Best way to turn $1000 into $1250?
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My crack at it is the player bet in baccarat, betting as follows, until we either reach our goal or go bust:
1. If bankroll >= $625, bet as much as is needed to reach $1250.
2. If bankroll < twice the table min (which I assume to be $25), then bet it all.
3. Otherwise, bet half the current bankroll.
My simulation shows this to have a 79.17% chance of success, which means each attempt would lose $10.38 on average. I might have erred, though, since intuitively it seems that the probability of winning should be <75%.
What other combination of games/betting methods would give a higher chance of winning? And can anyone verify or debunk my baccarat results?
Remember a casino undergoes a sort of Monte Carlo attack every night... if the casino has lasted the simulation must be wrong.
Please note the follow curiousities:
You have not proposed the most talked about game with the lowest house edge: Blackjack.
You have not proposed any of this fancy stuff like 0.6 percent craps at 100x.
You have chosen PLAYER which pays 1:1 rather than BANKER which pays 19:20.
Most important, you've chosen to do what all the mathematicians suggest about a bankroll and House Edge: Walk in and bet the wad!
If those first few bets don't prevail, we are are out of it.
Quote: FleaStiffRemember a casino undergoes a sort of Monte Carlo attack every night... if the casino has lasted the simulation must be wrong.
The simulation *does* show an average player loss. Did you miss that?
Quote:You have not proposed the most talked about game with the lowest house edge: Blackjack.
That's because the goal is to have the greatest chance of winning, not the lowest house edge. Those two are not the same. Your chances of winning any particular hand at BJ are much lower than with baccarat. For a method where you hope to finish after a few hands, you want the chances of winning any particular hand to be much closer to 50%.
Quote:You have not proposed any of this fancy stuff like 0.6 percent craps at 100x.
I'm assuming normal games, not esoteric ones. Besides, I don't know that 100x craps would be increase the chances of winning anyway. I certainly don't know how to devise a betting strategy to do so.
Quote:You have chosen PLAYER which pays 1:1 rather than BANKER which pays 19:20.
Player is easier because there are no commissions to figure. It would be cumbersome at best and impossible at worst to run the system while having to figure commissions if the sequence lasts more than a few hands. And I doubt it would increase the chances of winning, either.
I based the win probabilities off of an ask the Wizard post.
My calculations are a little different, because I didn't read your first post thoroughly. Anyhow, in my calcs, you will wager your whole bankroll if you are under $625, and not half of your bankroll.
FYI, I got my figure for the probability of winning a single hand from the the Wizard's baccarat page. I assumed 8 decks.
Anyway, do you have any ideas of game/betting-method combinations that would yield a chance of winning higher than 79.3%? I'm not the kind of person who usually comes up with the best way to do something like this.
Nice questionQuote: MichaelBluejayHere's an interesting puzzle: What's the best way to win $250 with a $1000 bankroll in the casino? That is, what combination of game and betting method gives the highest probability of winning?
The best it could be with a fair game is just 1000/1250 or 80%
This is Gambler's Ruin 101
under Bold Play
I feel there are a few threads here at WoV about this
(search should work)
The math can be done in a spreadsheet (I did mine in excel a few years ago)
or some code as in R
I get
Baccarat Banker:
> prob # Probability of reaching goal
[1] 0.7931378
> delta
[1] 7.372651e-34
Baccarat Player:
> prob # Probability of reaching goal
[1] 0.7926508
> delta
[1] 0
Craps Pass Line:
> prob # Probability of reaching goal
[1] 0.7923835
> delta
[1] 0
Don't Pass:
> prob # Probability of reaching goal
[1] 0.7924462
> delta
[1] 0
Not really much of a difference between the 4 bets IMO
Code by BruceZ below
(matches my excel results from a few years ago)
br = 1000 # Starting Bankroll
goal = 1250 # Target bankroll
odds = 1 # Payoff Odds
p = (244/495) # P(win bet) Pass Line Craps
max_bet = 5000 # Table max
prob = 0
p_fail = 1
count = 0
while ( (br >= 1) & (br < goal) & (count < 1000) ) {
# Compute probability of completing parlay up
bet = min(br,max_bet)
consec = 0
while (br + bet*odds <= goal) {
br = br + bet*odds
consec = consec + 1
bet = min(br,max_bet)
}
p_parlay = p^consec
# Compute probabilty of reaching goal
bets = 0
if (br < goal) {
while ( min(br,max_bet) >= ceiling((goal-br)/odds) ) {
br = br - ceiling((goal-br)/odds)
bets = bets + 1
}
}
p_goal = ifelse(bets, 1-(1-p)^bets, 1)
delta = p_fail*p_parlay*p_goal
prob = prob + delta
p_fail = p_fail*p_parlay*(1-p)^bets
count = count + 1
}
prob # Probability of reaching goal
delta
Math professor and 'all around good guy' Stewart N. Ethier has a paper (a fun read)
'Improving on Bold Play at Craps' also in his Doctrine of Chances book
that the don't pass bet in craps with odds to be nearly optimal way of hitting a win goal.
I have not duplicated that math because it assumes one can make fractional bets
maybe he wants to join in and show a layman's example with
real world wagers with his claim from 1987
(we were all so much younger then than now)
Good Luck
Anyway, I'm surprised that I apparently got the right answer on the first try.
Thanks 7craps and CrystalMath for the help!
(Bold Play also can minimize the number of games to play)
1/37
> prob # Probability of reaching goal
[1] 0.7799039
> delta
[1] 0
1/38
> prob # Probability of reaching goal
[1] 0.7708013
> delta
[1] 0
more thrill for your entertainment at 35 to 1 payoff
with just a small decrease in the probability of success
Plus one will over shoot the $1250 win goal this way with integer wagers
I appreciate all the math experts on here and they add so much to the forum and are always helpful when someone not as strong at math (like myself) asks for it. But I just dont agree with the header above telling forum members not to post if they "dont have the skills". Buy hey, its just my opinion and this then should be deleted because it really doesnt add to the thread.
Boz
Quote: BozMany of the threads end up off topic or with posts that dont exactly make sense or add anything to the discussion, but that is part of a forum. If this thread is something special or different, I can understand, but maybe you need another category that makes this clear from the start.
It's specific to this thread, certainly not the norm on the WoV forums. The thread was filling up with wild, ridiculous guesses from people who had no clue.
My feeling is that the OP (original poster) has the right to specify what kinds of posts they do and don't want in their thread. Whether other forum members will respect that, and whether admin(s) will agree to enforce it, are other matters, of course.
Everyone has a feeling or maybe even an educated guess when it comes to a gambling question.
There are public opinions held as truths that are totally against the proven truths.
Just with VP say JoB 9/6 (calculated)
turn $1k into $1.25k at $1 denom
at 1000 hands: prob of success is 29.24%
at 2000 hands: prob of success is 39.72%
at 3000 hands: prob of success is 44.98%
I would have to simulate higher # of hands played
(my program is limited)
This brings in the question of how long a wait (how many games are required) until target or ruin is achieved?
with Bold Play it is, with even money bets, just a few games most examples.
But the probability distribution takes a bit more effort to produce.
Maybe I fire up my Excel
or look into the R code when I have more time or interest
added:
for those so interested
The Baccarat Player Bold Play from first post
The first 10 hands played (ties excluded)
and the probabilities of being at a certain Bankroll level
v*p^1 = 1st hand played (formula used in Matrix Algebra Tool program)
It is possible to get stuck in an endless loop.
Lose first wager and second wager, win next 2 and repeat.
Of course the probabilities are very small for this to continue forever.
So the probability of success is really a limit if we want to be exact.
success: 0.792651
ruin: 0.207349
no snap of the matrix p for now
Quote: MichaelBluejayMy simulation shows this to have a 79.17% chance of success, which means each attempt would lose $10.38 on average. I might have erred, though, since intuitively it seems that the probability of winning should be <75%.
"Intuitively", you are about four times likely to reach the $250 win point as you are the $1000 loss point - so the "intuitive" target probability of winning is 4/5, or 80%, not 75%.
Quote: ThatDonGuy"Intuitively", you are about four times likely to reach the $250 win point as you are the $1000 loss point - so the "intuitive" target probability of winning is 4/5, or 80%, not 75%.
Yeah, I said that in a later post.
For example let's say you have $100 and want to double it to $200 (exactly), and you can only play Roulette. Assume standard single-zero roulette which has house edge of 1/37 = 2.70270...% on every bet.
The first idea would be to bet the whole $100 on a single even money bet (red/black, even/odd etc.). Usually betting it all in one round gives the best probability to reach target bankroll but in this case it turns out this is not the best strategy. Can you formulate a betting strategy that gives the best result?
Quote: onenickelmiracleWithout being able to prove it, the best way is a progressive slot machine only playing when you already know you are an advantage. You just have to do nothing until the opportunity arises. I suppose you could use the "must hit" games as just one example of a way to do it.
No, the odds on slot jackpots are so long that the chance of success would be extremely low -- regardless of whether the overall EV was positive or not.
#1 $270 Banker in Baccarat... lowest House Advantage with guaranteed one hand (no splits or doubles) 50.6% chance of a win
#2 $250 Player in Baccarat... 1.235% House Advantage 49.4% chance of a win
#3 $270 in PaiGow Poker... House Advantage is 2.7%, 48.9% chance of a win
#4 $216 on either 6 or 8 in Craps... 1.52% House Advantage 45.45% chance of a win
My personal choice is #3.
**EDIT** I had to look up some old QFit data... Winning 5 before losing 4 in Blackjack is on the order of 90%+. IIRC winning 6 before losing 5 was 97% given the following rules
S17, DA2, DAS, RS4, Aces Split once, Late Surr., six decks (Mohegan Sun Rules) At $250 per bet, win 5 before lose 4 is a good "chain bet". can I get some better numbers for this?, thanks
Sometimes I like to play something I call Double or Nothing. Here's how it works. When I'm in a casino with my friends and we have very limited time, we put a C note in a slot machine and play it on Max Bet until we double it to $200 or more or we lose. If either of those things happen, play stops and we move on. So far, I've done this on two different occasions with one other gambler. The first time, we both doubled up. The second time, I lost all and my friend doubled up. So we've had success 3 out of 4 times. I understand it's a tiny population from which to declare a trend but I think the chance of success is somewhat better than you may imagine.Quote: MichaelBluejayNo, the odds on slot jackpots are so long that the chance of success would be extremely low -- regardless of whether the overall EV was positive or not.
Until recently, I always thought if you flat bet every time and wanted to hit a certain win goal, it didn't really matter if you bet small or bet big since the house edge remained constant... it only affected your time at the table.
Quote: MichaelBluejay[EDIT: As I suspected, this thread quickly devolved. Clueless posts have been deleted. If you don't have the skills to actually calculate the result of your proposal...please don't post. You don't actually *have* to calculate it, but if you *can't* calculate it, then you probably can't answer the question. Thanks.]
Wow. Just wow.
Why not. There exists a house edge. IF you can find a single zero wheel, you expose your bankroll to it once and if you lose you go get drunk.Quote: Jufo81what is the mathematically best way to double your bankroll on Roulette?.
Usually betting it all in one round gives the best probability to reach target bankroll but in this case it turns out this is not the best strategy.
Quote: drjohnnyInteresting thread.
Until recently, I always thought if you flat bet every time and wanted to hit a certain win goal, it didn't really matter if you bet small or bet big since the house edge remained constant... it only affected your time at the table.
The house edge remains constant, a little nip on every bet. The more bets, the more it nips. The longer you bet, the more likely your variance will be in the negative.
Quote: KeyserSozeQuote: MichaelBluejay[EDIT: As I suspected, this thread quickly devolved. Clueless posts have been deleted. If you don't have the skills to actually calculate the result of your proposal...please don't post. You don't actually *have* to calculate it, but if you *can't* calculate it, then you probably can't answer the question. Thanks.]
Wow. Just wow.
Indeed.
Quote: Jufo81I recently thought about a closely related problem: what is the mathematically best way to double your bankroll on Roulette?
For example let's say you have $100 and want to double it to $200 (exactly), and you can only play Roulette. Assume standard single-zero roulette which has house edge of 1/37 = 2.70270...% on every bet.
The first idea would be to bet the whole $100 on a single even money bet (red/black, even/odd etc.). Usually betting it all in one round gives the best probability to reach target bankroll but in this case it turns out this is not the best strategy. Can you formulate a betting strategy that gives the best result?
You've already mentioned the best strategy, one bet spread equally over 18 numbers such as red/black etc.
Quote: FleaStiffWhy not. There exists a house edge. IF you can find a single zero wheel, you expose your bankroll to it once and if you lose you go get drunk.
In this case your expected loss is: $2.70 (expected loss = house edge*amount wagered). But you can turn $100 into $200 with smaller expected loss than this on Roulette. Can you think of how?
Quote: JeepsterYou've already mentioned the best strategy, one bet spread equally over 18 numbers such as red/black etc.
No, think again mate. This doesn't give the highest EV (smallest expected loss) to double your balance.
Quote: drjohnnyInteresting thread.
Until recently, I always thought if you flat bet every time and wanted to hit a certain win goal, it didn't really matter if you bet small or bet big since the house edge remained constant... it only affected your time at the table.
The simplest way to think about this is to know that the House Edge represents a theoretical amount won for the casino every time you make the bet. For example, if you bet $200 on an Even Money bet with the goal of winning $200, then you only expose $200 to the House Edge of the game in question. In contrast, if you flat bet $5, in all likelihood, you win some and lose some, but the result (win or lose) will see you exposing more than $200 in bets to the House Edge, the only extreme exception being if you do nothing but win or nothing but lose...because money won gets bet again.
Quote: MichaelBluejayOkay, I see now why the theoretical maximum is 80% rather than 75%. For some reason I was thinking of the inverse of 250/1000, rather than the inverse of 250/1250.
Since you know what the probability would be 80% with a zero house edge, the problem posed in the OP reduces to the simpler problem that you should bet in such a way to reduce the number of plays to as small as possible. That way you get as close to zero house edge as you can given the game.
Quote: pacomartinSince you know what the probability would be 80% with a zero house edge, the problem posed in the OP reduces to the simpler problem that you should bet in such a way to reduce the number of plays to as small as possible. That way you get as close to zero house edge as you can given the game.
Not necessarily. Please refer to the roulette example I posted in this thread. I posted it here for this very reason.
Why was this post not linked to it's own thread?Quote: Jufo81I recently thought about a closely related problem:
what is the mathematically best way to double your bankroll on Roulette?
For example let's say you have $100 and want to double it to $200 (exactly),
and you can only play Roulette. Assume standard single-zero roulette which has house edge of 1/37 = 2.70270...% on every bet.
The first idea would be to bet the whole $100 on a single even money bet (red/black, even/odd etc.). Usually betting it all in one round gives the best probability to reach target bankroll but in this case it turns out this is not the best strategy. Can you formulate a betting strategy that gives the best result?
Would make the trail easier to follow IMO
It should stand alone, a nice question,
and could end up lost or deleted in this thread.
Maybe that is what you wish?
Maybe I was in direct sunlight too long?
Quote: 7crapsWhy was this post not linked to it's own thread?
Would make the trail easier to follow IMO
It should stand alone, a nice question,
and could end up lost or deleted in this thread.
Maybe that is what you wish?
Maybe I was in direct sunlight too long?
I could have started a new thread for this question but I think it suits nicely to the topic of this existing thread. I believe that threads are supposed to evolve with new contributions and sometimes cover topics that extend beyond the original question by the OP. Starting a new thread for every single question or new idea is not practical IMO.
Strategy:
1.) I would say either betting $250 on the Don't Come after 4-5 rolls after the point is established for the same shooter
2.) Pass Bet $5 + ~$170 odds probability weighted such that the return on odds = $245 (guestimate) - assuming you hit the point, I would continue to reduce the odds by the weighted probability needed to hit your total return factoring in your prior winning.
3.) Place the 6 and 8 for 108 each.
If you're darksider, you like strategy 1 more than 2 and 3. =)
Quote: pewIt makes no difference what you play. It has nothing to do with mathematics. You peeps have to face the fact that there is only one way to do it. You need to get LUCKY! Period end of story. (sorry)
The question is how lucky you have to get, actually. It has everything to do with mathematics.
Step 1: Borrow $250 from Bob Dancer.
Step 2: Don't pay back Bob Dancer.
On a side note, I really don't have anything against Bob Dancer. I just think the situation is funny.
Quote: debitncreditAnswer:
Step 1: Borrow $250 from Bob Dancer.
Step 2: Don't pay back Bob Dancer.
Ladies and gentlemen- We have a winner!
Quote: debitncreditAnswer:
Step 1: Borrow $250 from Bob Dancer.
Step 2: Don't pay back Bob Dancer.
On a side note, I really don't have anything against Bob Dancer. I just think the situation is funny.
Winner!
Thank you Professor Ethier for joining in.Quote: MoivreI get the same result for betting on Player.
BTW (off topic - no need to answer)
Does any of your family members or close friends call you "Stewie"?
I can not help but chuckle (maybe giggle) when I see your name in print
Stewart Gilligan "Stewie" Griffin also comes to mind
If you are NOT SN Ethier
that is nice too
Find a stock that has options and buy some out of the money calls a few days before earnings are announced. Sell the call options right before the earnings are announced. I've been watching stocks lately with an eye toward the reporting date of earnings and this seems especially true with high profile stocks. The price of the stock runs up right before earnings are announced.Quote: MichaelBluejayBest way to turn $1000 into $1250?
Doesn't work every time but neither does dice influencing.
Quote: WizardLet it be known that MichaelBluejay is no longer a secret admin.
And this forum will be none the better because of his departure. Your integrity will be missed !
The logic runs as follows (using a spreadsheet)
Phase 1 - you try to bet enough to get to $1250 and get out winning.
(i) Start bankroll $1000, Bet $250 - Win Pr=P
(ii) Bankroll now $750, bet $500 - Win Pr=(1-P)P
Phase 2 - you place everything to get back to your starting capital
(iii) Bankroll now $250, bet $250 - Win Pr=(1-P)(1-P)P Lose and out (1-P)(1-P)(1-P)
(iv) Bankroll now $500, bet $500 - Win Pr=(1-P)(1-P)PP Lose and out (1-P)(1-P)P(1-P)
Now you've either
(a) Won = .743 128 592
(b) Lost = .194 394 695
(c) No result = .062 476 713
Phase 3 repeat above.
Clearly No result has the same chances as when you started over - so using recursion the p(winning)=.743/(.743+.194)
The other method, using similar logic, was just betting numbers (1/37) allowing fractional bets (the first 57 bets allowed you to reach your target) and if the 58th spin wins you have enough to make 5 more bets etc. In theory by the 63rd spin you've got to within 99.999% of the payback and has a 79.096% chance.
Quote: s2dbakerFind a stock that has options and buy some out of the money calls a few days before earnings are announced. Sell the call options right before the earnings are announced. I've been watching stocks lately with an eye toward the reporting date of earnings and this seems especially true with high profile stocks. The price of the stock runs up right before earnings are announced.
Doesn't work every time but neither does dice influencing.
NXPI: Superconductor that is being put into all smart phones to allow customers to use their phones to make purchases. It's huge in Europe, slowly gaining ground in America. This stock is a 1st mover. When I found the stock it was 25.00. Last time I checked it was 34.00. It's a buy/hold until 2015.
Through pimping this stock to the 2-3 hundred people here, I promise I will not make any money as I do not own any of the stock. I just felt like sharing.
It will take 15 years, but Saving Bonds will get you there. I really like watching the value of mine grow month to month.
New England over Buffalo: While the money line won't get you there you can pair it up with either the Colts or Denver for a 2 team parlay. I haven't done much research on the Colts or Bronco, but Buffalo doesn't have a quarterback at the moment. Therefor, Patriots over Buffalo is my play of the week in Week 1 of the NFL Season.
