Solve this math problem and win a prize (expired)

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The chances are 7/8 because sam (being an intelligent person) has his friends or has strangers invite him to a game where the only portal that hasn't been opened is the one he needs. So each time he uses a machine he has a 50% chance of getting the organ. He repeats three times so he only has a 1/8 chance of not getting the organ.
Brute force, of all possible outcomes. "Good" results in bold:

1st portal, correct, drops: 1/6

1st portal, correct, no drop: 1/6
- 2nd portal correct, drops: 1/6 * 1/3 * 1/2 = 1/36
- 2nd portal correct, no drop: 1/6 * 1/3 * 1/2 = 1/36
- - 3rd portal correct, drops: 1/36 * 1/3 * 1/2 = 1/216
- 2nd portal incorrect: 1/6 * 2/3 = 1/9
- - 3rd portal correct, drops, 1/9 * 1/2 * 1/2 = 1/36

1st portal incorrect : 2/3
- 2nd portal correct, drops: 2/3 * 1/2 * 1/2 = 1/6
- 2nd portal correct, no drop: 2/3 * 1/2 * 1/2 = 1/6
- - 3rd portal correct, drops: 1/6 * 1/3 * 1/2 = 1/36
- 2nd portal incorrect: 2/3 * 1/2 = 1/3
- - 3rd portal correct, drops: 1/3 * 1/2 = 1/6

1/6 + 1/36 * 1/216 + 1/36 + 1/6 + 1/36 + 1/6 = 127/216 = 58.8%

Er, you're not brute forcing an optimal strategy here. Actually, you're not brute forcing anything. I mean look at the middle block - correct portal, no drop. He is going to leave this game, he isn't going to check out the second and third portals and if they will drop. He doesn't care. They don't have the organ he wants. Adding the following chances is meaningless.


So, are you agreeing with the above math or disagreeing? What did you say the chance was in the end?

~Philoi.
07/22/2013 04:07 PMPosted by DrunkenPanda
The chances are 7/8 because sam (being an intelligent person) has his friends or has strangers invite him to a game where the only portal that hasn't been opened is the one he needs. So each time he uses a machine he has a 50% chance of getting the organ. He repeats three times so he only has a 1/8 chance of not getting the organ.


I know you are trying to be funny but this actually sounds like the best method so far. That is if you can do it and your friends/strangers are willing to help. But what are the chances of that? Did you take that into account? : P ~Philoi.
best odds? 100%...duh...thatd be your best odds...now actual odds? lol

Er, you're not brute forcing an optimal strategy here.


Brute force refers to the math, not the logic. The methodology used was totaling up each possible outcome, instead of using a more formulaic/generic approach.


Actually, you're not brute forcing anything. I mean look at the middle block - correct portal, no drop. He is going to leave this game, he isn't going to check out the second and third portals and if they will drop. He doesn't care. They don't have the organ he wants. Adding the following chances is meaningless.


And that is in the math. If you notice, the chances of getting the right portal in the second block is 1 in 3, since he started a new game. In the third block, when he is in the same game, it is 1 in 2.

Does anyone else hate it when wrong people call others retards?


Except he isn't wrong. Although right or wrong, you shouldn't call people retards, this isn't simple math :).
Except he isn't wrong. Although right or wrong, you shouldn't call people retards, this isn't simple math :).


I think it is simple math. But people still shouldn't call one another retards anyway.

I just didn't get that people would leave games and make new ones. That changes the problem entirely. I think Getch got it right or close to right.

~Philoi.
07/22/2013 12:58 PMPosted by DuckOfDeath
The math is right. Its just that you are using the wrong math. That formula is the odds of having EXACTLY 1 success in 3 tries. The odds of you getting any number of successes in 3 tries is a different formula and the outcome would be higher.


He stated the problem ambiguously.I took it as he needed 1 specific organ out of the portals.
So if its 50% chance to get AN organ from each portal it is actually a lower probability that he will get the one he wants.

If he was talking about need any one organ from any one portal then yes 50% is the correct answer. But if he needs the specific organ my answer is correct.

/thread

Can I have a consultation prize now
So, what are the odds that three Infernal Machines will be parlayed into the last required organ?


Everybody is wrong so far.

He is asking for odds, not probability.

For those not familiar, odds are occurring versus not occurring or vice versa.

Whereas probability, which is what everybody is giving, is occurring versus total or not occurring versus total.
If he needs a specific organ and he opens all three portals the answer is......
Its P(1)= (3 choose 1) * 1/2 ^1 * 1/2 ^2= 37.5%

if he needs ANY one organ then the answer is 50% because in this case each portal would be treated as a separate entity as one does not depend on the other and vise versa.

This problem is EXACTLY like a coin flip. 50% head and tails no matter how many coins you flip in a row.
But if you flip 3 coins at the same time (3 different portals in the same game) and each coin has a different heads representation (lets say black is tails for all, meaning no drops then coin 1 face is blue, coin 2 face is yellow, and coin 3 face is red) you end up with a 37.5% chance that you actually get the desired face color you are looking for (lets say red).
This is why EVERYONE rages about RNG. You don't have a mathematical grasp of how probability works. Its confusing if you aren't good with number. The best way is just to go color coins and flip them for 10 trials.....
Math noobs.....
07/22/2013 06:17 PMPosted by Sephirath
He is asking for odds, not probability.


He wins^ so 37.5% probability is 3/8 odds....
07/22/2013 06:01 PMPosted by shindog3
The math is right. Its just that you are using the wrong math. That formula is the odds of having EXACTLY 1 success in 3 tries. The odds of you getting any number of successes in 3 tries is a different formula and the outcome would be higher.


He stated the problem ambiguously.I took it as he needed 1 specific organ out of the portals.
So if its 50% chance to get AN organ from each portal it is actually a lower probability that he will get the one he wants.

If he was talking about need any one organ from any one portal then yes 50% is the correct answer. But if he needs the specific organ my answer is correct.

/thread

Can I have a consultation prize now


Seriously? You are going to use the excuse that he was ambiguous? Seemed pretty straightforward to me. Maybe you stats majors dont need to take English or something?

And yes he does need a specific organ and no you are still wrong. Basic logic will dictate that at worst he will have a 50% chance by opening up all three portals in one game. The only time it would be less than 50% is if he doesnt use his machines logically and opens up random portals in new games regardless of if its the right portal or not.

You can have a consolation prize just for making me laugh though.
I dare you to go collect the machines and try it. I'm sorry you can't comprehend it even when the math is laid out in front of you.
Correct answer is 127/216, as stated by others. The strategy is to keep opening portal(s) in the same game until/unless he opens the correct portal but does not get a drop (obviously).
If he needs a specific organ and he opens all three portals the answer is......
Its P(1)= (3 choose 1) * 1/2 ^1 * 1/2 ^2= 37.5%

if he needs ANY one organ then the answer is 50% because in this case each portal would be treated as a separate entity as one does not depend on the other and vise versa.

This problem is EXACTLY like a coin flip. 50% head and tails no matter how many coins you flip in a row.
But if you flip 3 coins at the same time (3 different portals in the same game) and each coin has a different heads representation (lets say black is tails for all, meaning no drops then coin 1 face is blue, coin 2 face is yellow, and coin 3 face is red) you end up with a 37.5% chance that you actually get the desired face color you are looking for (lets say red).
This is why EVERYONE rages about RNG. You don't have a mathematical grasp of how probability works. Its confusing if you aren't good with number. The best way is just to go color coins and flip them for 10 trials.....
Math noobs.....


If he heeds a specific organ and opens all three portals in the same game then his odds are 50%. This as I have said before should be his worst case scenerio unless he is dumb about opening portals in different games when he shouldnt.

This problem is not exactly like a coin flip because what happens with the second portal relies on what happened in the first. The outcome of the first portal affects how you open the second and thus affects the odds. A coin flip doesnt have memory but a set doors do. And everytime you open a door the odds for the other doors change because they are part of a static set.
07/22/2013 06:27 PMPosted by shindog3
I dare you to go collect the machines and try it. I'm sorry you can't comprehend it even when the math is laid out in front of you.


You dont have to collect the machines. And the math is in front of you and you are the one not comprehending it. Here I will try to show you why your math is wrong.

I have 3 keys. I need a specific organ (lets say the magda/king organ). And I am doing MP5. If I open all three portals in the same game regardless of which one I got first. Then my odds of getting that specific organ is 50%. Its 0% for the ghom/rak portal, 0% for the siege/kulle portal, and 50% for the magda/king portal. 0+0+50=50. That is the absolute worst case scenerio that can happen. Unless you are a total idiot opening random portals in new games you cannot get any lower than 50%. Logic should alert you that using a formula that tells you that your odds of getting the organ is less than 50% is wrong.
Uggg, I had a reply all written out and everything and went to post and BNET ate it. But, looking above, someone already beat me to it.

It's 58.80% or 0.58796296296298
I found a copy of it, just in case mine is easier to understand

Not sure if answered already but...

to make things nicer to read, I'll use some terminology.

C = correct portal
NC = not correct portal
D = drop
ND = no drop

C/NC/D/ND# = correct portal/not correct portal/drop/no drop for particular machine

[1/3 * 1/2] + <----- C1, D1

[1/3 * 1/2 * 1/3 * 1/2] + <----- C1, ND1, C2, D2

[1/3 * 1/2 * 1/3 * 1/2 * 1/3 * 1/2] + <------ C1, ND1, C2, ND2, C3, D3

[1/3 * 1/2 * 2/3 * 1/2 * 1/2] + <----- C1, ND1, NC2, C3, D3

[2/3 * 1/2 * 1/2] + <------ NC1, C2, D2

[2/3 * 1/2 * 1/2 * 1/3 * 1/2] + <------- NC1, C2, ND2, C3, D3

[2/3 * 1/2 * 1 * 1/2] <----- NC1, NC2, D3

Total probability =

0.16666666666667 +
0.02777777777778 +
0.00462962962963 +
0.02777777777778 +
0.16666666666667 +
0.02777777777778 +
0.16666666666667 =

0.58796296296298 -> 58.80% -> 127/216 chance to get correct organ with 3 machines on MP5.
07/22/2013 06:20 PMPosted by shindog3
He is asking for odds, not probability.


He wins^ so 37.5% probability is 3/8 odds....


He is arguing semantics. Its the same data just written in different ways. One is usually expressed as a % or fraction and the other is a ratio.
07/22/2013 06:49 PMPosted by DuckOfDeath
Logic should alert you that using a formula that tells you that your odds of getting the organ is less than 50% is wrong.


It should, but the guy calling people math noobs believes otherwise.
07/22/2013 07:13 AMPosted by Darkul
The first person to solve this problem will win


There is an algebraic solution. Show your work.


this is a forum about diablo gaming no math problems
if you want to be clever go to troll somewhere else

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