October 10th, 2011 at 6:06:07 AM
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Hi Guys,

I was wondering if someone could help me out with the following question/problem:

Following on from the progressive blackjack probabilities table at:

http://wizardofodds.com/blackjack/appendix8.html#progressivebj

What would these probabilities be if under exactly the same conditions (6 decks etc) you had seen an entire deck having been dealt out with no aces among those 52 cards?

i.e. you only entered this bet when conditions were more favorable

The payouts are different where I am (much more favorable), so it is just the probabilities I am after./

Thanks in advance for anyone who helps me out with this!

Regards!

I was wondering if someone could help me out with the following question/problem:

Following on from the progressive blackjack probabilities table at:

http://wizardofodds.com/blackjack/appendix8.html#progressivebj

What would these probabilities be if under exactly the same conditions (6 decks etc) you had seen an entire deck having been dealt out with no aces among those 52 cards?

i.e. you only entered this bet when conditions were more favorable

The payouts are different where I am (much more favorable), so it is just the probabilities I am after./

Thanks in advance for anyone who helps me out with this!

Regards!

October 10th, 2011 at 8:11:09 AM
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I see that the payout depends on how many consecutive aces the player gets. Here are my results for 5 of 6 decks remaining with no aces gone:

Here's a good question: What's the probability of not having any aces in the first deck of a 6-deck shoe?

Hand | Permutations | Probability |
---|---|---|

4 red/black aces | 23760 | 0.0000053 |

4 aces | 231264 | 0.0000518 |

3 suited aces | 113280 | 0.0000254 |

3 non-suited aces | 2752704 | 0.0006165 |

2 suited aces | 7278240 | 0.0016300 |

2 non-suited aces | 26201664 | 0.0058682 |

1 ace | 375557184 | 0.0841105 |

no aces | 4052887944 | 0.9076923 |

Total | 4465046040 | 1.0000000 |

Here's a good question: What's the probability of not having any aces in the first deck of a 6-deck shoe?

October 10th, 2011 at 10:15:18 AM
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Hmmm.. The odds of not getting an ace on the first card of a 6 deck shoe is (312-24)/312. The odds of not getting an ace on the second card of a 6 deck shoe is (311-24)/311... and so on until you get (261-24)/261.

Multiple them all together and you get.... 0.01043 (1.043%). Not impossible!

Multiple them all together and you get.... 0.01043 (1.043%). Not impossible!

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

October 10th, 2011 at 10:36:35 AM
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ChesterDog,

Thanks for your reply- very interesting/helpful to know...

I imagine the probability is too low to be feasible as an stand alone strategy!

As backdrop, I was curious as to the results as a reference point for using quick and dirty judgments in the game. For example, whilst the probability of that exact scenario might very low, there will be many more opportunities as a whole where a below statistically expected (not the correct phrase I am sure, but you get what I mean) distribution of aces have emerged in the first 52 cards (e.g. only 6 aces being dealt out of 104 cards etc). Assuming these can easily be kept track of (say putting a coin in one hand per ace) and the remaining cards can be estimated to a reasonable degree (using the discards, as counters using true counts should be skilled at doing to within 1/4 deck say) in theory it could give you more opportunities to play, if the progressive side bet is close to +ve expectation (before thinking about the potential change in house/player edge if the deck is ace heavy).

I don't have the skill to work these combinations out myself with conviction, so was just looking for a reference point between the odds as per a completely random un-dealt shoe (courtesy of the Wizard) and something more favorable, which I can then keep in the back of my mind.

thanks again!

Thanks for your reply- very interesting/helpful to know...

I imagine the probability is too low to be feasible as an stand alone strategy!

As backdrop, I was curious as to the results as a reference point for using quick and dirty judgments in the game. For example, whilst the probability of that exact scenario might very low, there will be many more opportunities as a whole where a below statistically expected (not the correct phrase I am sure, but you get what I mean) distribution of aces have emerged in the first 52 cards (e.g. only 6 aces being dealt out of 104 cards etc). Assuming these can easily be kept track of (say putting a coin in one hand per ace) and the remaining cards can be estimated to a reasonable degree (using the discards, as counters using true counts should be skilled at doing to within 1/4 deck say) in theory it could give you more opportunities to play, if the progressive side bet is close to +ve expectation (before thinking about the potential change in house/player edge if the deck is ace heavy).

I don't have the skill to work these combinations out myself with conviction, so was just looking for a reference point between the odds as per a completely random un-dealt shoe (courtesy of the Wizard) and something more favorable, which I can then keep in the back of my mind.

thanks again!

October 10th, 2011 at 11:00:28 AM
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The problem with this the HA without the jackpot is something on the order of 53%, and that really, no count of Aces would really push the bet into a favorable one.

You would hate NOT to bet this bet and have 4 Red Aces get dealt to you.

You would hate NOT to bet this bet and have 4 Red Aces get dealt to you.

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

October 10th, 2011 at 11:43:27 AM
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UK or Australia?

Or somewhere else where they say "maths" instead of "math"?

Or somewhere else where they say "maths" instead of "math"?

October 10th, 2011 at 5:58:12 PM
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UK- London!

October 10th, 2011 at 6:05:30 PM
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Another British Invader. WELCOME

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