Hi, so last time I learned about binomial probabilities and I've been trying to actively use them. The numbers were matching things in that very simple scenario, but now I'm learning the actual case has a few more stipulations. Namely, my population can be more than the number of events. Also, in the case an event fails, then another could succeed until either the whole population is exhausted or 6 events succeed.
Here's the situation:
I have a population of 7, but I can only choose 6. Each event has a 65% chance of success. So I did a normal calculation for every draw and got the following:
0- 0.064%
1- 0.836%
2- 4.66%
3- 14.42%
4- 26.78%
5- 29.84%
6- 18.47%
7- 4.9%
I can never get the last possibility since I can only choose 6, so how do I distribute that into a skew? Then, what's more aggravating is that the live data is closer to a 6 population spread (32 trials):
Live:
0- 0%
1- 3.125%
2- 6.25%
3- 15.625%
4- 40.625%
5- 28.125%
6- 6.25%
7- 0%
6-Pop spread:
0- 0.183%
1- 1.33%
2- 9.51%
3- 23.54%
4- 32.79%
5- 24.36%
6- 7.54%
The best I've done to try and reconcile it falls apart pretty quickly.
(7C6) 0.65^6 * 0.35^1 * (1C0) * 0.65^0 * 0.35^1 = 0.0646.
pretty close to the live data. But then for (7C5), it's so far off I'm inclined to believe I messed up.
Here's the situation:
I have a population of 7, but I can only choose 6. Each event has a 65% chance of success. So I did a normal calculation for every draw and got the following:
0- 0.064%
1- 0.836%
2- 4.66%
3- 14.42%
4- 26.78%
5- 29.84%
6- 18.47%
7- 4.9%
I can never get the last possibility since I can only choose 6, so how do I distribute that into a skew? Then, what's more aggravating is that the live data is closer to a 6 population spread (32 trials):
Live:
0- 0%
1- 3.125%
2- 6.25%
3- 15.625%
4- 40.625%
5- 28.125%
6- 6.25%
7- 0%
6-Pop spread:
0- 0.183%
1- 1.33%
2- 9.51%
3- 23.54%
4- 32.79%
5- 24.36%
6- 7.54%
The best I've done to try and reconcile it falls apart pretty quickly.
(7C6) 0.65^6 * 0.35^1 * (1C0) * 0.65^0 * 0.35^1 = 0.0646.
pretty close to the live data. But then for (7C5), it's so far off I'm inclined to believe I messed up.
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