TomG
TomG
Joined: Sep 26, 2010
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September 14th, 2021 at 6:35:46 PM permalink
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.
unJon
unJon
Joined: Jul 1, 2018
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September 14th, 2021 at 7:15:11 PM permalink
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.

  • link to original post



    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?
    The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
    chevy
    chevy
    Joined: Apr 15, 2011
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    September 14th, 2021 at 7:34:32 PM permalink
    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)
    FinsRule
    FinsRule
    Joined: Dec 23, 2009
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    Thanks for this post from:
    teliot
    September 14th, 2021 at 7:41:45 PM permalink
    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

  • link to original post



    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.
    chevy
    chevy
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    Thanks for this post from:
    teliot
    September 14th, 2021 at 9:08:23 PM permalink
    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

  • link to original post




    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.
    ChumpChange
    ChumpChange
    Joined: Jun 15, 2018
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    September 14th, 2021 at 9:16:19 PM permalink
    One player can lose 8 bets on one hand.
    The dealer can win 15-20+ times in a row.
    charliepatrick
    charliepatrick
    Joined: Jun 17, 2011
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    Thanks for this post from:
    unJon
    September 15th, 2021 at 1:40:35 AM permalink
    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
    Dieter
    Administrator
    Dieter
    Joined: Jul 23, 2014
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    September 15th, 2021 at 1:55:26 AM permalink
    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

  • link to original post



    Is standing on 12 a basic strategy play?
    May the cards fall in your favor.
    charliepatrick
    charliepatrick
    Joined: Jun 17, 2011
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    September 15th, 2021 at 2:06:18 AM permalink
    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

  • link to original post



    Is standing on 12 a basic strategy play?
  • link to original post

    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.
    Dieter
    Administrator
    Dieter
    Joined: Jul 23, 2014
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    September 15th, 2021 at 2:12:18 AM permalink
    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

  • link to original post



    Is standing on 12 a basic strategy play?
  • link to original post

    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.
  • link to original post



    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!
    May the cards fall in your favor.

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