TTBit
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August 26, 2010, 05:48:27 PM 

While thanking nelisky in a private message, a little math broke out. I'm not a math professor, and would like everyone's opinion. nelisky is running The Amazing Anonymous Bitcoin Lottery (TAABL), and can be found here: http://taabl.datlatec.com/It is a lottery where people bet on the outcome of a hash from certain blocks. Last 3 characters. Meaning 16^3 possibilities or 4096. 1 btc per ticket. My strategy: Bet on every outcome, in fact, it is positive expectation to do so (ignore the 2% charity). Is this a crazy strategy or smart strategy? I'll try to backup my statement that it is a smart strategy, but wanted to get opinions first.

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nelisky
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August 26, 2010, 05:56:22 PM 

My argument is that if a considerable number of people bet less than you, say 200 bets of 20, chances are you'll be dividing the amount you put in with them, thus effectively doing charity for all the winners As a side note, this was part of the decision of letting unclaimed prizes roll over to next draws, giving them higher jackpots or, alternatively, using the full bet pot on the current draw, blending in unclaimed prized with the for for those actually claimed. In the latter approach, betting all possibilities would, I believe, be even more harmful, as in the former at least there's an accumulated jackpot to add as incentive to your bet.




FreeMoney
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August 26, 2010, 07:56:46 PM 

Bet every uncovered number. Your profit comes from other peoples duplicates.

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nelisky
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August 26, 2010, 08:05:31 PM 

Bet every uncovered number. Your profit comes from other peoples duplicates.
Except there's statistical advantage in doing that, unless the number of of 'other betters' is very low, and that defeats the monetary advantage, right? I mean, if there are few uncovered numbers, your investment is small, payoff potentially big, chances of getting it small. If there are many uncovered numbers your investment is that much bigger, chances go up in accordance and the payoff just gets smaller In any case there is absolutely no guaranteed return of investment, mainly because you're not investing It's a bet and what you bet on, how much you bet is only one part of the equation. The other part, what other people do, affect you greatly and you have little or no control over... maybe you can influence other betters, but that's that.




FreeMoney
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August 26, 2010, 08:19:32 PM 

Sorry, I was not clear. You bet on uncovered numbers and profit from when people duplicate each other. If someone lands on you, you take your bet down.

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Insti


August 26, 2010, 08:26:34 PM 

My strategy: Bet on every outcome, in fact, it is positive expectation to do so (ignore the 2% charity).
What if 2 people did this? (You can make anything a positive expectation if you ignore all the downsides.)




nelisky
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August 26, 2010, 08:28:07 PM 

Sorry, I was not clear. You bet on uncovered numbers and profit from when people duplicate each other. If someone lands on you, you take your bet down.
Oh no, you were clear. It's just the unspoken assumption that if you bet on uncovered numbers you'll win (and thus profit) that I find somewhat misleading Fixing your sentence: You bet on uncovered numbers and if you happen to win, and the chances increase with the number of bets you cover profit from when people duplicate each other, knowing that the profit will be reduced and may actually be a loss even when you win as your number of bets increase. If someone lands on you, you take your bet down (while canceling lasts ).




TTBit
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August 26, 2010, 09:30:40 PM 

I'm not suggesting that there is a nolose scenario, but there is a positive expectation solution. Much like getting 3:1 on a coin toss, you might lose $10, but it is still a good bet when you make $30 50% of the time.
Take the simplest of lotteries, a coin toss. You bet 10 BTC on heads and 10 BTC on tails. No matter what someone else does, your EV can not go down. if someone bets 10 btc on heads (total purse 30btc), you now have a 50% chance to return all 30 and a 50% chance at splitting the pot and getting 15 coins. 50% of +10 and 50% of 5 = 2.5 btc edge.
Now take nelisky's lottery:
You bet all 4096 spaces in the lottery, 4096btc. Your opponents bet only half of the lottery, 2048 btc. Total purse: 6144btc
50% chance you win all btc = 6144 (3072 Expected value) 50% chance you split and only receive 3072 (1536 Expected value) Combine is 3072+1536 = 4608 expected coins on a 4096 bet, a 512 btc expected profit. .or. 50% you win the 2048 put in by others (+1024) and 50% you lose 1024 because of the split (512) = +512
You do lose value by sharing a bet, but that value is more than made up when transferred to other bets.
I think my math works out.

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nelisky
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August 26, 2010, 09:41:32 PM 

I think my math works out.
It does, noone is arguing that, I believe. It's your assumptions that are slightly off There was this statistics professor that started the first year class with the following pearl about averages (loosely translated): "Statistics are like a bikini, showing almost everything but hiding the most important... If you and me both go and eat together, I eat two chiken and you eat none, on average we each had one, but I'll be stuffed and you'll be hungry." The error, in my non educated opinion, is one of historical averages. Take the coin toss example: you throw a coin 10 times, 8 of them it falls on tails. What would you say your chances are of hitting tails the 11th throw? Amazingly, they are 50%. Every time 50%. Regardless of how many times you throw them. This is not true for a physical coin, as it may have unbalanced weight or whatever, but still the same applies. You are assuming that if you keep betting all numbers all the time, chances are that your losses and your gains will accumulate to void one another. Each time you bet is a different time, same possibilities regardless of past prize history. It's a gamble, no matter how you dice your numbers (pun obviously intended)... You can minimize you chances of loosing by betting less or not at all, but your chances of winning depend on all the betters, not just on your own actions.




FreeMoney
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August 26, 2010, 09:54:39 PM 

Sorry, I was not clear. You bet on uncovered numbers and profit from when people duplicate each other. If someone lands on you, you take your bet down.
Oh no, you were clear. It's just the unspoken assumption that if you bet on uncovered numbers you'll win (and thus profit) that I find somewhat misleading Fixing your sentence: You bet on uncovered numbers and if you happen to win, and the chances increase with the number of bets you cover profit from when people duplicate each other, knowing that the profit will be reduced and may actually be a loss even when you win as your number of bets increase. If someone lands on you, you take your bet down (while canceling lasts ). Well, you can cancel as long as people can place, so you just have to be quick. Obviously it isn't profitable every time. Is simply positive expected value. Sorry, I don't bother saying expected anymore since I play and post about poker all day. The variance all evens out and the expectation is all that remains after a few decades or whatever. So, yes, you are still gambling, it's just that you can do it with an edge instead of paying an edge. It isn't some magic about covering all numbers either. Every ticket you buy is a better value than the last. The first ticket gets you 1/4096 of winning 1BTC the second is twice as good a value, it gets you a 1/4096 chance at winning 2BTC!

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TTBit
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August 26, 2010, 10:21:00 PM 

I think my math works out.
It does, noone is arguing that, I believe. It's your assumptions that are slightly off There was this statistics professor that started the first year class with the following pearl about averages (loosely translated): "Statistics are like a bikini, showing almost everything but hiding the most important... If you and me both go and eat together, I eat two chiken and you eat none, on average we each had one, but I'll be stuffed and you'll be hungry." The error, in my non educated opinion, is one of historical averages. Take the coin toss example: you throw a coin 10 times, 8 of them it falls on tails. What would you say your chances are of hitting tails the 11th throw? Amazingly, they are 50%. Every time 50%. Regardless of how many times you throw them. This is not true for a physical coin, as it may have unbalanced weight or whatever, but still the same applies. You are assuming that if you keep betting all numbers all the time, chances are that your losses and your gains will accumulate to void one another. Each time you bet is a different time, same possibilities regardless of past prize history. It's a gamble, no matter how you dice your numbers (pun obviously intended)... You can minimize you chances of loosing by betting less or not at all, but your chances of winning depend on all the betters, not just on your own actions. I remember my statistics professor starting off with this well known problem now (before internet): A car is behind 1 of 3 doors. the other 2 doors contain goats. You select a door at random, say door #2. The host (who knows where the car is) shows you a goat behind door #1. He gives you the chance to switch. Is it to your advantage to switch if you want a car? Answer: yes, 2/3 chance car is behind door #3. At the time, I couldn't believe that was true, now I get it. I am assuming that the draws are completely random, and that over time, my gains will overcome my losses by betting every number. This is a nolose *long term* bet, given enough tries, I'll come out, at worst, break even if everyone employs the same strategy. As soon as someone 'takes a shot', I gain edge from him.

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nelisky
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August 26, 2010, 10:34:00 PM 

Well, you can cancel as long as people can place, so you just have to be quick.
Obviously it isn't profitable every time. Is simply positive expected value. Sorry, I don't bother saying expected anymore since I play and post about poker all day. The variance all evens out and the expectation is all that remains after a few decades or whatever.
I get all that, the tendency to balance. It just doesn't apply on a small timescale (within a few years small) when you get one draw per week. It might, and it might just have you loose every single time. You play poker, you know about probabilities, and yet it's not the math wizard that makes it to the high stakes tables, it's the guy who can read other people's behavior and uses that to get some leverage. So, yes, you are still gambling, it's just that you can do it with an edge instead of paying an edge. It isn't some magic about covering all numbers either. Every ticket you buy is a better value than the last. The first ticket gets you 1/4096 of winning 1BTC the second is twice as good a value, it gets you a 1/4096 chance at winning 2BTC!
That is just wrong. Your bets go to the pot in full, but you don't take the whole pot if you win. Poker? sure, but not lottery. The first ticket gives you a chance of winning, at best, 0.5BTC, and it did cost you 1BTC. The second? in a perfect world an extra 0.24, so you are not even making up for the initial bet. It only levels out because many people play and, if you were to make no duplicates, the bets would be spread so well that prize distribution would suck even more. But I get it, eventually you'll bet, as well as many others and only you will win, making up for all the other losses. I say go for it! The more the merrier...




FreeMoney
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August 26, 2010, 11:08:39 PM 

Look, I can work out the numbers, but it's a pain so I simplified to a winner take all lottery.
When you buy the first ticket you have a (1/4096 * .5) + (1/256 * .25) + (1/16 * .25) = 0.016723633
When someone buys a second ticket it is worth (1/4096 * 1) + (1/256 * .5) + (1/16 * .5) = 0.033447266
Once there is one ticket on every number (1/4096 * 2048) + (1/256 * 1024)/16 + (1/16 * 1024)/256 = 1
Now if someone buys a duplicate that doesn't match your ticket in the rightmost digit:
(1/4096 * 2048.5) + (1/256 * 1024.25)/16 + (1/16 * 1024.25)/256 = 1.000244141
And now we have positive expected value.
Each ticket now has that much value, so the order doesn't matter. But people may still be wary of buying the first very low value tickets. I'm not saying they won't find other reasons to do it, but they are essentially donating value to future draws.

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FreeMoney
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August 26, 2010, 11:09:34 PM 

That is just wrong. Your bets go to the pot in full, but you don't take the whole pot if you win. Poker? sure, but not lottery. The first ticket gives you a chance of winning, at best, 0.5BTC, and it did cost you 1BTC. The second? in a perfect world an extra 0.24, so you are not even making up for the initial bet. It only levels out because many people play and, if you were to make no duplicates, the bets would be spread so well that prize distribution would suck even more.
But I get it, eventually you'll bet, as well as many others and only you will win, making up for all the other losses. I say go for it! The more the merrier...
I'm well aware of the math involved in splitting pots, that does happen in poker.

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nelisky
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August 26, 2010, 11:26:23 PM 

Look, I can work out the numbers, but it's a pain so I simplified to a winner take all lottery.
When you buy the first ticket you have a (1/4096 * .5) + (1/256 * .25) + (1/16 * .25) = 0.016723633
When someone buys a second ticket it is worth (1/4096 * 1) + (1/256 * .5) + (1/16 * .5) = 0.033447266
Once there is one ticket on every number (1/4096 * 2048) + (1/256 * 1024)/16 + (1/16 * 1024)/256 = 1
Now if someone buys a duplicate that doesn't match your ticket in the rightmost digit:
(1/4096 * 2048.5) + (1/256 * 1024.25)/16 + (1/16 * 1024.25)/256 = 1.000244141
And now we have positive expected value.
Each ticket now has that much value, so the order doesn't matter. But people may still be wary of buying the first very low value tickets. I'm not saying they won't find other reasons to do it, but they are essentially donating value to future draws.
Don't get me wrong here, you obviously know way more than I do about this, but just indulge me for a little longer if you can. You say buy every ticket not taken so all dups are working for you, but then you show math where you buy *all* the tickets and only one other is bought. On top of that you assume the one duplicate ticket is not the winner. This is where I think things go crazy, as we're talking about optimal situations which are not, I believe, remotely real. If you don't buy every ticket then the chances of ever winning the prize get substantially smaller and not every ticket is the same. There's 1 change in 16 for everybody to share 25% of the pot, and 1 in 256 of sharing another 25%. So if I buy 256 tickets wisely chosen to cover all 2 digit combinations, I have secured 15 64BTC in 3rd prizes and 68BTC in 1 2nd. Not enough to cover my bet, so you actually make a profit unless I hit first prize, which I have a 4% chance of doing... I just proved your point, didn't I? Damn... So what can I do to deter this? add a number?




FreeMoney
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August 26, 2010, 11:35:08 PM 

Okay, I am not saying that you are guaranteed to profit buy buying all empty tickets. I am saying that it has positive expected value.
If I offer to flip a coin with you and pay you $10 and heads and take $1 from you on tails you will on average profit by $4.5 per flip. You will never actually win $4.5 in a flip, that is just the probability weighted average of the outcomes. The exact odds of having a losing streak of 100 flips, or of being down after 17 flips or whatever is not important here.
The "problem" with this is not that buying all the empties is profitable, it is that buying a few tickets in a mostly empty field is very unprofitable. The easiest way to fix this is to make sure all money is awarded each draw (or with a sudden death, which sounds fun). I know it kills the growing jackpot though, which would be fun to have.

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FreeMoney
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August 27, 2010, 12:18:58 AM 

My math is off because the winner doesn't get runner up prizes. This goes both ways though because when someone else wins you'll get extra on all of your runner ups because he doesn't get a share. It changes things, but not the main idea.

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Insti


August 27, 2010, 06:54:56 AM 

Look, I can work out the numbers, but it's a pain so I simplified to a winner take all lottery.
When you buy the first ticket you have a (1/4096 * .5) + (1/256 * .25) + (1/16 * .25) = 0.016723633
Can you explain this formula more, what are you computing? what are the 0.5 and 0.25 numbers ?




FreeMoney
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August 27, 2010, 07:32:08 AM 

Look, I can work out the numbers, but it's a pain so I simplified to a winner take all lottery.
When you buy the first ticket you have a (1/4096 * .5) + (1/256 * .25) + (1/16 * .25) = 0.016723633
Can you explain this formula more, what are you computing? what are the 0.5 and 0.25 numbers ? When the first ticket is bought the pot is 1 so matching all 3 wins you .5 matching the rightmost 2 gets you .25 and matching the right digit gets you .25. To be accurate I should have reduced the odds on matching 1 and 2 because in order to win those prizes you have to also not match all 3. (1/4096 * .5) + ((1/256  1/4096) * .25) + ((1/16  1/256) * .25) = .015686 is the expected value of the first ticket. The formula for future tickets gets even more complicated because of the stairstepping of dupes. The number of tickets with the same rightmost digit as the Nth ticket (including that ticket) is floor(x/16 + 1) Floor meaning the largest integer less than. so the whole formula for the value of the Nth ticket is: ((1/4096) * (N/2)) / floor(N/4096 + 1) + ((1/256) * (N/4)) / (floor(N/256 + 1)) + ((1/16) *(N/4)) / (floor(N/16 + 1)) Oh, and all the .25s should be .24s if you consider the bounty.

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semyazza


August 28, 2010, 08:27:11 PM 

Okay, I am not saying that you are guaranteed to profit buy buying all empty tickets. I am saying that it has positive expected value.
If I offer to flip a coin with you and pay you $10 and heads and take $1 from you on tails you will on average profit by $4.5 per flip. You will never actually win $4.5 in a flip, that is just the probability weighted average of the outcomes. The exact odds of having a losing streak of 100 flips, or of being down after 17 flips or whatever is not important here.
The "problem" with this is not that buying all the empties is profitable, it is that buying a few tickets in a mostly empty field is very unprofitable. The easiest way to fix this is to make sure all money is awarded each draw (or with a sudden death, which sounds fun). I know it kills the growing jackpot though, which would be fun to have.
Just to throw in my .02BTC I love skalansky :D... His book "The Theory of Poker" covers this scenario perfectly. Also "Gambling for a Living" and "Getting the Best of it".




