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FrozenGate by Avery

Raffle! Win Big Prizes! (And help fund LPF!)

Lasher said:
Guys lets not turn this great event into something heated.

This was done to help the forum. Everyone knew going into this that your money was a donation to keep this little corner of the interwebs up and running for the good of all our members.

It was not suppose to be about buying so many tickets you were sure to win big.

Don't ruin what was a great idea and a fun event by bickering.

Glen,

Thank you again for such a fun raffle I for one was glad to be a part of it and could have cared less if I won or not as long as this forum that has taught me so much is kept up and running.

Lasher
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^^^^I agree 100%....^^^^

The money put into the Raffle was to Support the Forum expenses...
and if you got something back (prize/s) for your Forum support you got
lucky... if not you still supported the Forum and you should feel great
(for what the Forum gives back to you)...

If on the other hand you bought Raffle tickets just to win prizes then
I feel sorry for whoever didn't win what they wanted...

What was that saying..... "The luck of the draw"...

Jerry
 
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twhite828 said:
I'm ordering a copy when my Spyro Gyra album comes in. That seems really interesting. :angel:
...but what I posted here I didn't make up. :beer:
-Trevor
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I didn't think you did. I don't actually care much about the math. EVERY "C" on my transcripts is from a math class. Despite being a "C" math student, I still understand probabilities.

Some gotta win. Some gotta lose.
Good time Charlie's got the blues

Peace,
dave
 
twhite828 said:
Found the theorem that backs up my method of determining combined probability...

Multiplication Rule of Probability

-Trevor
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That's one of the De Morgan laws. The example provided in that website refers to DICE throwing. This is a TICKET drawing where tickets are NOT put back into the bowl.
Venn's diagram can explain what is explained on the link much better, A^B is not the same as A U B.
 
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nikokapo said:
That's one of the De Morgan laws. The example provided in that website refers to DICE throwing. This is a TICKET drawing where tickets are NOT put back into the bowl.
Venn's diagram can explain what is explained on the link much better, A^B is not the same as A U B.
Click to expand...

The dice example proves that you can multiply the two probabilities of discrete occurrences happening consecutively to yield the probability of both occurring.

You multiply the probability of him losing the first draw (354/394) by the probability of losing the second draw (353/393) to yield the probability of him losing the first two draws. You can then multiply in the probability of losing the third draw (352/392) to yield the probability of losing the first three draws. It applies with changing probability too...

-Trevor
 
twhite828 said:
The dice example proves that you can multiply the two probabilities of discrete occurrences happening consecutively to yield the probability of both occurring.

You multiply the probability of him losing the first draw (354/394) by the probability of losing the second draw (353/393) to yield the probability of him losing the first two draws. You can then multiply in the probability of losing the third draw (352/392) to yield the probability of losing the first three draws. It applies with changing probability too...

-Trevor
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You said it, they are NOT independent. After the first ticket is drawn, the remaining tickets have their probabilities of being drawn changed.
It doesn't apply to changing probability "just like that", you need to apply Bayes' theorem for a conditional probability calculation.
 
nikokapo said:
You said it, they are NOT independent. After the first ticket is drawn, the remaining tickets have their probabilities of being drawn changed.
It doesn't apply to changing probability "just like that", you need to apply Bayes' theorem for a conditional probability calculation.
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He had an 89.8477% chance to lose the first - it's concrete. He had an 89.8219% of losing the second. Those are independent. If he loses the first drawing 89.8477% of the time, and the second drawing 89.8219% of the time, then those two line up 80.7029% of the time. That is 89.8219% of the 89.8477% of the first drawings lost.

-Trevor
 
twhite828 said:
He had an 89.8477% chance to lose the first - it's concrete. He had an 89.8219% of losing the second. Those are independent. If he loses the first drawing 89.8477% of the time, and the second drawing 89.8219% of the time, then those two line up 80.7029% of the time. That is 89.8219% of the 89.8477% of the first drawings lost.

-Trevor
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Again, they're not independent, they're conditional.

Look up "Venn's diagram". Or try doing a probability tree.
 
nikokapo said:
Again, they're not independent, they're conditional.

Look up "Venn's diagram". Or try doing a probability tree.
Click to expand...

I have one thousand angry ticks. I choose ten percent at random. Of the 100 ticks that I now foolishly have in my hand, I choose to keep only ten percent, leaving me still with ten angry ticks in my hand.

Each tick at the beginning therefore had a 1% chance of being chosen. 1% are left from the original thousand I had in my hand.

These drawings aren't conditional. We have this set of sixty different probabilities. We want to know the chance of all of them occurring. You have cereal for breakfast 89.8477% of mornings, and I have a 89.8219% of contracting Lyme disease from all those angry ticks - thus, there's an 80.7029% chance that I'll have Lyme disease while you're eating cereal this Monday.

EDIT: I attached the Excel sheet... download and change the extension to .xls.

-Trevor
 

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^^^^Yeah... I agree with T_J on that one...^^^^
This is the LPF Raffle Thread... not the Theory of Probability Thread...:whistle:

Jerry
 
Tech_Junkie said:
Cant you guys make a new thread for this? Or PM maybe? :horse:
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beating-a-dead-horse.gif
 
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