# Error in Harrington?

• Bronze
Joined: 15.06.2009
First of all, this post is not at all as horrible as you might think. I think Harringtons books are fine. (Well the post may still be very very bad ...) You probably need to have Harringtons volume 3 to be able to answer the question I have.

In volume 3, problems 37-42 discuss all-in situations on the bubble. I think the theoretical discussion is correct, but there may be an error or two in his practical calculations.

Question: Does anybody know for sure if the table on page 260 is correct or incorrect? I get slightly different numbers for the third and fourth places. Please see the tables below.

Harringtons numbers:
Player Stack First Second Third Fourth
A 5000 0.370 0.329 0.244 0.057
B 5000 0.370 0.329 0.244 0.057
C 2000 0.148 0.193 0.279 0.380
D 1500 0.111 0.150 0.232 0.507

My numbers:
Player Stack First Second Third Fourth
A 5000 0.370 0.329 0.215 0.086
B 5000 0.370 0.329 0.215 0.086
C 2000 0.148 0.193 0.313 0.346
D 1500 0.111 0.150 0.257 0.482

The numbers are supposed to add up to 1, both horizontally and vertically. They do in both tables when disregaring rounding errors.

You reasonably ask why I want to know this. I have translated the verbal theory to a mathematical formula for computing the probability of a player to finish in a given position given stack sizes. This formula is more ghastly than you might imagine if you haven't tried it yourself. (Even so if my particular formula happens to be incorrect.)

Then I have implemented the formula in a little computer program. (Some nontrivial issues arose there too.) My program spits out plausible numbers, but for Harringtons example they don't agree exactly.

Of course, I don't ask anyone to make the actual calculation. It's very painful to calculate the third column (or further columns) by hand. Bring in more players and it gets worse very very fast. Unless the formula can be simplified (mathematically) which I doubt it can, no human made computer can take on a large tournament with, say, 200 players left without making approximations.

So, if you for some reason know the answer to my question, please let me know. I'm sure that at least one other person than me has debugged Harrington

/Johan =
• 5 replies
• Bronze
Joined: 12.10.2010
Hi, YohanN7

I will move your thread to tournaments discussion forum where it belongs.

Best regards,

Kurrkabin
Joined: 19.04.2006
Hi YohanN7,

I can verify that your calculation is correct.

3rd collumn easily calculates the % on getting 1st place - no worries there

4th and 5th collumn calculate that if A gets first, X gets 2nd.
The same for B,C and D
Still we reach the same numbers as Harrington.

Now I calculated the probability that A gets 3rd.
This happens on 6 different "occurences", I named all 6 of them.
Now I took the numbers from X getting 1st and put them in one collumn.
Furthermore from the previous calculations I took the probabilities that the 2nd place goes to X when Y gets 1st (so from the overall 2nd calculations)
The third place is then simply calculated for "B gets 1st,C gets 2nd,A 3rd" with the chips of A and D => 5000/(1500+5000)
==>all lines have to be multiplied with probability of getting 1st and 2nd in the line. All the results added up give 21.5043% of getting 3rd for Player A

So harrington was wrong, your calculation correct.
(I estimated that the calculations for all others would be hopefully the same result,too)

Gratulations!
• Bronze
Joined: 19.01.2007
Hey Jonah,

Assuming that the calculations are based on ICM your results are indeed correct. (you can double check here http://www.icmpoker.com/Calculator.aspx) Or in Sven's detailed post

If this is not an error in Harringtons book, my guess is that his method of equity modelling was somewhat different from ICM, but i'm not sure what that would be, maybe it's in the book somewherre (i don't have it with me). But given that the first two places are the same it's likely just an error because using a different model (ex. malmuth-weitzman) would result in different numbers for all places.

Cheers,
FF
• Bronze
Joined: 15.06.2009
Thank you very much for your answers SvenBe and FishermansFriend. SvenBe, your post is wonderfully detailed, and I can use it to double check the numbers in a document I've produced alongside with my code implementation of the ICM formula.

Yes, the formula is intended to be the ICM formula, and I'm pretty sure Harrington describes ICM in the surrounding text, tough he doesn't explictly say ICM.

While I'm at it: My implementation now works nicely and produces the correct numbers (I am fairly sure by now), but there are interesting theoretical problems associated with it. Performance, for instance, is an issue for many players and, especially, many prizes paid.

One can probably use previously calculated numbers (e.g first place probabilities) when calculating deeper down (e.g second place probabilitis). (SvenBe's post indicates this too). Thus it will in the end be memory vs CPU. As it stands now I loop through everything and store nothing. I have a recursive version too which (while very elegant) performs like the loop version.

The interested nerd can find a recursive code snippet here:
http://www.holdemresources.net/hr/sngs/icm/icmjava.html
That is Java, but is easily translated to any language where recursion is fine. I use C++ myself from old habits.

/Johan =