An example for discussion of EV, Equity, and Edge

Think I got it

Need to exclude the impossible combination - 'advance from semis and win wildcard':

Advance from semis or win wildcard: 0.090909 + (1-0.090909)*(1/60) = 0.10606
Times final seat value: $1242.86 X 0.10606 = $131.82


And the same thing would presumably apply in the first version:
Total chance of making the final table: 0.09259 + (1-0.09259)*(1/60) = 0.107714

Value of a seat in the finals: $8700 / 7 = $1242.86
EV of round 1 using this method: $1242.86 * 0.107714 = $133.87

 

Ken:

Since your original post on this thread contained some errors that you noted, could you repost your calculations with the correct numbers and without referring to any corrections. The way your post now stands, I'm having a hard time following the flow of numbers.

 

It's very much ingrained in my mind that when people use the terms 'edge' or 'advantage', they are referring to skill-derived EV, over and above the average player. Might using them as you suggest here add to the ambiguity?

And do we actually need a third piece of terminology? If equity has a sign, positive or negative, then it can surely have a value. You could say that +0.405% is positive equity, and the magnitude of that equity is 0.405%. (That would make the term 100% equity, meaning zero, problematic, though.)

The slight problem I have with the whole idea that EV must only refer to the dollar value of a seat is that it masks the fact that, conceptually, and in terms of some of the ways calculations can be done, EV (in the broad sense that can be applied in any domain, not just TBJ) is precisely the right word for the thing that quantities such as the above 0.405% represent.

E.g., When we say that a single hand of BJ has an EV of -0.50%, we mean that for every dollar bet, a player expects to lose $0.005. In the same way, the +0.405% figure means that for every dollar of entry fee handed over, the player expects to win $0.00405.

 

The error is in the re-buys

Ken,

Thanks for re-posting your "cleaned up" calculations. It's raining pretty heavy here so I spent the better part of the last hour watching the rain (very relaxing if one in not in it) with your calculations in hand. Once the rain stopped and true to form with my signature line, I solved the reason why the "complicated" method did not match the "simple" method for calculating EV. When you did your calculations for the re-buys you assumed that all players that lost in the 1st round would re-buy. This is an error. Only 42 were able to re-buy not the 44 that were eliminated in Round 1. That difference of 2 players reduces the prize pool by $100 when using the "complicated" method which in turn reduces its EV. On the other hand, the "simple" method took into account that $100.

As proof, I have copied your last post below and noted my calculation changes in RED. The EVs for both the "simple" and "complicated" methods are now identical as you originally suspected they should be.

So I can draw 2 conclusions about the "simple" method as long as players are not added after the first round:
1) Since both methods produce the same EV, there is no need to go through complex and error prone procedures. The simple formula "Total Prize Pool / Number of players" will give the right answer every time. With a few "tweaks", the "simple" method can be used for a tournament where players are added after the first round and these "tweaks" will result in substantially less calculations than the "complicated" method.
2) Everything is automatically taken into account: including re-buys, wild-cards, how players advance, and rules of the game.
3) Can be used "on the fly" when at a tournament with 100% reliability.

 

I agree 100%. However, we are stuck with this "standard" terminology just as we are stuck with a way of playing BJTs which I personally consider AWFUL. So we have to live with it as we are powerless to change it - see my signature line.

 

Pp/np

You’re right toolman1.
As you said, as long as no new players are injected (paid or not paid) the EV for an average player is PP/NP assuming that they all have the same chances of playing, which if not possible is best represented by an average.
In Ken’s particular scenario we need to differentiate two groups of players in the subgroup of reentries: 42 that were unlucky to lose in the first round and has gotten to play in the rebuy and 2 players that theoretically wanted to play in rebuys but didn’t get to do it. But for a representative of all players it is as you wrote 2/3 x 1/3 x 42/44

S.Yama

 

That was more or less what I understood you to mean. Would it be legitimate terminology to call this result a 'conditional EV'? (i.e. it is your expectation, conditioned on 100% probability of being able to rebuy?)

It seems to me that what you have done is set up a framework, a procedure in which various variables are available to be tweaked.

Give them all their default values -

Probability of getting a rebuy = 42/44
Probability of advancing from a table = 1/3
Probability of drawing a wildcard = 1/60

- and the result is the same as the simple calculation.

Modify one or more, based on some specific information, turn the handle, and viola! - your personal EV for the tournament.
(You might wonder how the wildcard probability could change; maybe you notice that a proportion of losers routinely don't hang around for the drawing.)

 

I want to address the issue of how the number of available seats and number of players doing a re-buy affects the SV, if at all. For discussion purposes, I’ll continue to use the hypothetical tournament we have been referring to on this thread. Below is a partial reproduction of that tournament incorporating all of the calculation corrections we have previously noted with the re-buy formula highlighted in RED for ease of reference. I also took the liberty to change “EV” to “SV” (Seat Value). To save space, I stopped at the point where the re-buy calculation is made since the rest is not germane to my discussion.

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The hypothetical tournament:

Round 1: 66 players pay $100 each. Two players advance from each of 11 tables. The result is 22 semifinalists and 44 losers.

Re-buy Round: Only 42 re-buy seats are available for the 44 eligible.
42 re-buys pay $50 each to play 7 tables of 2 advance. That creates 14 more semifinalists.

Semifinals: 36 players, 6 tables of 6 with 1 advance.

Finals: 6 players plus one wildcard from all 66 round one entrants makes 7 finalists. The breakdown of prizes is unimportant.

Let's assume that the value of a seat in round 1 is to be called the SV of this event. What is the appropriate calculation?

What if we use the round by round calculation:
Chance of advancing from round 1: 1/3
Chance of advancing from the re-buy round: (2/3) * (1/3) * (42/44) = .21212​

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Now let’s concentrate on the re-buy formula:

If we take the re-buy formula of (2/3) * (1/3) * (42/44) = .21212 and multiply out the fractions on the left side to have just one fraction we come to 84/396 = .21212. Now, to make that fraction have meaningful numbers, we divide the numerator and denominator by 6 to arrive at the fraction “14/66”

So now the re-buy formula can simply be presented as: 14/66 = .21212
The number “14” (numerator) represents the number of players that will advance from the Re-Buy Round.
The number “66” (denominator) represents the number of players in Round 1.
Note: No numbers in this formula refer to the number of seats available to the players in the re-buy round which is 42. ​

My proposal: Use the fraction 14/66 = .21212 to represent the re-buy round calculation. The numerator will always be the number of players that will advance out of the re-buy round and the denominator is the total number of players in Round 1. What I find interesting here is that the same rule applies for determining a fraction for the Round 1 play. In this case the fraction is 1/3. That is the same as 22/66 with 22 being the number of players that will advance out of Round 1.

The probable objection to using this single fraction:
There is no provision for those who chose not to re-buy which reduces the players in the re-buy round.

No provision for that is necessary because the probability of advancing out of the re-buy round will always be .21212. Let's take a look at what happens if only 35 players decide to re-buy:

The original formula is: (2/3) * (1/3) * (42/44) = .21212
The (2/3) is the reduced fraction from 44/66
So the original formula is actually (44/66) * (1/3) * (42/44) = .21212
The number 44 (in each of 2 of fractions) represented the number of players that wanted to re-buy.
So let's assume only 35 of the players decide to re-buy.
The formula becomes: (35/66) * (1/3) * (42/35) = 1470 / 6930 = .21212​

So we are stuck with the number .21212. Then we say .21212 is the probability of winning the re-buy round. If that's the case then there is only one number that can be multiplied by .21212 to have 14 players emerge from the re-buy round and that number is 66. Therefore, the number .21212 represents the PORTION OF THE ORIGINAL PLAYER POOL of 66 THAT WILL ADVANCE OUT OF THE RE-BUY ROUND: 66 * .21212 = 14. No matter what we do, if the math logic is correct, that answer will always be .21212 or 14 players out of the original 66. So, the number .21212 can never change (for SV calculation purposes) regardless of the number of players in the re-buy round. One exception - if too few players re-buy, additional wild-cards for Round 2 may be needed (if the tournament rules permit) but that does not change the SV.​

So my conclusion is that the calculation for the re-buy round remains at 14/66 = .21212 regardless of the number of players that sign up for the re-buy round. In reality, this is because if the tournament starts with 66 players, SV is only affected when the estimated prize pool is adjusted by the re-buy money estimate - not because more or less players are trying to advance through the re-buy round. When play began we don't know the future. There are 66 players competing and the chance of winning, say 1st place, is always 1/66 at the beginning of play. Yes, a given player's probability constantly changes as the play progresses while that player wins or loses at various points but at the start, any given player's chance for first place is always 1/66. Remember, SV is an estimate for the average player at the beginning of play and most of the time, unless the Prize Pool is fixed, it will not equal an SV that is calculated after a tournament is run and all the numbers are certain.

NOTE:
I will be in and out of town over the next 10 days or so. If any member’s response needs a response from me, just hold tight, I will respond when I can.