Flipping Coins

[BaseJester 1]

LW -0.10WL -0.10WW +20.10LL -19.90WWL +9.899LLW -10.099= -0.2 (the other was a joke)

Head asplodes.

[BaseJester 1]

The scam is that once you take the bet, you're committed to doing all million flips before you can stop (unless you go broke first). So you are *really* betting on the probability that a sequence of random flips will never cross the -$1,000 line over the course of one million flips. The probability of this happening is miniscule.

The thing is that we're multiplying minuscule probabilities by really large payouts. It's tempting for us to approximate very unlikely as never, but more difficult to comprehend big numbers like we might win. Consider:I may have the shape of the left squiggle all wrong, but the red hump is very tall and skinny. The blue hump is very short and wide. Turn the thing on its side and forget what the axes represent.1/big is small, but big is big in a completely compensating way. This whole proposition is just a slightly profitable lottery.

[psujohn 0]

This whole proposition is just a slightly profitable lottery.

No it's not. There's a subtle part of the prop that makes it -EV that looks innocent but makes all the difference and that's "you lose if your br ever falls below $1". If you can handle infinitely large and small numbers (ie you don't have the rounding error a computer does) you'll never actually reach zero. That's because you can only lose 1% of your br at a time. But even allowing for computer rounding error if you change the simulation to require only that your br isn't 0 you'll see that you don't often go completely bust and it becomes +EV because of the bonus. The variance is still crazy in this though and in practice you'd only want to do it if you could play for a long time.Under the revised rules (br > 0) if you could do it as many times as you want it becomes massively +EV because you can never lose more the $1000 in a trial but you can get huge wins occasionally. Here's 10 runs under the br > 0 rule:remaining br = 3.39410372865 with bonus = 3.56380891508remaining br = 1.89486813283 with bonus = 1.98961153947you lose after 1000000 flips. Your max was 53673983.6515remaining br = 60510.9639413 with bonus = 63536.5121384remaining br = 39935404.1697 with bonus = 41932174.3782you lose after 1000000 flips. Your max was 4412.59391032remaining br = 618.844915365 with bonus = 649.787161134you lose after 1000000 flips. Your max was 3062.48166784remaining br = 470121.459061 with bonus = 493627.532014remaining br = 10114.3356513 with bonus = 10620.0524339The first 3 runs you lose money, the 4th you make it all back and get a huge profit and the 5th you retire to your own private island in the Caribbean.

[psujohn 0]

I take HTT and give you HHT and I give you .3% bounus (.003) when you win.

Let's run this say 1000 times for $1 each. And you don't need to give a bonus.(Hint: The "mark" picks first - you gave me HHT - then you take the first 2 and the opposite of the 2nd at the front - you want THH. Under the conditions you set you took HTT and gave me the ideal hand against you HHT. I win about 2/3 of the time. You can just ship me the $330 or so on tilt or stars.)Of course this means I was wrong. If the mark picks a duplicating sequence you can still pick a hand that beats him and not just one that breaks even. I was a bit surprised by that at first but if you think about it, it makes sense. By picking this way you ensure that once (opposite of mark's 2nd choice) comes up at all he can't win. Because once that comes up you'll hit your pattern before he can hit his. So if my pattern is HHT and yours is THH then I can never win if the first flip is T and I can only win if the first 2 flips are H which only happens 25% of the time. If the mark picks the first two the same HH or TT you'll win 75% of the time.If the mark picks the first two different HT or TH you'll win 66% of the time.

[BaseJester 1]

No it's not.

Yes it is.

There's a subtle part of the prop that makes it -EV that looks innocent but makes all the difference and that's "you lose if your br ever falls below $1".

I don't think this makes sense. Why should being forced to stop an activity make it -EV? If coin flipping at 1:1 was -EV, then being forced to stop doing it would be good for you. The only way stopping is bad is if you have positive expectation from now on. Why would you suddenly switch from a negative expectation to a positive expectation just because you have lost $1000? There is no tendency to bounce back up. That's the gambler's fallacy.

simulations

I think the details are clouding the concepts.E(X + Y ) = E(X) + E(Y)That's it.

[thebottomline 0]

I don't think this makes sense. Why should being forced to stop an activity make it -EV?

Because it's a high variance activity, if you're betting a % but being forced to stop at a low limit compared to the number of flips ($1000 - 1m flips), and with the fact that you lose slowly anyway means you'd eventually go broke, but you wouldn't if you weren't forced to stop and could still bet 1% of your small denominations of coins, but you can't.

[BaseJester 1]

Think about the Martingale Method.--------------------Suppose you double your bet at roulette every time until you win. You have a very large bankroll. You're guaranteed to win because the chance of losing over and over again is minuscule. In this example, we show up with $1,000,000 and 100% of the time walk away with a dollar profit. 100% out of 1 million trials. It's guaranteed. I proved it with this simulation. ---------------------Now if you think the proposition as originally posed is -EV, try to explain how your thinking is different than the justification for Martingale. (Or alternatively, go make some money at roulette.)Expected ValueMartingdale Method

<?php# Martingale Method$win = 0;$lose = 0;define ("STARTING_BANKROLL",1000000); define ("NUM_TRIALS",1000000); $goal = STARTING_BANKROLL + 1;for ($trial = 0; $trial < NUM_TRIALS; $trial++) {   $br = STARTING_BANKROLL;   $wager = 1;   while(1)    {	   $spin = rand(-1,36);  # American roulette has a 0 and 00	   # Let's bet on low	   if ($spin > 0 && $spin < 19  ) {		   $br = $br + $wager;	   }	   else {		   $br = $br - $wager;		   $wager = $wager * 2;	   }	   #echo $br."\n";	   if ($br >= $goal )	   {		$win++;			break;	   }	   if ($br < 0) 	   {		$lose++;			break;	   }   }   }echo "Used a starting bankroll of ".STARTING_BANKROLL."\n";echo "Ran ".NUM_TRIALS." trials.\n";echo "win: ".$win."\n";echo "lose: ".$lose."\n";echo "win%: ".(100*$win/($lose+$win))."\n";?>

C:\Program Files\PHP>[b]php flip[/b]Used a starting bankroll of 1000000Ran 1000000 trials.win: 1000000lose: 0win%: 100

[ROACHBABY 0]

You have $1,000. A friend says that he will pay you twice your bet amount, every time you flip heads. So, you bet $1, it comes up tails, you lose $1. You bet $1, it comes up heads, you get $2 from him (but lose your original $1). (EDIT: i.e. It's a 2 for 1 payout when you win).

Ok, I just woke up so maybe I am not thinking right, but.....If I bet $1 and win then I lose my original bet (see quote) and get $2 in winnings. Isn't that the same thing as getting 1:1 on my money? I think to be getting 2:1 you would have to be getting your original bet back and then the $2 in winnings. Thoughts?

[psujohn 0]

I don't think this makes sense. Why should being forced to stop an activity make it -EV? If coin flipping at 1:1 was -EV, then being forced to stop doing it would be good for you. The only way stopping is bad is if you have positive expectation from now on. Why would you suddenly switch from a negative expectation to a positive expectation just because you have lost $1000?

We're going to flip coins you and I. I'll give you heads and take tails. Everytime you win I pay you twice your wager (2:1 for real not the bs in the op).Clearly a +EV situation for you. But I have a few additional rules.You must bet your entire net worth on every flip.You cannot ever quit if you have money left.And yes I have an infinite amount of money.So even though an individual flip is +EV the additional rules make it -EV. That's because you can only stop when it's beneficial for me.The trick of the problem as posed it that it sounds like the br < $1 rule is unimportant but it's not. In a pure math world you will never get to $0 because you always have 99% of your br after each flip. $1 sounds like a tiny portion of your br but given that you need to bet %1 every time it only takes a relatively small number of losing flips to get under $1 regardless of how big your BR is at the start. And even if your BR grows it doesn't take many more losses (relatively) to get under $1.At $1000 it takes < 700 loses in a row to get under $1. At 10,000 it's 900 something.But just like those long runs of losses will come in a large enough sample you'll also get long runs of wins if you stay in the game. Thus the $1 limit makes all the difference in the world.

[BaseJester 1]

We're going to flip coins you and I. I'll give you heads and take tails. Everytime you win I pay you twice your wager (2:1 for real not the bs in the op).Clearly a +EV situation for you. But I have a few additional rules.You must bet your entire net worth on every flip.You cannot ever quit if you have money left.And yes I have an infinite amount of money.So even though an individual flip is +EV the additional rules make it -EV. That's because you can only stop when it's beneficial for me.

I think you're confusing having a positive expectation with being a good idea.If you're going to posit something like an infinite amount of money, why do you have a problem with the idea that something with an infinitesimal probability could still occur? Don't you agree that your amassing an infinite amount of wealth is ****ing improbable? So I'm telling you that the game you suggested has positive expectation for me because we might flip a coin infinity times and it comes up heads every time.Andy: Suppose Wolverine goes around robbing banks. No one could stop him. Then he would be incredibly wealthy.Bob: Well, yeah, unless Superman were there. He might thwart Wolverine's robbery.Andy: There's no such thing as Superman.Bob: [Whack!]

[psujohn 0]

Now if you think the proposition as originally posed is -EV, try to explain how your thinking is different than the justification for Martingale. (Or alternatively, go make some money at roulette.)

Martingale with an infinite bankroll is indeed guaranteed profitable. The $1 min br part of the OP is like saying you can do martingale but if you lose 3 times in a row you're not allowed to play any more. Or as happens in the casino you cap the maximum bet size. That's why I'm saying that the game is +EV without the $1 br limit but -EV with it.

I think you're confusing having a positive expectation with being a good idea.If you're going to posit something like an infinite amount of money, why do you have a problem with the idea that something with an infinitesimal probability could still occur? Don't you agree that your amassing an infinite amount of wealth is ****ing improbable? So I'm telling you that the game you suggested has positive expectation for me because we might flip a coin infinity times and it comes up heads every time.

You said that you can't take a +EV game and make it -EV by limiting how often you can do it. I proved that's not the case. Well under the assumption that I can require you to keep playing as long as I like but force you to quit at a certain point. If you don't get that you don't get that and there's no point in my explaining it further. It really has nothing to do with infinitesimal probabilities. The game in the OP is -EV because the odds of you getting below $1 at some point are so high that you will lose nearly every time you play. Sure it's possible to come up with a sequence where you make a fortune. If you remove the $1 limit it's very likely that you'll make a fortune even without the 5% bonus at the end.

[BaseJester 1]

Martingale with an infinite bankroll is indeed guaranteed profitable.

Rats. That's bad for my argument. Everything is finite in the original problem, though. Here's another thread about Martingdale on FCP.

You said that you can't take a +EV game and make it -EV by limiting how often you can do it. I proved that's not the case. Well under the assumption that I can require you to keep playing as long as I like but force you to quit at a certain point. If you don't get that you don't get that and there's no point in my explaining it further. It really has nothing to do with infinitesimal probabilities. The game in the OP is -EV because the odds of you getting below $1 at some point are so high that you will lose nearly every time you play.

Isn't saying a thing happens "nearly every time" the same thing as saying the probability of the thing not happening is infinitesimal? That's what I thought it meant, anyway.

[BaseJester 1]

We're going to flip coins you and I. I'll give you heads and take tails. Everytime you win I pay you twice your wager (2:1 for real not the bs in the op).Clearly a +EV situation for you. But I have a few additional rules.You must bet your entire net worth on every flip.You cannot ever quit if you have money left.And yes I have an infinite amount of money.So even though an individual flip is +EV the additional rules make it -EV. That's because you can only stop when it's beneficial for me.

I want to pose a question about this, and I don't know the answer to it. If you don't impose a requirement that I continue, how many times should I flip for maximum expected value?

[speedz99 145]

tree

[psujohn 0]

I want to pose a question about this, and I don't know the answer to it. If you don't impose a requirement that I continue, how many times should I flip for maximum expected value?

Even though it's +EV you shouldn't flip at all because 50% of the time you're going to be left penniless. If you can do it for some smaller amount of money and quit whenever you want then you should pre-determine ho much you want to win and play to that amount - say you have $1000 you can afford to lose but you want to buy a $15K car you'd pre-determine that you're going to flip 4 times looking to run your 1K to 16K. But this really has nothing to do with EV and is all about the utility of the $$$.

[psujohn 0]

That's bad for my argument. Everything is finite in the original problem, though.

Nothing is infinite in the original. There are a fixed number of flips and a requirement that you quit early if below $1.

[BaseJester 1]

I want to pose a question about this, and I don't know the answer to it. If you don't impose a requirement that I continue, how many times should I flip for maximum expected value?.

Even though it's +EV you shouldn't flip at all because 50% of the time you're going to be left penniless. If you can do it for some smaller amount of money and quit whenever you want then you should pre-determine ho much you want to win and play to that amount - say you have $1000 you can afford to lose but you want to buy a $15K car you'd pre-determine that you're going to flip 4 times looking to run your 1K to 16K. But this really has nothing to do with EV and is all about the utility of the $$$.

I think this is a thoughtful response, and I'm pleased that you're bringing up utility. It doesn't, however, address the question as posed. When you say the original game is -EV, are you really saying it's a bad idea due to utility or is it really -EV?

[BaseJester 1]

Everything is finite in the original problem, though.

Nothing is infinite in the original. There are a fixed number of flips and a requirement that you quit early if below $1.

Yeah, that's what I said. If the problem with Martingale is that our bankroll isn't infinite, then maybe that says something about this backwards Martingale method in the OP. (Maybe, it is also EV neutral.)

[psujohn 0]

Yeah the actual EV is difficult to determine. There is a non zero chance that you'll win but the amount you win when you win is unknown. Clearly if you run it infinitely you will eventually win but my intuition is that you won't win enough or often enough to overcome the fact that you lose nearly $1000 almost every time. Even the odds of you winning is very difficult to determine because there are so many ways for you to lose. At any point 700 or so bad flips in a row will cause you to lose but you don't need to have 700 in a row to lose. For example if you lose 3 out of 5 flips you'll go broke after 3300 or so flips.The OP also expressed that you'd be able to do those million flips in 24 hours and I'm pretty well certain that given the constraints of a normal human lifetime the odds are overwhelmingly against you winning even once if you played every day.At any rate whether the game is +EV or -EV as stated I'd be happy to take the house side if anyone wanted to play even though I'd only stand to win $1000 and could potentially lose everything if I lost.

[SilentButDeadly3 0]

Interesting question, but the real question is would you call on the first hand of the WSOP main event if you held AA and the other nine players went all in?

Herm Edwards would insta-call.

[KevinFKHS 0]

Maybe i can't think straight...but aren't you always gonna end up broke?!??I mean, even if you go on rushes, you risk more money...and in the long run the number of wins and losses is gonna be the same, so you'll be losing money!BR(WL) = 100 + 1 - 1.01 = 99.99BR(WWLL) = 100 + 1 + 1.01 - 1.0201 - 1.009899 = 99.980001BR(WWWLLL) = 100 + 1 + 1.01 + 1.0201 - 1.030301 - 1.01999799 - 1.0097980101 = 99,9700029999...and so on........................right?

[BaseJester 1]

Maybe i can't think straight...but aren't you always gonna end up broke?!??

No, just nearly always. So nearly that nobody wants to take the bet in practice.

I mean, even if you go on rushes, you risk more money...and in the long run the number of wins and losses is gonna be the same,

There's no guarantee of the number of wins and losses being equal. It's just the mean.

so you'll be losing money!BR(WL) = 100 + 1 - 1.01 = 99.99BR(WWLL) = 100 + 1 + 1.01 - 1.0201 - 1.009899 = 99.980001BR(WWWLLL) = 100 + 1 + 1.01 + 1.0201 - 1.030301 - 1.01999799 - 1.0097980101 = 99,9700029999...and so on........................right?

That's probably all right, but look what happens in the extremely unlikely case that you go on a string of 688. If the hero loses every flip, he's lost his $1000 bankroll. But if he wins every flip, he wins $940,000.Playing the house on this gamble is like selling home flood insurance for $1 a year to people living in the desert. If you should happen to lose, then look for a government bailout or fly to Venezuela.

[navybuttons 15]

I would bet on THH and you get THT For even $.

i didn't see this until tonight. if you're still interested in this bet i am willing to take this action. if you're still willing to take it let me know and we can begin to arrange it.

[oldmangrimis 0]

i didn't see this until tonight. if you're still interested in this bet i am willing to take this action. if you're still willing to take it let me know and we can begin to arrange it.

I am trying to PM you but your box is full.

[oldmangrimis 0]

i didn't see this until tonight. if you're still interested in this bet i am willing to take this action. if you're still willing to take it let me know and we can begin to arrange it.

THH will beat THT, I don't what you to take a bet where you are giving up that kind of edge to me. I just wanted work out the math to this bet. So I would say for you not to take the bet, look it over.