So I am trying to calculate the chances of something to happen, but some percentage are just gone...

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pineapplewithmouse

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This is in a game so I will simplify it:
There is a temple. In the temple, there are 6 monsters. Each monster has a 10% chance to drop a rose (It can drop or 0 or 1).
So, the chance that no one will drop a rose is 53.1% (because 0.9^6)
The chance that one monster will drop the rose is 5.9% (because 0.1*0.9^5)
The chance that 2 monsters will drop the rose is 0.6% (because 0.1^2*0.9^4)
The chance that 3 monsters will drop the rose is 0.0729% (because 0.1^3*0.9^3)
The chance that 4 monsters will drop the rose is 0.0081% (because 0.1^4*0.9^2)
The chance that 5 monsters will drop the rose is 0.0009% (because 0.1^5*0.9)
The chance that 6 monsters will drop the rose is 0.0001% (because 0.1^6)
So theoretically, the sum of all these odds should be 100% because there is a 100% that some of these events will happen.
But their sum is only 59.7%.
40.3% just gone for some reason.
Why? What I did wrong in my calculations?
 
pineapplewithmouse said:
The chance that one monster will drop the rose is 5.9% (because 0.1*0.9^5)
Click to expand...
That's the probability that Monster #1 drops a rose. But there are 6 monsters, so the probability that Monster #1, or Monster #2, etc. drops their rose is 6 times as much, right?

pineapplewithmouse said:
The chance that 2 monsters will drop the rose is 0.6% (because 0.1^2*0.9^4)
Click to expand...
But there are 15 pairs of monsters that might both drop their roses. You aren't interested only in one specific pair.

And so on.

This is a binomial probability distribution.

pineapplewithmouse said:
But their sum is only 59.7%.
40.3% just gone for some reason.
Why? What I did wrong in my calculations?
Click to expand...
This is a great observation, and it shows that you are really thinking! Good for you. Now see if you can fix it.
 
I did not understand, can u please explain more?
"That's the probability that Monster #1 drops a rose" How?
"But there are 15 pairs of monsters" How? Where?
 
pineapplewithmouse said:
This is in a game so I will simplify it:
There is a temple. In the temple, there are 6 monsters. Each monster has a 10% chance to drop a rose (It can drop or 0 or 1).
So, the chance that no one will drop a rose is 53.1% (because 0.9^6)
The chance that one monster will drop the rose is 5.9% (because 0.1*0.9^5)
Click to expand...
Can you tell us how much you have learned about probability? That will help us know what explanations will work for you.

But let's call the monsters A, B, C, D, E, F, and define 6 events, A, B, C, D, E, F, meaning that monster A drops its rose, etc..

The probability of A is 0.1. The probability that only A occurs is, as you indicate, (0.1)*(0.9)^5.

But the probability that only B occurs is the same, as are the probabilities for C, D, E, and F.

The event you are asking about here is that exactly one monster drops its rose; that is, exactly one of A, B, C, D, E, and F occurs. and

P(only A or only B or only C or only D or only E or only F) =​
P(only A) + P(only B) + P(only C) + P(only D) + P(only E) + P(only F) = 6*P(only A) =​
6*(0.1)*(0.9)^5.​

The point is, there is a big difference between "exactly one occurs" and "only this particular one occurs". You calculated the latter.

As for pairs of monsters, do you know anything about combinations, or Pascal's Triangle?
 
I am so sorry for being stupid but...
1. What are the events? Event A is when monster A will drop it's rose, event B is when monster B will drop it's rose and etc?
2. So, if only A will happen this is not the same like only B will happen? They have different meaning?
3. How do you know that the probability that one of these events to happen is equal to the sum of all the events to happen separately?
4. So if your calculation is right, the chance that one monster will drop it's rose is 35.42%. I still do not get it... how it jumped from 10% to 35.42%?
I still do not understand about the pairs of monsters... I did not learn about combinations but I did learn about Pascal's Triangle.
 
pineapplewithmouse said:
1. What are the events? Event A is when monster A will drop it's rose, event B is when monster B will drop it's rose and etc?
Click to expand...
Yes, that's what I said.

pineapplewithmouse said:
2. So, if only A will happen this is not the same like only B will happen? They have different meaning?
Click to expand...
Of course. They have the same probabilities, but if A dropped its rose, that doesn't imply that B also did.

pineapplewithmouse said:
3. How do you know that the probability that one of these events to happen is equal to the sum of all the events to happen separately?
Click to expand...
Have you learned about the probability of A or B? I asked you how much you have learned about probability, but you haven't answered.

If not, then do you see that the number of ways in which the event "exactly one of A, B, C, D, E, F occurs" is 6? Use that as your numerator.

pineapplewithmouse said:
4. So if your calculation is right, the chance that one monster will drop it's rose is 35.42%. I still do not get it... how it jumped from 10% to 35.42%?
Click to expand...
You are told the probability that any given monster drops its rose. That is not the same as any monster.

pineapplewithmouse said:
I still do not understand about the pairs of monsters... I did not learn about combinations but I did learn about Pascal's Triangle.
Click to expand...
Please tell us what you have learned. Have you learned anything about problems like "how many ways are there two choose two letters from ABCDEF"?
 
Idk how can I answer on how much I have learnt.
"how many ways are there two choose two letters from ABCDEF"
1. I will assume that you meant "too choose"
2. If I did it right, there is 15 ways to do that.
3. How this problem related to probability?

"Have you learned about the probability of A or B?"
Kinda?
I learnt how to calculate conditional probability...
You can ask if I learnt specific things, so you will get more clear picture...

"You are told the probability that any given monster drops its rose. That is not the same as any monster"
I am so confused...
My question is simple: "What is the chance to get a rose from only one monster?"
How do I calculate it?...
 
You are correct that the probability that no monster drops a rose is 53.1441%. Good thinking.

You are also correct that the probability that one specific monster (say monster A) will drop a rose and the other five will not is is 5.9094% or approximately 5.9%. But there are six monsters so you have neglected five other possibilities, namely that monster B is the one that drops the rose, monster C, and so on. So the probability that that exactly one of the six monsters is 6 times 5.9049% = 35.4294%.

You are also correct that the probability that monsters A and B drop roses and C, D, E, and F do not is indeed 0.6561%. But you have neglected many other possible pairs. The possible pairs are AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, and EF. 15 of them. The probability that any pair both drop a rose is 15 * 0.6561% = 9.8415%.

And the probability that monsters A, B, and C drop a rose, but the other three do not is indeed 0.0729%. But again you are neglecting that there are twenty possible triplets, namely ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF, BCD, BCE, BCF, BDE, BDF, BEF, CDE, CDF, CEF, and DEF

So the probability some triplet drops a rose and another triplet does not is 20 * 0.0729% = 1.4580%.

The probability that ABCD drop a rose and E and F do not is indeed 0.0081%. Once again however, you are ignoring the total number of such situations: ABCD, ABCE, ABCF, ABDE, ABDF, ABEF, ACDE, ACDF, ACEF, ADEF, BCDE, BCDF, BCEF, BDEF, and CDEF,

The probability that some four drop a rose is 15 * 0.0081% = 0.1215%.

The probability that A does not drop a rose, but the other five do is 0.009% as you say. But then you have not accounted for only B not dropping a rose or for only C not dropping a rose, or for only D not dropping a rose, or for only E not dropping a rose, or for only F not dropping a rose.

The probability that exactly five drop a rose is 6 * 0.009% = 0.0054%.

And you are right that the probability that all six drop a rose is 0.0001%.

Now add them up

53. 1441 + 35.4294 + 9.8415 + 1.4580 + 0.1215 + 0.0054 + 0.0001 = 100.0000.

By the way, you may not get exactly 100 if you round.
 
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I am so sorry... but I guess I have another mistake...
I wanted to calculate this thing to calculate another thing.
If the chances to get one rose is 35.42% and the chance to not get is 53.14%, so the chance to get rose every 3 runs is 10% (because 0.3542^2*0.5314). But... If I want that will happen 7 times (because I need 7)... the chances of this to happen is 1 in 10 million (because 0.1^7).
I have a strong feeling I did something wrong again. This chance is just too small...
 
I am not quite sure what you are asking.

First, the probability of getting AT LEAST one rose on a visit to the temple is 46.8559%. You have to add together the probabilities for exactly one, exactly two, exactly three, exactly five, and exactly six. So if you are wondering what is the probability that you get AT LEAST one rose in three trips, the answer is approximately 85%.
 
I did not think I am so lucky to get 2 roses in one visit, but I guess it will be right to calculate it like you said.

I do not understand how you got these 85%...
If I calculate the chance to get at least one rose in every visit in 3 visits, I need to do 0.4685^3 which is 0.10283 (approximately 10%)
I originally meant "What are the chances that in every 3 visits, in 2 of them I will not get roses, and in the other 1 I will get a least one rose?"
So I did 0.4685*0.5314^2 which is 0.1322 (approximately 13%)
And what are the chances that 3 visits like that will happen 7 times?
So I did 0.1322^7 which is 7 in 10 million... And these odds seems to small for me, so I asked if I did any mistakes here again?
 
pineapplewithmouse said:
I did not think I am so lucky to get 2 roses in one visit, but I guess it will be right to calculate it like you said.

I do not understand how you got these 85%...
If I calculate the chance to get at least one rose in every visit in 3 visits, I need to do 0.4685^3 which is 0.10283 (approximately 10%)
I originally meant "What are the chances that in every 3 visits, in 2 of them I will not get roses, and in the other 1 I will get a least one rose?"
So I did 0.4685*0.5314^2 which is 0.1322 (approximately 13%)
And what are the chances that 3 visits like that will happen 7 times?
So I did 0.1322^7 which is 7 in 10 million... And these odds seems to small for me, so I asked if I did any mistakes here again?
Click to expand...
No. You have to add up the probability that you get at least one rose on the first visit, plus the probability that you get no rose on the first visit but at least one on the second visit plus the probability that you get no rose on the first visit and no rose on the second visit but at least one on the third visit. That is approximately

[math]0.47 + 0.53 * 0.47 + 0.53 * 0.53 * 0.47 = 0.851123[/math]
The probability that you get at least one rose in three visits is quite decent.

If instead you are asking what is the probability that you get exactly one rose on each of three successive visits, then that is approximately (0.354^3) or a bit above 4%. In probability theory you need to be exact in what you ask.
 
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Wait.
I will try to make it more clear:
"What are the chances, that after 3 visits in the temple, I will get at least 1 rose?"

Ok, I think I understand why you did this drill...
So just to make sure I really understood:
So if I want to calculate the same, but with 5 visits, I need to do:
0.47+0.53*0.47+0.53*0.53*0.47+0.53*0.53*0.53*0.47+0.53*0.53*0.53*0.53*0.47=0.9581

Yessss, now this is make sense!
In my previous calculation, the more visits I did, the less chance was the find a rose, which is not making any sense...
But now the more visits I do, the more chance I have to find it!
Thank you so much, I really appreciate it
 
I did not check your arithmetic, but you now understand the logic.

I think that logic is the easiest to grasp, but the computation is burdensome. There is an alternative computation that is quicker to do.[

[math]1 - 0.53^5 \approx 95.82\%.[/math]
The logic of that computation is this.

If you visit the temple five times, you must get either no roses or else at least one. Those two “events” are “mutually exclusive and exhaustive.” Consequently their probabities sum to 1.

Prob(none) + Prob(at least one) = 1

Prob(at least one) = 1 - Prob(none)

And the probability of none in five visits is approximately

[math]0.53^5.[/math]
 
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