What constitutes 2/3 of the events (for annual points)

[froggy47]

If we have 12 events that's easy 2/3 * 12 = 8. You must run 8 events to be in annual points.

If we have 13 events it's 2/3 * 13 = 8.67. Do you still have to run 8 events or is it rounded up to 9?

If we have 14 events it's 2/3 * 14 = 9.33. Is that rounded down to 9 events required or round up to 10?

Just trying to plan the end of season schedule. Thanks if anyone knows. Not looking for "I think it should be this way or that way". I was wondering if some "old timer" knows for a FACT how it's calculated.

:)

[Steve35]

In the past we’ve rounded up. 9.33 = 10, 8.01 = 9, etc. I’m not sure what the current interpretation is.

[froggy47]

Thanks Steve.

:)

[MX5bob]

In the past we’ve rounded up. 9.33 = 10, 8.01 = 9, etc. I’m not sure what the current interpretation is.

Nothing like making up new rules for rounding up or down. :rolleyes:

[Bimota Guy]

Nothing like making up new rules for rounding up or down. :rolleyes:
Bob, This is not a math rule, it is a rule based on practicality. As the rule is presently written (it is silent on the issue of rounding) we must always round up in order to complete the minimum number of events. However, if you can figure out a reasonable way to, for example, compete in 8.01 events w/o competing in 9 events we will listen. :p

[MX5bob]

Bob, This is not a math rule, it is a rule based on practicality. As the rule is presently written (it is silent on the issue of rounding) we must always round up in order to complete the minimum number of events. However, if you can figure out a reasonable way to, for example, compete in 8.01 events w/o competing in 9 events we will listen. :p
I'd say that competing in 8 events as opposed to 8.01 events should be sufficient. 8.67 rounds to 9. In a stretch, you can round 9.33 to 10. Besides, two-thirds or 66.66% of any of the numbers mentioned does not equal 8.01. :D

OR, you can write a rule that specifically says:

7 out of 10
8 out of 11
8 out of 12
9 out of 13
10 out of 14

See, isn't that simple? And no math rules were harmed in the process.

[73STS]

I'd say that competing in 8 events as opposed to 8.01 events should be sufficient. 8.67 rounds to 9. In a stretch, you can round 9.33 to 10. Besides, two-thirds or 66.66% of any of the numbers mentioned does not equal 8.01. :D

OR, you can write a rule that specifically says:

7 out of 10
8 out of 11
8 out of 12
9 out of 13
10 out of 14

See, isn't that simple? And no math rules were harmed in the process.

Bob

You really need to find a job quick!:D

[MX5bob]

Bob

You really need to find a job quick!:D

You're not implying that I'm losing my, uh, uh, . Well anyway, I'm pretty sure I'll be employed soon.

[Bimota Guy]

Well, the final count is 13 championships for 2006. Therefore with 2/3 of 13 being approx. 8.7, you will need to participate in 9 events to get a trophy. Moreover Bob can sleep well. :D :p

-Steve

[MX5bob]

Well, the final count is 13 championships for 2006. Therefore with 2/3 of 13 being approx. 8.7, you will need to participate in 9 events to get a trophy. Moreover Bob can sleep well. :D :p

-Steve

Thank god, I've been worried sick. :p:D