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teukon
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October 09, 2011, 07:16:31 PM 

Excellent work. I have just invented a simple game for miners. I believe it clarifies the problem with the proportional reward system (at least for people who are more comfortable with words than with formulae). If you think that I've missed the point or made a subtle error then please correct me. Also, feel free to use/extend this as you see fit. This is a 4player game. There is a central pot into which each player must deposit a bitcoin to play. A fair coin is flipped repeatedly. If a tails comes up at any point then the game ends immediately and the 4 players split what is left in the pot evenly. If the coin comes up heads on the first flip then player 1 takes a bitcoin from the pot. After the second head, player 2 takes a bitcoin. After the third head, player 3 takes a bitcoin. After the fourth head, player 4 takes a bitcoin.This game comes with the question: If you were invited to play this game, which player would you like to be?Of course, the idea is that:  Honest 24/7 proportional miners will be happy as player 3 or 4.
 People who have an ethical problem with pool hopping will be disgusted with the game and refuse to have any part in it.
 Pool hoppers will ask to be player 1.




organofcorti
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October 11, 2011, 08:28:59 AM 

Excellent work. I have just invented a simple game for miners. I believe it clarifies the problem with the proportional reward system (at least for people who are more comfortable with words than with formulae). If you think that I've missed the point or made a subtle error then please correct me. Also, feel free to use/extend this as you see fit. This is a 4player game. There is a central pot into which each player must deposit a bitcoin to play. A fair coin is flipped repeatedly. If a tails comes up at any point then the game ends immediately and the 4 players split what is left in the pot evenly. If the coin comes up heads on the first flip then player 1 takes a bitcoin from the pot. After the second head, player 2 takes a bitcoin. After the third head, player 3 takes a bitcoin. After the fourth head, player 4 takes a bitcoin.This game comes with the question: If you were invited to play this game, which player would you like to be?Of course, the idea is that:  Honest 24/7 proportional miners will be happy as player 3 or 4.
 People who have an ethical problem with pool hopping will be disgusted with the game and refuse to have any part in it.
 Pool hoppers will ask to be player 1.
Thanks for posting your game teukon  it kept me busy for quite a while. I like the fact that it's easily understandable, and it's something that makes for a simple thought experiment  and it's clear to see that it's unfair. It's not a good analogy of proportional pool hopping vs full time mining since the sequential payout part = which I think you intend to mimic the skew toward smaller block sizes  has a much too large an effect. But generally it's a good introduction to proportional pool mining, moral judgements aside. It got me to thinking for the last day or so  how could I change your game to be a better analogy of prop pool mining and at the same time keep it a simple thought experiment? The short answer is that I couldn't. But your game did inspire me, and I extended it into a dice game that mimics proportional pool mining quite well. It's no longer a good thought experiment, but I think it would be a good teaching tool. DicecoinDicecoin is a dice game played at the Zero Sum Casino. It uses a 10 sided die, and there can be any number of players. Rules: 1. A die is thrown until a '10' appears. 2. Each throw costs a total of $10, which is split evenly among all players present for the throw. 3. When the '10' appears, the prize of $100 is split proportionally amongst all players, per throw played. 4. Players can leave or join a game at any time. 5. There must be more than one player for a game to continue. Assumptions: 1. Gamblers at the Zero Sum Casino are just as irrational about gambling as anyone else  they will stay in a game that lasts a long time and costing a lot because they want to recoup or at least offset their losses. They also jump into a game in progress because it has been going on for a while and "it's due for a '10' ". 2. More rational players also play to the end of a game because they know that at the Zero Sum Casino game rules are written so that on average their wins = losses. 3. There are never any shortage of players. Example: A game lasts for 20 throws. There are ten players and all players stay until the end of the game and no one else joins. Each throw costs each player: $10 divided by the number of players = $1, for a total cost of $20 per player. Each player wins $100 divided amongst the players proportional to the number of throws they played: $100/10/20 = $0.50 per player per throw played = $10 per player. The total profit = winnings  loss = $10. So for this example, the players lose $10 each. Question: is there a winning strategy for this gambling game? What is it? Extra credit questions: What is the percentage improvement in winnings (winnings for your strategy/winnings for no strategy), and what is your expected profits in dollars per game? Frame your answer in mathematical terms as per How to hop 6: Basic probability for bitcoin miners. So here's a competition for everyone: Post your strategy here, with your reasoning. If you can answer the first question, have the winning strategy and your reasoning is sound, you'll win 1 btc from me. The winner for the extra credit questions will be the poster with the simplest, clearest and most informative explanation, will win 4btc from me. Good chartporn will be helpful but not necessary. Rules:  The winner will be the first poster with a correct answer before midnight 19102011 UTC.
 You can post multiple answers if you think your first answer might be incorrect, but only the most recent answer will be considered.
 Feel free to discuss each other's answers.
I'll post hints here from time to time. A read of How to hop 5: Back to basics and How to hop 6: Basic probability for bitcoin miners might be helpful. Thanks again for such a great idea, teukon!




teukon
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October 11, 2011, 08:56:13 AM 

Cool! My idea was to try to distill the essential problem of the proportional reward system to the point where most people can quickly read it and see the problem as obviously as we do. Unfortunately it has warped so much that I feel most people will see only a loose connection with the proportional reward system. Dicecoin fills a big gap here. You really do have something which closely reflects the proportional reward system but have neatly stripped away talking about hashes, shares, and difficulty (which just confuse the issue). Unfortunately the game is quite a bit more complicated than mine but is certainly more accessible than the proportional reward system and likely will become more so with a little tidying up. I like the idea of phrasing it as a problem but it very much reminds me of 1st year undergraduate probability questions . Some thoughts: Point (2) begins with "Each each". I personally slightly prefer using a 6sided dice for it's ubiquity and for shorter games but I see that 10 allows for simpler costs and rewards. The "5" in $50 and $500 is just noise in my opinion and I'd be tempted to go with $10 and $100. Perhaps some of the wording can be cleaned up to make the game quicker to understand. I only suggest little things like replacing "goes" with "lasts" in "Example: A game goes for 20 throws." Your rewards for solving these problems are generous. I would instead promise to reward particularly elegant and well explained solutions as these have real value. This is however my opinion as a maths tutor and I'm known for being harsh . Best of luck with this. Feel free to use my game on your blog as a simple thought experiment if you like. There is no obligation of course, I merely wish to point out that it should be considered public domain information.




organofcorti
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October 11, 2011, 10:11:47 AM 

Thanks for the props, teukon. Some thoughts:
Point (2) begins with "Each each".
Ooops! Fixed. I personally slightly prefer using a 6sided dice for it's ubiquity and for shorter games but I see that 10 allows for simpler costs and rewards. The "5" in $50 and $500 is just noise in my opinion and I'd be tempted to go with $10 and $100.
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!). Perhaps some of the wording can be cleaned up to make the game quicker to understand. I only suggest little things like replacing "goes" with "lasts" in "Example: A game goes for 20 throws."
I agree. I'll change that. Your rewards for solving these problems are generous. I would instead promise to reward particularly elegant and well explained solutions as these have real value. This is however my opinion as a maths tutor and I'm known for being harsh . That well expressed. It's what I'd hoped for the 'extra credit' answers, but I think I'll use your phrasing. Best of luck with this. Feel free to use my game on your blog as a simple thought experiment if you like. There is no obligation of course, I merely wish to point out that it should be considered public domain information.
I'm already planning a post on mininglike games. Plus 'Dicecoin' is only an extension of your game, so i'm already using it  thanks for making the game public domain.




teukon
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October 11, 2011, 10:37:12 AM 

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.
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.




organofcorti
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October 11, 2011, 12:44:20 PM 

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.
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?




3phase
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October 11, 2011, 01:35:42 PM 

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=0Ar02QJTJDS8dG5wblRYUTVzeUVhaUMtQjlwUXJwWlE&hl=elExplanation 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 (RewardCost)*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. (3.641)/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 poolhopping 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 .




iopq


October 11, 2011, 02:57:12 PM 

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




3phase
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October 11, 2011, 03:50:37 PM 

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




iopq


October 11, 2011, 04:46:36 PM 

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




3phase
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October 11, 2011, 05:12:06 PM 

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
Please have a look again. Maybe my explanation is not good enough, but the spreadsheet shows this. 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 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 20throw round, does something change if I participate in throws 15 or if I participate in throws 1115 (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 1115 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 15). 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 perceptionwise would be to consider a more "strange" strategy whereby a player only plays on throws 1115 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)? Tablehopping 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.




organofcorti
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October 12, 2011, 01:42:40 AM 

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.




3phase
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October 12, 2011, 08:22:26 AM 

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. 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.




organofcorti
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October 12, 2011, 09:21:39 AM 

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.
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.




organofcorti
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October 12, 2011, 12:21:41 PM 

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. 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. 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. 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?




iopq


October 14, 2011, 01:56:52 AM 

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) + ... 50.5 + 9/10*(10*2/41) + (9/10)^2(10*2/51) + (9/10)^3(10*2/61) + ...
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 5050 and negative expected value vs. someone who throws only the first few




iopq


October 14, 2011, 05:16:46 AM 

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




iopq


October 14, 2011, 06:09:46 AM 

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




iopq


October 14, 2011, 06:24:35 AM 

A more faithful game design would be:
Each person pays $1 to throw two tensided 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?




