Can one overcome a -EV game? [Archive] - Two Plus Two Older Archives

yellowjack

10-18-2005, 08:28 PM

You can't. Below is a response I made in response to blackjack, but the same trend occurs.

[ QUOTE ]
WWhen you play for $1, it's -EV
When you play for $2, it's -EV

Even though you're doubling your bet each time, it's still -EV

I don't know any strategy for BJ, but I do know that surrendering, doubling down, etc. are only moderately +EV in comparison to the overall -EV in playing. To simplify this, I'll put the probability of winning a hand at 40%, even though I think it is a bit generous (esp. online)

Suppose you're playing a max of 4 hands, where you go from $1 to $2 to $4 to $8 if you lose, and you're stopping at $8 (we'll see later that the max amount doesn't matter, it's still -EV). Let's say you have a 40% chance of winning a hand. If you win at $1 you just play for $1 again.

Note that the win is always just +$1 since when you're playing for say $4, you have already lost $3. So your net gain is +$1.

EV = 0.40*(+$1) + 0.6 * [0.4*(+$1) + 0.6 * [0.4*(+$1) + 0.6 * [0.4*($1) + 0.6(-$15)]]]

EV = $0.40 + 0.6 * [$0.40 + 0.6 * [$0.40 + 0.6 * [$0.40 + (-$9)]]]

EV = -$1.0736 (OP should check this)

So in words, even though the chances of losing 4 bets in a row is small, when you do lose 4 bets in a row it costs a lot. It's worth noting that the more bets you're prepared to make, the worse your EV is.

Take for example just making $1 and $2 bets.

EV = 0.4(+$1) + 0.6[(0.4)(+$1) + 0.6(-$3)]
EV = -$0.44

[/ QUOTE ]

SheetWise

10-18-2005, 09:54 PM

You'll find no end of posters who literally believe (from The Doctrine of Chances) that "anything that can happen, will happen" -- and will use abstractions such as infinity to "prove" it. Since your question states "eventually ruin a Casino that has a finite bankroll", it could be proven with 99.999999... certainty.

pzhon

10-18-2005, 10:15 PM

[ QUOTE ]
I made a post on another forum in which I claimed that without betting caps on a game like Craps a double up betting system could be used to eventually ruin a Casino that has a finite bankroll.

[/ QUOTE ]
Your bankroll is also finite.

You have a probability of winning. That probability is less than 1. Your expected value is not positive.

[ QUOTE ]
A poster in response, called me an idiot (might be right), and said that no huge bankroll could overcome a -EV game? I believe he is wrong and I am right. I would like some help from you probability guys to decide this issue.

[/ QUOTE ]
Your friend is right that you can't produce a positive expectation by a betting system. You need to have a +EV game. This has been covered many times before.

Imagine your task is to write down numbers that add up to 100. You can be as creative as you want, you can even use a random number generator, but you can only write down negative numbers. This is impossible for essentially the same reason you can't beat a -EV game by a betting system. I hope that makes it more intuitive to you.

10-18-2005, 10:46 PM

Infinite (dream world) Yes, because it's never ending. All the money in the world today, NO! So it doesn't really matter but;

Consider this, there are world records for winning and losing streaks recorded in the history of gambling. For example, games like craps, roulette, BJ, etc. Think about how many times the games have been played. "IF" one could figure out how many hands of BJ have been played vs. the longest losing steak recorded should reflect what the odds say should happen. 1,000 hands= (1) 10 hand losing streak, 1,000,000 hands= (1) 20 hand losing streak, etc. Not exact but you get the point. Sooner or later the streak will happen if you play enough and will eventually be more than all the money in the world (not infinite) and the world will go broke except for one person. Sooner or later gambling winning streaks and losing streak records will be broken once enough events have happened which is never ending. Though it will take many life times to see this.

Goodluck

Vincent Lepore

10-18-2005, 11:56 PM

[ QUOTE ]
Without involving infinities it will still always be -EV

[/ QUOTE ]

Well the point is not to change a negative EV to a positive EV. The point is that involving infinites on the side of the -EV player one can take all of the Casino's finite bankroll. I do not know the math or the xact amount necessary but I would bet that one could come up with two reasonable finite numbers (one for the Casino and one for the opponents) would yield a > %99.999999 probability that the -EV player would win all of the Casino's money using a double up system

Vince

Vincent Lepore

10-19-2005, 12:21 AM

Suppose the Casino has a $1 billion dollar Bankroll. Suppose I have a 500 trillion dollar bank roll. Both a lot less than infinite. I bet my bankroll on the pass line of their dice table until one of us is broke. What is the probability that I break the Casino before they break me? I bet that is close enough to %100 in my favor as to be accepted as the expected result. For a bunch of mathematicians I don't understand how you guys cannot accept the power of large numbers. Don't forget the Casino has no Max on their bets so I just pick a number large enough in comparison to their bankroll that makes it a near certainty that I win. It doesn't have to be an infinite number. The casinos also know that there are people in this wolrd that have bankrolls that can cause extreme damage and even destroy them if they do not cap their bets.

Vince

10-19-2005, 12:50 AM

[ QUOTE ]
[ QUOTE ]
Without involving infinities it will still always be -EV

[/ QUOTE ]

Well the point is not to change a negative EV to a positive EV. The point is that involving infinites on the side of the -EV player one can take all of the Casino's finite bankroll. I do not know the math or the xact amount necessary but I would bet that one could come up with two reasonable finite numbers (one for the Casino and one for the opponents) would yield a > %99.999999 probability that the -EV player would win all of the Casino's money using a double up system

Vince

[/ QUOTE ]

See my post

10-19-2005, 12:56 AM

Hi Vincent-

I'm going to make a nice assumption about you and think you aren't a troll, despite a flood of silly posts today by you. Here's my take on the matter:

If Casinos didn't cap their bets, yes, theoretically a gambler with a giant bankroll could ruin them. Let's make this easy: Assume the game gives you a 40 % chance of winning, so it's -EV. If you have a bankroll of $1.023 trillion and the casino has $1 billion, you start by wagering $1 billion. If the casino wins, you wager $2 bil, then $4, etc, until they are broke or you are. The casino would have to beat you 10 times in a row to take your $1.023 trillion bankroll. (2^10) With a 60 % chance of beating you in any given game, the casino will succeed at this venture .6 % of the time, or 6.05 times out of 1,000. (.6^10) So, 6 times out of 1,000, the casino takes $1 trillion from you. 994 times out of 1,000, the casino gives you $1 billion. This is a net loss of $5.05 million for you, per hand of blackjack or whatever it is we're playing.

This seems like bad odds for you, and I'm sure billionaires would be lining up to bankroll the casino. This is why this theorem doesn't work. Casinos cap their bets not because it's profitable, but because they don't have a line of billionaires waiting to back them. But if someone set up an explicit deal such as yours, I'm sure there'd be a long line /images/graemlins/smile.gif

This, by the way, is why Negreanu's strategy of a massive NL buyin isn't fundamentally a bad thing for the players. Whether one should be playing against Negreanu is another story, but if he's making poor plays, then it's +EV for a good player to take him on, as long as the player keeps a good amount of their bankroll off the table at any given time.

10-19-2005, 01:03 AM

Your talking about doubling up, not just making any bet you want. After you lose 29 bets in a row starting with a $5 bet you will need to wager $2,684,354,560 to win $5 assuming it's a 0 EV gamble. OK, so you have $500 Trillion to crack the casino for $1 Billion. How long do you think it would take to crack the casion of $1 Billion by doing this???

10-19-2005, 01:12 AM

Exactly, the casinos have a BR in respect to the bets they allow the people to wager so there is hardly and ROR, just like us poker players. But Vincent was talking about the Martingale system not just trying to crack them in one shot.

Vincent Lepore

10-19-2005, 01:26 AM

Thank you for the post it was very well done. I accept your conclusion especially the one I was looking for:

[ QUOTE ]
If Casinos didn't cap their bets, yes, theoretically a gambler with a giant bankroll could ruin them.

[/ QUOTE ]

You may call my posts silly but anyone reading them should have gotten the point that my comments were based on "theory". But instead I got a rash of insults. You at least understood that theoretically large numbers are very powerful.

I reread Negreanu's comments on massive bankroll. I will agree that there is no inherent mathematical advantage to a massive comparative buy given the conditions he outlines But there are advantages that a skilled player can develop through proper use of his massive bankroll. As long as the skilled player is making good plays a skilled opponent will not gain anything but the massive stack will likely prevent a him from making marginal +EV calls. Mainly because when making a decision the more likely the call is close the less likely a good player will make it. That is an advantage but not really quantifiable.

Another factor that will give the massive stack an advantage is that there are players that rarely, if ever, quit when they double up once or even twice if they havent been playing for a long period. Now the massive stack will always have them covered and be able to turn an earlier loss into a win. Even the greatest players make mistakes and also lose close decision. The massive stack will be able to always be in a position to take all of there chipsvalmost without regard to how much they have built their stack when this happens. Negreanu does not give enough credit to the psychological part of poker and the thought of losing ones buy in in a single swoop. Negreanu claims that poker is a people game yet points to mathematics when it is to his advantage. Yes there is no way to quantify the psychological effect that facing losing everything one has in front of them has on their game but I bet it is a substantial consideration to most solid poker players.

Vince

Vincent Lepore

10-19-2005, 01:31 AM

[ QUOTE ]
But Vincent was talking about the Martingale system not just trying to crack them in one shot.

[/ QUOTE ]

Yes, I did mention Martingdale but I also said a double up system. I believe that if you allow an inifinite bankroll then the martingdale system will also work to bust a finite casino bankroll.

Vince

Vincent Lepore

10-19-2005, 01:33 AM

The point is not how long it takes. The point is whether or not it will happen.

Vince

10-19-2005, 01:40 AM

Hi Vince:

I understand the point you're trying to make. But you do understand that even if the casino can be busted, it doesn't mean the casino should shy away from the venture because of the .006 margin they have against a stack 1,000 times bigger (I know there's a formula we can make from this but I'm tired and have done enough probability today.)

In the same way, poker players against Daniel are going to be at a disadvantage if they don't have unlimited buyins. However, if we use the old 60 % number I keep dragging out because I'm lazy, a player buying in for $1,000 against Daniel's stack (let's say $128k instead of $146k or whatever as I'm still lazy) would successfully be able to take all of Daniel's money 2.7 percent of the time. This doesn't require players to cash out when they're ahead. It does mean if they keep on getting their money in on 60 % advantages, they will eventually be broken by Daniel 97.3 % of the time, but it's still a money-winning proposition for them in the long run, and illustrates why it's to Daniel's disadvantage to do this in the long term. (It might be entertaining enough to counter the -EV , though /images/graemlins/grin.gif )

Vincent Lepore

10-19-2005, 01:49 AM

[ QUOTE ]
keep on getting their money in on 60 % advantages

[/ QUOTE ]

Come on now, how likely is it that people, even pros, are going to consistently have this type of advantage against Daniel Negreanu or any other skilled pro? You know as well as I do that skilled poker players are not going to just blindly blow off their chips. Oh I get it, maybe Daniel started drinking again... He is not being fair with his comments. In the hands of a skilled NLH poker player a massive stack is a horrible thing to face.

Well thanks for answering anyway.

Vince

10-19-2005, 01:56 AM

Well, I read his blog entry the other day and it seemed to me he was making OK but sometimes reckless plays.

[ QUOTE ]
The first hand I played was the Ac 10c and I made it 50. One player called and the button re-raised to 200. I came to play so I called the raise as did the other player.

The flop came Ah Qc 10h and it was checked to the button who bet 650. He only had about 2000 more so I figured, "Why not, he might have A-K," and made it $47,000 to go!

Nope, it wasn't AK, my opponent had a set of queens and I was stuck $2800 right off the bat.

I continued to raise every pot. At one point, two players limped in for $10 and I said,"10? I'm in for 10," and threw out two $5000 chips with the Q-3 off suit. They all folded naturally.

[/ QUOTE ]

I really like Daniel and think he's a great player. But when he describes his play as such, I think a reasonable player could have 60 % or higher with most decisions - just factor in what two cards he's raising with and play really tight. (This is not an evaluation of his normal play, simply the situation he brought up in the blog, which is what I thought we were discussing.)

Vincent Lepore

10-19-2005, 05:59 AM

I thought we were discussing the advantage or disadadvantage of making a massive buy-in. If not I guess this discussion is over with. That in fact was my whole purpose for all of these posts that you found silly. After all the issue is not what DN does or doesn't do. It's the massive buy in that is the issue. At least that's how I see it.

Vince

pzhon

10-19-2005, 08:25 AM

[ QUOTE ]
A poster in response, called me an idiot (might be right), and said that no huge bankroll could overcome a -EV game?

[/ QUOTE ]
You are now being told on multiple forums that you don't understand the martingale.

<ul type="square">"Your understanding of the mathematics behind the Martingale system is off. These topics have been covered in detail in the Other Gambling forum. You should go read them to understand why the Martingale system doesn't work."

"Are you seriously this stupid? No bankroll is BIG enough to beat a negative ev game. "Super Huge" is not "close" to infinity. "[/list]
The people who are telling you that you are wrong are quite well-respected, and many of your statements have been specifically refuted. Why are you wasting everyone's time with these old fallacies? Do you enjoy being wrong publicly, or can you not believe that you are wrong despite the overwhelming consensus against you?

I'm willing to bet a lot of money, at 10:1 odds, that your statements about being able to overcome a -EV game by having a large bankroll and playing a martingale are wrong. If you think you have 1 chance in 11 of being correct (we can let any 2+2 author referee), please accept this wager.

If you don't think you have 1 chance in 11 of being right, please don't repeat yourself. Retract your statements and apologize. If you don't, I will ask the moderators to change your title from addict to troll.

10-19-2005, 10:38 AM

Because I don't have a massive buyin, I think it's more useful in discussing the advantages of playing AGAINST a massive buyin. Interesting discussion though, Vince /images/graemlins/smile.gif

Vincent Lepore

10-19-2005, 01:51 PM

[ QUOTE ]
I'm willing to bet a lot of money, at 10:1 odds, that your statements about being able to overcome a -EV game by having a large bankroll and playing a martingale are wrong. If you think you have 1 chance in 11 of being correct (we can let any 2+2 author referee), please accept this wager.

[/ QUOTE ]

O.K. I made the statement that if a player had an infinite bankroll and played against a casino with a finite bankroll and no betting cap that employing the martingale system would work. I used the pass line of the craps table. Under those circumstances I will take your 10-1 bet.

By the way I'm pretty sure that one would be very close to a certainty to win with a bankroll less than infinite but large enough in relation to the casinos but I don't know what that relationship needs to be. Those fellows that said I was wrong are well...wrong.

Vince

pzhon

10-19-2005, 02:41 PM

[ QUOTE ]
[ QUOTE ]
I'm willing to bet a lot of money, at 10:1 odds, that your statements about being able to overcome a -EV game by having a large [not infinite] bankroll and playing a martingale are wrong. If you think you have 1 chance in 11 of being correct (we can let any 2+2 author referee), please accept this wager.

[/ QUOTE ]

O.K. I made the statement that if a player had an infinite bankroll and played against a casino with a finite bankroll and no betting cap that employing the martingale system would work. I used the pass line of the craps table. Under those circumstances I will take your 10-1 bet.

[/ QUOTE ]
As has been pointed out many times, you do not have an infinite bankroll. You claimed that a martingale would allow someone with a large finite bankroll to "beat" a casino:

<ul type="square">While it is true that there is no one with an infinite bankroll it is also true that there are people with bankrolls large enough to beat a casino with a doubling up system that is given a large number of double ups. (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=3707339&page=2&view=e xpanded&sb=5&o=14&vc=1) [/list]
I'd like to bet on this statement of yours. I'd like to bet on whether Bill Gates, with a bankroll of $100B, can use a martingale on craps to win money on average against a casino with a bankroll of $1. Or do you think the bankroll disparity needs to be greater than 10^11, which would allow the player to double up 36 times?

Vincent Lepore

10-19-2005, 03:48 PM

You are not serious are you? A casino with $1 against $100B with no betting cap? Now that is not a certainty for the player but the probability that the Casino will go broke before the player is pretty close to 1. Your "on average" statement is the kicker here. You can show that if you run this scenario out to infinity that eventually the casino will win but your problem is that now you must use an infinite series of events to prove your point of "on average". That's interesting since you won't let me use an infinite bankroll. But if you would like to give me 10 to 1 on a one time computer simulation of this scenario I'll take the bet that Gates wins. What do you say?

Vince

pzhon

10-19-2005, 04:56 PM

[ QUOTE ]
You can show that if you run this scenario out to infinity that eventually the casino will win but your problem is that now you must use an infinite series of events to prove your point of "on average".

[/ QUOTE ]
No, there is no requirement that something be infinitely repeatable for the average value to make sense. The casino wins on average even if there is only one trial. That's basic mathematics.

[ QUOTE ]
That's interesting since you won't let me use an infinite bankroll.

[/ QUOTE ]
You didn't restrict your comments to the ridiculous (and trivial) case of an infinite bankroll.

<ul type="square">"While it is true that there is no one with an infinite bankroll it is also true that there are people with bankrolls large enough to beat a casino with a doubling up system that is given a large number of double ups." [/list]
Do you stand by your words or not?

Vincent Lepore

10-19-2005, 06:39 PM

[ QUOTE ]
Do you stand by your words or not?....

[/ QUOTE ]

Yes. Anytime you would like to run a one time trial simulation and give me 10 to 1 I'll agree. I'll take the $100B bankroll.

Do you stand by your words?

[ QUOTE ]
The casino wins on average even if there is only one trial

[/ QUOTE ]

Vince

TomCollins

10-19-2005, 09:50 PM

If you lose 500 trillion dollars 1/10000th of the time, its still EV for the casino to play against you. Define what it means to win. If it is busting the casino more than 50% of the time, you have a point. But this is a lot different than having a positive expectation.

pzhon

10-19-2005, 11:23 PM

[ QUOTE ]
Anytime you would like to run a one time trial simulation and give me 10 to 1 I'll agree. I'll take the $100B bankroll.

[/ QUOTE ]
You can't think that resembles my offer. You aren't fooling anyone. You are trying to pretend the issue is something that is both trivial and not in dispute. You have been called an idiot not because you posted correct trivialities, but because you are insisting on things that are wrong.

You are trying to back away from your statement,

<ul type="square">"While it is true that there is no one with an infinite bankroll it is also true that there are people with bankrolls large enough to beat a casino with a doubling up system that is given a large number of double ups."[/list]
and my offer,
<ul type="square">"I'm willing to bet a lot of money, at 10:1 odds, that your statements about being able to overcome a -EV game by having a large bankroll and playing a martingale are wrong. If you think you have 1 chance in 11 of being correct (we can let any 2+2 author referee), please accept this wager." [/list]
Is it really that hard for you to admit that you were wrong?

Vincent Lepore

10-20-2005, 02:50 AM

In fact I was not wrong. Being called an idiot and being an idiot are not the same thing as you so appropriately can attest. No one called you an idiot. See how wrong they are. I stand by what I said. Anytime you would like to wager and perform A trial to prove your point let me know. As for this discussion with you I leave it so you may have the pleasure of conversing with yourself since you adamantly refuse to face the truth.

Bye

TomCollins

10-20-2005, 09:55 AM

The problem is Vince and "the reasonable" people simply have a different definition of "beating" the casino. Vince defines "beating the casino" as taking their entire bankroll a large percent of the time. The reasonable folks think that beating the casino means having a positive expectation.

With Vince's definition, he is in fact "beating" the casino. However, he is not using the same definition as everyone else. So its quite clear why there is a debate on this trivial issue.

Vince, if I am wrong about what you think it means to "beat" a casino, please clarify. The more exact terms the better.

pzhon

10-20-2005, 10:49 AM

[ QUOTE ]
In fact I was not wrong.

[/ QUOTE ]
Then why do you refuse to bet on the whether your statements are correct? Not when you say something like 2+2=4, but the statements with which people disagree, such as the one I keep quoting and you keep deleting.

If you restricted your statements to the things you were willing to bet on, you wouldn't be having arguments on this in multiple forums. Or, I'd be richer.

<ul type="square">While it is true that there is no one with an infinite bankroll it is also true that there are people with bankrolls large enough to beat a casino with a doubling up system that is given a large number of double ups. (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=3707339&page=2&view=e xpanded&sb=5&o=14&vc=1) [/list]
Whether you believe it or not, the martingale does not beat the game for any finite bankroll. 7 double-ups is not enough; 36 double-ups (Gates versus a casino with $1) is not enough. The player loses on average. The casino wins on average. No betting system will allow you to double your money with a probability greater than 50% on a -EV game.

[ QUOTE ]
Anytime you would like to wager and perform A trial to prove your point let me know.

[/ QUOTE ]
If you think being a 10:1 favorite to end a session ahead means you have a good deal, you have a definition of good deal considered absurd by serious +EV gamblers. I'd be happy to be the casino against you, and let you place bets on 35 out of 38 of the numbers in roulette (and quit after one trial, if you want). Are you trying to say that this is a way to beat roulette, because you can win 92% of the time with this system? I'd be annoyed to lose $1000 92% of the time, but the line of people willing to be the casino against you for one trial each would be very long.

Now, if someone had only $1000 or a bit more, they might not like being the casino against you. The risk of ruin might be too high for them. Someone else with a tougher stomach might not mind betting their whole bankroll with an 8% chance to multiply their bankroll by 35. However, no serious +EV gambler would think that you are getting a good deal, even though you win 92% of the time. The martingale is the same. Whether you think you might be too risk-averse to be the casino or not, it is never a good deal for the player, even if the player has the bankroll of Bill Gates.

A probability of winning close to 1 does not allow you to conclude the gamble is good, since the player is risking far more than the house. See the definition of EV.

[ QUOTE ]
you adamantly refuse to face the truth.

[/ QUOTE ]
Then why am I the one willing to bet on whether the statement is true? <font color="white">(Alex/Mugaaz also offered to bet.)</font>

You had a disagreement in the Mid-High NL forum. You came here and asked who was right. You have been told.

Vincent Lepore

10-20-2005, 11:15 AM

[ QUOTE ]
If you think being a 10:1 favorite to end a session ahead means you have a good deal,

[/ QUOTE ]

I have a buddy that whenever he feels he is losing a debate he changes or rearranges the issue to suit (support) his point. I have never and still never calim anything accept that the betting max in a Casino is to prevent ruin. A winning "session" never entered the equation until you mentioned it. The point as you well know is that an infinite bankroll would destroy a Casino and there are finite numbers that would make it so close to certain (mathematically) that a Casino with a relatively inconsequential bankroll would go broke without betting caps. Now that was the arguement, is the arguement and is a correct statement. That's all.

Vince

Yads

10-20-2005, 11:44 AM

[ QUOTE ]
The problem is Vince and "the reasonable" people simply have a different definition of "beating" the casino. Vince defines "beating the casino" as taking their entire bankroll a large percent of the time. The reasonable folks think that beating the casino means having a positive expectation.

[/ QUOTE ]

I think this is what this disagreement is all about. Vince is right, that given a large enough bankroll, a player could bust the casino w/o even using martingale (in fact he would have a better chance w/o using a martingale system,) a large portion of the time. This does not make the game +EV, however, and that's not his point.

SheetWise

10-20-2005, 12:56 PM

[ QUOTE ]
The problem is Vince and "the reasonable" people simply have a different definition of "beating" the casino.

[/ QUOTE ]
We should all agree that given any finite series of trials in a game with a known expectation, the probability of any series occurring will always be > 0. In the martingale (for this argument) the "house edge" is only realized in the occurrence of this single series. Even though it's mathematically correct to calculate the EV of every trial incorporating this single negative outcome, there are times that it's just plain silly to do so. If we believe that "everything that can happen will happen", then we have to believe that every insurance company must eventually go broke. Which makes them a bad investment. If we factor into our decisions occurrences that are less frequent than being hit by a meteorite -- then how do we factor the meteorite risk into our decisions? I think Vince is incorporating reasonable risk-tolerance in his belief that the "infinite monkey" represents zero risk, while pzhon holds to the reality that the number is something > 0 and that the risk has to burden the entire process.

Pzhon has a better chance of proving his position logically, but I don't think even he would suggest logic should always dictate action.

pzhon

10-20-2005, 01:55 PM

[ QUOTE ]

I have a buddy that whenever he feels he is losing a debate he changes or rearranges the issue to suit (support) his point.

[/ QUOTE ]
You must have taken lessons from him.

It is as though you said, "2+2=4 and the square of a negative number is negative." When people point out that you are wrong, and offer to bet that the square of a negative number is positive, you say you stand by your words, and are willing to bet that 2+2=4.

[ QUOTE ]

I have never and still never calim anything accept that the betting max in a Casino is to prevent ruin.

[/ QUOTE ]
Obviously, that is wrong. You have claimed many other things, and you said as much, "I thought we were discussing the advantage or disadadvantage of making a massive buy-in." Here are a few of the incorrect statements you have made in this discussion on the original thread and this one:

/images/graemlins/frown.gif "The cap on betting is the reason that the martingdale system cannot work." (No player has an infinite bankroll. Playing a martingale loses on average with any finite bankroll, and in fact, loses more money on average as your bankroll increases.)

/images/graemlins/frown.gif "There is a built in mathematical advantage that large numbers have over small numbers when it come to probability." (I have no idea what you meant, but the same equations apply to large numbers as to small numbers, even if you find the large numbers exciting and powerful.)

/images/graemlins/frown.gif "While it is true that there is no one with an infinite bankroll it is also true that there are people with bankrolls large enough to beat a casino with a doubling up system that is given a large number of double ups." (No finite bankroll can make a martingale win on average. Even if the casino faces a high risk of ruin, it is not a good gamble for the player.)

/images/graemlins/frown.gif "Whatever you might want to believe you will find that the resulting numbers here will not be overcome by some -EV casino game unless the game is a %100 certainty." (Of course that was wrong. Changing the numbers without making the game a lock could change the player's expected loss from .1% of his bankroll to 99.9% of his bankroll. The player can't expect to win without making the game +EV.)

/images/graemlins/frown.gif "The fact is that the only way to clasiify a game as a negative EV game is to do so by using an infiite number of Trials...when the pass line is evaluated in the Game of craps that to show that it is a -EV bet one must use an infinite number of trials." (Sometimes the idea of repeated trials is used to help explain probability or expected value. However, that is not part of the mathematical definition. Probability and expected value are defined for events that happen only once.)

/images/graemlins/frown.gif "that is close enough to %100 in my favor as to be accepted as the expected result." (That a probability is close to 1 does not make it 1. That something happens with high probability does not mean you can say it is the mathematically expected result. This is particularly important in the context of gambling. See the definition of expected value.)

/images/graemlins/frown.gif "Now the massive stack will always have them covered and be able to turn an earlier loss into a win." (It could happen, but on average, there is no restoring force without a skill advantage. Having a massive stack does not give you a skill advantage.)

/images/graemlins/frown.gif "In the hands of a skilled NLH poker player a massive stack is a horrible thing to face." (It's a problem if you have a deep stack, too. It doesn't matter if you don't have a large stack yourself. In fact, it can be profitable to play a short stack in a table of good players with deep stacks. See KaneKungFu123's posts.)

If you stuck with correct statements, you would not have so many respected posters disagreeing with you.

From your comments, my guess is that you feel it is an advantage to buy in for a huge amount in NL so that you can keep setting smaller stacks in until you bust them. However, you don't have an infinite number of chips, the short stacks don't have to stay at the table until you win, and the short stacks may buy in multiple times, taking many shots at doubling up repeatedly against you.

This reminds me of how I cleared the Bet365 September bonus. I bought in short on several NL 100 tables since I only had $100 on the site and didn't feel like depositing more. On one table, my pushes with AK, AJ, 99, QQ, etc. were called by KK (oops), K5, 22, AT, etc. and I busted out 3-4 times. One opponent cackled, "How many leaves are there on that money tree?" On another table, I bought in for $10 and ran it up to $140. Net gain: $173 before the bonus, and by my estimates I was slightly unlucky. When I started buying in short, my variance dropped tremendously, but to my surprise, my winrate (over 10 PTBB/100 at NL 100) has not fallen.

It is a common fallacy that it is always an advantage to have more chips in NL. As has been pointed out individually by many of the Mid-High NL players, it is tremendously profitable to play a short stack when people with deep stacks are playing badly and think no short stack can hurt them, that their stack size means they don't have to play +EV poker. Sometimes the optimal way to exploit this behavior with a short stack.

Vincent Lepore

10-20-2005, 02:04 PM

[ QUOTE ]
that given a large enough bankroll, a player could bust the casino w/o even using martingale (in fact he would have a better chance w/o using a martingale system,) a large portion of the time. This does not make the game +EV, however, and that's not his point.

[/ QUOTE ]

Thank you. In fact once I realized that I did not need the Martingale I changed to a simple double up strategy. Believe it or not there is an important point here. That is the potential strength of a large bankroll. It is very evident in tournaments. I was trying to apply this same idea to cash games. I realize that there is no inherent (mathematical) advantage to a massive buy-in when compared to the average buy-in by the rest of one's opponents. But if a player was to buy-in comparatively small and not quit until he or the massive bankroll played a significant number of all-in hands against each other I believe that the massive buy-in would have an advantage. The concept of getting the best of it is in direct conflict with this way of thinking and is, in my opinion, what Daniel is saying. He's right of course. When you gamble you want to get your money in with the best of it. But you also want to do it with the concept of "gambler's" ruin in the back of your mind. That is the point. If there is an advantage to massive buy-in it may be an intimidation factor that might not be quantifiable.

Vince

TomCollins

10-20-2005, 02:17 PM

Vince, I'm going to ask you one more time (you must have me ignored).

What is your definition of "advantage" in this case.

Does it mean you will bust the other player a very large percentage of the time?

Or does it mean having +EV?

Vincent Lepore

10-20-2005, 02:17 PM

[ QUOTE ]
"The cap on betting is the reason that the martingdale system cannot work."

[/ QUOTE ]

This is absolutely true. You claim it is untrue and back it up with the statement that no player has an infinite bankroll. An infinite bankroll vs a finite bankroll is part and parcel to my statement. You know that. You refuse to acknowledge that. You must know then that given an infinite bankroll and no betting cap the Martingdale will break a casino with a finite bankroll. I'll wait until you honestly answer this before continuing.

Vince

10-20-2005, 02:36 PM

[ QUOTE ]
[ QUOTE ]
"The cap on betting is the reason that the martingdale system cannot work."

[/ QUOTE ]

This is absolutely true. You claim it is untrue and back it up with the statement that no player has an infinite bankroll. An infinite bankroll vs a finite bankroll is part and parcel to my statement. You know that. You refuse to acknowledge that. You must know then that given an infinite bankroll and no betting cap the Martingdale will break a casino with a finite bankroll. I'll wait until you honestly answer this before continuing.

Vince

[/ QUOTE ]

The -EV game is still -EV game even though the ROR of the casino is 1 or close to one. I thought that originally you meant that an infinite bankroll with a Martingale betting system would turn a -EV game to a +EV game. Now I'm not sure what do you mean when you say that Martingale would work.

pzhon

10-20-2005, 03:02 PM

[ QUOTE ]
"The cap on betting is the reason that the martingdale system cannot work."

This is absolutely true. You claim it is untrue and back it up with the statement that no player has an infinite bankroll.

[/ QUOTE ]
For a martingale to succeed with probability 1, you need both ingredients:

/images/graemlins/diamond.gif An infinite bankroll.
/images/graemlins/diamond.gif No cap on the bets.

With either ingredient alone, a martingale would not succeed with probability 1. In particular, if you have a finite bankroll, and a casino lets you play a martingale with no cap on the betting limit, you will not win with probability 1, and you will lose money on average. The average amount of money you lose is greater when your bankroll is greater.

Unless there are people with infinite bankrolls (not just so large as to impress you, but infinite), your statement is wrong. Since no one has an infinite bankroll, no one could make the martingale succeed with probability 1, even if the casinos removed the betting limits.

I'm willing to bet that a 2+2 author would say your statement above is wrong, even though you say it is absolutely correct. Are you willing to bet on it?

By the way, your theory that the reason casinos have betting limits is to protect themselves from whales who would bet $20 billion is ridiculous. It doesn't explain why the betting limit for a red chip game is often $500 or less rather than $50k or more. By your theory, the betting limit should be the same on all tables, and should vary with the financial health of the casino. The explanations others have given make much more sense.

[ QUOTE ]
An infinite bankroll vs a finite bankroll is part and parcel to my statement. You know that. You refuse to acknowledge that.

[/ QUOTE ]
I'm not sure what you think you just said.

Your statements about infinite bankrolls are completely trivial, when correct. No one is arguing with them.

You also made statements about people with finite bankrolls. All of the applications involve people with finite bankrolls, and buying in for a finite number of chips. Your assumption that the case of a large finite bankroll resembles what happens with an infinite bankroll is wrong, and your statements about the finite case and real world are wrong.

pzhon

10-20-2005, 03:24 PM

[ QUOTE ]
if a player was to buy-in comparatively small and not quit until he or the massive bankroll played a significant number of all-in hands against each other I believe that the massive buy-in would have an advantage. The concept of getting the best of it is in direct conflict with this way of thinking

[/ QUOTE ]
Please elaborate on what advantage you see that is more important than getting the best of it. It looks like a common, easily exploitable weakness that I (and other NL players) exploit regularly when a fish gets a big stack, but maybe you mean something else.

10-20-2005, 04:08 PM

What's the point of bringing Martingale to the equation anyway? If you have an infinite bankroll and infinite time, you could ruin the casino by flat betting. The series of events needed for that is quite unlikely, but you have enough time and money to wait for the win streak to happen.

pzhon

10-20-2005, 04:36 PM

[ QUOTE ]
If you have an infinite bankroll and infinite time, you could ruin the casino by flat betting.

[/ QUOTE ]
No, the probability would be less than 1. See this post (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Board=&Number=683150&page=0& view=expanded&sb=5&o=14&fpart=) by BruceZ on the risk of ruin of a +EV gambler with a finite bankroll.

[ QUOTE ]
The series of events needed for that is quite unlikely, but you have enough time and money to wait for the win streak to happen.

[/ QUOTE ]
It's not just a fixed sequence. If the casino starts with 10 units, it would immediately lose if you hit a sequence of 10 wins. However, by the time that happens, the casino might have 100 units, and the losses would not bankrupt it.

10-20-2005, 04:49 PM

Alright, I thought you could have an infinite series of wins if you had infinite money and time.

bobman0330

10-20-2005, 05:49 PM

Pzhon, I used to think you were an arrogant jerk as a result of a disagreement we got into a few months ago. I still think you're a little arrogant, but this quote:
[ QUOTE ]
/images/graemlins/frown.gif"There is a built in mathematical advantage that large numbers have over small numbers when it come to probability." (I have no idea what you meant, but the same equations apply to large numbers as to small numbers, even if you find the large numbers exciting and powerful.)

[/ QUOTE ]

has forced me to conclude that you're also hilarious. I think the use of the frown as a bullet has something to do with it.

Vincent Lepore

10-20-2005, 06:52 PM

[ QUOTE ]
what do you mean when you say that Martingale would work.

[/ QUOTE ]

I mean the same thing that I've always meant that a player with an infinite bankroll using the Martingale system against a casino with no betting cap would eventually take all of the Casinos Bankroll. I can't believe that this is even being argued. It's understood. I never cliamed it would turn a -EV game into a +EV game. I started this thread to try and determine if the same could be said for a player with a finite bankroll that was massive in relationship to the Casino's if the Casino allowed no limit on betting.

Vince

Vincent Lepore

10-20-2005, 06:59 PM

[ QUOTE ]
Alright, I thought you could have an infinite series of wins if you had infinite money and time.

[/ QUOTE ]

Yes you can and will.

Vince

10-20-2005, 07:08 PM

Right. And the answer you received was "Yes, the overwhelming amount of time a player with a proportionally gigantic bankroll compared to the casino's will bust the casino time and time again." We all agree on this. I think, however, that the point most people are trying to make at this point is that even if you could get ahold of a trillion dollars to bust the casino's billion-dollar bankroll, it's not a good investment and is -EV even if it works most the time. But you already know that. /images/graemlins/grin.gif

Vincent Lepore

10-20-2005, 07:39 PM

[ QUOTE ]
"There is a built in mathematical advantage that large numbers have over small numbers when it come to probability."

[/ QUOTE ]

What I meant by a built in mathematical advantage is that even with finite numbers a relationship can be established between a big vs small number. The relationship can be such that the disparity is so great that if the two sides tossed an unbiased coin the small stack taking tails and the big stack taking heads and bet their bankrolls on each toss the big stack that it is a certainty (close enough) that the big stack breaks the small stack. I believe that if you slightly bias the coin in tails favor say 1% the results will be the same.

Vince

TomCollins

10-20-2005, 07:47 PM

[ QUOTE ]
[ QUOTE ]
"There is a built in mathematical advantage that large numbers have over small numbers when it come to probability."

[/ QUOTE ]

What I meant by a built in mathematical advantage is that even with finite numbers a relationship can be established between a big vs small number. The relationship can be such that the disparity is so great that if the two sides tossed an unbiased coin the small stack taking tails and the big stack taking heads and bet their bankrolls on each toss the big stack that it is a certainty (close enough) that the big stack breaks the small stack. I believe that if you slightly bias the coin in tails favor say 1% the results will be the same.

Vince

[/ QUOTE ]

Very rigorous proof Vince. You should apply for the Nobel Prize and beat Sklansky to it.

TomCollins

10-20-2005, 07:49 PM

What you are arguing is not debated. A big stack will bust a small stack a large percent of the time even if the small stack has an edge. This percent depends on betting patterns, bankroll differences, and edge in the game. Will you beat the casino with this? No. Do you have a large proabability of "breaking the bank"? Depends on those three variables.

pzhon

10-20-2005, 09:28 PM

[ QUOTE ]
[ QUOTE ]
Alright, I thought you could have an infinite series of wins if you had infinite money and time.

[/ QUOTE ]

Yes you can and will.

[/ QUOTE ]
No, that is another false statement by you. The probability of an infinite streak of wins is 0. However, there will be streaks of wins of every finite length with probability 1.

This is really easy to prove.

TomCollins

10-20-2005, 09:38 PM

Its sad how few people understand the concept of infinity /images/graemlins/frown.gif

SheetWise

10-21-2005, 12:02 AM

[ QUOTE ]
Its sad how few people understand the concept of infinity.

[/ QUOTE ]
And sad how many use it in the same way a drunk uses a lamp post -- for support rather than illumination.

Vincent Lepore

10-27-2005, 01:15 AM

[ QUOTE ]
The probability of an infinite streak of wins is 0.
.....This is really easy to prove.

[/ QUOTE ]

O.K. Prove it. I'm open to learning.

Vince

Vincent Lepore

10-27-2005, 01:35 AM

I believe the following applies.

[ QUOTE ]
Theorem 4. The cardinality of the power set of an arbitrary set has a greater cardinality than the original arbitrary set, or |*A| > |A|.
This is called simply Cantor's Theorem. It generalizes the previous theorem, in which we proved that the power set of a particular set, N, had a greater cardinality than the original. The present theorem is trivial for finite sets, but is fundamental for infinite sets.
Proof. Let A be an arbitrary set of any cardinality, finite or infinite. Again we supply a negative proof, and assume that the members of A can be put into one-to-one correspondence with the subsets of A. Take any one of the supposed ways of pairing off the members of A with the subsets of A. Let us say that if a member of A is paired with a subset of A of which it happens to be a member, then it is happy; otherwise it is sad. Let S be the set of sad members of A. Clearly S is one of the subsets of A. Therefore S is paired off with one of the members of A, say, x. Is x happy or sad? If x is happy, then x is a member of the set to which it is paired, which is S, but that would make it sad. If x is sad, then x is not a member of the set to which it is paired, which is S, but that would make it happy. So if x is happy, then it is sad, and if it is sad, then it is happy. Our assumption implies a contradiction and is therefore false. So the members of A cannot be put into one-to-one correspondence with the subsets of A.
But if A and *A cannot be put into one-to-one correspondence, then they cannot have the same cardinality. If so, then the larger one must be *A, for A can be put into one-to-one correspondence with a proper subset *A. For example, if the members of A are A1, A2, A3..., then they can be put into one-to-one correspondence with this subset of *A: {{A1}, {A2}, {A3}...}.

Many profound consequences follow directly from Cantor's theorem. But we make the most important of them explicit in the next theorem.

Theorem 5. If S is a set of any infinite cardinality, then its power set has a greater infinite cardinality, or |*S| > |S|.
This follows directly from Cantor's Theorem (Theorem 4). Cantor's theorem applies equally to finite and infinite sets; this corollary focuses on the important consequence for infinite sets.
If we follow the notation for finite sets, and say that a set of cardinality a has a power set of cardinality 2a, then this theorem asserts that 2a > a, for each transfinite cardinal a.
This theorem asserts that for any infinite cardinality, there is a larger infinite cardinality, namely, the cardinality of its power set. Hence, there is an infinite series of infinite cardinal numbers. We will meet some of the infinite cardinals larger than N shortly.
This theorem also implies that, for every set, there is a greater set. It follows that there is no set of all sets, or no set of everything.

[/ QUOTE ]

I believe that an infinite set of wins is a subset of the infinite set of events of wins and losses.

Vince

jason1990

10-27-2005, 09:26 AM

Let p be the probability of winning the game and assume p<1. Let P_k be the probability that we win every game after the k-th game. If we can show that P_k=0 for all k, then the result is proven.

Let k be an arbitrary integer and let x be an arbitrary positive real number. Since p<1, the limit as n goes to infinity of p^n is 0. Therefore, there exists and integer n such that p^n<x. Let P_{k,n} be the probability that we win all the games from game number k+1 to game number k+n. Then

P_k < P_{k,n} = p^n < x.

So P_k<x for all positive numbers x. Hence, P_k=0. Since k was arbitrary, P_k=0 for all k, and this completes the proof.

jason1990

10-27-2005, 09:28 AM

Cardinality and measure are two different ways of describing the "size" of set. The concept which is relevant in probability is measure, not cardinality.

Vincent Lepore

10-27-2005, 06:55 PM

[ QUOTE ]

Let p be the probability of winning the game and assume p<1

[/ QUOTE ]

I take it that you mean p<1 and >0. You also mean the probability of having an infinite streak of wins and not winning ALL games. Right?

Vince

Vincent Lepore

10-27-2005, 07:08 PM

If you toss a fair coin 2 times the probability distribution will be as follows.

2 heads: .25
2 tails: .25
head:tail:.25
tail:head:.25

If you toss the coin an infinite number of times the probabiliy distribution will be similar. One of the distributions will be an infinite series of heads and one an infinite series of tails, etc. No? Or are you through has mathematics able to eliminate a potential result? I'm sure common sense plays no part in this discussion but even common sense tells us that one possibility is to continuously toss a head.

http://www.netnam.vn/unescocourse/statistics/52.htm

Vince

college kid

10-28-2005, 05:00 AM

[ QUOTE ]
I made a post on another forum in which I claimed that without betting caps on a game like Craps a double up betting system could be used to eventually ruin a Casino that has a finite bankroll.

A poster in response, called me an idiot (might be right), and said that no huge bankroll could overcome a -EV game? I believe he is wrong and I am right. I would like some help from you probability guys to decide this issue.

Vince

[/ QUOTE ]

First, that's what table limits are for. Second, the house's bankroll is bigger than your. Third, even with a huge bankroll, you will eventually hit a streak large enough that you can't afford the double up. And so and and so forth.

This post reminds me of all crap money management systems people talk about with the stock market and think they are such hot [censored] when the market goes up and they are temporarily playing a +EV game. It has nothing to do with the way you are betting. You must get down to the core of how the game works. Either the payoff odss must be changed, or the physical aspects of the game must be altered so the payoffs are better (read: dice control).

Now many people on this site think I'm a damn idiot, but I see so many people who just don't fundamentally understand where the expectation comes from and the uselessness of betting systems. How can they not get it!?

TomCollins

10-28-2005, 10:55 AM

The probability of an infinite run of heads or tails is 0. The probability of having a run of X heads or tails where X is any number, given infinite time and infinite flips is 1.

TomCollins

10-28-2005, 10:59 AM

The odd numbers are a subset of all integers. But the set of all odd numbers and the set of all integers have the same Cardinality.

jason1990

10-28-2005, 11:03 AM

The number p is the probability of winning one game. I was proving that the probability of having an infinite streak of wins is 0. Having an infinite streak of wins is the same as saying there exists k such that you win all games after the k-th game. I'm not requiring p>0. If p=0, then the probability of winning one game is zero, so the probability of an infinite streak of wins is zero.

jason1990

10-28-2005, 11:09 AM

If X is the sequence of all wins and losses, then X is a random variable taking values in the space of all sequences of 0's and 1's. Obviously, X is not a discrete random variable, so the link you provided is not relevant. In fact, the range of X is uncountable, which implies there are uncountably many sequences a such that P(X=a)=0. In fact, if 0<p<1, then P(X=a)=0 for all sequences a. For instance, the probability that every odd flip is heads and every even flip is tails is zero. This is the same phenomenon that occurs with continuous random variables. If X is a uniform or normal random variable, for example, then P(X=a)=0 for all real numbers a. In order to get a non-zero probability, you have to ask for the probability that X lies in some interval.

Vincent Lepore

10-28-2005, 06:06 PM

[ QUOTE ]
Since p<1, the limit as n goes to infinity of p^n is 0.

[/ QUOTE ]

This sounds like a self serving proof. Why does p^n = 0?

Vince

Vincent Lepore

10-28-2005, 06:33 PM

O.K. Jason, I agree with your summation of infinite sequences. Since you know more about these things than I do it is my belief that I do not need to use an infinite sequence to prove that I am correct. I am confident that you can provide proof that the Casino will go broke using finite numbers for each (the casino and the opponent) given no cap on bettings. Correct?

Vince

Komodo

10-28-2005, 07:25 PM

Ok, infinite losing streaks is impossible which is proven.
But its also trivial to prove that infinite money is impossible too, which kills any martingale formula both in practice and in theory.
I dont see how you can allow infinite money and finite losing streaks in the same mathematical formula. It just doesnt make any sense.

AlphaWice

10-28-2005, 09:59 PM

Remember, the initial discussion was supposed to be centered around "why do casinos not have betting caps"?

Suppose you have a casino, raking in $100M a year, with a $10B roll. Investors, backed by $100B, decide to stupidly "martingale" you, in a game of craps/bacarrat. They succeed some % of the time greater than 50% (because their money is much greater than yours). So do you let them bet, knowing they ruin you - or stand pat with guaranteed income?

TomCollins

10-28-2005, 10:23 PM

[ QUOTE ]
[ QUOTE ]
Since p<1, the limit as n goes to infinity of p^n is 0.

[/ QUOTE ]

This sounds like a self serving proof. Why does p^n = 0?

Vince

[/ QUOTE ]

p^n doesn't equal 0. It's limit is zero. This means that you can pick any positive number, as small as you want. Call this number x. For each x you pick, I can pick an n such that p^n < x. So if you make x 1/1 billion, I can pick an n that p^b < 1/1 billion.

jason1990

10-29-2005, 12:05 AM

[ QUOTE ]
O.K. Jason, I agree with your summation of infinite sequences. Since you know more about these things than I do it is my belief that I do not need to use an infinite sequence to prove that I am correct. I am confident that you can provide proof that the Casino will go broke using finite numbers for each (the casino and the opponent) given no cap on bettings. Correct?

Vince

[/ QUOTE ]
The first step toward making a proof is formulating a provable (or disprovable) statement. I don't know exactly what it is you want to prove, but I can try to work with you to create a mathematical formulation of your claim. You said, "the Casino will go broke using finite numbers for each (the casino and the opponent) given no cap on bettings." If I take this literally, it seems to mean

A. Suppose the "Casino" and the "Player" play a game that pays n to 1, the Player's chance of winning is p<1/(n+1), and there are no betting limits. Then there exist finite numbers A and B such that if A and B are the Player's and Casino's bankrolls, respectively, and the Player uses the "martingale strategy," then the probability the Casino goes broke is 1.

Is this what you mean? Somehow, I think it might not be. I'm guessing you mean

B. Let q be an arbitrary number strictly less than 1 (such as 0.99999, for example). Suppose the "Casino" and the "Player" play a game that pays n to 1, the Player's chance of winning is p<1/(n+1), and there are no betting limits. Then there exist finite numbers A and B such that if A and B are the Player's and Casino's bankrolls, respectively, and the Player uses the "martingale strategy," then the probability the Casino goes broke is greater than q.

Is this it? Or is it something else entirely?

Alex/Mugaaz

10-29-2005, 03:57 AM

Vince. Why don't you understand that someone can be the most likely to get all the money on the table but have the worst of it? These ideas aren't contradictory.

Here is a horrible example. You vs 100 ten year olds in a fight to the death. You are definitely not a favorite to be alive at the end of the battle, but none of the ten year olds are the favorite to beat you, each one who fights you is most likely going to lose.

Think about this until you get it. If you can understand this then you can understand why you can't beat a -ev game.

skiier04

10-29-2005, 11:39 AM

[ QUOTE ]
If you toss a fair coin 2 times the probability distribution will be as follows.

2 heads: .25
2 tails: .25
head:tail:.25
tail:head:.25

If you toss the coin an infinite number of times the probabiliy distribution will be similar. One of the distributions will be an infinite series of heads and one an infinite series of tails, etc. No? Or are you through has mathematics able to eliminate a potential result? I'm sure common sense plays no part in this discussion but even common sense tells us that one possibility is to continuously toss a head.

http://www.netnam.vn/unescocourse/statistics/52.htm

Vince

[/ QUOTE ]

You cant flip a coin an infinate number of times.

Vincent Lepore

10-29-2005, 02:00 PM

[ QUOTE ]
Is this it? Or is it something else entirely?

[/ QUOTE ]

Hmmm...Yes with one minor exception. Martingale might be interesting but a simple bet the whole bankroll on each toss of the coin (%1 bias for the casino) will do.

I'm pretty sure the following applies, especially when one uses bet the whole bankroll on each toss.

Oh, one other thing. Thank you for taking the time.

Vince

[ QUOTE ]
Gambler's Ruin

Let two players each have a finite number of pennies (say, for player one and for player two). Now, flip one of the pennies (from either player), with each player having 50% probability of winning, and transfer a penny from the loser to the winner. Now repeat the process until one player has all the pennies.

If the process is repeated indefinitely, the probability that one of the two player will eventually lose all his pennies must be 100%. In fact, the chances and that players one and two, respectively, will be rendered penniless are

(1) P1 = N2/N1 + N2
(2) P2 = N1/N1 + N2

i.e., your chances of going bankrupt are equal to the ratio of pennies your opponent starts out to the total number of pennies.

Therefore, the player starting out with the smallest number of pennies has the greatest chance of going bankrupt. Even with equal odds, the longer you gamble, the greater the chance that the player starting out with the most pennies wins. Since casinos have more pennies than their individual patrons, this principle allows casinos to always come out ahead in the long run. And the common practice of playing games with odds skewed in favor of the house makes this outcome just that much quicker.

[/ QUOTE ]

Vince