Check my math on this fantasy hand
March 1st, 2015 at 12:48:58 PM
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This is borne out of a casual conversation I had with a customer recently and I was curious.
In blackjack side games including Match the Dealer, 21+3 and Double Action Blackjack, a nice payout is available when you're dealt a pair, and that rank is identical to the dealer's up card. It could be a Double-Non, A Suited-Non, A Double-Suited, a Trips, or a Suited Trips depending on the game you're playing, but they all offer up nice sized payouts.
The other day, a customer idly asked me "Hey! What's the odds that all 7 of us would get a pair and match the dealer?"
I gave him the cop-out answer: A snowball in hell, but I have to admit I was curious, so I went home and ran the math on it. I'd like you to check it for me if you would.
This is the chance that, in a six-deck shoe, 15 cards of the same rank would be clumped together in the shoe regardless of suit, AND that the first of those 15 cards lands in first base's first card (A 1-in-16 shot since there are 7 players plus the dealer).
1 * 23/311 * 22/310 * 21/309 * 20/308 * 19/307 * 18/306 * 17/305 * 16/304 * 15/303 * 14/302 * 13/301 * 12/300 * 11/299 * 10/298
= 0.0000000000000000012107 (15 cards of the same rank will be together in a 6-deck shoe about once every 826 quadrillion hands)
0.0000000000000000012107 / 16
= 0.0000000000000000000757 (7 trips will appear on the table once every 13.2 quintillion hands)
Now since I'm counting down from 312, I guess this is assuming the shoe will open up with these odds. Is it a good accurate average for any point in the shoe though? Obviously conditions may improve or worsen the deeper into the shoe you go.
Thanks for helping me satisfy my sick curiosity!
In blackjack side games including Match the Dealer, 21+3 and Double Action Blackjack, a nice payout is available when you're dealt a pair, and that rank is identical to the dealer's up card. It could be a Double-Non, A Suited-Non, A Double-Suited, a Trips, or a Suited Trips depending on the game you're playing, but they all offer up nice sized payouts.
The other day, a customer idly asked me "Hey! What's the odds that all 7 of us would get a pair and match the dealer?"
I gave him the cop-out answer: A snowball in hell, but I have to admit I was curious, so I went home and ran the math on it. I'd like you to check it for me if you would.
This is the chance that, in a six-deck shoe, 15 cards of the same rank would be clumped together in the shoe regardless of suit, AND that the first of those 15 cards lands in first base's first card (A 1-in-16 shot since there are 7 players plus the dealer).
1 * 23/311 * 22/310 * 21/309 * 20/308 * 19/307 * 18/306 * 17/305 * 16/304 * 15/303 * 14/302 * 13/301 * 12/300 * 11/299 * 10/298
= 0.0000000000000000012107 (15 cards of the same rank will be together in a 6-deck shoe about once every 826 quadrillion hands)
0.0000000000000000012107 / 16
= 0.0000000000000000000757 (7 trips will appear on the table once every 13.2 quintillion hands)
Now since I'm counting down from 312, I guess this is assuming the shoe will open up with these odds. Is it a good accurate average for any point in the shoe though? Obviously conditions may improve or worsen the deeper into the shoe you go.
Thanks for helping me satisfy my sick curiosity!
Casinos are not your friends, they want your money. But so does Disneyland.
And there is no chance in hell that you will go to Disneyland and come back with more money than you went with.
- AxelWolf and Mickeycrimm
March 1st, 2015 at 12:55:18 PM
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Last month I saw a table with one player and two boxes, both had JJ and the dealer J - and yes the sixth card was also a Jack.
btw There's a Jackpot bet in the UK run by one chain where you win the Jackpot if you get 4 suited Aces and lower prizes for fewer or non-suited Aces down to 5/1 for one Ace ( http://www.ukcasinotablegames.info/blackjackaces.html ).
btw There's a Jackpot bet in the UK run by one chain where you win the Jackpot if you get 4 suited Aces and lower prizes for fewer or non-suited Aces down to 5/1 for one Ace ( http://www.ukcasinotablegames.info/blackjackaces.html ).
March 1st, 2015 at 12:56:32 PM
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You don't need to divide by 16. You can only really figure this out from the start of the shoe, as you have no idea what cards have been dealt otherwise.
The answer is 1 * 23/311 * 22/310 * 21/309 * 20/308 * 19/307 * 18/306 * 17/305 * 16/304 * 15/303 * 14/302 * 13/301 * 12/300 * 11/299 * 10/298 = about 1 in 826 quadrillion.
The answer is 1 * 23/311 * 22/310 * 21/309 * 20/308 * 19/307 * 18/306 * 17/305 * 16/304 * 15/303 * 14/302 * 13/301 * 12/300 * 11/299 * 10/298 = about 1 in 826 quadrillion.
March 1st, 2015 at 12:57:31 PM
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Technically this is countable. In order to make a bet at improved odds (albeit slightly), you'd want to watch for decks in which many cards of the same rank have already been played. In fact, given that you wanted to play this side-bet on the first hand of the shoe, you'd want to play again on the second hand of the shoe if you won on the first hand.
