When is 50 to 1 suited tie profitabe?
June 21st, 2023 at 11:34:07 AM
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Sorry if this was posted already, I was looking for an answer for a while but I can't find one.
So, in Blackjack you often have a "perfect pair" side bet, which pays 25 to 1 if it's suited and 8 to 1 off-suit. In Dragon/Tiger you have a 50 to 1 payout for a suited tie and 11 to 1 for off-suit. How many cards have to be drawn and how many "same" cards have to stay in the shoe to reach profitable odds?
For example, I'm playing Dragon/Tiger, I notice that we're 70% trough the shoe and not a single Ace of Spades has been drawn. That means out of 416 cards, about 250 have been drawn and there's still 8 Ace of Spades left inside. Should I start betting for the suited tie in every game now? What is the exact edge? What about the odds on a perfect pair BJ bet with 25:1 suited + 8:1 off-suit payouts? Thanks in advance for answers :)
So, in Blackjack you often have a "perfect pair" side bet, which pays 25 to 1 if it's suited and 8 to 1 off-suit. In Dragon/Tiger you have a 50 to 1 payout for a suited tie and 11 to 1 for off-suit. How many cards have to be drawn and how many "same" cards have to stay in the shoe to reach profitable odds?
For example, I'm playing Dragon/Tiger, I notice that we're 70% trough the shoe and not a single Ace of Spades has been drawn. That means out of 416 cards, about 250 have been drawn and there's still 8 Ace of Spades left inside. Should I start betting for the suited tie in every game now? What is the exact edge? What about the odds on a perfect pair BJ bet with 25:1 suited + 8:1 off-suit payouts? Thanks in advance for answers :)
July 2nd, 2023 at 10:13:12 AM
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For the Dragon/Tiger problem (Suited Pair with 166 cards remaining, 8 of which are Ace of Spades), if you limit it to dealing just two Aces of Spades, the answer is No.
Suppose there are only 54 cards left in the shoe, and all eight Aces of Spades are still there, but each of the other 46 cards is different.
There are 54 x 53 / 2 = 1431 pairs of cards that can be dealt, of which 8 x 7 / 2 = 28 are both Aces of Spades.
Each of the 28 pairs of Aces of Spades is worth +50, for a total of 28 x 50 = 1400
Each of the other 1403 pairs is worth -1, for a total of -1403.
You cannot calculate the "exact" edge of your problem as you need to know how many of each of the 52 cards remain. If, in addition to there being 8 Aces of Spades, there are 6 3s of Hearts and 5 Queens of Diamonds, this increases your edge considerably.
Suppose there are only 54 cards left in the shoe, and all eight Aces of Spades are still there, but each of the other 46 cards is different.
There are 54 x 53 / 2 = 1431 pairs of cards that can be dealt, of which 8 x 7 / 2 = 28 are both Aces of Spades.
Each of the 28 pairs of Aces of Spades is worth +50, for a total of 28 x 50 = 1400
Each of the other 1403 pairs is worth -1, for a total of -1403.
You cannot calculate the "exact" edge of your problem as you need to know how many of each of the 52 cards remain. If, in addition to there being 8 Aces of Spades, there are 6 3s of Hearts and 5 Queens of Diamonds, this increases your edge considerably.
