Playing Break-Even

SleightOfHand

Well-Known Member
#1
Lets say that you want to play a break even game. If I am correct, this means a RoR of 100%. How do you decide what your betting units will be?
 

UncrownedKing

Well-Known Member
#2
SleightOfHand said:
Lets say that you want to play a break even game. If I am correct, this means a RoR of 100%. How do you decide what your betting units will be?
A break even game has a 0% HE. Its like betting on coin flips. If you flat bet, half the time you will win, half the time you won't. But after 100 million plays, you will wind up exactly where you started.

The highest RoR you would have would be if you bet everything on one flip, which gives you 50% RoR. Which is equivalent to a 1 unit BR.

Someone correct me if I'm wrong, but you should be able to apply the following formula for break even RoR(for the coin flipping example, I don't think it would be accurate for break even BJ spread):

(.5)^BR in units
 
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sagefr0g

Well-Known Member
#3
SleightOfHand said:
Lets say that you want to play a break even game. If I am correct, this means a RoR of 100%. How do you decide what your betting units will be?
i don't know the answer to this question but it is exactly what i've been trying to figure out, of late.
what i'm finding from looking at cvcx sims for play all and changing bet spreads for a given bankroll is that it appears your limited to how close you can get to break even by the table minimum allowed and lowest denomination chips, i guess. but yeah the ROR goes way up there as you try and adjust the spread down towards playing a break even game.
like the sim below i got the dollars per hour down to zero in the custom column and the ROR is only 99.3% lol.
not sure if this sort of stuff is what your getting at though.:confused::whip:
edit: if you can raise your bankroll way up there (in this case from six grand to nineteen grand) with this spread and game you can get the ROR down to around 26.4% but your gonna make around 35 cents per hour then. lol.
 

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SleightOfHand

Well-Known Member
#4
sagefr0g said:
i don't know the answer to this question but it is exactly what i've been trying to figure out, of late.
what i'm finding from looking at cvcx sims for play all and changing bet spreads for a given bankroll is that it appears your limited to how close you can get to break even by the table minimum allowed and lowest denomination chips, i guess. but yeah the ROR goes way up there as you try and adjust the spread down towards playing a break even game.
like the sim below i got the dollars per hour down to zero in the custom column and the ROR is only 99.3% lol.
not sure if this sort of stuff is what your getting at though.:confused::whip:
edit: if you can raise your bankroll way up there (in this case from six grand to nineteen grand) with this spread and game you can get the ROR down to around 26.4% but your gonna make around 35 cents per hour then. lol.
I would think that the decrease in RoR is from the change in advantage from .001% to .0046%. My question is: since a break-even game is 100% RoR, what kind of betting units/ramp would one choose? Perhaps one that generates a low trip RoR? And if that is the case, what kind of parameters (hours played/trip BR) would be recommended?
 
#5
Hi SleightOfHand,

This is pretty easy. If you insist on playing, you should bet and play as little as possible. I recommend betting $0 and playing 0 hands, unless your objective is something other than winning money.
 

jimbiggs

Well-Known Member
#6
SleightOfHand said:
Lets say that you want to play a break even game. If I am correct, this means a RoR of 100%. How do you decide what your betting units will be?
I don't know what the answer is for a break even game, but if I wanted to reduce heat and play a slightly better than break even game, then I would just take my normal bet size and reduce the spread. For example, instead of spreading 1-8 on a DD game, I would only spread 1-4.
 

DonR

Well-Known Member
#8
non-self-weighter said:
Hi SleightOfHand,

This is pretty easy. If you insist on playing, you should bet and play as little as possible. I recommend betting $0 and playing 0 hands, unless your objective is something other than winning money.
While I agree that the objective should be winning money, you can still get a few comps along the road while playing the break even game that SleightOfHand is talking about. You are still a winner, in a way. I know it's not much, but it is certainly better than just playing BS and losing. If you are limited to a few casinos only, where everybody knows you, this may be a good way of avoiding heat and having a few freebies, while still having fun.
 

SleightOfHand

Well-Known Member
#9
non-self-weighter said:
Hi SleightOfHand,

This is pretty easy. If you insist on playing, you should bet and play as little as possible. I recommend betting $0 and playing 0 hands, unless your objective is something other than winning money.
Well, obviously the objective is something other to win money as I am playing a break-even game. Perhaps have a unit/ramp that will make my standard deviation a bit less than that of a standard ramp?
 

sagefr0g

Well-Known Member
#10
SleightOfHand said:
I would think that the decrease in RoR is from the change in advantage from .001% to .0046%. My question is: since a break-even game is 100% RoR, what kind of betting units/ramp would one choose? Perhaps one that generates a low trip RoR? And if that is the case, what kind of parameters (hours played/trip BR) would be recommended?
yeah, i screwed that up letting the program change the ramp to get the .0046% . sorry i didn't notice that. well anyway below is that ramp giving the .0046% in the custom bets for a six grand roll and shows a higher ROR than for the nineteen grand roll. 65.7% ror for the six grand roll and 26.4% for the nineteen grand roll.

whatever, i'm not even sure we're talkin about the same thing. like when you say break even game i'm not sure what you mean. and as far as a 100% ror for a break even game, well that just doesn't sound right to me. might be right but to me if you say i'm playing a break even game then your not losing money, not winning any either, so to me that shouldn't end up being a 100% ror.

i mean usually if your just playing basic strategy (excepting for some good single deck game) then your not playing a break even game, it's a losing game, sort of a slow bleed, with a 100% ror, lol.
but to me, it would seem theoretically at least that a counter could play a break even game to where he doesn't lose or doesn't win money, just stays even. but i think table min's and chip denomination size is a good bit of what gets in the way of a counter being able to bet size and ramp so as he can just play a break even survival sort of game.

anyway that's the sort of thing i'm currently curious about. remember the chat last evening, where i was asking about me and the wife sitting at the same table just hoping to play a break even game? you told me about BJFB.:)

like me playing alone, i can play a near break even game or better with my measly six grand roll. thing is when my wife and i sit and play together then it's like me playing two hands all the time. and if it's play all then it's like having to bet two units at zero and negative counts instead of one unit like i'd be doing if i was playing solo. turns out to be very costly betting those two units in zero and negative counts instead of just one unit.

i guess the only really solution for us is to have a much larger bankroll in order to get that ror down if we are going to play like that. :(
 

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1357111317

Well-Known Member
#11
I think the reason why playing a $0.00 Expectation game is that at some point in time you will hit a one in a million losing streak that will break your bankroll no matter how big it is. I might be a one in a million or one in a billion chance that you hit that losing streak but at some point if you play infinitely you will hit that streak. I think for a breakeven game you would have to calculate the odds of hitting that ridiculous losing streak. That is probably what you are looking for.
 

nightspirit

Well-Known Member
#12
sagefr0g said:
like me playing alone, i can play a near break even game or better with my measly six grand roll. thing is when my wife and i sit and play together then it's like me playing two hands all the time. and if it's play all then it's like having to bet two units at zero and negative counts instead of one unit like i'd be doing if i was playing solo. turns out to be very costly betting those two units in zero and negative counts instead of just one unit.

i guess the only really solution for us is to have a much larger bankroll in order to get that ror down if we are going to play like that. :(
Wait, when you adjust the size of the units for your 2 hands then your ROR shouldn't change.
 

sagefr0g

Well-Known Member
#14
nightspirit said:
Wait, when you adjust the size of the units for your 2 hands then your ROR shouldn't change.
thing is i'm looking at it for the same bankroll, sort of thing.
a six grand bankroll for playing one hand play all ....ror 19.8%
and
a six grand bankroll for playing two hands, play all .... ror 30.7%
so i guess when you go to the two hands, play all it's just not enough bank to keep the ror lower.:confused::whip:
note: in the two hand sim i had to fudge the custom bet at tc=1 to get it similar to the one hand sim.
 

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Pelerus

Well-Known Member
#16
sagefr0g said:
when you say break even game i'm not sure what you mean. and as far as a 100% ror for a break even game, well that just doesn't sound right to me. might be right but to me if you say i'm playing a break even game then your not losing money, not winning any either, so to me that shouldn't end up being a 100% ror.
From p. 112 of Blackjack Attack,

RUIN = [(1 - w/sd) / (1 + w/sd)] ^ bank/sd,

where w = win rate. Thus if the win rate is equal to zero, as it is in the case of a break-even game, the ruin must equal 1, or 100%.

I believe that Mr. 1357111317 is correct in the following:

1357111317 said:
I think the reason why playing a $0.00 Expectation game [has a 100% RoR] is that at some point in time you will hit a one in a million losing streak that will break your bankroll no matter how big it is.
Regardless of the starting size of one's bankroll, playing a break-even game against an opponent with an infinite bankroll will inevitably result in ruin, provided that the game has any element of variance whatsoever.

That being said, the ruin formula as presented above is tailored to an AP audience (for whom win rate > 0), so it doesn't provide much information about the case of win rate = 0, or for that matter, win rate < 0. The RoR in both of those two cases is the same (100%), but for our purposes that is not enough information - it would not be correct for us to say that the case of a player with a $0 EV is the same as that of a player with a -$20 EV, because given the same starting bankroll, the latter player will reach his ruin much more quickly than the former.
 

nightspirit

Well-Known Member
#17
sagefr0g said:
thing is i'm looking at it for the same bankroll, sort of thing.
a six grand bankroll for playing one hand play all ....ror 19.8%
and
a six grand bankroll for playing two hands, play all .... ror 30.7%
so i guess when you go to the two hands, play all it's just not enough bank to keep the ror lower.:confused::whip:
note: in the two hand sim i had to fudge the custom bet at tc=1 to get it similar to the one hand sim.
I see, you are using the same unit size. Yes, if your bankroll would allow you to play a $10 unit then playing two times $5 would surely keep your ROR lower.
 

stophon

Well-Known Member
#18
Well it's already been said, but if you play a break even game forever your RoR is 100% because your opponents bankroll is infinite and yours is not.

But there is also an equal chance you will double your bankroll as there is that you will lose all of it. So this means that your RoR is only 50% if your playing until you double. Your trip RoR will fluctuate largely with the variance of the game your playing.

And answering some of the earlier posts, there is no way to have a 100% RoR unless you play for an infinite amount of time.

The RoR for betting your whole BR every time is (1-.5^n), where n is the number of times that you play. So even though your RoR rapidly approaches 100% as you keep playing it will never reach it (it would be an asymptote if you were to graph it).
 

sagefr0g

Well-Known Member
#19
Pelerus said:
From p. 112 of Blackjack Attack,

RUIN = [(1 - w/sd) / (1 + w/sd)] ^ bank/sd,

where w = win rate. Thus if the win rate is equal to zero, as it is in the case of a break-even game, the ruin must equal 1, or 100%.

I believe that Mr. 1357111317 is correct in the following:



Regardless of the starting size of one's bankroll, playing a break-even game against an opponent with an infinite bankroll will inevitably result in ruin, provided that the game has any element of variance whatsoever.

That being said, the ruin formula as presented above is tailored to an AP audience (for whom win rate > 0), so it doesn't provide much information about the case of win rate = 0, or for that matter, win rate < 0. The RoR in both of those two cases is the same (100%), but for our purposes that is not enough information - it would not be correct for us to say that the case of a player with a $0 EV is the same as that of a player with a -$20 EV, because given the same starting bankroll, the latter player will reach his ruin much more quickly than the former.
so really two things i guess one can say.
the risk of ruin math for AP's is a bit more informative for winning games than break even or losing games.
and it would seem that practically speaking if one is playing a break even game that it's more the great 'tsunami' or series of lesser 'tsunami' standard deviation events in the negative sense that will ruin the player more than anything else otherwise it would seem the break even player just might be able to play a long happy recreational life time, lol.
so i guess some one playing a break even game has a lot less room for error than an optimal betting player, sort of thing.

but anyway, i guess this idea of the danger of standard deviation dragging one down the tubes when playing a break even game makes sense, since there is the affect of the ror lowering some just by raising ones bank roll size. just gives one a bit more slack in the hang man's noose i guess, lmao.:eek:

i guess this is getting a bit voodoo to mention here, but the idea of playing a break even game sort of thing begs the question of if the idea of 'regression to the mean' or the 'law of averages' sort of ideas might become more meaningful for the survival minded player.:rolleyes::whip:
 

Pro21

Well-Known Member
#20
What is the question you are really asking?

If you want to play a break even game I am guessing that you plan to make it up elsewhere like in comps. You spread say 1-3 or 1-4 on a 6 deck game you are about break even and you shouldn't get any heat. But then the question should be what do I have to bet to get the comps I want, and then can I afford to bet that amount.

If you could clarify what you want to know you might get a better answer.

By the way, this idea that a game with an expectation of zero suddenly becomes expectation of -1 because of bankroll size is ludicrous.
 
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