Headaches With Probability :) ... What Chance Will Monster(s) Drop Treasure?

Lance_Botelle · November 23, 2020, 12:51pm

Hi All,

I decided to post this as a little puzzle, in case others are interested. I am not great with understanding this sort of thing, although I think I have the answer now … so offer this for answers from others who know the solution to this sort of thing.

Anyway, here was my query …

You have five monsters, which the PCs will kill one after the other. The first four monsters have 10% chance of dropping a gold coin and the last one has 60% chance.

My questions for your answer (and please show your working), is what is the probability of the party collecting:-

A) One gold coin.
B) Two gold coins.
C) Three gold coins.
D) Four gold coins.
E) Five gold coins.

As I say, I believe I have my solution now and will post it a little later after others have had the opportunity to post their own answers.

Thanks all, Lance.

4760 · November 23, 2020, 5:56pm

Let’s put it this way: the first four monsters are similar probability-wise, so we’ll use combinations (p elements amongst n).

For one gold coin, the possibilities are:

only the fifth monster and none of the first four
or any of the first four and not the fifth.
This can be calculated as follows:
60% x 90% x 90% x 90% x 90% = 39,366%
Cnp(1;4) x 40% x 10% x 90% x 90% x 90% = 4 x 2,916% = 11,664%

=> P(1 gp) = 0,5103 (51,03%)

For two gold coins, the same principle applies:

either the fifth monster and only one of the other four, i.e. 60% x 10% x 90% x 90% x 90% x Cnp(1;4) = 4,374% x 4 = 17,496%
or two of the first four and not the fifth i.e. 40% x 10% x 10% x 90% x 90% x Cnp(2;4) = 0,324% x 6 = 1,944%

=> P(2 gp) = 0,1944 (19,44%)

etc…

P(3 gp) = 0,0306 (3,06%)
P(4 gp) = 0,0022 (0,22%)
P(5 gp) = 0,00006 (0,006%)

And the probability to get 0 gp is when neither monster drops a gold coin, i.e. 40% x 90% x 90% x 90% x 90% = 26,244%

As expected, P(0 gp) + P(1 gp) + P(2 gp) + P(3 gp) + P(4 gp) + P(5 gp) = 1

Lance_Botelle · November 23, 2020, 6:21pm

Oh my!

NB: My answer below is assumed to be wrong.

That appeared far more complex than I was expecting … I said I was bad at this. So, OK, here is what I thought the answer might be, and how I worked it out. Maybe you can explain it to me some more.

To begin with and for clarification, the PCs meet each of these monsters, kill them, examine the body, collect any coin that may have fallen, and then move to the next. Kill, examine, and collect again (if dropped) until they have killed all five monsters.

So, I thought it worked out like this …

I thought we used an average likelihood based upon the individual drops chances, so … (*)

(*) I know this is where I go wrong. (Mixing “apples with oranges”?)

10 + 10 + 10 + 10 + 60 = 100 (divided by 5 for five monsters) gives an average of 20% chance of drop.

Therefore, I thought the probability of gaining one coin was 20%. (or 0.2)
Then, the probability to gain a 2nd coin (2 coins total) was 0.2 x 0.2 = 0.04 (or 4%).
The probability to gain a 3rd coin (3 coins total) was 0.2 x 0.2 x 0.2 = 0.008 (or 0.8%).
The probability to gain a 4th coin (4 coins total) was 0.2 x 0.2 x 0.2 x 0.2 = 0.0016 (or 0.16%).
Finally, to gain a 5th coin (5 coins total) was 0.2 x 0.2 x 0.2 x 0.2 x 0.2 = 0.00032 (or 0.032%).

Just looking at the figures, I can see that there should be more chance to drop 2 coins than I have it (I would have thought anyway), which your calculations demonstrate.

I think I am still struggling to grasp this.

Why is this not 60% x 10% x 10% x 10% x 10%?

EDIT: Oh … I see it is the probability of them NOT also dropping alongside the 60% chance of a drop. So, we end up with the probability of a single drop alone.

4760 · November 23, 2020, 7:05pm

Because you want to get one gold coin, and one gold coin only. In the above calculation, only the fifth monster drops a coin (60% chance) and the four others don’t (90% each). The probabilities are independent of the result of the previous encounters, so the probability to get only the coin from the fifth monster is 90% (I don’t get a coin from the first) x 90% (I don’t get a coin from the second) x 90% (no coin from the third) x 90% (no coin from the fourth) x 60% (I do get a coin from the fifth creature)

With 60% x 10% x 10% x 10% x 10%, you get 5 gold coins (probability of each monster dropping their gold coin).

Exactly

Lance_Botelle · November 23, 2020, 7:06pm

This turned out to be far more complicated than I first realised …

Thanks for educating me!

EDIT: I have copied and pasted your answer to my “NOTES” section of my module to keep me informed of how to look at this again.

4760 · November 23, 2020, 7:14pm

That’s the fun with probabilities!
For instance, it’s sometimes much easier to calculate the probability NOT to get the result and then get the probability to get the result as 1-P(not result)

4760 · November 23, 2020, 7:16pm

In that case, Cnp (p elements chosen among n without any order) is calculated as n!/[p!(n-p)!] with n! = n x (n-1) x (n-2) x … x 2 x 1

Lance_Botelle · November 23, 2020, 7:26pm

I think that is what I “intuitively” did to realise my answer must be wrong. I mean I figured surely 20% must be wrong for a single coin drop … or to put it another way, how could there be 80% of no drop?

That’s far too complicated for someone like me … I’m much too old and out of touch to grasp that … Maybe if I was to be taught from the beginning, then maybe I might grasp it. However, I can just about get my head around your last example having looked at it about ten times.

When I was doing some internet background reading about probability, I found it extremely difficult to get my thinking around probability of a coin toss always being 50/50 even after it had been tossed nine heads in a row. I now have grasped the concept that the individual probability is independent of the “group” probability. However, it still messes with my head when I imagine this scenario …

A man has ten coins in a row and tosses the first nine, which all end up heads. Even though, the probability of the last coin toss ending up heads is still 50%, I cannot help but think the chances of rolling another heads (for the tenth in a row) is a lot slimmer for him. One minute I think I understand, then I don’t.

EDIT: I suppose the difference is that the probability of rolling all heads is applied across the total coin tosses and so it can “break” at any coin toss, where the probability of a likely fail is based on the number of coins involved. i.e. The weight of probability applies over all the coin tosses as opposed to any one individual coin toss.

4760 · November 23, 2020, 10:24pm

Think about it this way: the probability for any single coin to end up heads is 50%, even if it’s the last one in a series of ten, because the result of the last toss is independent of the previous nine.
But the chances to have all ten coins ending up heads is (1/2)^10, since as you figured in this case we’re not looking at each coin independently but consider the whole series, and not getting heads will stop the tosses.

Tarot_Redhand · November 23, 2020, 10:32pm

It’s been a long time since I touched on probability. I seem to recall that it was something to do with someone called Bayes (I converted a Bayesian Inferencing engine from one language to another) but all the formula I can find are different to the one I remember. The one I remember was quite simple being P(e) where P is probability and e is an event and it should be read as the probability of something, P, given that event e has occurred. Whereas the formulas I see online are all P(A | B). Cest la vie.

TR

Lance_Botelle · November 24, 2020, 12:27am

@4760

OK, I looked over this again, and I do not understand why we multiply by 4 on the first calculation, and 6 on the second one?

EDIT: OK, I think it is because of the possible combinations … i.e. We can have 4 permutations of the 60% with each of the 10% (of which there are 4), but a possible 6 permutations of any two of the four. If that is the case, is there any obvious formula to use to help determine how many permutations we are using? E.g. Is it something like a factorial of the (total - 1)? i.e. If total = 4, -1 = 3. 3 factorial is 3 x 2 x 1 = 6?

Can you explain this a little more for me. Thanks!
Lance.

EDIT: I was trying to apply your working to another test issue …

Roll a d4, a d6 and a d8. What is the probability of one die rolling a one, two dice rolling ones and all three rolling ones?

I was going to try to show my working using your example, but lost my way again … OK … maybe I have found my way back …

All three I guess would be 0.25 x 0.1666 x 0.125 = 0.005206 = 0.52%

OK, here is my calculations for the dice scenario …

DICE (D4 + D6 + d8) Roll 1 COMBINATION PROBABILITIES

USING 25% 16.6% 12.5%

ANY JUST ROLLS A 1

0.25 x 0.8334 x 0.875 = 0.1823 = 18.23%
0.75 x 0.1666 x 0.875 = 0.1093 = 10.93%
0.75 x 0.8334 x 0.125 = 0.0781 = 7.81%

TOTAL = 36.97%

ANY TWO ROLL A 1

OPT 1 (D4 & D6 roll a 1): 0.25 x 0.1666 x 0.875 = 0.03644 = 3.64%
OPT 2 (D4 & D8 roll a 1): 0.25 x 0.8334 x 0.125 = 0.02604 = 2.60%
OPT 3 (D6 & D8 roll a 1): 0.75 x 0.1666 x 0.125 = 0.01561 = 1.56%

TOTAL = 7.8%

ALL ROLL A 1

0.25 x 0.1666 x 0.125 = 0.005206 = 0.52%

TOTAL = 0.52%

Any chance of a verification?

4760 · November 24, 2020, 12:49pm

@Lance_Botelle
Exactly…

Permutations of n elements: n!
Explanation: you have n possibilities for the first element, (n-1) for the second, … until there’s only 1 element left for the n-th: in total you have factorial of n ways of organising your n elements.

Arrangements (not sure if it’s the term in English) of p elements chosen among n: n! / (n-p)!
Explanation: you have n possibilities for the first element, (n-1) for the second, … (n-p+1) for the p-th element.

Combinations of p elements chosen among n: n! / (p! (n-p)!)
Explanation: it’s the same principle as above, but in this case the order of the selection is not important, which means for any given selection of p elements, we have p! possibities (see “permutations” above) that give the same result => we divide the total by p! so we get the total number of selections of p elements among n, independently of the order they are selected.

In the context: choosing 1 creature among 4 can be done in 4!/(1! 3!) = 4 ways.
Choosing 2 creatures among 4 can be done in 4! / (2! (4-2)!) ) = 24 / (2 x 2) = 6 different ways

4760 · November 24, 2020, 1:11pm

Probability of rolling a 1 on a n-sided die: 1/n
Probability of NOT rolling a 1: (n-1)/n

Probability of getting either event A or event B: P(A U B) = P(A) + P(B) [A and B independent, which is the case here]

Probability of getting both event A and event B: P(AB) = P(A) x P(B)

Your calculations are correct (except for the second line, where you get 10.27% although it should be 10.93%).

=> Probability of rolling three 1: 1/4 x 1/6 x 1/8 = 1/192 = 0,00520833…

=> Probability of rolling only one 1:
d4 and not d6 and not d8: 1/4 x 5/6 x 7/8 = 35/192
d6 and not d4 and not d8: 3/4 x 1/6 x 7/8 = 21/192
d8 and not d4 and not d6: 3/4 x 5/6 x 1/8 = 15/192
P(only one 1) = 71/192 = 0,3698

=> Probability of rolling only two 1:
d4 and d6 and not d8: 1/4 x 1/6 x 7/8 = 7/192
d4 and d8 and not d6: 1/4 x 5/6 x 1/8 = 5/192
d6 and d8 and not d4: 3/4 x 1/6 x 1/8 = 3/192
P(exactly two 1) = 15/192 = 0,078

Lance_Botelle · November 24, 2020, 2:00pm

This is useful to know, although my head is still wrestling to fully comprehend how to always apply it.

I was trying to apply it to a “simple” coin toss scenario, where a coin is tossed 5 times, and then giving the probability of seeing only one head, two heads, three heads, four heads or five heads. (I realised as I wrote this I am making sure I do not say “at least” x many heads, as I can see how that changes the probabilities - I assume you would just add the answers to each up?)

HERE IS MY CURRENT UNDERSTANDING …

Where have I gone wrong?

Using … Combinations of p elements chosen among n: n! / (p! (n-p)!)

HEADS (ELEMENT?) REQUIRED AMONG N?

1 = 5!/(1! x (5-1)!) = 120/(1 x 24) = 120/24 = 5.
2 = 5!/(2! x (5-2)!) = 120/(2 x 6) = 120/12 = 10.
3 = 5!/(3! x (5-3)!) = 120/(6 x 2) = 120/12 = 10.
4 = 5!/(4! x (5-4)!) = 120/(24 x 1) = 120/24 = 5.
5 = 5!/(5! x (5-5)!) = 120/(120 x 0) = 120/0 = 0 … ?

1 HEAD: 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.03125‬ x 5 = 0.15625‬ or 15.625%
2 HEADS: 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.03125‬ x 10 = 0.3125 or 31.25%
3 HEADS: 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.03125‬ x 10 = 0.3125 or 31.25%
4 HEADS: 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.03125‬ x 5 = 0.15625‬ or 15.625%
5 HEADS: 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.03125‬ x 0 = ? (6.25% remaining?)

Corrected - Thanks!

EDIT: I think my last calculation has to use x1, which would make 5 heads:

5 HEADS: 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.03125‬ x 1 = 0.03125 or 3.125%

4760 · November 24, 2020, 2:43pm

Don’t forget the case 0 head

Then you’ll get the following:
P(0) = 0.03125
P(1) = 0.15625
P(2) = 0.3125
P(3) = 0.3125
P(4) = 0.15625
P(5) = 0.03125

Yes, by definition 0! = 1

Lance_Botelle · November 24, 2020, 3:32pm

@4760,

Ah yes, of course!

Obvious really! Thanks for all your help on this topic.

Hopefully, I have a better grasp now.

Take care, Lance.

jimdad55 · November 24, 2020, 5:25pm

Could I just say I’m totally impressed by the crunching of numbers in this thread? Don’t understand a word of it and I don’t even have NWN2 but I’m hooked, in a weird, living vicariously, underling sort of way.

Lance_Botelle · November 24, 2020, 7:38pm

@jimdad55,

It has certainly been a re-education for me - It has been years since I looked at the basics of this, but @4760 helped me to understand some more again.

If I recall correctly, one for your class of students perhaps?

Take care, Lance.