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onenickelmiracle
onenickelmiracle
Joined: Jan 26, 2012
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September 13th, 2014 at 7:27:07 PM permalink
Some casinos won't allow large bets on roulette. Years ago I remember Turning Stone having $2000 allowed outside and only $25 inside.

Played $40 yesterday on roulette losing but close(#30 not #9), then won $250 from a $20 buy in. Was absolutely wasted and even accidentally bet the first ten on the pass line after the point was made. 7 out the first throw and was shocked I lost.

A few 7-11s, doubling pass line wagers, then $10 pass lines minimum with odds on 4 to get into comfort. Skipped one odds on 6 point bet early because the dice landed 45° angle against the wall-voodoo maybe thinking 6 wasn't supposed to be the number. Nothing happened too much in succession though.
I was the hot shooter for 20-30 minutes though, so the probability was probably about 40%. Lol
The most I took was 5x odds by the way.

Somehow say you go $200 to $7000, I don't want to hold so much as $3200 and even risk it to get there.
I am a robot.
Ahigh
Ahigh
Joined: May 19, 2010
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September 13th, 2014 at 7:28:22 PM permalink
Quote: Wizard

Earlier this week there was a show on the Discovery Channel called Beating the House. Part of the show featured Andy Bloch making the claim that the best way to turn $30 into $1000 was to bet the full $30 on a single number in double-zero roulette.

The probability of success with that strategy is 1/38=2.63%.

An obviously superior strategy would be to bet just $27 on a single number. If you win, you'll be at $975. If you lose you'll be at $3. Either way, make the smallest bet possible that will put you at exactly $1000 on any of the available bets, or your full bankroll if you don't have enough to reach $1000 in one bet. If such optimal bet is not an integer, then round down. Keep repeating until you get to $1000 or $0. I find the probability of success is 2.81%.

However, an even better strategy of betting 1/6 of your bankroll on the pass line in craps, followed by max odds (assume 3-4-5x allowed). If this would put you over $1,000 or cause a bet that isn't evenly divisible by $1, or cause a win that isn't divisible by $1, then round the bet down. I find such a strategy has a probability of success of 2.91%.

I'm going to look at the don't pass plus laying odds next.

The question for the poll is what would be your first bet if the goal was to maximize your chance of turning $30 into $1000? Let's assume the standard casino games available in any medium-sized casino.



Vegas 2047 will allow you to bet $30 to win $1000 in a single bet. Our entire mission is GIVING ALL PLAYERS THE BEST POSSIBLE CHANCE TO WIN.
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Wizard
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Wizard
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September 13th, 2014 at 7:48:21 PM permalink
Well blow me down.

I just did a don't pass plus odds simulation and the probability of reaching the $1,000 goal using the same kind of grinding technique, careful to never exceed the goal or lose anything to rounding, is 2.92%.

It's not whether you win or lose; it's whether or not you had a good bet.
Mission146
Mission146
Joined: May 15, 2012
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September 13th, 2014 at 8:06:44 PM permalink
Quote: Wizard

Earlier this week there was a show on the Discovery Channel called Beating the House. Part of the show featured Andy Bloch making the claim that the best way to turn $30 into $1000 was to bet the full $30 on a single number in double-zero roulette.

The probability of success with that strategy is 1/38=2.63%.



I don't necessarily know what the best strategy is, but I can do better than this Andy Bloch without even thinking about it very hard.

First off, take the full $30 and Reverse-Martingale at Craps, five wins in a row goes ($60-$120-$240-$480-$960) and has probability:

(.4929)^5 = 0.02909337405

At this point, you need $40 to reach a total of $1,000, so you're going to Martingale as follows:

$40 ($920 left), $80 ($840 left), $160 ($680 left), $320 ($360 left).

Okay, so you take the probability of initial success of 0.02909337405 and then you only have to win one out of four to be at exactly $1,000, and the probability of losing four in a row is:

(.5071)^4 = 0.06612633336

Now, you have $360 left, so really you should bet it all to $720 and then bet $280, but all of those possibilities are going to take half of forever, so I'm just going to go for two wins in a row and not be heart-broken by the fact that I have $1440 instead of $1,000, if successful.

Okay, so the ways for me to lose after getting to the $960 are to either lose five in a row, or to lose four in a row, win one, and then lose again.

(.5071)^5 + ((.5071)^4 * .4929 * .5071) = 0.05006091356

In order for that 5.006091356% event to matter in the first place, I must first have won the initial five in a row, so I have to multiply the probability of winning the five in a row by the probability of the subsequent events (Losing five in a row thereafter OR losing four, winning one, and then losing) occurring and then subtract that from the probability of winning five in a row in the first place. If my methodology is sound, then the result of the formula will be the probability of success given this strategy:

0.02909337405 - (0.02909337405 * 0.05006091356) = 0.02763693316

This reflects a probability of success of 2.763693316%, which is higher than the 2.63%, and moreover, if I went as far as to bet $280 when at the $720 level and then, if that failed, tried to double $440 into $880 and then tried to bet $120 followed by $240 followed by $480 and then to double the remaining $40....etc....etc....etc....the probability of success would be higher.

For example, the series where the player wins five, loses four, wins one, loses one, wins one, loses one, wins one and then wins again:

(60-120-240-480-960-920-840-680-360-720-440-880-1000 ---$120 is the last bet) adds:

(.4929 * .4929 * 4.929 * .4929 * .4929 * .5071 * .5071 * .5071 * .5071 * .4929 * .5071 * .4929 * .5071 * .4929 * .4929) = 0.00029200549 is added to the probability of winning, so:

0.02763693316 goes up to 0.02763693316 + 0.00029200549 = 0.02792893865 is the new probability of winning when you add that possible series.

If we add all other possible series' of wins, then it would probably end up close to 2.9%.

From a practical standpoint, my advice is: If you Reverse Martingale $30 into $960 by winning five PL bets in a row at Craps, the other $40 isn't really THAT important, now is it?
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
mustangsally
mustangsally
Joined: Mar 29, 2011
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September 13th, 2014 at 8:10:18 PM permalink
Quote: Wizard

Earlier this week there was a show on the Discovery Channel called Beating the House. Part of the show featured Andy Bloch making the claim that the best way to turn $30 into $1000 was to bet the full $30 on a single number in double-zero roulette.

I love Andy! Ahhhhhhhhh

He and my husband are from the same town

Do you have exactly the words he said?

I knew the dpass/odds might be the top
Bold Play on the Don't Pass was a paper from the 1980s by 3 well known math guys, including Peter Griffin

The Banker at Baccarat even with the 5% comm is at 2.81% rounded up

how about $30 Banker, after a win, to Roulette
or
craps

dinner and dancing and craps tonight!
Sally
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Mission146
Mission146
Joined: May 15, 2012
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September 13th, 2014 at 8:11:27 PM permalink
Mike,

As long as you're running simulations, do you mind determining the probability of winning with my PL Reverse Martingale never to exceed a total of $1,000 but always Reverse Martingaling if the total is under $500, and maybe the same for Don't Pass? I'm just curious, and it would take most of my life to do every possible (and, likely, practically irrelevant) series by hand.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
mustangsally
mustangsally
Joined: Mar 29, 2011
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September 13th, 2014 at 8:19:38 PM permalink
Quote: Mission146

If we add all other possible series' of wins, then it would probably end up close to 2.9%.

nice try!

I show just the pass line =
0.02785283
(244/495)^5 * 0.9570755

> p <- pdoubb(960,1000,50000,244/495,1,0.000000001)
> p
[1] 0.9570755

lower than the Banker Baccarat method
but very close
Sally
I Heart Vi Hart
Mission146
Mission146
Joined: May 15, 2012
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September 13th, 2014 at 8:27:55 PM permalink
Quote: mustangsally

nice try!

I show just the pass line =
0.02785283
(244/495)^5 * 0.9570755

> p <- pdoubb(960,1000,50000,244/495,1,0.000000001)
> p
[1] 0.9570755

lower than the Banker Baccarat method
but very close
Sally



Thanks!

I figured there would be other methods superior to mine (and there are probably more than that), but I immediately recognized that my method is superior to Bloch's suggestion in terms of probability...00 Roulette is simply the wrong game for anything like this, regardless of the strategy.

I'm also not greedy, I'd take that $960 and run for the door...well...after checking the UX machines.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
mustangsally
mustangsally
Joined: Mar 29, 2011
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September 13th, 2014 at 8:31:12 PM permalink
Quote: Mission146

do you mind determining the probability of winning with my PL Reverse Martingale never to exceed a total of $1,000 but always Reverse Martingaling if the total is under $500, and maybe the same for Don't Pass? I'm just curious, and it would take most of my life to do every possible (and, likely, practically irrelevant) series by hand.

my values are calculated using 2 different methods

i have not completed the Wizard's pass/dpass yet
but soon will

yours is easy to do, just Bold Play
> p <- pdoubb(30,1000,50000,244/495,1,0.000000001)
> p
[1] 0.02785283

Sally
more later
I Heart Vi Hart
teddys
teddys
Joined: Nov 14, 2009
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September 13th, 2014 at 9:03:19 PM permalink
Quote: Wizard

Earlier this week there was a show on the Discovery Channel called Beating the House. Part of the show featured Andy Bloch making the claim that the best way to turn $30 into $1000 was to bet the full $30 on a single number in double-zero roulette.

The probability of success with that strategy is 1/38=2.63%.

An obviously superior strategy would be to bet just $27 on a single number. If you win, you'll be at $975. If you lose you'll be at $3. Either way, make the smallest bet possible that will put you at exactly $1000 on any of the available bets, or your full bankroll if you don't have enough to reach $1000 in one bet. If such optimal bet is not an integer, then round down. Keep repeating until you get to $1000 or $0. I find the probability of success is 2.81%.

However, an even better strategy of betting 1/6 of your bankroll on the pass line in craps, followed by max odds (assume 3-4-5x allowed). If this would put you over $1,000 or cause a bet that isn't evenly divisible by $1, or cause a win that isn't divisible by $1, then round the bet down. I find such a strategy has a probability of success of 2.91%.

I'm going to look at the don't pass plus laying odds next.

The question for the poll is what would be your first bet if the goal was to maximize your chance of turning $30 into $1000? Let's assume the standard casino games available in any medium-sized casino.

Didn't you do this already with the reality TV show and the frat guys and concluded the best bet was the don't pass plus max odds?
"Dice, verily, are armed with goads and driving-hooks, deceiving and tormenting, causing grievous woe." -Rig Veda 10.34.4

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