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Eaglesnest
Eaglesnest
Joined: Jul 29, 2014
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August 28th, 2014 at 8:23:16 PM permalink
Quote: RS

The proper play is to bet $13 ($500/38) on each number, straight up, on roulette. It is the most +EV, assuming you get only one shot and are using someone else's money.



No matter how you arrange your bets at roulette, the EV will be the same, as long as you avoid that five-number bet which I think is called the "basket."
NowTheSerpent
NowTheSerpent
Joined: Sep 30, 2011
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August 28th, 2014 at 11:20:13 PM permalink
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NowTheSerpent
NowTheSerpent
Joined: Sep 30, 2011
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August 28th, 2014 at 11:26:27 PM permalink
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AxelWolf
AxelWolf
Joined: Oct 10, 2012
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August 28th, 2014 at 11:26:34 PM permalink
Quote: NowTheSerpent

There you go, Buzzard, insulting hats yet again. LOL!

I thought I said that?
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
RS
RS
Joined: Feb 11, 2014
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August 29th, 2014 at 4:00:37 AM permalink
Quote: Eaglesnest

No matter how you arrange your bets at roulette, the EV will be the same, as long as you avoid that five-number bet which I think is called the "basket."



Not quite true.

Say you have $38. If you had one shot and bet every number straight up, you'd get paid $35 when a number hit and you'd lose the original wager (ie: one shot). If you bet each of the numbers $2 at a time on the in-between-line (ie: $2 on 0/00 split), you'd get paid $34 and lose the original wager.

If you do get to keep the chips, then playing the lowest HE game would probably be best.
NowTheSerpent
NowTheSerpent
Joined: Sep 30, 2011
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August 29th, 2014 at 4:09:01 AM permalink
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FatGeezus
FatGeezus
Joined: Jun 12, 2010
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August 29th, 2014 at 7:15:22 AM permalink
In an AC casino I would bet $250 on Red and $250 on Black.

If the 0 or 00 comes up, you would only lose half you bet ($125) on the Black and ($125) on the Red.

You are guaranteed to walk away with $250 no matter what number comes up
Romes
Romes
Joined: Jul 22, 2014
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August 29th, 2014 at 10:37:53 AM permalink
Depends what your goal is...

1) Lowest House Edge for Maximum Value: Blackjack, hands down. Lowest house edge (~.5%) you can find with basic strategy. I'd be prepared to put my own money up on the small chance you get a split/double situation. They generally only give you promotional chips in accordance to your level of play. Thus, a $500 chip you should easily be able to match $500 "if needed." The majority of the time BS will suffice for this bet.

2) Highest guaranteed amount back - Roulette $13 per number straight up including the zero's. This will return $455.

3) If you're not allowed to break it up like that, I'd probably go Don't Pass in craps, which would return the full $500 but at a 1.36% HE. Or if you really wanted 'guaranteed' money you could match your $500 with $500 and play the Pass & Don't Pass.

I think in the situation you're referring to we're assuming a $500 promo chip that CAN'T be broken down and you get to keep the winnings. In this case I'd back count a blackjack shoe and throw it in on the highest true count I could find =P. If that wasn't an option, I'd probably still play blackjack, baccarat, or craps, in that order.
Playing it correctly means you've already won.
Romes
Romes
Joined: Jul 22, 2014
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August 29th, 2014 at 10:38:45 AM permalink
post != edit... =/
Playing it correctly means you've already won.
RS
RS
Joined: Feb 11, 2014
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August 29th, 2014 at 1:28:01 PM permalink
Quote: NowTheSerpent

You would win $35 and lose $37, so the HE = -2/38 = -1/19 = -5.26% as usual. If you bet $2 across 0/00, only one can hit for $35, the other would lose. In 38 results, you'd lose $2 36 times and win $34 twice, so your return would be ($34*2 - $2*36)/($2*38) = -4/76 = -1/19 = -5.26% once again.



You're using someone else's money....

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