November 20th, 2009 at 5:33:19 AM
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Hi there,

I would like to run a promo and I am no math whizz, so here goes. We use 5 decks for the Black Jack shoe, and the idea is to allow the punter to use an Ace (as a voucher promo) for a single hand, the punter can choose when he/she would like to add it as the first card to the table game and is only allowed one card thereafter/no splits if another ace is dealt (making the total 12). The Ace is removed after the hand is dealt. I would like to ask how do I work out the cost implication, this means I need to look at the probabilities. How do I go about working out the cost?

Calculations thus far (very basic) based on one box/player and the dealer

260 cards/shoe

80 picture cards therefore one picture card every 3.25 hands

30.76%

Black Jack pays 3 to 2

100 x $100 (Aces/Vouchers)- 30 will achieve a Black Jack = $ 4 500

Number of cards which total allow a total of 17 or higher

20 x 9 (value of cards) Ace + 9 = 20

20 x 8 cards - Ace + 8 = 19

20 x 7 cards - Ace + 7 = 18

20 x 6 cards - Ace + 6 = 17

20/260 = 7.69% x 4 = 30.76% ($ 3 000)

Therefore $ 4 500 + $ 3 000 = $ 7 000

I understand that the player may still win on the smaller cards against the dealer, but it is this probability that I would like to further understand, if I am on the right track at all. Are the above probability assumptions correct? How do I go about working out the cost? Does the additional Ace affect the probability making the shoe 261 cards for one hand?

I am asking for you help, please!

Please assist

vix1308

I would like to run a promo and I am no math whizz, so here goes. We use 5 decks for the Black Jack shoe, and the idea is to allow the punter to use an Ace (as a voucher promo) for a single hand, the punter can choose when he/she would like to add it as the first card to the table game and is only allowed one card thereafter/no splits if another ace is dealt (making the total 12). The Ace is removed after the hand is dealt. I would like to ask how do I work out the cost implication, this means I need to look at the probabilities. How do I go about working out the cost?

Calculations thus far (very basic) based on one box/player and the dealer

260 cards/shoe

80 picture cards therefore one picture card every 3.25 hands

30.76%

Black Jack pays 3 to 2

100 x $100 (Aces/Vouchers)- 30 will achieve a Black Jack = $ 4 500

Number of cards which total allow a total of 17 or higher

20 x 9 (value of cards) Ace + 9 = 20

20 x 8 cards - Ace + 8 = 19

20 x 7 cards - Ace + 7 = 18

20 x 6 cards - Ace + 6 = 17

20/260 = 7.69% x 4 = 30.76% ($ 3 000)

Therefore $ 4 500 + $ 3 000 = $ 7 000

I understand that the player may still win on the smaller cards against the dealer, but it is this probability that I would like to further understand, if I am on the right track at all. Are the above probability assumptions correct? How do I go about working out the cost? Does the additional Ace affect the probability making the shoe 261 cards for one hand?

I am asking for you help, please!

Please assist

vix1308

November 20th, 2009 at 6:49:05 AM
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From the Wizard Of Odds Site under promotional chips:

Free Ace

The free ace coupon may be used as an ace, in lieu of the first card dealt, in blackjack. According to my blackjack appendix 14, the expected value of an ace as the first card is 50.4% of the amount bet, assuming liberal six-deck rules (dealer stands on soft 17, double after split allowed, re-splitting aces allowed). These statistics are on a per hand basis, and include pushes. The probability of a push in blackjack is 8.5%. I have not calculated the conditional probability of a tie, given the first player card is an ace. Assuming the push probability is still 8.5%, and the player keeps the coupon on a push, the value of a free ace is 55.1% of face value, under the same liberal rule assumptions.

Free Ace

The free ace coupon may be used as an ace, in lieu of the first card dealt, in blackjack. According to my blackjack appendix 14, the expected value of an ace as the first card is 50.4% of the amount bet, assuming liberal six-deck rules (dealer stands on soft 17, double after split allowed, re-splitting aces allowed). These statistics are on a per hand basis, and include pushes. The probability of a push in blackjack is 8.5%. I have not calculated the conditional probability of a tie, given the first player card is an ace. Assuming the push probability is still 8.5%, and the player keeps the coupon on a push, the value of a free ace is 55.1% of face value, under the same liberal rule assumptions.

November 20th, 2009 at 8:53:43 AM
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I did a bit of analysis and came up with a player advantage of .4807 x the player bet.

I would limit the value of the promotion by stating that the maximum bet can only be $25 or you could limit the bet to what the player normally plays. This would eliminate the losses due to counting. This is fairly similar to the match play coupon.

I assumed that the dealer stood on soft 17.

I used the Wizard's table for infinite decks to get the expected values based on A-x vs the player hands. I did not want to use the 5 deck table because I could not calculate the effect of using an undealt Ace in the calculation.

Because you specfied that you could not split Aces, I then used the Wizard's analysis for 5 decks for the A-A combination using the available strategy (which is to hit every thing except to double when the dealer has a six).

In the end, I created a table of the expected values of the dealer hands vs the player's hand and multiplied them by the probability of each hand occurring to get the result above.

I would limit the value of the promotion by stating that the maximum bet can only be $25 or you could limit the bet to what the player normally plays. This would eliminate the losses due to counting. This is fairly similar to the match play coupon.

I assumed that the dealer stood on soft 17.

I used the Wizard's table for infinite decks to get the expected values based on A-x vs the player hands. I did not want to use the 5 deck table because I could not calculate the effect of using an undealt Ace in the calculation.

Because you specfied that you could not split Aces, I then used the Wizard's analysis for 5 decks for the A-A combination using the available strategy (which is to hit every thing except to double when the dealer has a six).

In the end, I created a table of the expected values of the dealer hands vs the player's hand and multiplied them by the probability of each hand occurring to get the result above.

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You want the truth! You can't handle the truth!

November 25th, 2009 at 10:53:31 AM
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Thank you Lucky13 and Boymimbo! I really appreciated your input, it helped me work through the heady calculations as well as a greater understanding of the probability to lead to an estimated costing. In the end, it was decided that we will allow for the standard, liberal rules for the player, in spite of the increase in probability this will bring. I am not entirely sure of the impact on the costing itself, we will know soon enough...

Over and Out.

vix1308

Over and Out.

vix1308

November 25th, 2009 at 12:30:00 PM
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Quote:vix1308Thank you Lucky13 and Boymimbo! I really appreciated your input, it helped me work through the heady calculations as well as a greater understanding of the probability to lead to an estimated costing. In the end, it was decided that we will allow for the standard, liberal rules for the player, in spite of the increase in probability this will bring. I am not entirely sure of the impact on the costing itself, we will know soon enough...

Over and Out.

vix1308

Send me a pack of coupons and I'll be right there :)

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You want the truth! You can't handle the truth!