Betfred are currently running a promotion on their Blackjack game where players can win up to £10 million as a bonus for hitting 13 cards and not going bust in their Black Jack game. The rules of the game are as follows:

The opportunity to win a Charlie Hand prize is triggered when:

the player’s hand stems from an initial (2 card) deal; and

the player attains a hand comprising 6 to 13 cards without busting (exceeding a total of 21 points in the base game)

regardless of whether the final total of the player’s hand is better than the total of the dealer’s hand or not.

For the avoidance of doubt, the outcome of the base game has no bearing upon the Charlie Hand prize, other than when the player busts or the player’s hand comprises 6 to 13 cards stemming from a split of an initial (2 card) deal.

There are no staking requirements other than the initial base game stake required to play the game

The Betfred Blackjack game is played with 6 decks of cards.

The Betfred Blackjack game will allow the player to draw on a soft 21 (i.e. where the player has 21 points and there is an Ace in the player’s hand valued as 11 points).

The Betfred Blackjack game will not allow the player to draw more than 13 cards in any individual hand regardless of the hand total.

The Charlie Hand prizes can be won on any individual hand/box played.

I did try and post a link to the game rules but I can't as I'm a newbie.

I was wondering if anyone knew what the odds of hitting 13 cards without going bust were under the above game conditions please? I'm assuming its extremely unlikely but I'd be interested to know how unlikely.

Thanks in advance for any advice!

Quote:dafcno1

The Betfred Blackjack game will allow the player to draw on a soft 21 (i.e. where the player has 21 points and there is an Ace in the player’s hand valued as 11 points).

doubling on my blackjack at a favorable count...? bad idea?

It looks like about 171 billion to 1 using only Aces and Two's.

See the value of all the "charlies" below:

Charlie Hand / Prize

13 Cards / 1,000,000 x Initial Stake

12 Cards / 100,000 x Initial Stake

11 Cards / 10,000 x Initial Stake

10 Cards / 1,000 x Initial Stake

9 Cards / 100 x Initial Stake

8 Cards / 10 x Initial Stake

7 Cards / 5 x Initial Stake

6 Cards / 1 x Initial Stake

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I am not good enough to work out the probabilities for this type of thing, but there are plenty of people on this site that should be able to help.

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Edit/Update (about 325 pm)

The prizes for the 6 and 7 card charlie would shave about 0.21%*** off the base game house edge (?)

I used the WoO page about charlies, which said something like:

***: A six-card charlie is worth about 0.16%, and a seven-card charlie is worth about 0.01% (when it pays "even money").

see WoO links below:

https://wizardofodds.com/games/blackjack/rule-variations/

https://wizardofodds.com/games/blackjack/rule-variations/automatic-winner-charlie-rule/

I wouldn't know where to even start trying to calculate the odds so I was hoping someone cleverer than me (not difficult) could help!

If you have a six card 12, you would aim for the seven card charlie (see proof).

Reasoning/Proof: Cards that will bust a Player 12, 16/52 (so 36/52 is the chance of "improving" to a "7 card charlie", in this case).

Note: 36/52 is about 69.23% and 16/52 is about 30.77%

The value of the 6-card charlie in this hand is 100%.

The value of "aiming" for the 7- card charlie for this hand is:

( "chance of winning" x 5 ) - ("chance of busting" x 1)

=(about 69.23% x 5) - (about 30.77% x 1)

"roughly =" 346.15% - 30.77%

= about 315.38%

315.38% is better value than 100%, so therefore you would hit a six card 12 (based on the "charlie bonus table" for this game).

Note: I used "infinite deck" to make the maths easier for me (since there are more "bust cards in the shoe", the real value of hitting this hand would be a bit less than the figure I gave above).

Note 2: This is just the proof for one scenario, you would probably hit up to a "six card 16" , since that would be worth about 130.77% (using "infinite decK").

Quote:DRichWow, 13 cards under 22 must be huge odds. I would guess you could only do it with just two's and aces and nothing higher.

It looks like about 171 billion to 1 using only Aces and Two's.

My top amount of cards under 21 is 8 and i got 20, and im sure that specific hand can be guessed with some thought but it consisted of mostly aces

Average card value is 85 / 13 = 6.54, standard deviation 3.15.

3.15 * 13^.5 is 11.37.

For 13 cards, expected value is 85 +/- 11.37.

21.5 is 5.586 SDs below expectations. So the result will be <= 21.5 about 1 in 86 million trials. This is a rough estimate due to small sample size and large deviation from mean.

I would have expected a "stand on 20-21" strategy to work better, as, if you hit a 19, you have about a 2/13 probability of multiplying your winnings by 10, but that's about 97.9%.

Quote:ThatDonGuyI simulate a house edge of about 97.5% using a "stand on 18-21 with 6 or more cards" strategy. This also assumes that all bets are "to 1" - i.e. a 6-card hand pays even money, as opposed to being a push; if odds are "for 1," the strategy is the same, but the house edge is about 98.4%.(snip)

Just to clarify, the promotion says "4. There are no staking requirements other than the initial base game stake required to play the game".

Since you don't have to pay any extra for the "charlie", the game should be a lot better than 97.5% house edge ?

To the OP, ThatDonGuy is one of the people on this forum I would want to help me with this problem or similar (if he disagrees with my post, I would go with whatever he says).

Quote:ThatDonGuy(snip)I would have expected a "stand on 20-21" strategy to work better, as, if you hit a 19, you have about a 2/13 probability of multiplying your winnings by 10, but that's about 97.9%.

I would have expected a "stand on 20-21" too, for when the "charlie payouts multiply by 10".