+3) - 2.5 = 22.5
toss i(=n+1) profit: 100*1(1+3i) - 2.5 = 100*1(1+3+3n) - 2.5
EV = 2.25 + Sum[(9/10)^n((10*1/(1+3+3n)) - 1/4),{n, 1, Infinity}] ~= 4.76

two toss strategy
toss 1 profit: 100*1/(1+3) - 2.5 = 22.5
toss 2 profit: 100*2/(2+3+3) - 5 = 20
toss i profit: 100*2(2+3+3n) - 5
EV = 2.25 + 1.8 + Sum[(9/10)^n((10*2/(2+3+3n)) - 2/4),{n, 2, Infinity}] ~= 6.55

three toss strategy
toss 1 profit: 100*1/(1+3) - 2.5 = 22.5
toss 2 profit: 100*2/(2+3+3) - 5 = 20
toss 3 profit: 100*3/(3+3+3*2) - 7.5 = 17.5
toss i profit: 100*3(3+3+3n) - 7.5
EV = 2.25 + 1.8 + 1.4175 + Sum[(9/10)^n((10*3/(3+3+3n)) - 3/4),{n, 3, Infinity}] ~= 7.29

four toss strategy
toss 1 profit: 100*1/(1+3) - 2.5 = 22.5
toss 2 profit: 100*2/(2+3+3) - 5 = 20
toss 3 profit: 100*3/(3+3+3*2) - 7.5 = 17.5
toss 4 profit: 100*4/(4+3+3*3) - 10 = 15
EV = 2.25 + 1.8 + 1.4175 + 1.0935 + Sum[(9/10)^n((10*4/(4+3+3n)) - 4/4),{n, 4, Infinity}] ~= 7.51

five toss strategy
toss 1 profit: 100*1/(1+3) - 2.5 = 22.5
toss 2 profit: 100*2/(2+3+3) - 5 = 20
toss 3 profit: 100*3/(3+3+3*2) - 7.5 = 17.5
toss 4 profit: 100*4/(4+3+3*3) - 10 = 15
toss 5 profit: 100*5/(5+3+3*4) - 12.5 = 12.5
EV = 2.25 + 1.8 + 1.4175 + 1.0935 + 0.820125 + Sum[(9/10)^n((10*5/(5+3+3n)) - 5/4),{n, 5, Infinity}] ~= 7.43

again, the result is four toss strategy is best, but again our edge is lower, and we pay less so we win less
first toss is 190% efficiency, second toss is 172% efficiency, third toss is 130% efficiency, fourth toss is 122% efficiency, fifth toss is 97% efficiency

let's analyze what the players who enter the game "when it's due for a 10" do for you
they actually just act to decrease the contribution of the "play from beginning to end" players because they will pay less part of the fee for the round, but they also get a share so they cut into your profits if you're no longer putting any more money in while not "helping" to get a 10
this is different from late hoppers in bitcoin mining because those people actually let the pool find a block faster

now let's calculate the same for a three player game because there's a constraint that there has to be more than one player for the game to continue

player 1 pays $3.(3)
player 2 pays $3.(3)
player 3 pays $3.(3)

let's calculate the EV of the one toss strategy by player 1

toss 1: paid 3.(3), wins 33.(3), profit 30
toss n: paid 3.(3), profit 100/(1+2+2n) - 3.(3)

EV = 3 + Sum[(9/10)^n((10*1/(1+2+2n)) - 1/3),{n, 1, Infinity}] ~= 6.85

two toss strategy
toss 1: paid 3.(3), wins 33.(3), profit 30
toss 2: paid 6.(6), wins 33.(3), profit 26.(6)
toss n: paid 6.(6), profit 100/(2+2+2n) - 6.(6)

EV = 3 + 2.4 + Sum[(9/10)^n((10*2/(2+2+2n)) - 2/3),{n, 2, Infinity}] ~= 9.32

three toss strategy
toss 1: paid 3.(3), wins 33.(3), profit 30
toss 2: paid 6.(6), wins 33.(3), profit 26.(6)
toss 3: paid 10, wins 33.(3), profit 23.(3)
toss n: paid 10, profit 100*3/(3+2+2n) - 10

EV = 3 + 2.4 + 1.89 + Sum[(9/10)^n((10*3/(3+2+2n)) - 1),{n, 3, Infinity}] ~= 10.3

four toss strategy
toss 1: paid 3.(3), wins 33.(3), profit 30
toss 2: paid 6.(6), wins 33.(3), profit 26.(6)
toss 3: paid 10, wins 33.(3), profit 23.(3)
toss 4: paid 13.(3), wins 33.(3), profit 20
toss n: paid 13.(3), profit 100*4/(4+2+2n) - 13.(3)

EV = 3 + 2.4 + 1.89 + 1.458 + Sum[(9/10)^n((10*4/(4+2+2n)) - 4/3),{n, 4, Infinity}] ~= 10.5

five toss
toss 1: paid 3.(3), wins 33.(3), profit 30
toss 2: paid 6.(6), wins 33.(3), profit 26.(6)
toss 3: paid 10, wins 33.(3), profit 23.(3)
toss 4: paid 13.(3), wins 33.(3), profit 20
toss 5: paid 16.(6), wins 33.(3), profit 16.(6)
toss n: paid 16.(6), profit 100*5/(5+2+2n) - 16.(6)
EV = 3 + 2.4 + 1.89 + 1.458 + 1.0935 + Sum[(9/10)^n((10*5/(5+2+2n)) - 5/3),{n, 5, Infinity}] ~=10.4
which is less than the previous strategy, so four throws is better

first throw has 206% efficiency, second throw has 174% efficiency, third throw has 129% efficiency, fourth throw has 106% efficiency, fifth has below 100%

notice that it costs us less to play, but the three player scenario is worse because we put in less money overall and because our edges are smaller

with two players the game goes as following:

player 1 pays $5
player 2 pays $5

then player 1 leaves the game, player 2 keeps playing
if the game ends on the...

first toss
player 1 gets 1/2, pays $5, so wins $50
same for player 2

second toss
player 1 gets 1/3, pays $5, wins $33.(3)
player 2 pays $15, gets 2/3, wins 66.(6)

nth toss
player 1 gets 1/(n+1), pays $5, wins 100/(n+1)
player 2 pays $10n + $5, gets n/(n+1), wins 100n(n+1)

there's 10 outcomes, 1 of which is a win, and 9 of which is wait until next toss
player 1 - win for the nth toss

win(n) =  (100/(n+1) + 9*win(n+1))/10 = 10/(n+1) + 9/10*(win(n+1))
n starts at 1

so win(1) = 5 + 9/10*(win(2))
win(2) = 3.(3) + 9/10*(win(3))

10/2 + 9/10*(10/3) + (9/10)^2 * (10/4)
5 + 3 + 2.25 + ...

win(n) = (9/10)^n * (10/n+2)
Sum[(9/10)^n * (10/(n+2)),{n, 0, Infinity}] = 100/81 (-9+10 log(10)) ~= 17.3

so he profits a little over 12.3 dollars every time he plays if he only plays the first game which is 246% profit!

if he plays the first two:
toss 1: pays 5, wins 50
toss 2: pays 10, wins 50
toss 3: paid 10, wins 2/5 * 100 = 40
toss 4: paid 10, wins 2/6 * 100 = 33.3
toss n: paid 10 if n=>2, win 2/(n+2) * 100

but now let's include profit calculations into the series because the first time we only paid 5 but second time ten
1/10(50 - 5) + (9/10)*1/10 (50 - 10) + (9/10)^2 * 1/10 (40 -10) + ...
5-0.5 + 9/10*(10*2/4-1) + (9/10)^2(10*2/5-1) + (9/10)^3(10*2/6-1) + ...

profit(n) = 4.5 + (9/10)^n(10*2/(n+3) - 1)
Sum[(9/10)^n(10*2/(n+3) - 1),{n, 1, Infinity}] ~= 11.7
11.7 + 4.5 = 16.2 which is higher than 12.3, but we had to invest more money on the subsequent throws, but still a higher EV, so throw at least twice

first three:
toss 1: paid 5, wins 50, profit 45
toss 2: paid 10, wins 50, profit 40
toss 3: paid 15, wins 50, profit 35
toss 4: paid 15, wins (3/7 * 100), profit 3/7 * 100 - 15

4.5 + 9/10(4) + (9/10)^2(3.5) + (9/10)^3(10 * 3/7 - 1.5) + ...
n>=3, profit(n) = (9/10)^n(10*3/(n+4) - 1.5) + 4.5 + 9/10(4) + (9/10)^2(3.5)
Sum[(9/10)^n(10*3/(n+4) - 1.5),{n, 3, Infinity}] + 4.5 + 9/10(4) + (9/10)^2(3.5) ~= 17.6

which is higher than 16.2, so we will keep going
first four:
toss 1: paid 5, wins 50, profit 45
toss 2: paid 10, wins 50, profit 40
toss 3: paid 15, wins 50, profit 35
toss 4: paid 20, wins 50, profit 30
toss 5: paid 20, wins (4/9 * 100), profit 4/9 * 100 - 20

4.5 + 9/10(4) + (9/10)^2(3.5) + (9/10)^3(3) + Sum[(9/10)^n(10*4/(n+5) - 2),{n,4, Infinity}] = 17.7

keep going

first five:
4.5 + 9/10(4) + (9/10)^2(3.5) + (9/10)^3(3) + (9/10)^4(2.5) + Sum[(9/10)^n(10*5/(n+6) - 2.5),{n,5, Infinity}] ~= 17.3

we could keep going, but 5 is worse than 4, so the solution is stay for the first 4 tosses vs. one player who keeps on playing

so we get the EVs here:
1 toss: 12.3
2 tosses: 16.2
3 tosses: 17.6
4 tosses: 17.7
so the first throw has the expectation of 17.3 (346% expected), second throw has the expectation of 8.9 (178% expected), third throw has the expectation of 6.8 (136% expected), fourth throw has the expectation of 5.1 (102% expected) and all throws afterwards have below 100%

throwing every time gives you 0 expected value vs. an every time thrower as you just split the winnings 50-50 and negative expected value vs. someone who throws only the first few

You did have a good idea about how to solve it though, only your expected value calculations were wrong. 'How to hop 7' will be done soon and might be helpful.


The solution is actually very interesting. Check out How to hop part 1: Slush's pool, How to hop part 2: More on score., and How to hop 4.

Now, back to the competition - ready for another try?

Try using this online dice generator to generate data. Or this webpage shows you how to generate a bunch of geometrically distributed random numbers in Excel. Give it a try. 50 games worth of data will be enough to allow you check the accuracy of your strategy to your own satisfaction.

It only took half an hour with Excel. My probability 101 class was 25 years ago, so I don't remember much, but I can still understand certain things.  And yes, a simulator would be best, but it is beyond my means.

Thanks.

We have our first entrant! Looks like you did a whole lot of hard work there 3phase.

I have clarified the rules of the competition, see the competition post.

2nd hint for the competition - if you don't have the probability theory chops to to answer the first question and if you have some time, a 10 sided die(or online dice generator), a pencil and lots of paper you should be able to answer this question by playing the game and recording the results.

I would probably just generate lots of rolls and then calculate results for different strategies you think might work. Post your best strategy and the results and working out and see if you win!

If you do have the enough knowledge of probability to calculate results, I would probably check your conclusions using dice and a pencil, or a simulator if you can make one.

EDIT: I will leave the post as is, as it might be useful for discussion, but I just want to say that my conclusion is wrong. iopq is correct. If you enter in the 11th throw, the probability of the next throw rolling a 10 is 1/10 as stated, but your expected reward is much smaller, because 10 throws have already been played.

This says to me that for pure prop pools, only hopping at the beginning of the round is beneficial. It also however points to a curious tweak with score based pools such as Slush, where the value of the previous shares is gradually diminished to zero over time, and might provide an opportunity to hop there, even later than the start of the round. I am currently thinking on this.

I also edited my first post to remove the mistaken conclusion

ORIGINAL POST:

Please consider the simple question for this simplified game that was proposed:

I play for 5 throws in a round with 20 throws. I invest 5 units of money. I receive as a reward (10/20)*5=2.5 units according to my initial assumption. Of course if the round has only 10 throws to roll a 10, I will receive 5 units (as much as I had invested).

But on the 20-throw round, does something change if I participate in throws 1-5 or if I participate in throws 11-15 (after having seen the first 10 unsuccessful throws)? No, it doesn't. The reward is the same.

My point is that if you get to the gaming table during a round which already has 10 unsuccessful throws, this fact does not affect the probabilities of the next throws at all due to the independence of events. So you are perfectly fine to follow the proposed strategy (playing for the next 5 throws, throws 11-15 that is) and if the round ends with 20 throws you receive the same amount of reward as someone who followed the same strategy from the beginning of the round (playing in throws 1-5).

You cannot know in advance the number of throws that will be needed. For the first two throws, the probability of rolling a 10 is 0.09.

If there have already been 10 unsuccessful throws in a round, the probability of rolling a 10 in the next two throws (throw 11 and throw 12) is again 0.09.

What might also help perception-wise would be to consider a more "strange" strategy whereby a player only plays on throws 11-15 every round. This strategy still has positive expected value, but it would be pointless, as they will miss many rounds (65% of them actually) which will be shorter than 10 throws, meaning that they don't exploit the positive expectancy of the strategy fully. But what if there were another 20 tables around at different stages of the round (in terms of number of throws)? Table-hopping would enable a player to fully exploit the positive expectancy if his strategy, regardless of his starting throw number.

Sorry, I give up, I can't explain this in English any better.

after 10 throws total the person with 5 throws gets a better share of the pot than at 20, so your conclusion that 5 any consecutive trials is best cannot be possibly true

Please have a look again. Maybe my explanation is not good enough, but the spreadsheet shows this.

3phase: you misunderstood the game
the prize is distributed per amount of throws you play

so after 20 throws, the people who played 20 will get a larger share than the person who played 5

Winning strategy:

Play five throws per game only, starting at the beginning of a round and then stop until a 10 is rolled. Repeat.

Assumptions:

A. 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)

B. Based on the first assumption, the reward is 100/N and your cost per round is 10/N, therefore the reward is 10 times your cost per round.

In the link below, you will find an spreadsheet detailing the results that a player would expect to get if they would stay in for a number of throws (1 to 10 throws) and then leave and the results they would get if they would stay and play throughout the game without leaving for any throw. There are therefore 11 sheets (for 1 throw, for 2 throws, ..., for 10 throws, for all throws) which are 11 different strategies.

https://docs.google.com/spreadsheet/ccc?key=0Ar02Q-JTJDS8dG5wblRYUTVzeUVhaUMtQjlwUXJwWlE&hl=el

Explanation for the sheet :

1. Column 1 is the number of throws until a 10 is rolled, distinct possibilities. I have put numbers up to 50 throws, the reason will become clear in column 3

2. Column 2 is the probabiltity of this number of throws happening. The probability of rolling a 10 after 1 throw is 1/10. The probability of rolling a 10 after exactly 2 throws is (9/10)*(1/10) that is, anything else but a 10 is rolled in the first throw and a 10 is rolled in the second. The probability of rolling a 10 after exactly 3 throws is ((9/10)^2)*(1/10) and so on

3. Column 3 is the accumulated probability of rolling a 10 with UP TO the specified number of throws. So, the accumulated probability of throwing a 10 after at most 2 throws is (1/10)+((9/10)*(1/10)) which is the probabilty of the first throw rolling a 10 and the second throw rolling a 10, as calculated in column 2. As seen on the tables, the probability of rolling a 10 after at most 50 throws is about 99.5%, so this is why I stopped it there and did not calculate further results for a greater number of throws

4. Column 4 is the cost to the player if they play only for the number of throws which is referred in the Sheet name. So for the "1 throw" sheet, the player only plays during one throw. Notice for example that for the "4 throws" sheet, the cost increases by 1 unit up to the 4th throw and then remains the same. The cost of one throw is calculated as 1 unit, because, given the assumption B above, the reward can be very easily calculated as 10 units, regardless of the actual amount and the number of players present at the game.

5. Again according to assumption B above, the reward is calculated as follows: (Total reward of 10 units for each player) divided by (number of throws until a 10 is rolled) and then multiplied by the number of throws in which the player has participated (which is exactly his cost for these number of throws in cost units).

6. Finally in column 6 (or more specifically in the Sum of column 6 which is found above the table in each sheet) the truth is revealed: Multiplying the probability of a number of throws until a 10 is rolled by the difference between Reward and Cost for this number of throws and this particular strategy gives the expected value in pure profit of each outcome. In short (Reward-Cost)*Probability. Obviously for a round which takes 10 throws to roll a 10 the expected value is 0 which means everybody gets exactly their money back. For a round greater than 10 throws the expected value is negative, and for a round with less than 10 throws the expected value is positive.

Now look at the number at the top right of each sheet which sums up all the expected values of the possible outcomes (1 to 50 throws to roll a 10). For the player participating in all throws the expected value converges to 0 after a long number of rounds (if I had calculated the results up to 200 throws it would be more obvious).

The best number is given by the strategy of playing 5 throws and then leaving. In this case the expected value for a large number of rounds is 2.64 which means that for every unit of betting currency you get 2.64 units as a profit.

To answer the extra credit question, with typical calculations:

A player with no strategy that just stays in for all throws and all rounds until he gets bored, can expect to receive 1 unit for every 1 unit that they play with. 0% return

A player playing the optimal strategy can expect to receive 3.64 units for every 1 unit that they play with. (3.64-1)/1=2.64 = 264% return. In other words, he gets an improvement of 264% over having no strategy.

Really enjoyed the challenge and the analysis, which proved the basic pool-hopping premise also to myself. I suppose that taking this analysis to the Bitcoin mining conditions, a similar result can be made plain, with much larger numbers of course.



Feel free to use the spreadsheet if you find it useful. And thanks for the very informative posts you've done so far.

I just hope I win this competition, as I have little luck mining lately  .

Interesting idea. I think it's doable - I already have multipool sims that can calculate efficiency of a hopped round, and the difference in D (at current nmc D) isn't a problem.  It would have to be vanilla, because like you say the extraneous stuff is just annoying and it changes from time to time so I'd just ignore it.

At the current namecoin D the 1.0 efficiency share should still be at 0.43xD. The only difference is the expected value of a share. The number of rounds and reward per round would be built into the share value. Then you have to make a function relating btc share value to nmc share value which would allow you to hop at the right time. I already made a function for determining the best time to hop from prop to slush and back, so i think a similar function for nmc to btc might work.

But I don't really know enough about it yet, and there's bound to be complications and issues I haven't thought of. If enough people want a strategy for merged mining I'll spend some time on it - unless you come up with an answer first?

Ah! I see.

One thing I've been wondering about this morning is pool hopping with merged mining.  Calculating when it is profitable to mine in a proportional pool taking into account the Bitcoin and Namecoin blocks is an interesting extension.  One could also try factoring in fees and promotions but I don't think this would be as much fun.

Thanks for the props, teukon.

Ooops! Fixed.


I actually started with a six sided die but I saw that there were so many online dice generators I figured people wouldn't have to brave their local D&D store to try some experiments if I used a 10 sided die. I used ten sided because, as you say, it makes working with the mean values easier. The $500 vs $50 started as the 50 btc reward we're used to, but I had to make the game seem a bit more enticing so I just increased it by an order of magnitude. $100 vs $10 would be better, and might distract people less - important since the actual reward doesn't even matter for the first question (hint here, folks!).


I agree. I'll change that.

That well expressed. It's what I'd hoped for the 'extra credit' answers, but I think I'll use your phrasing.


I'm already planning a post on mining-like games. Plus 'Dicecoin' is only an extension of your game, so i'm already using it - thanks for making the game public domain.