piroukay
piroukay
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December 25th, 2017 at 5:37:06 PM permalink
Quote: noy2222

I propose naming this side bet "Stupid Motherf$$ker".
It costs a dollar per round and pays off when you double and win on a natural 20 or a natural 6.



Casino 5 card poker could claim that title. odds of getting a flush are 1 in 508. Can the player's odds doubling on hard 20 on BJ be worse than this?
beachbumbabs
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beachbumbabs
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December 26th, 2017 at 7:02:33 AM permalink
Quote: piroukay

Casino 5 card poker could claim that title. odds of getting a flush are 1 in 508. Can the player's odds doubling on hard 20 on BJ be worse than this?



Kind of apples-to-oranges comparison, right?

Unless..

96/312 x 95/311 chance of getting a hard 20 in the first place, x 24/310 chance of then getting an ace, for 6 deck. 1 in 137.4267 hands is what I get, apples to apples.

But once you already HAVE 2 tens (6 deck), you have a 7.742% chance of getting an ace to go with it, or 1 in 12.99 hands. Which I would think you knew (1 in 13 cards being an ace).
If the House lost every hand, they wouldn't deal the game.
Dalex64
Dalex64
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Thanks for this post from:
piroukay
December 26th, 2017 at 7:52:38 AM permalink
It could be a free bet.

If a player has a 20, offer to pay some multiple of his existing wager if he hits and gets an ace.

If he busts, he just loses his existing wager.

As a free bet, and with a sufficiently high multiplier, you could get lots of suckers to fall for it while maintaining an obnoxiously high house advantage.
piroukay
piroukay
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December 26th, 2017 at 1:54:58 PM permalink
Is it 1 in 13 still with six decks?
Doc
Doc
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December 26th, 2017 at 7:07:19 PM permalink
At the point of possibly hitting a hard 20, the odds of an ace are not exactly 1 in 13, because the only two known cards already in play are both 10-vale cards; i.e., non-Aces. (Or rather at least two non-Ace cards totalling 20? I suppose your hard 20 could include an Ace, but I haven't been thinking of the problem that way.) I think that should make the odds very slightly higher than 1 in 13, perhaps the 1 in 12.99 suggested by BBB (if I am remembering her post correctly without looking back.)
onenickelmiracle
onenickelmiracle
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December 27th, 2017 at 1:25:05 AM permalink
It sounds like a waste of space to me. Someone has a good hand and they want to pay extra to destroy it, I don't think so. Sounds like it would slow down the game. There are better ideas than this to investigate.
I am a robot.
MaxPen
MaxPen
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December 27th, 2017 at 1:35:26 AM permalink
Seems Iike this would appeal to someone that plays blackjack like poker. i.e. If they can't raise they fold. There is maybe a few players out there that want to and will double everything. Saw one regularly in the Cannery couple years ago. I think he hated money.😳
Joeman
Joeman
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December 27th, 2017 at 6:12:19 AM permalink
Assuming a fresh shuffle and the only exposed cards are 2 ten-value cards and the dealer's upcard (which we'll assume is not an ace), there are 24 Aces and 285 non-Ace cards unseen in a 6-deck shoe. So, odds for an Ace = 24 in 309, or 1 in 12.88.

Further, if the dealer's upcard is an Ace (or if your 20 consists of A-9), then your odds would be 1 in 13.43. And if you had A-9, and the dealer also had an Ace, the odds would be 1 in 14.05.

Of course, the odds will change with each card dealt. I'm no counter, but I would think this type of bet would be easily vulnerable to someone just counting aces.
"Dealer has 'rock'... Pay 'paper!'"
Kellynbnf
Kellynbnf
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December 27th, 2017 at 8:38:32 AM permalink
The only way you can make a 2-card hard 20 is with two 10-value cards. Since most casinos do not allow doubling on 3+ cards there are no other possible combinations for this scenario under normal rules (and A+9 is a soft 20, which although still a bad play to double is much safer than doubling on a hard 20).
piroukay
piroukay
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May 11th, 2018 at 2:13:58 PM permalink
Thank you, everybody, for all your help, guidance and amusement.

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