Solve this math problem and win a prize (expired)

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The first person to solve this problem will win a decent Uhkapian Serpent (5 million gold value), a nice wand (5 million gold value), and 1 million in gold.

Sam wants to craft a Hellfire ring. He has three Infernal Machines, and only needs one more organ to craft it.

At MP5 (50%) drop-rate, what are the chances that he will get that particular drop?

There is obviously three possible portals, and a 50/50 drop rate once the correct portal opens - so even if he ends up using all three machines (the last used opening the correct portal), the worst the odds can be is 50/50. But the correct portal could theoretically be opened three times (all on the first try, albeit in different games).

So, what are the odds that three Infernal Machines will be parlayed into the last required organ?

There is an algebraic solution. Show your work.
07/22/2013 07:15 AMPosted by Garthandal

No. Not even close. If he had a single Infernal Machine the odds would be:

1/3 x 1/2 = 1/6

But he has three, so at worst the odds are...

1 x 1/2 = 1/2

What are his best odds? Anyone?
For 1 portal:

1-((1-0.5)^1) = 0.5
50% Chance

For 2 portals:

1-((1-0.5)^2) = 0.75
75% Chance

For 3 portals:

1-((1-0.5)^3) = 0.875
87.5% Chance
I can't believe no one has gotten it yet.
I will give a hint - the answer has already been given. The person failed to show their work.

Another poster had the formula right but the answer wrong.

Edit: 5.5% was not the right answer. I was thinking .55% (which was close but still wrong).
if I am reading the question right it is probably a 1 in 9 chance of the portal opening which comprises of the first 2 1 in 3 chances failing and the last 1 in 3 chance succeeding
U asked for the best chance. rolling the dice 3 times in the same game means he has a chance to win of 50%. rolling the dice in 3 different games means he has a chance to get it by a chance of 1/6
This is a trick question.
Sam's gear is too bad to do Ubers, and so he effectively has a 0% chance.
i probably had a miss calculating in the worst chance. this is (1/3)^3*1/2. but im still sure the best chance is 50% by doing it all in 1 game --> 50%
hmm i think i see how u look at it

doing it in 1 Portal he has a chance of 50%.

doing it in 2 Portal he has a chance of 50% in the first portal, and 25% (50% of a 50% chance) in for the second portal. so -> 75%

doing it in 3 Portals means: 50% + 25% + 12.5%. so the one with 87.5% was right
Come pm for your prize Azaroth
1 game three portals: 1/2 = 50%

3 games 1 portal: 3*(1/3)/2 = 50%
but then you dont calculat in the chance to get all the right portals in the first try?
07/22/2013 07:46 AMPosted by Azaroth
but then you dont calculat in the chance to get all the right portals in the first try?

I thought it was 66% at first.

.33 / .50 = .66%

But I figured Azaroth put so much time into the equation I would go ahead and give him the prize.

It may actually be .66% - or it may be .50%...I don't know myself anymore!

*note to self, think your questions through a little more before creating threads*

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