Free Bet Blackjack Mohegan Sun

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I read on your web site that you recently had a chance to visit Tunica, Mississippi. I play very often at the Grand. I am sure you have heard of the side bet known as Triple 7s. I don't consider myself a good card counter, or even an average one, but say no 7s had been played at all in six hands would the odds not be in my favor to play the side bet?

Yes, I've seen that bet in Tunica. I address it in my blackjack appendix 8. I agree that side bet would seem very countable.

Mr. Wizard, I just recently played on a Casino Boat that had a Bust Bet at the Blackjack table. You could place this even money bet at anytime after seeing the dealers’ up card. Is this a bad bet and what might the odds be? Thank You.
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This is a sucker bet. The most likely time the dealer will bust is with a 6 up. However even then the dealer will only bust 42% of the time, give or take depending on the exact rules, for a house edge of 16%.

The Hollywood casino in Tunica started offering a FREE progressive side bet on two $5 tables. The rules are six decks and dealer hits a soft 17, all other rules are standard. They swear there are no rule changes to the game (Dealer hits, DAS, 4 splits, 6 decks). Suited sevens of diamonds gets the progressive, which starts at $1,000. All other triple sevens pay $50. So how high would the progressive need to be to get to breakeven?

The probability of three seven of diamonds is combin(6,3)/combin(312,3) = 0.00000398937. The probability of three unsuited sevens is (combin(24,3)-combin(6,3))/combin(312,3) = 0.000399735. According to my blackjack calculator the house edge is 0.6233%. The expected loss on a $5 bet would be 3.12 cents. Just the value of the $50 for three unsuited sevens is $50*0.000399735=2.00 cents. To make up the other 1.12 cents the meter would need to reach $2802.

Wheel of Madness is another one of those silly blackjack side bets. It is not included in your ever growing list of exotics. Is it proprietary information that keeps you from its analysis?

The wheel in Wheel of Madness is weighted. In other words it is designed to stop on the lower prizes more often. Without knowing the exact weights I can’t analyze it. I have tried to get the weights from casinos and the manufacturer but alas have had no luck.

Spirit Mountain Casino in Grand Ronde Oregon added a side bet in the last 24 hours called 'Field Gold 21.' It resolves before the rest of the hand begins and concerns the first two cards dealt to a player. The side bet can be between 1 and 25 dollars. The pay table follows.
  • Ace, Jack Suited = 25 - 1
  • 2 Aces = 10 - 1
  • 3 or 4 Total = 3 - 1
  • 9 or 10 Total = 2 - 1
  • 11 or 12 Total = 1 - 1
  • Any Blackjack = 3 - 2

Aces always count as 1 and 10’s and faces count as 10. What is the house advantage? If I keep an Aces and Fives count is there a positive count where the possible remaining aces make the bet a positive proposition? Would counting remaining aces divided by remaining decks be better?

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You didn’t tell me the number of decks, but assuming six the house edge is 5.66%. Here is the return table.

Field of Gold — Six Decks

EventPaysPermutationsProbabilityReturn
Ace/jack suited251440.0029680.074202
Two aces102760.0056890.056888
3 or 4 total314280.0294340.088301
9 or 10 total248840.1006680.201336
Any other blackjack1.521600.0445210.066782
11 to 12 total166120.1362850.136285
All other-1330120.680435-0.680435
Total485161-0.056641
Blackjack

Just eyeballing it, I would say aces would be the best card to track, betting into an ace-rich deck. My advice would be to count aces as −12 and all other cards as +1.

In San Diego casinos Super Fun 21 has a $1 side bet that on the first hand of single deck a diamond suited blackjack pays $300. What are the correct odds for getting this with 6 players and you are sitting at 1st base?

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There is one way to get the ace and four ways to get the 10-point card, for a total of 1*4=4 winning combinations. There are combin(52,2)=1,326 ways to choose 2 cards out of 52. So the probability of winning is 4/1326 = 0.30%. Fair odds would be 330.5 to one. The expected return is 0.0030*300 + 0.9970*-1 = -0.0920. So the house edge is 9.2%.

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The reason they limit this bet to the first hand after the shuffle is a card counter could take advantage it otherwise. Without tracking the cards, you can assume the house edge is 9.2% all the time.

I play at the Mohegan Sun casino, which has the ’Match the Dealer’ side bet, at many of its blackjack tables. When using a high/low count, do the odds on the side bet get better or worse as the count goes up/down? Thanks very much.

For the benefit of other readers, the Match the Dealer side bet pays when either of the player’s first two cards match the dealer’s up card. A traditional count is not going to be useful against this side bet. Rather, the odds would swing to the player’s advantage if the distribution of cards by rank were unusually unbalanced. It isn’t going to be practical to keep track of 13 different suits. The Big Book of Blackjack by Arnold Snyder, which I highly recommend, has a short chapter on how to beat a similar bet, the ’Royal Match.’ With only four suits to worry about, this side bet is vulnerable to the method described in that book in a single-deck game.

I’m a novice counter, using the hi-lo system. Knowing the count, do my odds increase significantly enough to make a side bet of 'over 13, under 13' worthwhile?

Yes! That side bet is extremely vulnerable to card counters. As long as the minimum is not too low, you should be using another strategy to exploit it, one that treats aces as a low card. Arnold Synder presents such a strategy in The Big Book of Blackjack. Otherwise, if you are using a standard hi-lo count, Synder says to only make the Over bet in very high counts.

I know the commandment to not make side bets. However, I have seen a side bet in blackjack that pays 11 to 1, if the player has a pair in his first two cards. Would it be possible using a count system to gain an advantage?

It sounds like you are talking about Lucky Pairs, a side bet that wins if the player’s first two cards are a pair. Many baccarat tables also offer this bet. As I show in my baccarat page, the house edge is 10.36%, assuming eight decks. In either game, you would pretty much need to eliminate all cards of at least one rank to have an advantage. To know that, you would need to keep 13 different counts. In baccarat, this could be done, since you are allowed to take notes while you play. However, based on some very extensive analysis, profitable opportunities don’t happen often enough for this to be a practical use of time.

If the Lucky Lady side bet is only played when the true count is, say, 10 or above, will one beat the house edge? If so, what is the minimum true count at which one can beat the house edge with this side bet?

I have not studied the effect of card counting of that bet for myself. However, Arnold Snyder has, and his results can be found in his Big Book of Blackjack. There he says you should make the bet in a six-deck game if it is the last two decks, and the count is +10 or greater, using the Red Sevens count. In a double-deck game he says to bet in the last deck, and a count of +6 or greater.

I ran into a new sidebet in a blackjack game and was wondering what the house edge was. In addition to the house edge, what would the optimum count have to be to lessen the edge and gain an advantage, if there is one? The player wins if the rank of the dealer’s up card falls between the ranks of the player’s first two cards. The fewer ranks between the player’s two cards, the more a win will pay. A win with a one-rank gap pays 12 to 1, two ranks pays 6 to 1, three ranks pays 4 to 1, and four or more ranks pays 1 to 1. A three of a kind pays 30 to 1. Any insight or help would be appreciated.

Based on six decks, I get a house edge of 3.40%. I show all my math in my blackjack appendix 8. An extremely high or low count would indicate the ranks in the remaining cards are clumped together, which would lower the house edge, but I don’t think it would be enough to warrant bothering with.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

In Casino Vienna, there is a side bet in blackjack. The game uses six decks and the dealer stands on soft 17. It pays 5 to 2 if the dealer busts, regardless of the outcome on the original bet.

My blackjack appendix 2B was created for questions like this. It shows the probability the dealer will bust is 28.19% under those rules. That would make the house edge 1.33%.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

At the Lake Elsinore Hotel and Casino in California there is a blackjack side bet called the Red Flex. It pays according to the number of consecutive red cards in the dealer's hand, starting with the first card. The pay table is as follows:
  • Seven or more reds pays 200 to 1
  • Six reds pays 100 to 1
  • Five reds pays 50 to 1
  • Four reds pays 10 to 1
  • Three reds pays 5 to 1
  • Two reds pays 1 to 1

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If the dealer busts, or isn't required to draw cards because all the players busted, the dealer will still draw cards as necessary to adjudicate the side bet.

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What are the odds?

I show my analysis of the Red Flex in my blackjack appendix 8.

This question is discussed in my forum at Wizard of Vegas.

I found a blackjack dealer who exposes his hole card when peeking for blackjack. He offers insurance after peeking. So, when the dealer has an ace up and I can see a 10, I take insurance. How much does this reduce the house edge of the overall game?

Let's assume six decks. The probability of the up card being an ace is 1/13. Next, the probability the hole card is a 10-point card is 96/311. So, this opportunity will occur 2.37% of hands.

The insurance bet I assume will be half a unit and insurance pays 2 to 1. So, every time this happens, you can expect to gain a unit. Since it happens with probability 2.37% of the time, that is what it is worth to you. Subtract the house edge under whatever the blackjack rules are to get the overall player advantage of the game.

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