Harrington on Hold 'em Workbook

BlondeFishBlondeFish
in General Poker Chat
I just finished Harrington on Hold 'em, Volume III: The Workbook. I would rank all three Harrington volumes as my top three books on tournament poker.

Now that forum members have shared our scores for The Donkey Test, how about sharing our scores for the Workbook? My score is 384, out of a maximum ~591 points possible.

"400 or more: A very good player who should show a solid profit in big tournaments.
300 or more: A player with a solid base of skills to build on."

Comments

  • compueasecompuease Admin
    I also did Vol III the workbook a few months ago, don't recall my score but one thing to be cautious of when using evaluations such as this to judge how one should fair in real life situations. This only gives a guidline, it cannot tell you how all the intangibles will effect your results.. such as....

    1/ how predictable your opponent is or isn't.
    or perhaps
    2/ how well your opponent is reading YOU.

    I'm sure there are many more but you get the idea. It's not difficult to pick out an opponent who is playing by the "book" and adjust so be careful....
    Scores can be deceiving and the proof is in the pudding, success in real multi player tournaments over a period of time.... and I'm not referring to free or low level buy-ins...
  • blackmagiczblackmagicz
    I found that book very good (I did the test as well...but don't remember my score)

    You will find your hand reading ability will increase dramatically (at least it did for me)

    Understanding betting patterns and player thinking is a big key in playing tournies and this book I highly recommend.
  • jpajamasjpajamas
    HoH VIII is an excellent exercise. All three Harrington volumes should be in any serious tournament player's library. I would also suggest Phil Gordon's Little Green Book, and Little Blue Book. I received the PG trilogy for Christmas (Little Black Book - recreational read), and combining the 2 authors makes for some very powerful philosophy on poker...
    Scored 422 the first time I did the wookbook, and did it again about 4 months later, after reading PG, and actually disagreed with some answers! Ballsy, maybe, but just goes to show that poker is a game of situations, and everyone has their own interpretation.
  • BlondeFishBlondeFish
    My favourite problem is #39 on page 257.

    Situation: Online SNG tournament. The tournament pays 3 players in the standard ratio of 50% 1st, 30% 2nd, 20% 3rd. 4 players remain. You are one of the smaller stacks with $2,000 chips.

    Your hand: ??

    Action to you: Player A with $5,000 goes all-in, Player B with $5,000 folds as does the small blind with $1,500 chips. You think Player A has a pushing range of: any pair, any A, KJ+, QJ+.

    Which range of hands would you call with?

    Unlike the five other single-table bubble problems (#37-42) where Harrington neglects to show how the reader can arrive at the same solution as him, I get a better idea with his nine pages of analysis for #39. I used an "Independent Chip Model" calculator to try to verify Harrington's calculations, but the $ equities are different.

    Player . Harrington . ICM
    A . . . . $332.63 . . $326.81
    B . . . . $332.63 . . $326.81
    C . . . . $187.71 . . $194.48
    D . . . . $147.02 . . $151.91

    Has anybody been gone through and verified Harrington's calculations? I think his assumptions and calculations are better than ICM. I remember pkrfce9 trying to solve a similar problem.
  • pkrfce9pkrfce9
    hmmm yes... i did do some spreadsheet work on that. i'll go back and have a look to compare what i have against dan's numbers. give me a reminder in a week or so.
  • BlondeFishBlondeFish
    For the probability of player A finishing third, I calculated 18.0%, but Harrington gave an answer of 24%. I can see only six possible combinations for player A to finish third.

    1) B 1st & C 2nd: 5000/13,500 * 0.19281 * 5000/6500 = 0.0549
    2) B 1st & D 2nd = 0.040
    3) C 1st & B 2nd = 0.037
    4) C 1st & D 2nd = 0.011
    5) D 1st & B 2nd = 0.026
    6) D 1st & C 2nd = 0.011.

    0.05 + 0.04 + 0.037 + 0.011 + 0.026 + 0.011 = 0.180. Did anybody come up with Harrington's answer or mine?
    pkrfce9 wrote: »
    i'll go back and have a look to compare what i have against dan's numbers. give me a reminder in a week or so.
  • pkrfce9pkrfce9
    BlondeFish wrote: »
    For the probability of player A finishing third, I calculated 18.0%, but Harrington gave an answer of 24%. I can see only six possible combinations for player A to finish third.

    1) B 1st & C 2nd: 5000/13,500 * 0.19281 * 5000/6500 = 0.0549
    2) B 1st & D 2nd = 0.040
    3) C 1st & B 2nd = 0.037
    4) C 1st & D 2nd = 0.011
    5) D 1st & B 2nd = 0.026
    6) D 1st & C 2nd = 0.011.

    0.05 + 0.04 + 0.037 + 0.011 + 0.026 + 0.011 = 0.180. Did anybody come up with Harrington's answer or mine?
    i checked with harrington's chip counts. calculating 3rd place, i've got a slight rounding problem since these are dependant on the 1st 2 calcs but i come up with an answer very close to harrington's numbers and EV calcs.

    to calculate the odds of getting third, i summed up the odds of everyone else getting 4th multiplied by the others 2 getting 1st and 2nd, same calc involved just combining your 6 cases into 3.

    B getting 4th approx 2.5% (your cases 4 and 6 = 2.2%)
    C getting 4th approx 12.5% (your cases 2 and 5 = 6.6%)
    D getting 4th approx 9% (your cases 1 and 3 = 9.2%)
    total approx 24% (your total = 18%)

    you seem way off on your calcs for C. you tell me why...
  • BlondeFishBlondeFish
    I have verified my calculations for cases 2 and 5.
    (2) (Probability of B finishing 1st) * (probability of D finishing 2nd) * (probability of A finishing 3rd)
    = 5000/13,500 * 0.15004 * 5000/7000
    = 0.03969

    (5) P(D 1st) * P(B 2nd) * P(A 3rd)
    = 1500/13,500 * 0.329 * 5000/7000
    = 0.02611

    This totals to 6.6%, which seems to make more intuitive sense than the 12.5% you got for your case "C getting 4th". The probability of C getting 4th (your 12.5%) should be LOWER than D getting 4th (your 9%), i.e., D with the fewest chips should be the most likely to finish last.
    pkrfce9 wrote: »
    C getting 4th approx 12.5% (your cases 2 and 5 = 6.6%)
    D getting 4th approx 9% (your cases 1 and 3 = 9.2%)
    you seem way off on your calcs for C. you tell me why...
    pkrfce9 wrote: »
    to calculate the odds of getting third, i summed up the odds of everyone else getting 4th multiplied by the others 2 getting 1st and 2nd, same calc involved just combining your 6 cases into 3.
    How were you able to calculate the "odds of everyone else getting 4th" without first calculating the probabilities of everyone finishing third?
    In the words of Vinnie Barbarino, I'm soooooooooooooooooooo confused! :)
  • pkrfce9pkrfce9
    BlondeFish wrote: »
    The probability of C getting 4th (your 12.5%) should be LOWER than D getting 4th (your 9%), i.e., D with the fewest chips should be the most likely to finish last.

    How were you able to calculate the "odds of everyone else getting 4th" without first calculating the probabilities of everyone finishing third?
    In the words of Vinnie Barbarino, I'm soooooooooooooooooooo confused! :)
    maybe i have the players mixed up? i don't have the book handy and am going from your post. i have the stacks as follows:
    sb C 1500
    bb D 2000
    utg A 5000
    button B 5000

    and of course, when i wrote this up originally, i did everything as SB, BB, UTG and BUTTON so I may be mixing up my ABC's here.

    as far as how to calculate the odds for 4th, i spoke to dan about this way back. i got the sense he's uncomfortable with too much of this stuff being made public.
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