Let's say 2 specific players are in a BJT with 162 players. The final table has 6 players and each table leading to the final table has 6 players with 2 advancing per table. The 2 specific players we are interested in are John Q. Average and Peter Pro. Mr. Average is an average player (surprise) and Mr. Pro thinks he is twice as good as Mr. Average. So, Mr. Average will advance from his table 1/3 of the time or 33.33% (6 players, 2 advance = 1/3). Assuming Mr. Average and Mr. Pro play at different tables, the question is: Assuming Mr. Pro is indeed twice as good as Mr. Average, how often (what percent of the time) will Mr. Pro advance from his table?
Tough One Good to see you posting again Toolman! It's hard to define twice as good. There are a number of ways to look at that. You could say twice as good means having twice as many units of strength, we'll call them bananas, as another player. But then how many bananas does an average player have? We could assign an arbitrary baseline of ten bananas for an average player. A player with zero bananas makes the worst possible decision every time. An average player makes good decisions often enough to be given ten bananas. Then a player who is twice as good has twenty bananas. But we still haven't even begun to define how much more often bananas translate to wins. You could state that a player who is twice as good wins twice as often as the average expectation. That would mean in a two from six table his expectation is to win two-thirds of the time. That's a lot. I don't think the best player in the world would win that often in tables full of novices. And what about tables in which three from five advance? To be twice as good under this definition he would have to advance six times in every five of these games he plays. Even a ploppy will tell you that's impossible. You could state that a player who is twice as good as average would lose half as often. In our two from six table he would advance two-thirds of the time. But in a one from five table he would advance three-fifths of the time by this definition. Again, not going to happen. We could define an arbitrary percentage of chances of losing, but how would we prove it? For example, we could arbitrarily define being twice as good as being 75% as likely to lose a table as an average player. By this definition our good player would advance in two-from-six tables half the time. But how could you prove that means twice as good? You can't. A player being twice as good means that "goodness" is quantifiable in defined units of measure. We don't have that in tournament blackjack, poker, golf, baseball, or any other sport. I believe the most meaningful way to quantify a player's relative strength is to record his historical percentage of games he lost versus the theoretical average. For example, if he lost 60% of the time in the past year and 75% of the players at the tables he played at lost, his percentage would be 60%/75% = 80%. He has an 80% chance of losing versus the theoretical average. In our two-from-six table his chance of losing is 80% times 2/3 = 8/15. His probability of advancing is 47%. In the one-from-five table stated earlier his chance of advancing is 36%. In the three-from-five table his chance of advancing is 68%.
wondering If bananas improve I.Q. which I believe to be true, based on Monkey's math skills. How many of Peter Pro's extra bananas could he afford to eat and still win as offen or more often than before? tgun
We want more bananas, we want more bananas! This is before I gorge on bananas! The edge over average player and estimating value of tournaments, especially compared to the average player is a very thick chapter in my “understanding tournament blackjack” book, but I will try to simplifying it AMAP. To say that someone is twice as good as the average one should be able to prove that she or he wins twice the prize value offered to the average player. Basic and easily quantified unit useful for comparisons is frequency of advancement per round. There will be lots of conditional assumptions, which I write after the main body of the post. The average player in a round where 2 of 6 advance succeeds 1/3 of times (33.33%), so if he is given a chance to play 100 tournament rounds he should on average advance 33 or 34 times. Consider someone who advances about 40 times out of 100 played rounds. We can say that she advances 20% more often than the average player. (40%-33.33=6.667 6.667/33.33=20%) In the 40 second rounds (of initial 100 first round tries) she executes the same edge over her opponents, thus making third round (semifinal) 16 times, and enjoys being in the finals 6 or 7 times (to be statistically exact it is 6.4 versus average of 3.7). In the finals she still performs at the same superiority level (by that 20%) and cashes about twice as much as Mr. Average. Here goes your twice as good player. He or she needs to advance from played rounds on average 20% more often than others do. Now, the partial list of conditional assumptions is: We are talking about tournament with exactly 4 rounds 2 of 6 advance. The skill level of competitions in higher rounds does not improve. Prize distribution is structured in a way that our player has similar edge. We are omitting all kind of variances. The most influencing is “the streakiness”, for example, a player who advances only 6 times out of 100 vs. average of 33, but wins 3 consecutive rounds in only two tournaments will make final twice and perhaps win more than the average. Gotta go, casino management just brought into my suit a basketful of bananas... S. Yama
I used the 2 advance, 4 rounds as a starting point - to provide something to calculate from. However, what I'm really after is a general "rule of thumb" to judge the abilities of another player. Sure, they can say they are good and they think they should be cashing in more tournaments than they actually do, but are they over-rating their true skills? Seems to me, from the responses so far, that if Mr. Pro has an overall advance rate from a 6 player 2 advance table of about 40%, then that player is substantially better than Mr. Average. Mr. Pro would need to advance from the lower rounds of a 6 player 2 advance table more than 40% of the time (not an easy task) because he will advance less than 40% of the time from the upper rounds only because the competition level will increase as one advances to each successive round. Mr. Average will experience the same up and down percentage wins as Mr. Pro but he will average 33.33% advances. So if Mr. Average advances 33.33% of the time from the first round then he is actually below average because generally he is competing against less quality players (on average) in that first round. So if Mr. Pro experiences an overall average 40% table advances when 2 advance from 6 players, then most players I know overrate their skills as few reach that level. They may be confusing luck with skill or just plain wishful thinking. The only way to know how often one advances is to keep a detailed log, preferably on an Excel spreadsheet, to determine the rate of wins - memory don't cut it. I know the "40%" is not a hard number but as a "rule of thumb" it seems pretty good to me - bananas excluded of course.
The Law of Averages That's a pretty good way to look at it. Someone who's twice as "good" advances to the finale twice as often as the average expectation. But then, because of the variances Yama stated, we also have to add one more caveat. He has to advance to the finale twice as often in a theoretical average tournament. That would mean average number of players, average number of elimination tables, average amount of carnival/skill-friendly features in the rules, average opposition, etc. If Toolman's tournament were considered average then a player who's twice as good as average would advance to the finale 2/27 of the time, versus 1/27 for an average player.
Only money counts The only statistic that means anything is ... do you win money. It is pretty easy to calculate expected winnings for a tournament. I track each "Try". Each buy-in, rebuy, or wild card is a "Try". If the format is 2 of 6 advance, then, 2 of 6 advance, then, 3 of 6 advance, the probablility of getting to the final table that Try, for an average player, is 1/3*1/3*1/2 = 0.0556. If you wild card to the semi, then the probability of getting to the final table that Try is, of course, 1/2 = 0.50. An average player can expect average results at the final table, so if 6 players at final table, and $24,000 prize pool, an average player at that table can expect to win, on average, $4,000. With a 0.0556 chance of getting there, the expected winnings are 0.0556*$4,000 = $222.40. Total your expected winnings for all Tries, and compare to your actual winnings. If your actual winnings are $40,000 while your expected winnings are $20,000; then you are twice as good as the average player in the tournaments you play. Your "Money Edge" is 100%. That's it, as far as I am concerned. If you want or need a "Table Edge", just calculate the average number of tables you actually play per Try (including any final tables made), and use the formula TE=(AW/EW)^(1/ATPT) Where TE is table edge, AW is actual winnings, EW is expected winnings, and ATPT is average tables per try. ^ is the sign for expotentiation. Nothing counts but the money.
