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EVE Corporation - Page 1071

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https://discord.gg/c8jHgQpMSY

mity hat tree discord if you care
Firebolt145
Profile Blog Joined May 2010
Lalalaland34491 Posts
April 18 2012 20:16 GMT
#21401
wtf has this thread become
Moderator
Impervious
Profile Blog Joined March 2009
Canada4200 Posts
April 18 2012 20:17 GMT
#21402
On April 19 2012 05:04 KwarK wrote:
Show nested quote +
On April 19 2012 04:51 Impervious wrote:
On April 19 2012 04:15 KwarK wrote:
On April 19 2012 02:27 abominare wrote:
Ignoring payout value differences (you take a nice cut for cash or a smaller cut liquidating yourself via npc taxes). if you bought 9 tickets you would have about a 56% chance of turning 378m into 525m. So in theory if you had an infinite amount of capital to run an infinite amount of runs, you would technically make money of the long run. Of course you don't have infinite money so you always run the risk of simply hitting a string values in which you would lose all your capital.

-_-
I invested in you. Now I feel bad.

Why? He's right.

Here's an example showing how you could always come ahead if you had infinite money:

At a roulette table, you can bet on black or red. Neither of the options has a 50% chance of winning (since 0 is green), however, the payout for you being right is 2 times your bet. In the long run, it sounds like it is impossible to win money, right?

Well, try this. Put a 1 dollar bet on red. If you win, you collect the 1 dollar and start again with 1 dollar.

If you lose, place a new bet on red for 2 dollars. If you win, you collect 2 dollars. Subtracting the loss from the first bet (1 dollar), you have a net gain of 1 dollar.

If you lose again, place a new bet on red for 4 dollars. If you win, you collect 4 dollars. Subtracting the losses from the first two bets (1 and 2 dollars respectively), you get a net gain of 1 dollar.

If you lose again, place a new bet on red for 8 dollars. If you win, you collect 8 dollars. Subtracting the losses from the first three bets (1, 2, and 4 dollars respectively), you get a net gain of 1 dollar.

If you lose again, place a new bet on red for 16 dollars. If you win, you collect 16 dollars. Subtracting the losses from the first four bets (1, 2, 4, and 8 dollars respectively), you get a net gain of 1 dollar.

If you lose again, place a new bet on red for 32 dollars. If you win, you collect 32 dollars. Subtracting the losses from the first five bets (1, 2, 4, 8, and 16 dollars respectively), you get a net gain of 1 dollar.

Repeat as many times as it takes to win. As long as you have infinite cash backing you, you actually can come out ahead. There is more complicated mathematics that can show this as well, but this is a simple way of explaining it.

So, while theoretically it is a fucking terrible idea to gamble, since the odds are stacked against you, if you have an infinite source of cash backing you, you can beat the odds. There is mathematics that actually backs this up in a more precise way (I studied it for a semester), but I forget a lot of that stuff already..... Large numbers can be really, really weird :/

Premise
That if you have infinite bets then after gambling enough money at a negative expected value you will still have infinite money.
Problem
All you have effectively proved is that 1/10th infinity is still infinity

The bet is always negative EV, each time you take it you lose EV. What you've done is thrown infinity into the equation to counter this but it in no way changes the fact the infinity money you get out is of a lower order than the infinity you started with.

EV has absolutely nothing to do with the example I just gave.

You could be rolling a 20 sided dice each time, with a 1:1 payout plan, yet still eventually come out ahead.
~ \(ˌ)im-ˈpər-vē-əs\ : not capable of being damaged or harmed.
ghost_403
Profile Joined September 2010
United States1825 Posts
Last Edited: 2012-04-18 20:33:08
April 18 2012 20:25 GMT
#21403
Holy hell, you guys are retarded.

In this scenario, each ticket is (effectively*) independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

edit: Proving that it's effectively independent:

+ Show Spoiler +

n = number of tickets you buy
N0 = total tickets for one drawing
P0 = cost of one ticket
eta = efficiency of blink

cost of playing : P0*n
payout : N0*P0*eta
chance of winning : n/N0

average winnings for one ticket: 1/N0*N0*Po*eta - P0 = P0*eta - P0

average winnings : (chance of winning)*(payout) - (cost of playing)
: (n/N0) * (N0*P0*eta) - n*P0
: n*P0*eta - n*P0 = n*(P0*eta-P0)
: n*(average winnings for one ticket), which implies independence
They say great science is built on the shoulders of giants. Not here. At Aperture, we do all our science from scratch, no hand holding. Step aside, REAL SCIENCE coming through.
KwarK
Profile Blog Joined July 2006
United States42691 Posts
April 18 2012 20:30 GMT
#21404
On April 19 2012 05:25 ghost_403 wrote:
Holy hell, you guys are retarded.

In this scenario, each ticket is independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

I believe they get around this by escalating the amount bet on each subsequent lottery by a huge amount, thus rendering every previous bet completely irrelevant should they win. Of course if there was a maximum bet, which there is, then this would be impossible. The maximum bet on blink is a lottery on the highest value item they offer and once you hit that and lose you can't do a lottery on two of them to try and make it back.
Once you can no longer escalate the odds and render the past history irrelevant then you need a streak good enough to undo all the previous losses. If you take the "but infinity" argument here and make the case that after infinite time you'll get a really long winning streak then you would be correct but you'd also have been losing EV for the infinite time before then. You still lose.
ModeratorThe angels have the phone box
Ame
Profile Joined October 2009
United States246 Posts
Last Edited: 2012-04-18 20:43:02
April 18 2012 20:31 GMT
#21405
I thought casinos didn't care about it since they have betting limits :x

If still unconvinced about things or curious and enjoy reading wiki,
http://en.wikipedia.org/wiki/Martingale_(betting_system)

Assuming that one trusts wiki,
"It is only with unbounded wealth, bets and time that the martingale strategy can succeed."

And oops,
+ Show Spoiler +
"Contrary to popular belief, table limits are not designed to limit players from exploiting a Martingale strategy. Instead, table limits exist to reduce the variance for the casino. For example, a casino which wins an average of $1000 a day on a given roulette table might not accept a $7000 bet on black at that table. While that bet would represent a positive expectation of over $368 (10/19 * 7000 - 18/38 * 7000 = 368.42) to the casino, it would also have a 47.37% chance of negating an entire week's profit."


More derail:
You can have an positive infinite expected value (I think you can use one of them crazy nonconverging sums), but can still lose finite-length runs (aka go broke).
Impervious
Profile Blog Joined March 2009
Canada4200 Posts
April 18 2012 20:37 GMT
#21406
On April 19 2012 05:30 KwarK wrote:
Show nested quote +
On April 19 2012 05:25 ghost_403 wrote:
Holy hell, you guys are retarded.

In this scenario, each ticket is independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

I believe they get around this by escalating the amount bet on each subsequent lottery by a huge amount, thus rendering every previous bet completely irrelevant should they win. Of course if there was a maximum bet, which there is, then this would be impossible. The maximum bet on blink is a lottery on the highest value item they offer and once you hit that and lose you can't do a lottery on two of them to try and make it back.
Once you can no longer escalate the odds and render the past history irrelevant then you need a streak good enough to undo all the previous losses. If you take the "but infinity" argument here and make the case that after infinite time you'll get a really long winning streak then you would be correct but you'd also have been losing EV for the infinite time before then. You still lose.

I'm not denying that betting in the lottery is a fucking terrible idea in reality. But if you had infinite backing, you could make money.

Just because the EV is negative does not mean that it is impossible for you to ever come across a string of bets where the EV over the course of all of those bets is actually positive. In fact, it's impossible for there to not be a string of bets where the EV of the string of bets is not positive at that point, regardless of the design of the lottery.

What you need to do is say "I'm going to stop betting when I am ahead by 100 billion isk". And if you have an infinite amount of bets, eventually you will come across a point in time where you are actually ahead by 100 billion isk. If you stop right then, you have made money, not lost it. But the only way to guarantee the ability to get to that point is to have an infinite amount of isk backing this endeavour.....
~ \(ˌ)im-ˈpər-vē-əs\ : not capable of being damaged or harmed.
ghost_403
Profile Joined September 2010
United States1825 Posts
April 18 2012 20:39 GMT
#21407
Whatever. Back on subject.

I'm moving into a wormhole. Based on empirical evidence, the people that live there are even dumber than the people who live in lowsec. I spent a good ten minutes scanning down exit holes last night in a C2, and all five of the people there running sites completely ignored me. Their insolence could only be tolerated for so long, so I jumped out and came back in a combat ship, and promptly killed a myrm running a site. Another myrm jumped in to defend his friend, but I failed to get a point on him before he warped off.

I think I'm going to move into a C3, run sites for isk and kill everyone in anything smaller than a carrier in the surrounding holes when that gets boring. Who's in?
They say great science is built on the shoulders of giants. Not here. At Aperture, we do all our science from scratch, no hand holding. Step aside, REAL SCIENCE coming through.
KwarK
Profile Blog Joined July 2006
United States42691 Posts
April 18 2012 20:46 GMT
#21408
On April 19 2012 05:37 Impervious wrote:
Show nested quote +
On April 19 2012 05:30 KwarK wrote:
On April 19 2012 05:25 ghost_403 wrote:
Holy hell, you guys are retarded.

In this scenario, each ticket is independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

I believe they get around this by escalating the amount bet on each subsequent lottery by a huge amount, thus rendering every previous bet completely irrelevant should they win. Of course if there was a maximum bet, which there is, then this would be impossible. The maximum bet on blink is a lottery on the highest value item they offer and once you hit that and lose you can't do a lottery on two of them to try and make it back.
Once you can no longer escalate the odds and render the past history irrelevant then you need a streak good enough to undo all the previous losses. If you take the "but infinity" argument here and make the case that after infinite time you'll get a really long winning streak then you would be correct but you'd also have been losing EV for the infinite time before then. You still lose.

I'm not denying that betting in the lottery is a fucking terrible idea in reality. But if you had infinite backing, you could make money.

Just because the EV is negative does not mean that it is impossible for you to ever come across a string of bets where the EV over the course of all of those bets is actually positive. In fact, it's impossible for there to not be a string of bets where the EV of the string of bets is not positive at that point, regardless of the design of the lottery.

What you need to do is say "I'm going to stop betting when I am ahead by 100 billion isk". And if you have an infinite amount of bets, eventually you will come across a point in time where you are actually ahead by 100 billion isk. If you stop right then, you have made money, not lost it. But the only way to guarantee the ability to get to that point is to have an infinite amount of isk backing this endeavour.....

The point of the infinity argument is that you can go "well although I need a streak of X in a row it doesn't matter because with infinity it will eventually happen if I keep trying for infinite time". The problem with infinity applied to a negative EV situation here with non escalating stakes is that each failure renders the chance of an eventual profit less likely. Yes, you will eventually win 100 in a row but by the time you do 100 in a row is very unlikely to be enough. By the time you win 200 in a row, you'll need 400. By the time you win 400, you'll need 800. While the chances of a streak that breaks you even are made higher by saying "but infinity" the length of that streak is also subject to the same infinity. Yes, eventually you'll make infinity money because infinity but by then you'll need a higher order of infinity to break even.
ModeratorThe angels have the phone box
tofucake
Profile Blog Joined October 2009
Hyrule19057 Posts
April 18 2012 20:52 GMT
#21409
On April 19 2012 05:39 ghost_403 wrote:
Whatever. Back on subject.

I'm moving into a wormhole. Based on empirical evidence, the people that live there are even dumber than the people who live in lowsec. I spent a good ten minutes scanning down exit holes last night in a C2, and all five of the people there running sites completely ignored me. Their insolence could only be tolerated for so long, so I jumped out and came back in a combat ship, and promptly killed a myrm running a site. Another myrm jumped in to defend his friend, but I failed to get a point on him before he warped off.

I think I'm going to move into a C3, run sites for isk and kill everyone in anything smaller than a carrier in the surrounding holes when that gets boring. Who's in?

oddly enough jovian just finished training itty 5 (and by oddly I mean "I was planning something like this myself")

count me in
Liquipediaasante sana squash banana
Body_Shield
Profile Blog Joined February 2011
Canada3368 Posts
April 18 2012 20:53 GMT
#21410
http://community.eveonline.com/devblog.asp?a=blog&nbid=28603


So, five-card stud, nothing wild... and the sky's the limit
Impervious
Profile Blog Joined March 2009
Canada4200 Posts
April 18 2012 21:28 GMT
#21411
On April 19 2012 05:30 KwarK wrote:
Show nested quote +
On April 19 2012 05:25 ghost_403 wrote:
Holy hell, you guys are retarded.

In this scenario, each ticket is independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

I believe they get around this by escalating the amount bet on each subsequent lottery by a huge amount, thus rendering every previous bet completely irrelevant should they win. Of course if there was a maximum bet, which there is, then this would be impossible. The maximum bet on blink is a lottery on the highest value item they offer and once you hit that and lose you can't do a lottery on two of them to try and make it back.
Once you can no longer escalate the odds and render the past history irrelevant then you need a streak good enough to undo all the previous losses. If you take the "but infinity" argument here and make the case that after infinite time you'll get a really long winning streak then you would be correct but you'd also have been losing EV for the infinite time before then. You still lose.

Yes, it is impractical to try to beat the system. Even with the escalating bet scenario I posted earlier: it would take an infinite number of bets on average to earn 1 dollar..... It's a fucking terrible return on investment.

But the same holds true for any situation where there can be a positive return, even when the EV is negative. And when you're dealing with a string of bets like this lottery, there will eventually be a point where the return is positive if you bet on it long enough. It might happen on the first bet, it might happen after 100 bets, it might happen after 100 million bets, or it might happen after many more than that, but it will happen.

Abom isn't dumb when he says that you can theoretically make money if you have an infinite bankroll.....
~ \(ˌ)im-ˈpər-vē-əs\ : not capable of being damaged or harmed.
KwarK
Profile Blog Joined July 2006
United States42691 Posts
April 18 2012 21:43 GMT
#21412
On April 19 2012 06:28 Impervious wrote:
Show nested quote +
On April 19 2012 05:30 KwarK wrote:
On April 19 2012 05:25 ghost_403 wrote:
Holy hell, you guys are retarded.

In this scenario, each ticket is independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

I believe they get around this by escalating the amount bet on each subsequent lottery by a huge amount, thus rendering every previous bet completely irrelevant should they win. Of course if there was a maximum bet, which there is, then this would be impossible. The maximum bet on blink is a lottery on the highest value item they offer and once you hit that and lose you can't do a lottery on two of them to try and make it back.
Once you can no longer escalate the odds and render the past history irrelevant then you need a streak good enough to undo all the previous losses. If you take the "but infinity" argument here and make the case that after infinite time you'll get a really long winning streak then you would be correct but you'd also have been losing EV for the infinite time before then. You still lose.

Yes, it is impractical to try to beat the system. Even with the escalating bet scenario I posted earlier: it would take an infinite number of bets on average to earn 1 dollar..... It's a fucking terrible return on investment.

But the same holds true for any situation where there can be a positive return, even when the EV is negative. And when you're dealing with a string of bets like this lottery, there will eventually be a point where the return is positive if you bet on it long enough. It might happen on the first bet, it might happen after 100 bets, it might happen after 100 million bets, or it might happen after many more than that, but it will happen.

Abom isn't dumb when he says that you can theoretically make money if you have an infinite bankroll.....

The chance of the streak that wins goes down at a faster rate than the number of attempts you have goes up. After infinite lotteries it'll be infinitely unlikely. You're treating the number of possible lotteries as infinite because you have an infinite bankroll while at the same time treating the amount you'll have to win as a finite number. This is a mistake, after infinite attempts and losses it will be infinitely unlikely to get the infinitely long streak you need to make it all back.
ModeratorThe angels have the phone box
Mysticus
Profile Joined April 2011
298 Posts
Last Edited: 2012-04-18 21:53:20
April 18 2012 21:52 GMT
#21413
If you play, play for fun expecting to lose up to your limit, set your limit, and if you are ever up; you stop.

^^

Advice for all gambling.
Impervious
Profile Blog Joined March 2009
Canada4200 Posts
April 18 2012 22:15 GMT
#21414
On April 19 2012 06:43 KwarK wrote:
Show nested quote +
On April 19 2012 06:28 Impervious wrote:
On April 19 2012 05:30 KwarK wrote:
On April 19 2012 05:25 ghost_403 wrote:
Holy hell, you guys are retarded.

In this scenario, each ticket is independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

I believe they get around this by escalating the amount bet on each subsequent lottery by a huge amount, thus rendering every previous bet completely irrelevant should they win. Of course if there was a maximum bet, which there is, then this would be impossible. The maximum bet on blink is a lottery on the highest value item they offer and once you hit that and lose you can't do a lottery on two of them to try and make it back.
Once you can no longer escalate the odds and render the past history irrelevant then you need a streak good enough to undo all the previous losses. If you take the "but infinity" argument here and make the case that after infinite time you'll get a really long winning streak then you would be correct but you'd also have been losing EV for the infinite time before then. You still lose.

Yes, it is impractical to try to beat the system. Even with the escalating bet scenario I posted earlier: it would take an infinite number of bets on average to earn 1 dollar..... It's a fucking terrible return on investment.

But the same holds true for any situation where there can be a positive return, even when the EV is negative. And when you're dealing with a string of bets like this lottery, there will eventually be a point where the return is positive if you bet on it long enough. It might happen on the first bet, it might happen after 100 bets, it might happen after 100 million bets, or it might happen after many more than that, but it will happen.

Abom isn't dumb when he says that you can theoretically make money if you have an infinite bankroll.....

The chance of the streak that wins goes down at a faster rate than the number of attempts you have goes up. After infinite lotteries it'll be infinitely unlikely. You're treating the number of possible lotteries as infinite because you have an infinite bankroll while at the same time treating the amount you'll have to win as a finite number. This is a mistake, after infinite attempts and losses it will be infinitely unlikely to get the infinitely long streak you need to make it all back.

Ok, lets try this:

We both bet 1 dollar on a coin flip, and we keep going on forever until I decide to stop. Eventually, there will be a point where I won 1 more than you did. That can happen on the first flip, on the 3rd flip, on the 5th flip, etc. If we keep going forever, it will eventually happen.

There will also be a point where I won 2 more than you did. That can happen on the 2nd flip, on the 4th flip, on the 6th flip, etc.

If we keep going, there will eventually be a point where I have an infinite number of wins more than you.

If I am the one who chooses when we stop, and I have an infinite bankroll, I can guarantee that I make money, as well as how much. It just takes a lot of time, and will take proportionally more depending on how much I want to win.

And it doesn't have to be a coinflip (although it is a lot easier to comprehend when dealing with 50/50 odds). It can be rolling a dice, where you win 5/6 times. It doesn't matter. It will eventually happen. What the odds and EV determine is how long it can be expected to take before the result you desire happens.

You aren't thinking about this properly.....
~ \(ˌ)im-ˈpər-vē-əs\ : not capable of being damaged or harmed.
Impervious
Profile Blog Joined March 2009
Canada4200 Posts
April 18 2012 22:15 GMT
#21415
On April 19 2012 06:52 Mysticus wrote:
If you play, play for fun expecting to lose up to your limit, set your limit, and if you are ever up; you stop.

^^

Advice for all gambling.

Or just don't gamble and buy a beer with the money you would have spent?
~ \(ˌ)im-ˈpər-vē-əs\ : not capable of being damaged or harmed.
abominare
Profile Joined March 2010
United States1216 Posts
April 18 2012 22:25 GMT
#21416
On April 19 2012 06:28 Impervious wrote:
Show nested quote +
On April 19 2012 05:30 KwarK wrote:
On April 19 2012 05:25 ghost_403 wrote:
Holy hell, you guys are retarded.

In this scenario, each ticket is independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

I believe they get around this by escalating the amount bet on each subsequent lottery by a huge amount, thus rendering every previous bet completely irrelevant should they win. Of course if there was a maximum bet, which there is, then this would be impossible. The maximum bet on blink is a lottery on the highest value item they offer and once you hit that and lose you can't do a lottery on two of them to try and make it back.
Once you can no longer escalate the odds and render the past history irrelevant then you need a streak good enough to undo all the previous losses. If you take the "but infinity" argument here and make the case that after infinite time you'll get a really long winning streak then you would be correct but you'd also have been losing EV for the infinite time before then. You still lose.

Yes, it is impractical to try to beat the system. Even with the escalating bet scenario I posted earlier: it would take an infinite number of bets on average to earn 1 dollar..... It's a fucking terrible return on investment.

But the same holds true for any situation where there can be a positive return, even when the EV is negative. And when you're dealing with a string of bets like this lottery, there will eventually be a point where the return is positive if you bet on it long enough. It might happen on the first bet, it might happen after 100 bets, it might happen after 100 million bets, or it might happen after many more than that, but it will happen.

Abom isn't dumb when he says that you can theoretically make money if you have an infinite bankroll.....


Wait who called me dumb?
Impervious
Profile Blog Joined March 2009
Canada4200 Posts
April 18 2012 22:27 GMT
#21417
On April 19 2012 07:25 abominare wrote:
Show nested quote +
On April 19 2012 06:28 Impervious wrote:
On April 19 2012 05:30 KwarK wrote:
On April 19 2012 05:25 ghost_403 wrote:
Holy hell, you guys are retarded.

In this scenario, each ticket is independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

I believe they get around this by escalating the amount bet on each subsequent lottery by a huge amount, thus rendering every previous bet completely irrelevant should they win. Of course if there was a maximum bet, which there is, then this would be impossible. The maximum bet on blink is a lottery on the highest value item they offer and once you hit that and lose you can't do a lottery on two of them to try and make it back.
Once you can no longer escalate the odds and render the past history irrelevant then you need a streak good enough to undo all the previous losses. If you take the "but infinity" argument here and make the case that after infinite time you'll get a really long winning streak then you would be correct but you'd also have been losing EV for the infinite time before then. You still lose.

Yes, it is impractical to try to beat the system. Even with the escalating bet scenario I posted earlier: it would take an infinite number of bets on average to earn 1 dollar..... It's a fucking terrible return on investment.

But the same holds true for any situation where there can be a positive return, even when the EV is negative. And when you're dealing with a string of bets like this lottery, there will eventually be a point where the return is positive if you bet on it long enough. It might happen on the first bet, it might happen after 100 bets, it might happen after 100 million bets, or it might happen after many more than that, but it will happen.

Abom isn't dumb when he says that you can theoretically make money if you have an infinite bankroll.....


Wait who called me dumb?

Kwark's response implied it.....
~ \(ˌ)im-ˈpər-vē-əs\ : not capable of being damaged or harmed.
KwarK
Profile Blog Joined July 2006
United States42691 Posts
April 18 2012 22:27 GMT
#21418
On April 19 2012 07:15 Impervious wrote:
Show nested quote +
On April 19 2012 06:43 KwarK wrote:
On April 19 2012 06:28 Impervious wrote:
On April 19 2012 05:30 KwarK wrote:
On April 19 2012 05:25 ghost_403 wrote:
Holy hell, you guys are retarded.

In this scenario, each ticket is independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

I believe they get around this by escalating the amount bet on each subsequent lottery by a huge amount, thus rendering every previous bet completely irrelevant should they win. Of course if there was a maximum bet, which there is, then this would be impossible. The maximum bet on blink is a lottery on the highest value item they offer and once you hit that and lose you can't do a lottery on two of them to try and make it back.
Once you can no longer escalate the odds and render the past history irrelevant then you need a streak good enough to undo all the previous losses. If you take the "but infinity" argument here and make the case that after infinite time you'll get a really long winning streak then you would be correct but you'd also have been losing EV for the infinite time before then. You still lose.

Yes, it is impractical to try to beat the system. Even with the escalating bet scenario I posted earlier: it would take an infinite number of bets on average to earn 1 dollar..... It's a fucking terrible return on investment.

But the same holds true for any situation where there can be a positive return, even when the EV is negative. And when you're dealing with a string of bets like this lottery, there will eventually be a point where the return is positive if you bet on it long enough. It might happen on the first bet, it might happen after 100 bets, it might happen after 100 million bets, or it might happen after many more than that, but it will happen.

Abom isn't dumb when he says that you can theoretically make money if you have an infinite bankroll.....

The chance of the streak that wins goes down at a faster rate than the number of attempts you have goes up. After infinite lotteries it'll be infinitely unlikely. You're treating the number of possible lotteries as infinite because you have an infinite bankroll while at the same time treating the amount you'll have to win as a finite number. This is a mistake, after infinite attempts and losses it will be infinitely unlikely to get the infinitely long streak you need to make it all back.

Ok, lets try this:

We both bet 1 dollar on a coin flip, and we keep going on forever until I decide to stop. Eventually, there will be a point where I won 1 more than you did. That can happen on the first flip, on the 3rd flip, on the 5th flip, etc. If we keep going forever, it will eventually happen.

There will also be a point where I won 2 more than you did. That can happen on the 2nd flip, on the 4th flip, on the 6th flip, etc.

If we keep going, there will eventually be a point where I have an infinite number of wins more than you.

If I am the one who chooses when we stop, and I have an infinite bankroll, I can guarantee that I make money, as well as how much. It just takes a lot of time, and will take proportionally more depending on how much I want to win.

And it doesn't have to be a coinflip (although it is a lot easier to comprehend when dealing with 50/50 odds). It can be rolling a dice, where you win 5/6 times. It doesn't matter. It will eventually happen. What the odds and EV determine is how long it can be expected to take before the result you desire happens.

You aren't thinking about this properly.....

The infinity rolls of the dice argument applies to both the odds of your streak eventually happening and the length the streak needs to be in a negative EV situation. The chance of the streak that puts you ahead hits 1 over infinity dice rolls but the length that streak needs to be hits a higher order of infinity. You're ignoring the impact of infinite dice rolls on the liklihood of getting the streak, it becomes infinitely unlikely faster than the odds of getting it become infinitely likely.
ModeratorThe angels have the phone box
Impervious
Profile Blog Joined March 2009
Canada4200 Posts
April 18 2012 23:04 GMT
#21419
On April 19 2012 07:27 KwarK wrote:
Show nested quote +
On April 19 2012 07:15 Impervious wrote:
On April 19 2012 06:43 KwarK wrote:
On April 19 2012 06:28 Impervious wrote:
On April 19 2012 05:30 KwarK wrote:
On April 19 2012 05:25 ghost_403 wrote:
Holy hell, you guys are retarded.

In this scenario, each ticket is independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

I believe they get around this by escalating the amount bet on each subsequent lottery by a huge amount, thus rendering every previous bet completely irrelevant should they win. Of course if there was a maximum bet, which there is, then this would be impossible. The maximum bet on blink is a lottery on the highest value item they offer and once you hit that and lose you can't do a lottery on two of them to try and make it back.
Once you can no longer escalate the odds and render the past history irrelevant then you need a streak good enough to undo all the previous losses. If you take the "but infinity" argument here and make the case that after infinite time you'll get a really long winning streak then you would be correct but you'd also have been losing EV for the infinite time before then. You still lose.

Yes, it is impractical to try to beat the system. Even with the escalating bet scenario I posted earlier: it would take an infinite number of bets on average to earn 1 dollar..... It's a fucking terrible return on investment.

But the same holds true for any situation where there can be a positive return, even when the EV is negative. And when you're dealing with a string of bets like this lottery, there will eventually be a point where the return is positive if you bet on it long enough. It might happen on the first bet, it might happen after 100 bets, it might happen after 100 million bets, or it might happen after many more than that, but it will happen.

Abom isn't dumb when he says that you can theoretically make money if you have an infinite bankroll.....

The chance of the streak that wins goes down at a faster rate than the number of attempts you have goes up. After infinite lotteries it'll be infinitely unlikely. You're treating the number of possible lotteries as infinite because you have an infinite bankroll while at the same time treating the amount you'll have to win as a finite number. This is a mistake, after infinite attempts and losses it will be infinitely unlikely to get the infinitely long streak you need to make it all back.

Ok, lets try this:

We both bet 1 dollar on a coin flip, and we keep going on forever until I decide to stop. Eventually, there will be a point where I won 1 more than you did. That can happen on the first flip, on the 3rd flip, on the 5th flip, etc. If we keep going forever, it will eventually happen.

There will also be a point where I won 2 more than you did. That can happen on the 2nd flip, on the 4th flip, on the 6th flip, etc.

If we keep going, there will eventually be a point where I have an infinite number of wins more than you.

If I am the one who chooses when we stop, and I have an infinite bankroll, I can guarantee that I make money, as well as how much. It just takes a lot of time, and will take proportionally more depending on how much I want to win.

And it doesn't have to be a coinflip (although it is a lot easier to comprehend when dealing with 50/50 odds). It can be rolling a dice, where you win 5/6 times. It doesn't matter. It will eventually happen. What the odds and EV determine is how long it can be expected to take before the result you desire happens.

You aren't thinking about this properly.....

The infinity rolls of the dice argument applies to both the odds of your streak eventually happening and the length the streak needs to be in a negative EV situation. The chance of the streak that puts you ahead hits 1 over infinity dice rolls but the length that streak needs to be hits a higher order of infinity. You're ignoring the impact of infinite dice rolls on the liklihood of getting the streak, it becomes infinitely unlikely faster than the odds of getting it become infinitely likely.

The chance that there is never a situation where I am 1 ahead is an even more infinitely unlikely situation. You can't look at the end result and not consider all of the permutations prior to that point.....

For the first flip, there is a 50% chance that I win. Let's say that I win with a result of H.

Assuming I lost the first flip, lets consider the results of the next 2 flips together. There are 4 results, HH, HT, TH, and TT. Of those 4 situations, there is 1 situation where I win. This occurs 1/4 of the time that I lose the first flip.

This means that if I choose when we stop, there are 2 situations where I win - a 1/2 from the first flip, and a 1/2 (the chance that you won) * 1/4 (the chance that the next two are HH) from the 2nd and 3rd flip, or a total of 5/8, or 62.5%.

Of the situations where I have not won after the first 3 flips, there are 2 different variants: Either I am down by 3, or I am down by 1.

Considering the situation where I am down by 3, there is no way I can win in the next 2 flips, so I'll ignore it for right now (although it needs to be considered if you continue to run the numbers). Considering the situation where I am down by 1, we are back into the same situation as the 2nd and 3rd flips.

If I am down by 1 at this point, which happens 1/4 of the time in the first 3 flips, I can come back to win it with a result of HH, which occurs 1/4 of the time. This means that the chance of me winning on flip 4 or 5 to be 1/4 (the chance that we have gone 1:2 to this point) * 1/4 (the chance that the next two are HH).

So, the chance of me having 1 win more than you after the first 5 flips, at some point, is 1/2 (first flip) + 1/8 (2nd and 3rd flips) + 1/16 (4th and 5th flips) = 11/16, or 68.75%.

If you keep going, this will actually converge to 1 (meaning 100%). I will eventually win at some point. The same principle applies regardless of the chance of winning an individual flip, it just takes a considerably longer time to start to appear to converge at 1. Hell, even this one takes a long fucking time to converge.....

The same thing happens if you want to have 2 wins, 100 wins, or even an infinite number of wins more than your opponent.
~ \(ˌ)im-ˈpər-vē-əs\ : not capable of being damaged or harmed.
KwarK
Profile Blog Joined July 2006
United States42691 Posts
April 18 2012 23:06 GMT
#21420
On April 19 2012 08:04 Impervious wrote:
Show nested quote +
On April 19 2012 07:27 KwarK wrote:
On April 19 2012 07:15 Impervious wrote:
On April 19 2012 06:43 KwarK wrote:
On April 19 2012 06:28 Impervious wrote:
On April 19 2012 05:30 KwarK wrote:
On April 19 2012 05:25 ghost_403 wrote:
Holy hell, you guys are retarded.

In this scenario, each ticket is independent. Buying four tickets means that your probability of winning is four times the probability of winning with one ticket. Since, on average, you lose, buying more tickets means, on average, you lose more.

You are going to lose money on this no matter how many tickets you buy (up to and including infinity).

Math is here.

I believe they get around this by escalating the amount bet on each subsequent lottery by a huge amount, thus rendering every previous bet completely irrelevant should they win. Of course if there was a maximum bet, which there is, then this would be impossible. The maximum bet on blink is a lottery on the highest value item they offer and once you hit that and lose you can't do a lottery on two of them to try and make it back.
Once you can no longer escalate the odds and render the past history irrelevant then you need a streak good enough to undo all the previous losses. If you take the "but infinity" argument here and make the case that after infinite time you'll get a really long winning streak then you would be correct but you'd also have been losing EV for the infinite time before then. You still lose.

Yes, it is impractical to try to beat the system. Even with the escalating bet scenario I posted earlier: it would take an infinite number of bets on average to earn 1 dollar..... It's a fucking terrible return on investment.

But the same holds true for any situation where there can be a positive return, even when the EV is negative. And when you're dealing with a string of bets like this lottery, there will eventually be a point where the return is positive if you bet on it long enough. It might happen on the first bet, it might happen after 100 bets, it might happen after 100 million bets, or it might happen after many more than that, but it will happen.

Abom isn't dumb when he says that you can theoretically make money if you have an infinite bankroll.....

The chance of the streak that wins goes down at a faster rate than the number of attempts you have goes up. After infinite lotteries it'll be infinitely unlikely. You're treating the number of possible lotteries as infinite because you have an infinite bankroll while at the same time treating the amount you'll have to win as a finite number. This is a mistake, after infinite attempts and losses it will be infinitely unlikely to get the infinitely long streak you need to make it all back.

Ok, lets try this:

We both bet 1 dollar on a coin flip, and we keep going on forever until I decide to stop. Eventually, there will be a point where I won 1 more than you did. That can happen on the first flip, on the 3rd flip, on the 5th flip, etc. If we keep going forever, it will eventually happen.

There will also be a point where I won 2 more than you did. That can happen on the 2nd flip, on the 4th flip, on the 6th flip, etc.

If we keep going, there will eventually be a point where I have an infinite number of wins more than you.

If I am the one who chooses when we stop, and I have an infinite bankroll, I can guarantee that I make money, as well as how much. It just takes a lot of time, and will take proportionally more depending on how much I want to win.

And it doesn't have to be a coinflip (although it is a lot easier to comprehend when dealing with 50/50 odds). It can be rolling a dice, where you win 5/6 times. It doesn't matter. It will eventually happen. What the odds and EV determine is how long it can be expected to take before the result you desire happens.

You aren't thinking about this properly.....

The infinity rolls of the dice argument applies to both the odds of your streak eventually happening and the length the streak needs to be in a negative EV situation. The chance of the streak that puts you ahead hits 1 over infinity dice rolls but the length that streak needs to be hits a higher order of infinity. You're ignoring the impact of infinite dice rolls on the liklihood of getting the streak, it becomes infinitely unlikely faster than the odds of getting it become infinitely likely.

The chance that there is never a situation where I am 1 ahead is an even more infinitely unlikely situation. You can't look at the end result and not consider all of the permutations prior to that point.....

For the first flip, there is a 50% chance that I win. Let's say that I win with a result of H.

Assuming I lost the first flip, lets consider the results of the next 2 flips together. There are 4 results, HH, HT, TH, and TT. Of those 4 situations, there is 1 situation where I win. This occurs 1/4 of the time that I lose the first flip.

This means that if I choose when we stop, there are 2 situations where I win - a 1/2 from the first flip, and a 1/2 (the chance that you won) * 1/4 (the chance that the next two are HH) from the 2nd and 3rd flip, or a total of 5/8, or 62.5%.

Of the situations where I have not won after the first 3 flips, there are 2 different variants: Either I am down by 3, or I am down by 1.

Considering the situation where I am down by 3, there is no way I can win in the next 2 flips, so I'll ignore it for right now (although it needs to be considered if you continue to run the numbers). Considering the situation where I am down by 1, we are back into the same situation as the 2nd and 3rd flips.

If I am down by 1 at this point, which happens 1/4 of the time in the first 3 flips, I can come back to win it with a result of HH, which occurs 1/4 of the time. This means that the chance of me winning on flip 4 or 5 to be 1/4 (the chance that we have gone 1:2 to this point) * 1/4 (the chance that the next two are HH).

So, the chance of me having 1 win more than you after the first 5 flips, at some point, is 1/2 (first flip) + 1/8 (2nd and 3rd flips) + 1/16 (4th and 5th flips) = 11/16, or 68.75%.

If you keep going, this will actually converge to 1 (meaning 100%). I will eventually win at some point. The same principle applies regardless of the chance of winning an individual flip, it just takes a considerably longer time to start to appear to converge at 1. Hell, even this one takes a long fucking time to converge.....

The same thing happens if you want to have 2 wins, 100 wins, or even an infinite number of wins more than your opponent.

You keep using coinflips to respond to my argument based upon a negative EV situation. Are you sure you understand what a negative EV situation is?
ModeratorThe angels have the phone box
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