fpod
fpod
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June 23rd, 2015 at 8:15:16 AM permalink
I would like to know the exact language governing video blackjack. i have quotes stating there is a law relating to fair odds, fairly shuffled, rng etc. but these examples all have different wording and the gaming commission site i couldn't navigate. specifically IGT's game king multi-game blackjack. single deck-shuffled after each hand-dealer stands 17-bj pays 2 for 1even money)double on10/11 only-split like pairs no re-split -no double after split-6 card no bust automatic win(for player)I would also like opinions/reviews of this game by those of you with experience and knowledge of the game as I have...I 'll say concerns. many thanks in advance, dave.
rsactuary
rsactuary
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June 23rd, 2015 at 9:13:26 AM permalink
Quote: fpod

why is the first 5, 0-00-1-2-3 considered the worst bet and why the higher house edge and advised against betting?



Because to make that bet consistent with all other bets on the table, it needs to return 6.2 chips for every chip bet. Since they can't do that, they round down to 6. (and I should note that if you bet 5 chips, they still only pay 30, instead of 31).

Payout = 36/5 = 7.2 - 1 (for the chip left on the table) = 6.2

Note that the payout is based off of 36 numbers.... not 38.
rsactuary
rsactuary
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June 23rd, 2015 at 9:17:48 AM permalink
Quote: fpod

second question, why bet a double street(6 numbers) that pays 5 to 1, when one could split his wager and bet half on two single streets of the same six numbers at 11 to 1? I am not a numbers or math person. that bet seems...I don't know that is why I am here.



It's the same... consider....

A bet of 2 on a double street... if a losing bet you are -2. If a winning bet you keep the 2 chips on the table, plus 10 chips for the win which gives you a total of 12 chips.


A bet of two single streets of one chip each... if losing bets, you are -2. if a winning bet, you will be paid 11 from the house, plus you keep the one chip that was a winner, for a total of 12 chips.

See? it's the same.
rsactuary
rsactuary
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June 23rd, 2015 at 9:32:28 AM permalink
Quote: fpod

Third question, when betting a color(red in this example) red and RED having more ODD numbers(1-3-5-7-9-19-21-23-25-27)why doesn't placing a wager on ODD increase ones chances? or wager BLACK and EVEN (having 10 even black numbers) or betting a color RED plus EVEN so one would have 28 numbers? or black with ODD getting 28 numbers? I know for some this is simple. however, I have thought about it and just don't seem to get it.



I too have noticed this discrepancy, but at the end of the day... they pay you like there's 36 numbers on the wheel, there are 38. Which means a house edge of 1-36/38 = 5.26%.
ThatDonGuy
ThatDonGuy
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June 23rd, 2015 at 10:12:38 AM permalink
Quote: fpod

Third question, when betting a color(red in this example) red and RED having more ODD numbers (1-3-5-7-9-19-21-23-25-27) why doesn't placing a wager on ODD increase ones chances? or wager BLACK and EVEN (having 10 even black numbers) or betting a color RED plus EVEN so one would have 28 numbers? or black with ODD getting 28 numbers?


Because when you bet on both, say, Red and Odd, you are making two bets, and you have to allow for the fact that if it lands on one of the ten numbers that are both Black and Even, or on 0 or 00, then you lose both bets. It is not a coincidence that the number of Odd Red numbers equals the number of Even Black numbers; in fact, if there are 18 Red and 18 Black numbers, then this will be true regardless of which 18 are Red and which 18 are Black.


Let RE be the number of Red Even numbers, RO the number of Red Odd numbers, BE the number of Black Even numbers, and BO the number of Black Odd numbers.
The total number of Red numbers = RE + RO = 18
The total number of Even numbers = RE + BE = 18
RE + RO = 18 = RE + BE, so RO = 18 - RE = BE; the number of Red Odd = the number of Black Even

The total number of Odd numbers = RO + BO = 18
RE + RO = 18 = RO + BO, so RE = 18 - RO = BO; the number of Red Even = the number of Black Odd

yay math

Specifically, when you bet on Red and Odd, one of five things is going to happen:
10/38 of the time, the number is Red and Odd (1, 3, 5, 7, 9, 19, 21, 23, 25, 27), and you win both bets.
8/38 of the time, the number is Black and Odd (11, 13, 15, 17, 29, 31, 33, 35); you win your Odd bet but lose your Red bet, so you are even.
8/38 of the time, the number is Red and Even (12, 14, 16, 18, 30, 32, 34, 36); you win your Red bet but lose your Odd bet, so you are even.
10/38 of the time, the number is Black and Even (2, 4, 6, 8, 10, 20, 22, 24, 26, 28), and you lose both bets.
2/38 of the time, the number is Green (0, 00), and you lose both bets.
10/38 of the time, you win 2; 12/38 of the time, you lose 2; 16/38 of the time, you break even.
fpod
fpod
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June 23rd, 2015 at 10:43:02 AM permalink
thank you much for your time and explaining it in simple terms. don't know who you are but I do appreciate it. thanx
fpod
fpod
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June 23rd, 2015 at 11:17:46 AM permalink
if you're the same person who answered my previous questions thanx again. if not, I really appreciate you doing so. and, it wasn't because I didn't believe you that I clicked the link just curious is all and I am glad I did as it gave me a different idea or approach to work on. not sure why/how but I understand (at least I think I do) your example and definition(s). in no way is this a criticism; i think you meant 29 where you put 19-3rd line down. thank you
ThatDonGuy
ThatDonGuy
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June 23rd, 2015 at 12:38:39 PM permalink
Quote: fpod

in no way is this a criticism; i think you meant 29 where you put 19-3rd line down. thank you


Yes, I did - and I also left out 19 in the Odd Red list. That post has been fixed.
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