What is the max times in a row you can lose at one deck blackjack?

Quote: ThatDonGuy

In theory, you can. It's pretty much impossible, but it "can" happen.

I'm having trouble understanding this. Are there really streaks that may or may not happen after an infinite amount of hands? If a streak "can" happen, but has not yet happened, shouldn't that mean there was not an infinite amount of hands played.

Quote: TomG

I'm having trouble understanding this. Are there really streaks that may or may not happen after an infinite amount of hands? If a streak "can" happen, but has not yet happened, shouldn't that mean there was not an infinite amount of hands played.

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The problem is that infinity is not a well defined concept.

Infinite consecutive wins
Infinite consecutive losses
Infinite alternating wins and losses

Can all three of those things happen in an infinite series?

EDIT:******This is so horribly wrong.......completely forgot dealer has to take cards.....please IGNORE.......**********

I think you can lose all 13 hands (though I admit to not being 100% on BS, so just relying on Wizard's charts)

Player - Dealer (up card first).....player always stands (Edit: per BS chart)

1) 87 - 6 10...Stand
2) 87 - 6 10...Stand
3) 87 - 6 10...Stand
4) 87 - 6 10...Stand

5) J2 - 4 Q....Stand
6) J2 - 4 Q....Stand
7) J2 - 4 Q....Stand
8) J2 - 4 Q....Stand

9) 93 - 5 A....Stand
10) 93 - 5 A....Stand
11) 93 - 5 A....Stand

12) 93 - 5 K....Stand

13) KK - K A....Stand.......I don't see a line for Splitting KK in chart, nor a surrender listing for 20.....so I think that is Stand (that's how I play it)

Quote: teliot

I get 11, but I can't get 12.

2-2 A-K
2-2 A-K
3-3 A-K
3-3 A-K
4-4-9 Q-8
4-4-9 Q-8
T-7 Q-8
T-7 Q-8
6-7-9 7-5
5-6-6 J-J
9-J J-T

Leftover = 556T

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I know this is what the thank you button is for, but this is amazing. I now wish I would have spent time thinking about this before just handing it off.

I think the Wizard would like a trivia question like this.

Quote: teliot

I get 11, but I can't get 12.

2-2 A-K
2-2 A-K
3-3 A-K
3-3 A-K
4-4-9 Q-8
4-4-9 Q-8
T-7 Q-8
T-7 Q-8
6-7-9 7-5
5-6-6 J-J
9-J J-T

Leftover = 556T

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Since my first attempt was so horribly, seriously, terribly, awfully, embarrassingly wrong, let me start with a proven commodity courtesy of teliot and ask can I get a 12th one......possibly by making dealer have to draw in some instances.

** are my modifications **

2-2 A-K
2-2 A-K
3-3 A-K
3-3 A-K
** Q-4 5-6-6
** Q-4 5-4-8
T-7 Q-8
T-7 Q-8
6-7-9 7-5
** 4-8 6-5-J
9-J J-T

** 9-9 J-T

Anyway, My guess is I am missing something....a split or something.

One player can lose 8 bets on one hand.
The dealer can win 15-20+ times in a row.

I get 12 using the following
2x 2 2 < A K
2x 3 3 < A K
2x 9 8 < Q Q
2x 9 8 < J J
4x 7 5 < 6 4 T

Quote: charliepatrick

I get 12 using the following
2x 2 2 < A K
2x 3 3 < A K
2x 9 8 < Q Q
2x 9 8 < J J
4x 7 5 < 6 4 T

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Is standing on 12 a basic strategy play?

Quote: Dieter

Quote: charliepatrick

I get 12 using the following
2x 2 2 < A K
2x 3 3 < A K
2x 9 8 < Q Q
2x 9 8 < J J
4x 7 5 < 6 4 T

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Is standing on 12 a basic strategy play?

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If the dealer has an up card of 6 or 4 then yes. btw you can always change the 9 8 vs Q Q to Q7 vs Q8 leaving 9 5 vs 6 4 T to get the same result.

Quote: charliepatrick

Quote: Dieter

Quote: charliepatrick

I get 12 using the following
2x 2 2 < A K
2x 3 3 < A K
2x 9 8 < Q Q
2x 9 8 < J J
4x 7 5 < 6 4 T

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Is standing on 12 a basic strategy play?

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If the dealer has an up card of 6 or 4 then yes. btw you can always change the 9 8 vs Q Q to Q7 vs Q8 leaving 9 5 vs 6 4 T to get the same result.

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Got it. Just looked a little funny for some reason.
Even if the player hits, they bust, so I suppose the hands per deck stays the same, with 1 player plus a dealer.
Nice solve!