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Wizard Math 101

AuthorMessage
Defender
Apr 27, 2009
127
Come on in class, quiet down, and take a seat please.

Ok, for your daily question, answer me the following:

A Frost Giant has a base damage of 475. But wait, you added a Colossus enchantment: that's 750 base damage now! You have one ice blade equipped: that's 40% more. The enemy has one elemental shield at 25% and one 30% ice trap. Oh no! The enemy put on a 25% weakness spell on you. You better calculate that as well! Yes! You have the pips, you have the buffs, now cast the spell! *Fizzle*. Oh no! You fizzled! You have 60 cards in your deck, you have discarded 20 and that was your only Frost Giant. First, calculate the probability that a Frost Giant spell will appear in the next 2 turns and calculate the amount of damage that the Frost Giant will do when you finally cast him!

~I used this problem with my daughter today. She got it right!

Mastermind
May 02, 2009
356
Explorer
Mar 27, 2011
58
nlhf1995 wrote:
Come on in class, quiet down, and take a seat please.

Ok, for your daily question, answer me the following:

A Frost Giant has a base damage of 475. But wait, you added a Colossus enchantment: ...



Sun spell name is colossal, colossus is one of ice damage spells 8)

(Pin the teacher! )

Hero
Jun 11, 2010
729
Wait! That was your only Frost Giant!?!?! That means the probability of getting a Frost Giant is 0/40! If you were to cast him (via treasure card or via reshuffle) then it would do... um... Hold on... carry the two... um... that's umm... dang, I need some paper.... try this again.... let's see, that's about 1500 damage according to my math. Ouch...

Defender
Dec 23, 2009
137
for the first question, it'll do approximately 768 damage if the enemy doesnt have any boosts or resistances. The damage will also be altered if you have an ice attack boost or chance of a critical. for the second question, it depends on how many spells you cast or discard, but assuming it's 1 per round, 5%. It isn't 0 since spells that fizzled are placed back in your deck.

Adherent
Mar 18, 2009
2737
My guess is:

27% chance of drawing the Frost Giant again.

768 damage (rounded up).

Astrologist
Jun 04, 2010
1008
nlhf1995 wrote:
Come on in class, quiet down, and take a seat please.

Ok, for your daily question, answer me the following:

A Frost Giant has a base damage of 475. But wait, you added a Colossus enchantment: that's 750 base damage now! You have one ice blade equipped: that's 40% more. The enemy has one elemental shield at 25% and one 30% ice trap. Oh no! The enemy put on a 25% weakness spell on you. You better calculate that as well! Yes! You have the pips, you have the buffs, now cast the spell! *Fizzle*. Oh no! You fizzled! You have 60 cards in your deck, you have discarded 20 and that was your only Frost Giant. First, calculate the probability that a Frost Giant spell will appear in the next 2 turns and calculate the amount of damage that the Frost Giant will do when you finally cast him!

~I used this problem with my daughter today. She got it right!


So, before I answer... should I adjust the elemental shield to -50% instead of -25%? Also, do I calculate the odds based on discarding or assuming that I'm only casting or discarding one spell per round for the next 2 rounds? Finally, since I have access to the Colossal Sun enchantment card, I should assume I have access to some pretty good gear. Should I calculate any gear damage boosts and if so, in what amount?

Mastermind
Sep 11, 2010
369
(note: calculations done with more decimals than appearing)

840 damage, by my calculations, to take the "easier" part first.

Now, I assume that you are desperate to get hold of that Frost Giant, and discard every single card that isn't a Frost Giant, and that you didn't discard any other cards when you cast the Frost Giant.

That means in the first round, we have a 5.13% chance of the Frost Giant appearing in the "available" card slots (as Colossal counts as a second card slot and as a used card - meaning that there's technically only 39 cards left for the game to pick between). In other words, 2/39.

No Giant, so all 7 cards are discarded, leading to the second round. Here, we have 32 cards left to pick between, meaning a base 3.13% for each card to be the Frost Giant (1/32). Or, you have about 21.9%'s chance of getting the Giant on hand in your second round (7/32).

Just for the lulz: Third round would give you a 28%'s chance of any card being Frost Giant (7/25); fourth round would be 38.9% (7/18), fifth would be 63.4% (7/11 per card), while the sixth round would guarentee that you'd find that Giant, seeing as you'd only have 4 cards left in your deck by then.

Adherent
Mar 18, 2009
2737
kingurz wrote:
My guess is:

27% chance of drawing the Frost Giant again.

768 damage (rounded up).


Whoops, I meant 37%. (for some reason I was thinking you only draw 5 cards, not 7).

Hero
Jul 27, 2009
755
ummm errrrr ahhhh (edges toward the door)

i think I'll skip class today....
my brain hurts from all those numbers...

(steps out the door )

um Bye~~~~~~~~~~~~~~~~~~

Champion
Jul 30, 2010
441
nlhf1995 wrote:
Come on in class, quiet down, and take a seat please.

Ok, for your daily question, answer me the following:

A Frost Giant has a base damage of 475. But wait, you added a Colossus enchantment: that's 750 base damage now! You have one ice blade equipped: that's 40% more. The enemy has one elemental shield at 25% and one 30% ice trap. Oh no! The enemy put on a 25% weakness spell on you. You better calculate that as well! Yes! You have the pips, you have the buffs, now cast the spell! *Fizzle*. Oh no! You fizzled! You have 60 cards in your deck, you have discarded 20 and that was your only Frost Giant. First, calculate the probability that a Frost Giant spell will appear in the next 2 turns and calculate the amount of damage that the Frost Giant will do when you finally cast him!

~I used this problem with my daughter today. She got it right!

depends on how many cards you discard in the next 2 turns? all? none? but the damage would be 1004.0625 if you discard all the cards in you hand for 2 turn the probability is 14 out of 26 or 7/13 or a 0.538 chance.

Astrologist
Jun 04, 2010
1008
nlhf1995 wrote:
Come on in class, quiet down, and take a seat please.

Ok, for your daily question, answer me the following:

A Frost Giant has a base damage of 475. But wait, you added a Colossus enchantment: that's 750 base damage now! You have one ice blade equipped: that's 40% more. The enemy has one elemental shield at 25% and one 30% ice trap. Oh no! The enemy put on a 25% weakness spell on you. You better calculate that as well! Yes! You have the pips, you have the buffs, now cast the spell! *Fizzle*. Oh no! You fizzled! You have 60 cards in your deck, you have discarded 20 and that was your only Frost Giant. First, calculate the probability that a Frost Giant spell will appear in the next 2 turns and calculate the amount of damage that the Frost Giant will do when you finally cast him!

~I used this problem with my daughter today. She got it right!


Ok, I like math too much to wait for the answers to my queries. :D I'm assuming no gear damage boost and I'll leave the elemental shield at 25% despite the card's value being 50%. The damage would break down to approximately 768 (rounded up) to all enemies assuming charm and ward processing in the order you have laid out. An exact number reported in game is extremely difficult to predict with multiple charms and wards because KI has never published exactly how they calculate and where they round in the process. However, it has been my experience that calculating the multiplier first and then applying that to the base damage (rounding the result) is consistently closer than starting with the damage and adjusting it by each multiplier as you go (thereby underscoring the importance of understanding the rules regarding rounding and operator processing).

In regards to the odds, in the round immediately following the fizzle... There is a 1 in 34 chance (approximately 2.9%) of Frost Giant coming up, based on straight numbers where the card draw(s) each round are random. This assumes that you only cast Frost Giant on the round in which it fizzled with no discarding. The numbers come from the fact that 6 of the 40 cards in your hand would remain in place with only 1 new draw from the remaining 34 in the pile. If it does not appear that first round and you discard or otherwise use your entire hand, there is a 7 in 33 chance (approximately 21%) that it will show up on the following round which assumes again a random card draw.

The random card draw issue is one I can't currently quantify because I have not personally tested card processing on a fizzle. Since this game is meant to be a card game "come to life", it is possible (even likely) that KI calculates the cards in your hand as a "stack" when you first enter combat. In other words, it performs a virtual "shuffle" thereby sealing the order of the cards you will draw at the very beginning of combat as would mimic how you would do it in real life. If the rules of the game on fizzle were to place the "fizzled" card at the bottom of your deck (again common in real life card games) then the odds of drawing Frost Giant in the next two hands are zero unless you have a "reshuffle" which you can cast on the round immediately following the fizzle. However, that could prevent you having the necessary pips to actually cast Frost Giant on the following round assuming it comes up (of which there is a 1 in 60 chance based on the statement "that was your only Frost Giant." from above).

How did I do?