The Peek Effect

Discussion in 'Blackjack Tournament Strategy' started by BlueLight, Aug 9, 2007.

  1. BlueLight

    BlueLight Active Member

    This is a teaser.
    We can all go to some web site and find dealer probabilities for different dealer upcards and decks with certain cards removed from the deck.
    For this teaser the problem is to find player probabilities. Two problems are given below.

    Problem (1) It's a one deck game with 2 players.

    Player #1.......3..Q..A..4... = ..18

    Player #2.......K..6

    Dealer............9

    When the players get to act on their hands, Player #1 will stand and Player #2 plans on taking only one card. The questions for this simple example is:
    1) What is the probability that Player #2 will be delt an A ?
    2) What is the probability that Player #2 will be delt a 4 ?


    Problem (2) Again it's a one deck game but the cards were delt out slightly different. The dealer 9 and the Player #1 Q have been switched to as follows:

    Player #1......3..9..A..4.. = ..17

    Player #2......K..6

    Dealer...........Q

    Now when the players get to act on their hands, Player #1 stands as before and Player #2 again plans on taking one card. Again the questions are:
    1) What is the probability that Player #2 will be delt an A ?
    2) What is the probability that Player #2 will be delt a 4 ?


    The answer to problem (2) is not as easy as some might think.


    ..............................BlueLight
     
  2. KenSmith

    KenSmith Administrator Staff Member

    I'll sit tight on this one until someone else answers.
     
  3. arlalik

    arlalik Member

    I'll try to answer it short

    Problem(1)
    1) 1/15
    2) 1/15

    Problem(2)
    1) 3/44
    2) 1/15
     
  4. London Colin

    London Colin Top Member

    Nice Teaser

    But you may have given too much of a hint with the title. :)

    I agree with arlalik's first three answers, but not the last one.

    We know the following -

    • The dealer has peeked, so the hole card is not an ace.
    • Deck + hole = 45 cards
    • Deck = 44 cards; contains 3 aces.
    • Hole card is one of 42 possible non-aces (i.e. 45 - 3)

    Therefore,
    P(ace) = 3/44

    To work out the probability of any other card, you need to consider the probability that the hole card itself may or may not be this same card.
    P(hole card = 4) = 3/42
    P(hole card <> 4) = 39/42

    So, P(4) = (3/42 * 2/44) + (39/42 * 3/44) = 0.0660

    This is just slightly less than 1/15, which reflects the fact that the next card is more likely to be an ace, once the dealer has peeked, and so all the other probabilities must be reduced accordingly.
     
    Last edited: Aug 10, 2007
  5. tirle_bj

    tirle_bj Member

    I agree with Colin

    Excellent job, Colin. I have nothing to add. The only minor correction:
    (3/42 x 2/44) + (39/42 x 3/44) = (2 x 3 + 39 x 3) / (42 x 44) = 123/1,848 =
    = 0.06656 instead of 0.0660. Probably this difference comes from rounding.
    Now in the first case we have 1/15 = 0.06666. So with the corrected number we have only 1/10000 of a difference.
     
  6. London Colin

    London Colin Top Member

    oops

    Alas, no. It seems I entered a parallel universe where 2 * 3 = 5. :D My answer was the result of 122/1848.
     
  7. BlueLight

    BlueLight Active Member

    Last Thoughts

    London Colin had the right formula - just jiggled the slide rule before writing down the numbers.

    Summing up : When the cards are delt out as was shown with the dealer having the Q up the probabilities are as follows:

    ..............................................Before Peek...............After Peek
    A on top of hand held pack.......... 0.0666667 .............. 0.0681818
    4 on top of hand held pack.......... 0.0666667 .............. 0.0665584

    There is also a "Peek Effect" for when the dealer has an A up. The results are a little different since there cannot be one of the many 10 value cards for the hole card.

    Years ago when analyzing the probabilities of getting a certain card at various point counts I ran into trouble when the dealer had an A or a 10 up; my sum of all the possibilities was greater than 1! This was because I didn't fully understand the peek effect.

    ..........................BlueLight
     
  8. London Colin

    London Colin Top Member

    More thoughts

    I stumbled upon the phenomenon a while ago, while thinking through the modifications needed to add ENHC support to some existing software.

    It also set me thinking about ways of generalising the idea to cover other cases where you have partial information about cards that have been dealt.

    E.g.,
    • If you were to have a face-down DD, like UBT, but with bust hands having to be revealed, then you would know which cards the DD card could not be.
    • If all cards are dealt face down and a player stands with a hand of three cards, the hand can contain no more than two from the set [10, 9, 8], etc.
    I didn't get very far with this, but added it to my growing to-do list, which is really more of an unlikely-to-ever-get-done list. :)
     
    PlayHunter likes this.

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