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bbo speedball suggestion for bbo speedball

#1 User is offline   nathan2008 

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Posted 2012-May-10, 07:16

I want to talk about the bbo speedball. I am sorry to say that all of you must admit that bbo speedball is not as successful as you expected even after the biggest prize u gave us :(. Here is my analyse:

1. For those who play with GIB, in general they are "professional" GIB players. If they want to make money, they can play robot reward, if they want to win masterpoints, they can play robot duplicate or bingo race etc.
2. For those who play regularily ACBL speedball, they are USA players and they want to win ACBL points which are useful for them.

As i am a big fan of bbo and an amateur in economy, i give u my suggestions :). I think the main problem now is that the demand is too low. So even you increase the prize to 55%, (it means u increase the supply) it is not that successful. So you should try to increase the demand, but how?

The first step is to decrease the prize i think. (In economy, if the prize decreases, the demand will increase...) Let's look at the robot duplicate MP. Why is it so popular on BBO? It is even more popular than robot rebate 55%. Every 7 mins, we have robot duplicate and every time, we get 20+ players. But robot rebate 55% is once every 15 mins, still difficult to get 10+ players. The pb is about the entry fee i think. Assume a player can get average 55% (it means he is already above average.) As the result follows "approximately" a Gaussian distribution, a player with 55% doesn't mean he can get 1 win 1 lose in robot rebate.(we can do calculation for the exact number, i would say he will get 1 win 2 lose in general.) Also those who play robot rebate are in general better players. So the player with 55% will go to play robot duplicate to get more points and "lose" less. I think maybe same idea can be followed for bbo speedball. My suggestion is that you can put the entry fee as 0.5 but no REWARD. we can do some calculations, if we have 50 pairs. with 0.5$, bbo will get 50*2*0.5=50. With rebate 55%, assume 20 pairs get 55%+, bbo get 50*2*1-1.5*20*2=40. BBO gets more lol. (Of course if 0.5 still cannot attract players, you can put it 0.25. Anyway having players is always better than no player :) )

If the first step still can't attract players, i think your goal should try to attract those who don't have any bb$. We know that a huge amount of people don't have bb$ on bbo. So you should let them have some at first. My suggestion is that every week, have one free bbo speedball to win bb$. (ex: players with 55%+ can get 1 bb$. You can limit that tourney to 100 pairs, then you will give 40*1*2=80 bb$) When people have money, they will play more bbo speedball i think.

These are my humble ideas. I have many others.... Of course you can ignore all this if you think i am very wrong :).
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#2 User is offline   barmar 

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Posted 2012-May-10, 09:19

In the latest changes to the BBO speedballs, you can no longer play with robots.

#3 User is offline   mgoetze 

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Posted 2012-May-10, 13:09

View Postnathan2008, on 2012-May-10, 07:16, said:

Gaussian distribution, [...] if we have 50 pairs. [...] assume 20 pairs get 55%+

LOL.

BTW, based on the standard deviation I have observed, you need to average about 53.5% in Robot Rebate in order to pay less per board than playing Robot Duplicate. If you want to actually break even, you need to average 58.5%.
"One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision"
    -- Bertrand Russell
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#4 User is offline   ArtK78 

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Posted 2012-May-10, 13:37

View Postmgoetze, on 2012-May-10, 13:09, said:

LOL.

BTW, based on the standard deviation I have observed, you need to average about 53.5% in Robot Rebate in order to pay less per board than playing Robot Duplicate. If you want to actually break even, you need to average 58.5%.

That cannot be right. If I understand correctly, you get paid 150% of your entry fee if you score 55% or better. Therefore, you need to cash 2 times out of 3 to break even. Hypothetically, a player who scores 55.1% twice in every three games and scores 10% in the third of every three games breaks even with an average score of about 40%. More realistically, a player who scores 51%, 56% and 61% every three games breaks even with an average score of 56%.

So, you will have to explain to me how you came up with the 58.5% figure.
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#5 User is offline   mgoetze 

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Posted 2012-May-10, 14:48

View PostArtK78, on 2012-May-10, 13:37, said:

That cannot be right. If I understand correctly, you get paid 150% of your entry fee if you score 55% or better. Therefore, you need to cash 2 times out of 3 to break even. Hypothetically, a player who scores 55.1% twice in every three games and scores 10% in the third of every three games breaks even with an average score of about 40%. More realistically, a player who scores 51%, 56% and 61% every three games breaks even with an average score of 56%.

So, you will have to explain to me how you came up with the 58.5% figure.

Easy, I changed the average score in my Gnumeric (but you can use Excel) sheet until the expected value was 0. If your mean score is in A1 and your standard deviation is in B1, then your expected value is

=normdist(0.55,A1,B1,1)*-1+(1-normdist(0.55,A1,B1,1))*0.5

My empirical standard deviation for robot rebate is 8.16%. If you are more consistent than me then you can probably make do with a slightly lower average, 'tis true.

This, of course, is predicated on the assumption that your results are normally distributed. That it is of course not quite the case, but it is a lot closer than you seem to think (my estimated theoretical EV and my empirical EV are within 1 cent of each other).
"One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision"
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#6 User is offline   nathan2008 

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Posted 2012-May-10, 15:27

View Postmgoetze, on 2012-May-10, 14:48, said:


=normdist(0.55,A1,B1,1)*-1+(1-normdist(0.55,A1,B1,1))*0.5



Are u informatician ? :) You are so cleverrrrr
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