Weighted Probability comparison?

jrlewis96

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Jun 5, 2018
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I'm trying to find the math to understand the relationships between possibilities.

if there are 36 iterations and every iteration has 2 attempts to produce a value of 9 (for a max value of 18 in any one iteration) called W


and


within the same 36 iterations, an iteration has 1 attempt to produce a value of 8 called X


and


within the same 36 iterations, an iteration has 1 attempt to produce a value of 7 called Y


and


within the same 36 iterations, 6 of those iterations have 1 attempt to produce a value of 1 called Z


How would i figure out the ratio values of z:y z:x x:y z:w w:x w:y

I think I'm trying to figure out the probability of Z having a value of 1 is in what ratio to say W having a value of 18.
 
I'm trying to find the math to understand the relationships between possibilities.

if there are 36 iterations and every iteration has 2 attempts to produce a value of 9 (for a max value of 18 in any one iteration) called W


and


within the same 36 iterations, an iteration has 1 attempt to produce a value of 8 called X


and


within the same 36 iterations, an iteration has 1 attempt to produce a value of 7 called Y


and


within the same 36 iterations, 6 of those iterations have 1 attempt to produce a value of 1 called Z


How would i figure out the ratio values of z:y z:x x:y z:w w:x w:y

I think I'm trying to figure out the probability of Z having a value of 1 is in what ratio to say W having a value of 18.
You WILL have to specify the distribution. What is the Range of possible values? How are the values determined? Die rolls? Rabbit out of a hat? Ping pong balls from a cage? One important consideration is the Independence of the trials. For example, if you roll a 10-sided die, to see if you get 9, 8, 7, or 1, do you roll one die for each of these examinations or do you roll it once and check for all four outcomes? This is very important. If you roll only one die and check for all four, then you cannot get a 9 and an 8, for example. If you roll one for each examination, you can achieve 9, 8, 7, and 1 on the same iteration. So, what's an "iteration"?
 
You WILL have to specify the distribution. What is the Range of possible values? How are the values determined? Die rolls? Rabbit out of a hat? Ping pong balls from a cage? One important consideration is the Independence of the trials. For example, if you roll a 10-sided die, to see if you get 9, 8, 7, or 1, do you roll one die for each of these examinations or do you roll it once and check for all four outcomes? This is very important. If you roll only one die and check for all four, then you cannot get a 9 and an 8, for example. If you roll one for each examination, you can achieve 9, 8, 7, and 1 on the same iteration. So, what's an "iteration"?

Ok so every time there's an iteration all 4 possibilities are checked. It's a binary outcome either it happens or it doesn't. If it does happen then the assigned value happens, if it doesn't the value remains at 0. You flip a coin for every possible chance of a value happening and it either does or does not.
 
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