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You are playing a blackjack game vs. a casino where all cards of value 10's are removed. (This is single deck blackjack) That means no Jacks, Queens, Kings and 10's. This is the very first game.

Casino hits on soft 17 or below, and stays on hard 17.

Dealer has 9 showing, and you are dealt two 7's.

Which is the best option?

1. Stay

2. Hit

3. Double Down

4. Split

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double down. Your current total is 14. There are only 11 cards that he can have that beat that total without further action (the four 8s, the three 9s and the four As). You have only 7 cards that can hurt you (make you go bust -- the 8s and 9s). It is far more likely that your total will be improved to a likely winning amount.

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This is a very difficult puzzle, only solveable through recursive functions using computer programming.

However, I'll provide the answer with the percentages.

Deck: A,A,A,A,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5,6,6,6,6,7,7,8,8,8,8,9,9,9 = 33 Cards total.

Dealer has 9 Showing...

Chance of hitting Soft 20: 12.121212121212%

Hard 17: 23.03301166760566%

Hard 18: 19.656737404234622%

Hard 19: 10.159733357731131%

Hard 20: 10.142273682318176%

Hard 21: 8.1274367703733663%

Busting: 16.759594996524914%

Therefore, if you stay, you will lose 100% - Bust% = 83.24040500% of the time.

Chances are you would want to either hit, double down, or split.

When you have pair of 7's, you have 7 cards that will bust you out of 33, total of 21.

Your chance of winning increase and losing decrease comparing to staying, regardless your % of winning does not overcome your % of losing.

Doubling down will cost you more money. So you do not want to double down.

Now you are left with a choice between hit and split.

This ultimately leads to the best choice being a split by a narrow margin.

Number of calculations that proves this is millions and millions of calculations.

Sorry for such hard one, I'll do lot easier one next time.

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COOL! B)) Got that one with a guess! Guess I'll go to Vegas! :lol:

(That's about the only game I play at casinos.)

It's about all that's with in my scope, am gonna throw 5k when I reach 65, may leap at it at 60, may use 1k on the wheel too - stays near the top of my todo list but back to london for that
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It's about all that's with in my scope, am gonna throw 5k when I reach 65, may leap at it at 60, may use 1k on the wheel too - stays near the top of my todo list but back to london for that

Oh, Yea! That's the other one I throw my money at! :D Too Much Fun!

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There's a 21% chance (7/33) that the dealer will have to stand, but the same 21% chance (7/33) that you will bust. However, with no 10's in the deck, the odds are in favor of the dealer not busting, so you're best option is to hit once (you have about a 49% chance of getting 17-21).

You are playing a blackjack game vs. a casino where all cards of value 10's are removed. (This is single deck blackjack) That means no Jacks, Queens, Kings and 10's. This is the very first game.

Casino hits on soft 17 or below, and stays on hard 17.

Dealer has 9 showing, and you are dealt two 7's.

Which is the best option?

1. Stay

2. Hit

3. Double Down

4. Split

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This is a very difficult puzzle, only solveable through recursive functions using computer programming...

...This ultimately leads to the best choice being a split by a narrow margin.

Number of calculations that proves this is millions and millions of calculations.

Sorry for such hard one, I'll do lot easier one next time.

Yeah...calculating the odds on BlackJack is pretty complicated...I know of a math student did their graduate project on showing that if you count cards you have a 51% chance of winning...

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