Mohegan Sun Blackjack Payout
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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.
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 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.
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?
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
| Event | Pays | Permutations | Probability | Return |
| Ace/jack suited | 25 | 144 | 0.002968 | 0.074202 |
| Two aces | 10 | 276 | 0.005689 | 0.056888 |
| 3 or 4 total | 3 | 1428 | 0.029434 | 0.088301 |
| 9 or 10 total | 2 | 4884 | 0.100668 | 0.201336 |
| Any other blackjack | 1.5 | 2160 | 0.044521 | 0.066782 |
| 11 to 12 total | 1 | 6612 | 0.136285 | 0.136285 |
| All other | -1 | 33012 | 0.680435 | -0.680435 |
| Total | 48516 | 1 | -0.056641 |
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.
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%.
Mohegan Sun Poker Payouts
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.
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.
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.
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.
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.
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.
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:
Mohegan Sun Blackjack Payouts
- 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
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.
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.