Theorycraft 101: 5.4 Trinkets

Blizzard’s pattern of trying to make more interesting raid trinkets in MoP has provided a lot of interesting theorycraft material. As in past Theorycraft 101 posts, the goal is here both to give theorycrafters some useful conclusions and equations to save them the trouble of reinventing them independently, and to give everyone some general information that helps them evaluate these new bonuses when selecting items.

Item Budgets

In general, the amount of any stats budgeted to a certain slot scales like:

V \cdot Q^I

  • V is the budget value controlling how many stats that item slot gets compared to other slots.
  • Q is a constant equal to the 15th root of 1.15.
  • I is the item’s ilvl.

Often a more convenient way to think of this, especially when dealing with trinkets, is to use ilvl 463 as a baseline. The reason is that most trinket procs are coded into the spelldata with ilvl 463 values (which is the pre-raid baseline for MoP), and then scaled from there on any individual item based on its ilvl. The Int proc on Nazgrim’s Burnished Insignia, for example, is this this spell. So in general, since looking up the ilvl 463 value for any proc/bonus on Wowhead is easy, you can think of the value at higher ilvls as:

V_{463} \cdot Q^{(I-463)}


V_{463} \cdot 1.15^\frac{I-463}{15}

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Hearthstone Probabilities and the Monty Hall Effect

The Monty Hall Problem

A certain probability puzzle is well-known in math circles for its unusual ability to cause people to refuse to believe the answer when it is explained to them. It’s usually known as the “Monty Hall problem” (after the host of Let’s Make a Deal):

Monty Hall has given you a choice of three identical doors. Behind one is a car and behind the other two are goats. You choose a door, but before it’s opened to reveal your prize, Monty adds a twist. He opens one of the other doors to reveal a goat (he always does this to add to the suspense). He then offers you the choice of staying with the door you chose, or switching to the remaining unopened door. Should you stay, or should you switch, and what’s your chance of winning in either case?

The answer is surprising to most: switching doubles your odds of winning the car (2/3 chance of winning, as opposed to 1/3 if you stay). The key fact is that Monty’s knowledge of which of the other doors (if any) was a car causes him to always remove a goat from the prize pool. The chance that the initial door you chose contained a car was 1/3 to start, and it’s unchanged by Monty’s ritual. But if the car was behind either of the other two (2/3 probability in total), Monty will remove the losing door and leave the winning one, and switching will win.

(If you don’t buy that the probability is anything other than 50% when everything started out equal and there are now two doors remaining, there are myriad sources on the internet trying to explain in different ways).

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