What to do?? What to do??

Discussion in 'Blackjack Tournament Strategy' started by Blackjack Rebel Too, May 24, 2006.

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  1. Blackjack Rebel Too

    Blackjack Rebel Too Member

    Final hand of round One advances Min bet $5 Max bet $200

    BR1 has $750
    BR2 has $725
    BR3 has $650
    no other players
    BR1 has the button

    BR2 and BR3 have bet max bets on two previous hands and seem certain to do so again. (chips already cut out of their stack)

    Betting first, BR1 bets max of $200, followed by max bets from BR2 (betting second) and BR3 (betting third)

    BR1 gets JJ
    BR2 gets 67
    Br3 gets 96
    Dealer's upcard is an 8

    BR2 is an experienced player who will certainly double. BR3 is a young guy who will probably double.

    My first question: Should BR1 now stand or split?

    My second question: What alternative bet should BR1 have made?

    Thanks in advance
     
    Last edited: May 24, 2006
    Blackjack Rebel Too, May 24, 2006
    #1
  2. elyssez

    elyssez Member

    What's the dealers' upcard?
     
    elyssez, May 24, 2006
    #2
  3. Blackjack Rebel Too

    Blackjack Rebel Too Member

    I just edited the first post to show that the dealer's upcard was an 8.
     
    Blackjack Rebel Too, May 24, 2006
    #3
  4. KenSmith

    KenSmith Administrator Staff Member

    I'll have to work it out to be sure, but my "at-the-table" answer is stand.
     
    KenSmith, May 24, 2006
    #4
  5. Blackjack Rebel Too

    Blackjack Rebel Too Member

    Thanks Ken. That is what I did in real time. The way the hand played out was:

    BR2 doubled and busted
    BR3 doubled and caught a 5 for a 20, matching my total
    I stood on my 20

    BR3 passed me by $100 and advanced on to the final round. (This was a semi-final table)
     
    Blackjack Rebel Too, May 29, 2006
    #5
  6. TXtourplayer

    TXtourplayer Executive Member

    Great hand, bad luck...

    You had two betting choises:

    #1. Bet the $200 max. and hope for a winning hand and no DD for either BR2 or BR3.

    #2. Bet $20 and hope the dealer win's or pushes agaisnt BR2 and BR3.

    I perfer #1 the best over all for your bet.

    As far as after the cards were out, I like standing. Both BR2 and BR3 need to DD on bustable hands to beat you.
    Put the monkey on their backs and make them beat you, I wouldn't open yourself up and possible give your lead away by spliting.

    With any luck you win this table with both BR2 and BR3 busting out on their DD attempts. As hard as it is, we see things like this happen all the time.

    Better luck next time.
     
    TXtourplayer, May 29, 2006
    #6
  7. KenSmith

    KenSmith Administrator Staff Member

    I believe I was mistaken, and splitting is better than standing here.
    Once again, I don't yet trust my numbers, but I'm getting stand=53% and split=61%. It looks like hit to 17 or higher on the first split hand is appropriate, and the strategy for the second split hand depends on the first hand total.
     
    KenSmith, May 30, 2006
    #7
  8. Reachy

    Reachy New Member

    My gut tells me to split as well

    Wouldn't know how to calculate split probabilities at this stage (can only imagine it involves a massive spreadsheet!!) but the feeling I'm getting from some of the recent posts is that you need to match the bets by splitting. If you split you still have the low and high and they are likely to bust anyway. If either of them win you have no means of beating them unless you match their bets.

    Incidentally if you had split you would have got probably a 19 for your 1st hand (guessing that BR2 must have at least got a 9 to bust a 13) and I guess you would have stood on that so your next hand would have been a 15. What would you have done then?

    Cheers

    Reachy

    (*sound of excel loading...)
     
    Reachy, May 30, 2006
    #8
  9. TXtourplayer

    TXtourplayer Executive Member

     
    TXtourplayer, May 30, 2006
    #9
  10. KenSmith

    KenSmith Administrator Staff Member

    Tx is right. I'm assuming that the players will play optimally behind you after you split. If you make poor hands, or bust, their options change. I'll have more to post after I've completed my work, but it definitely looks like I would have misplayed this at the table.
     
    KenSmith, May 30, 2006
    #10
  11. Reachy

    Reachy New Member

    I knew I shouldn't have made such a rash comment!

    I know, I would have stood on the t,t as well. It just seems that alot of the recent teasers have thrown up some unusual plays (like DD hard 20) and I got the feeling that this might be a similar scenario. I realise my post was slightly flippant but I've just got over a virus and my brain isn't working very well at present. However if we assume that both the other players double regardless (which is what I was doing initially) then BR1 will have the low if he splits won't he? i.e. BR1 - 350, BR2 - 325 and BR3 - 250. Just checking that my maths is correct on that one. I realise that the other players may change their play after a split of course and not double.

    Incidentally if BR1 split and hit 15 & 16 should he have stood on those totals?

    Cheers

    Reachy
     
    Reachy, May 30, 2006
    #11
  12. Walt

    Walt New Member

    Split?

    If I am BR2 and stiff, I'd be very happy if the leader gave up his 20 and the low.
     
    Walt, May 30, 2006
    #12
  13. TXtourplayer

    TXtourplayer Executive Member

    Stand on both

    Once you split you can't afford to bust either hand. You have to just ride it out and hope the dealer busts or even better that BR2 and BR3 do DD and both bust.

    Don't worry about messing up a hand, we all do it and with only seconds to decide it is hard to come up with the best possible play and even the highest percentage play dosen't always win.

    Luck is needed more than any other thing, without it you can have first high and low on the last hand only to get swung, it happens all the time.

    Now that is when you want to go throw-up!...LOL
     
    TXtourplayer, May 30, 2006
    #13
  14. Reachy

    Reachy New Member

    can you verify this?

    I just did some quick maths using Ken's Double Down outcome tables on BJI and calculated that there is nearly a 48% chance of at least one of the other players being successful in their doubles. Is this right?

    Cheers

    reachy
     
    Reachy, May 30, 2006
    #14
  15. TXtourplayer

    TXtourplayer Executive Member

    Off the top of my head...

    Just off the top of my head I would say the chances of either BR2's 13 or BR3's 15 wouldn't have that high of a percentage of hitting a winning hand.

    And if it was 48% that they could hit the DD that would mean 52% of the time they wouldn't, which is in your favor as BR1 to stand on your 20.
     
    TXtourplayer, May 30, 2006
    #15
  16. Reachy

    Reachy New Member

    Seperately they have odds of about 29% and 25% but is it not the case that the odds of at least one of them being successful is more than the highest individual odds? I figured it out as the odds of A and B winning + the odds of A winning and B pushing and so on. I'm probably on the wrong track?

    It also sort of ties in with Ken's preliminary calculations i.e. that standing has 53% chance of winning, but his split calculations seem to say so far (and I don't mean to pre-empt him) that there is a 61% chance of winning if you split. I guess we must await his results...

    Cheers

    Reachy
     
    Reachy, May 30, 2006
    #16
  17. TXtourplayer

    TXtourplayer Executive Member

    Another "If" play...

    What if you are BR1, but you have last bet? Now what would you do?

    With BR2 busting in front of you and BR3 hitting 5 for 20, now splitting looks like the best play.

    My point is the same hand, but a different scenario makes the play a lot easier to figure out.

    You just need to play a BJ tournament like a chess game. Think a few moves (or I should say hands) ahead so you know what to do and what the other players can do to in different situations to win.
     
    TXtourplayer, May 30, 2006
    #17
  18. KenSmith

    KenSmith Administrator Staff Member

    Unfortunately, there's no direct way to use the doubling charts to answer the question. Since the two hands are directly correlated to the dealer outcome, you really have to back-figure from there.

    If we assume both players will double, you could figure the chance of at least one succeeding with this process.

    Start with a dealer bust, 8 up, 6 decks, p = 24.37%

    When the dealer busts, hard 13's double will lose 5/13 of the time. Hard 15's double will lose 7/13 of the time. The chance that neither will win is (5/13) * (7/13). The chance that at least one will succeed is 1 - [(5/13)*(7/13)] =
    1 - 35/169 = 134/169

    So, for this line item. 24.37% * (134/169) = 19.32%

    Dealer 17: p(h13 loses or pushes) = 9/13, p (h15 loses or pushes) = 9/13.
    p(one or both win) = 1 - 81/169 = 88/169
    For dealer 17, 12.89% * 88/169 = 6.71%

    Notice that the chances of h13 and h15 are equal whenever the dealer doesn't bust. We'll use that to shorten our formula a bit.

    Dealer 18: p(one or both win) = 1 - (10/13)^2 = 69/169
    For dealer 18, 36.00% * 69/169 = 14.70%

    Dealer 19: p(one or both win) = 1 - (11/13)^2 = 48/169
    For dealer 19, 12.87% * 48/169 = 3.66%

    Dealer 20: p(one or both win) = 1 - (12/13)^2 = 25/169
    For dealer 20, 6.92% * 25/169 = 1.02%

    Dealer 21, they can't win, so ignore it.

    Add up the total, and we see the chance that at least one of the doublers will win is 45.41%.

    Since BR1 stood on 20, he'll advance 100-45.41 = 54.59%

    Note that this number for BR1's chances are higher than I stated above (53%). A small part of the difference is because I used infinite deck dealer outcomes in the previous calc, and 6 deck here.

    The rest is because BR3's best strategy is not to always double. Consider what happens if BR2 doubles and makes 20. Now, BR3 should hit to 21 and not double. My "BR1 stands = 53%" number takes into account pure optimal play by his opponents.
     
    KenSmith, May 30, 2006
    #18
  19. KenSmith

    KenSmith Administrator Staff Member

    When I saw a 1% discrepancy in the 53% and 54% answers, I was wondering exactly what BR3's optimal strategy is.
    BR3 should double with only one exception, and that's the one I mentioned before. If BR2's double yields a 20, BR3 should try to hit to 21 instead of double.

    I knew that small strategy change couldn't account for a whole 1% in difference, so I figured I had an error above. I found it.

    If the dealer makes 21, BR2 or BR3 can advance with a push. I left that out in the previous post.

    Add: For dealer 21, 6.95% * 25/169 = 1.03%.

    After one mistake, I wanted to double-check my exact numbers with infinite decks, so I redid the calculations using infinite decks.

    Dealer Bust: 24.47% * 134/169 = 19.40225%
    Dealer 17: 12.86% * 88/169 = 6.696331%
    Dealer 18: 35.93% * 69/169 = 14.66964%
    Dealer 19: 12.86% * 48/169 = 3.652544%
    Dealer 20: 6.94% * 25/169 = 1.026627%
    Dealer 21: 6.94% * 25/169 = 1.026627%
    Total = 46.47%
    BR1 advances 100-46.47 = 53.53% (if both BR2 & BR3 double)

    If BR3 uses the optimal strategy (hit to 21 if BR2 gets a 20), then BR1's chances drop ever so slightly to 53.49%.
     
    KenSmith, May 30, 2006
    #19
  20. Blackjack Rebel Too

    Blackjack Rebel Too Member

    While at the table, I did all those calculations but I came up with a 55.2345% chance of me winning. I must have made a miscalculation.:D :laugh::p
     
    Blackjack Rebel Too, May 31, 2006
    #20
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