Old 06-3-2005, 03:22 PM   #1
ToshX
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Default Help me with math

I am great at it, but neither me nor any of my friends could answer this question in which I have to ask:

Object A has a 12% chance of happening and Object B has a 13% chance of happening. These are two entirely different percentages, they do not correspond with each other. For example, you could not put these two together on a Pie Graph. Now then.

What is the chance that EITHER A will work OR B will work OR both A AND B will work, all in one percentage? Also, what formula(e) did you use to get this answer?
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Old 06-3-2005, 04:23 PM   #2
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Default RE: Help me with math

This is actually surprisingly simple. First of all, A has a 12% chance of happening, so 2 of the 3 possibilities are met. Next, we can find the odds of both happening at once. This is simply finding 13% of the 12% percent that A will happen. This gives us 1.56%. Then you just add the 13% and 12%, minus the overlap of 1.56%, to get 23.44%.
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Old 06-4-2005, 07:08 AM   #3
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Default RE: Help me with math

Nice! Thank you! Now theres a harder one that is no longer an example.

11% Chance To Cast Level 18 Confuse On Striking
9% Chance To Cast Level 11 Frozen Orb On Striking
11% Chance To Cast Level 9 Charged Bolt On Striking
9% Chance To Cast Level 11 Frozen Orb On Striking
11% Chance To Cast Level 9 Charged Bolt On Striking
12% Chance To Cast Level 15 Hydra On Striking

They are all different percentages, have fun :P I would do it myself but I don't know how to do it with that many.
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Old 06-4-2005, 08:30 AM   #4
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OMG LOL...i just burst out laughing.

DIABLO 2!!!

im too stupid to actually answer it, sorry.
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Old 06-4-2005, 08:49 AM   #5
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Tosh... you realize that with 6 variables, we could calculate thousands of different answers between the 2, 3, 4, 5, and 6 variable answers....

but, the simple way to figure this out... 11/100 * 11/100 * 11/100 * 12/100 * 9/100 * 9/100 = chance of all 6 working in 1 hit.
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Old 06-4-2005, 11:51 AM   #6
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Are you asking what are the odds of anything happening? Because if you are, it's still easy to figure out. I noticed you put a couple of the effects down twice, so let's just ignore those ones for now. So our 4 variables are:

12% Chance To Cast Level 15 Hydra On Striking, which I will call A
11% Chance To Cast Level 18 Confuse On Striking, or B
11% Chance To Cast Level 9 Charged Bolt On Striking, C
9% Chance To Cast Level 11 Frozen Orb On Striking, D

What we basically have to do is eliminate each percentage one at a time. We know that there is a 12% chance of A, so now we need to find the percentage of chances that result in B but not A. This is simply 11%-12%, which is 9.68%. This tells us that there is a 21.68% chance of A and/or B happening. We then repeat this for C, which gives us 8.6152% chance. This means that there is a 30.2952% chance of A and/or B and/or C happening. Finally, you need to find D, which is 9%-30.2952%, or 6.273432%. Adding it all up, the final total is 36.568632% chance of anything happening.
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Old 06-5-2005, 09:24 AM   #7
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That's because some of the effects can happen twice in different percentages :P Heres the fun part. Everything can be combined. Thus, we can get:
A+B, A+C, and all the rest.
A+B+C, A+B+D, and all the rest.
A+B+C+D, A+B+C+E, and all the rest.
A+B+C+D+E and all the rest.
AND all of them can work at once.
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This is why I need your help, to get this all into one big percentage. I know it seems ridiculously hard, but you people are smart.
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Old 06-5-2005, 05:25 PM   #8
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All you've done is make this question more annoying, not harder. The interesting thing about this problem is that there are so many ways to approach it. The first we can do is find the odds of E and/or F (which are 11% and 9%). This comes out to 19.01%. Using a slightly different approach from last time, we can find the percentage to add on based on the remander of the other percentage. In other words, there is a 63.431368% chance that nothing (besides E and F) will happen. 19.01% of 63.431368% is 12.05830306% chance of just E and/or F happening. Adding that to the previous percent, the total is 48.62693506% of anything happening. (this might be off as my calculator is only 10 digits.)

One last thing: If you add even one more percentage I will hunt down and kill you for making me do the rest by hand

(Off Topic: does anyone know how to take care of that virus that unselects your browser and installs party poker on your desktop)
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Old 06-6-2005, 03:33 AM   #9
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Quote:
Originally Posted by Matthew4444
All you've done is make this question more annoying, not harder. The interesting thing about this problem is that there are so many ways to approach it. The first we can do is find the odds of E and/or F (which are 11% and 9%). This comes out to 19.01%. Using a slightly different approach from last time, we can find the percentage to add on based on the remander of the other percentage. In other words, there is a 63.431368% chance that nothing (besides E and F) will happen. 19.01% of 63.431368% is 12.05830306% chance of just E and/or F happening. Adding that to the previous percent, the total is 48.62693506% of anything happening. (this might be off as my calculator is only 10 digits.)

One last thing: If you add even one more percentage I will hunt down and kill you for making me do the rest by hand

(Off Topic: does anyone know how to take care of that virus that unselects your browser and installs party poker on your desktop)
Thanks, this was exactly what I needed.
This is what you need when you can files, which you should do when you download a strange-lookin file. It's a shiatload of virus scanners all thrown in one thing, it only does files smaller than 10MB.
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Old 06-8-2005, 09:34 PM   #10
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There might be another, much simpler approach to this problem. Correct me if I'm wrong.

You want to find the chances of ANYTHING happening, right?

Well, then just find the chances of NOTHING happening, and subtract that from 1.

In your example, the probabilities are:

A. 11% Chance To Cast Level 18 Confuse On Striking
B. 9% Chance To Cast Level 11 Frozen Orb On Striking
C. 11% Chance To Cast Level 9 Charged Bolt On Striking
D. 9% Chance To Cast Level 11 Frozen Orb On Striking
E. 11% Chance To Cast Level 9 Charged Bolt On Striking
F. 12% Chance To Cast Level 15 Hydra On Striking

The probability of A NOT happening is 89%. The probability of B NOT happening is 91%. Then 89%, 91%, 89%, 88%.

So now you have all these probabilities of each individual event NOT happening.

To find the probability that they ALL do NOT happen, simply multiply the percentages together.

This number is the probability that all the events do NOT happen (nothing happens).

To find the probability that at least one thing will happen, simply subtract from 1 (100%).

The formula is: 1 - (.89 * .91 * .89 * .91 * .89 * .88) To add effects, simply multiply in the parentheses by the probability that this new event will NOT happen.

This answer agrees with Matthew's.
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Old 07-5-2005, 01:18 PM   #11
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42
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Old 07-5-2005, 03:05 PM   #12
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Quote:
Originally Posted by psychic25
There might be another, much simpler approach to this problem. Correct me if I'm wrong.

You want to find the chances of ANYTHING happening, right?

Well, then just find the chances of NOTHING happening, and subtract that from 1.

In your example, the probabilities are:

A. 11% Chance To Cast Level 18 Confuse On Striking
B. 9% Chance To Cast Level 11 Frozen Orb On Striking
C. 11% Chance To Cast Level 9 Charged Bolt On Striking
D. 9% Chance To Cast Level 11 Frozen Orb On Striking
E. 11% Chance To Cast Level 9 Charged Bolt On Striking
F. 12% Chance To Cast Level 15 Hydra On Striking

The probability of A NOT happening is 89%. The probability of B NOT happening is 91%. Then 89%, 91%, 89%, 88%.

So now you have all these probabilities of each individual event NOT happening.

To find the probability that they ALL do NOT happen, simply multiply the percentages together.

This number is the probability that all the events do NOT happen (nothing happens).

To find the probability that at least one thing will happen, simply subtract from 1 (100%).

The formula is: 1 - (.89 * .91 * .89 * .91 * .89 * .88) To add effects, simply multiply in the parentheses by the probability that this new event will NOT happen.

This answer agrees with Matthew's.
This is the same way i approached this problem but then realised that It had been done and there way no reason for me to repost it..







and as for the 42 quote.. thats an important number if your into Enders Game..
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Old 07-7-2005, 11:58 AM   #13
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Quote:
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and as for the 42 quote.. thats an important number if your into Enders Game..
or the hitchhikers guide
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Old 07-7-2005, 12:26 PM   #14
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Quote:
Originally Posted by Sera13
and as for the 42 quote.. thats an important number if your into Enders Game..
or the hitchhikers guide
lol thats the one i was thinking about.. scratch the enders game thing, your right it is hitchhikers lol.. BLONDE MOMENT
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