PSA: 10-pack has a 1/3 chance of having the featured char

Marvel Puzzle Quest MPQ General Discussion
1

This needs to be said clearly and loudly, and everyone needs to hear it and understand what it means. If you buy a Shulk 10-pack, you have a 1 in 3 chance of actually getting a single she-hulk cover. The math:

You have a 4.4% chance per draw of getting a She-Hulk, and therefore a 95.6% chance of getting no She-Hulk. Therefore the odds of you getting no She-Hulk 10 times in a row is 95.6% ^ 10, or approximately 63.76%. In other words, you have a 36.24% chance of getting even a single She-Hulk cover.

Let me repeat that, loud and clear.

The odds of getting a single She-Hulk cover in a 10-pack is 36.24%, or approximately 1 in 3.

In the doom event, they raised it back to 5.8%. So let’s crunch the numbers on that, shall we:
You have a 5.8% chance per draw of getting a Dr. Doom, and therefore a 94.2% chance of getting no She-Hulk. Therefore the odds of you getting no Dr. Doom 10 times in a row is 94.2% ^ 10, or approximately 55%. In other words, you have a 45% chance of getting even a single Dr. Doom cover. That’s slightly worse than 1 in 2.

Now keep in mind that a 10-pack costs approximately $25.

This is simple math that anyone should be able to do. Anyone buying a 10-pack at this point either has way too much money to blow, is absolutely desperate to do well in a PvE event, or is just bad at math. If you’re considering spending money on the game, they are a gigantic rip-off.

2

You need to look at the draws as individuals and not cumulative. Random numbers don’t have a memory and the chances aren’t cumulative. So the chance remains at 4% per draw. Sorry.

3

Budget Player Cadet is correct. Basic statistical analysis.

4

That is correct.

Basic probability states the following logic:

– Given: The odds of drawing a She-Hulk on ANY pull of a 10-pack is 4.4% (0.044).
– Given: All of the pulls are independent of each other.
– Therefore, the odds of NOT pulling a She-Hulk in a given pull is (100% - 4.4%) = 95.6% (0.956).

– In a sequence of 10 independent events, where the desired outcome is NOT pulling She-Hulk in ANY of the 10 pulls, the resulting calculation is (0.956)^10 = 0.637645 (63.7645%).

– Thus the odds of this chain of events NOT occurring (10 consecutive She-Hulk misses on token pulls), is (100% - 63.7645%) = 36.2355%, rounding to the 36.24% in the OP by BCP; 1 in 2.76 10-packs.

5

The way statistics work in this regard is like this: If you want to find the odds of something happening at least once, you instead find the odds of it never ever ever happening, and subtract that from 100%. So, OP did the correct thing. He looked at the odds of not pulling a featured character 10 times in a row (which is the equivalent of never pulling the character at all), and just took the inverse of that. Boom, odds of pulling the featured character once.

6

The most egregious aspect of this is that you need 8-10 covers of a character to even make them effective.

So, even if you ‘get lucky’ and pull a She Hulk, you have to repeat the feat 8 more times to get a character that you can use outside of their special event.

In other games where you have a ‘gacha pull’ mechanic, you immediately have access to that character and all their special powers. 9/10 times a rare (or super rare) pull is instantly useable and probably better than 90 percent of your collection.

7

…You seriously want to challenge a CompSci major on high-school math? icon_lol.gif
I did look at them individually. There’s no fun way of calculating the combined odds for the countless events that make up “at least 1 of something” in a random distribution with multiple pulls, so instead what you do is, as others pointed out, calculate the odds that in 10 pulls, you get nothing. Let’s simplify this from something like 30 objects to just 2 - a coin flip. If you flip a coin 3 times, and want to see the odds that you get at least one heads, you can build a tree, which will look a little like this:


a                                     
                                          START
                             H                               T
                       H          T                      H        T
                    H   T       H  T                  H  T    H  T
              0.125  0.125 0.125 0.125   0.125 0.125 0.125 0.125

…And so on, and so forth. (The tree would look similar for MPQ, except that instead of two children per node, you’d have 30-something). This is for independent events - each “flip” is independent of the last; it’s a 50-50 chance regardless of what the previous path was. Now, you could in theory, if you wanted to check the probability of getting all heads, simply add up all columns where a heads shows up. But for large graphs, this becomes incredibly unwieldy incredibly fast. It’s far, far easier to simply check “what are the odds that I don’t get heads at all?”. That’s just one branch, and that’s easy to calculate - you flip 3 times, each time you have a 50% chance of tails, so from the outset the odds that you get no heads can be calculated as 0.5^3 = 0.125.

The same applies to MPQ - we don’t care about anything else in the 10-pack. We only care about She-Hulk. So we can simplify it from 30 nodes to 2: She-Hulk or Not She-Hulk. In 3 consecutive pulls, you end up with a tree that looks like this:

a

                                                                  START
                                   S                                                                  NS
                       S                    NS                                           S                     NS
               S           NS          S          NS                             S          NS          S        NS

Once again, same issue - we could count all the paths that have SHulk, or we could just calculate the path which consists solely of NS nodes. That’s what I did in the OP. Statistical analysis 101 when dealing with multiple independent random events. icon_e_smile.gif

8

At first I was was all “that seems low.”

Then: 10 trials p=0.044 - Wolfram|Alpha

9

Now we need someone to analyze this with the Coupon Collector Problem to figure out how many 10 packs are needed to get 13 covers, with no more than 5 of any one color.

10

Bet it’ll run you more than my rent.

11

Has the % chance of getting any 3* ever gone up? I haven’t really followed it too closely I just assume they add a new 3* and then divide the rate over the new number of 3* characters which we know is growing unwieldy. Almost seems like it would be better suited to them is they lumped all the 3* together and posted that rate as apposed to showing us the putrid .07 rates. So yes piss on us and tell us its raining pls.

12

Hey, math! I like math. I’ll take a stab at it.

First, the easy stuff: since there is a .044 chance of getting the cover you want, in a 10 pack you expect to get .44 covers per pack. Yuck.

I tried thinking about the expected number of covers needed to get 13 if you can have no more than 5 of any one color, but I’m apparently not smart enough. Thankfully, I do know how to program (sort of), so I threw together a little Python Monte Carlo simulation:

[spoiler]```

import random

def trial():
count = [0, 0 ,0]
counter = 0

while count[0] + count[1] +	count[2] < 13:
cover = random.randrange (0, 3)
    if count[cover] < 5:
        count[cover] += 1
    counter += 1

print str(counter) + " covers required to get 13"

return counter

def Main():
trials = 1000000.0

counter = 0
total = 0.0
while counter < trials:
total += trial()
    counter += 1
print str (total/trials)

if name == “main”:
Main()


We can do a little bit better by buying 42-packs instead of 10-packs; the probability is the same, so you expect 1.848 covers per 42-pack. Thus you need 8.3 42-packs, or 131,314 hero points, saving us $22, for a total of $660 with 86 hero points left over.

But hang on, we can do MUCH better if we only buy packs until we get all three covers! According to the [wikipedia entry for the Coupon Collector Problem](http://en.wikipedia.org/wiki/Coupon_collector's_problem#Answer), we expect it to take 6 tries to get 3 unique coupons; 6 / .44 = 13.63 ten-packs to get all three covers (51,794 hp) or 3.25 42-packs (51,350 hp). At that point we'll have 6 covers, so need 7 more to max out; buying these directly with hero points gives us 7 * 1250 = 8750 hp, for a total of 60,100, or $302 with 100 hero points left over.

[quote="Budget Player Cadet"]
Bet it'll run you more than my rent.
[/quote]
Based on Budget Player Cadet's profile and the numbers [here](http://www.numbeo.com/cost-of-living/country_result.jsp?country=Germany), I'd stay pretty clear of that bet.
13

I also like math, so here is my advice. Spend 3x1250=3750 HP and buy 3 covers. That’s 100% to get 3 covers you want and you have 50 HP change for a healing pack. That’s the only way to play the game. Anything else means you are a gambler, but even casinos offer better chances so you might as well try your luck there.

14

But you can’t buy covers until you get your first one, so if you don’t have the character at all, you have to buy packs to get started.

15

Wait so 10-packs are a good deal then?

16

Hello, I am an AP Statistics teacher. If anyone reading this has any doubt in whether or not the original post is correct, read on. Or if you’re just a math nerd who loves numbers, read on.

If I were grading a paper, and came across BPC’s exact argument as he put down, I would give him a 70%. His math is correct - make no mistake about that. However he did not write the complete formula. The actual formula he should have used is the Complementary Binomial Probability Formula: 1-((n!/((n-X)!X!))•p^X•q^(n-X)) wherein n = the number of trials (10), X = the number of successes (0), p = the probability of success (.044), and q = the probability of failure (.956).

Now, you will notice that everything between the first parenthesis and the q works out to equal 1. So the answer derived from this formula is the same as the answer that BPC arrived at. I am willing to give him the benefit of the doubt and assume that he knew that and simplified the formula for the sake of clarity. I would have deducted anywhere from 5% to 10% for that, depending on whether this was a mistake he was prone to and whether or not I truly believed the elimination of the multiplicative identity portion of the formula was intentional.

The bulk of the points would be lost because his conclusion statement, which he repeated loud and clear, is wrong. BPC, I’m sorry to have to get all teacher-y on you, but you made a rookie Stat mistake. Your conclusion should have read “The odds of getting at least a single She-Hulk cover in a 10-pack is 36.24%, or approximately 1 in 3.” The odds of getting exactly a single cover, as your statement claims, are 29.35%. Always remember, if you’re dealing with Complementary Events, then chances are that your conclusion should include an “at least” in there somewhere.

TL;DR: If you don’t believe the CompSci major when it comes to statistical analyses, then believe the AP Statistics teacher when he tells you to listen to the CompSci major. icon_e_ugeek.gif

17

This means your odds of getting Spooderbag are 51%, and a 15% chance of getting two Spooderbags.

18

D’oh! Didn’t caveat my statement well enough. icon_lol.gif
Yay statistics.

Also, mischiefmaker, my rent is 250€. So… really not that far off the mark. icon_e_biggrin.gif

19

Great thread.

107×155 4.65 KB

NCSTL must feel a bit sheepish now.

20

Honestly you don’t need very fancy statistics models to tell you that buying this stuff is a bad idea. Statistics just tells you how bad it is to buy it.