# What are the odds of getting the same character in a row?

Laserjunge
Posts:

**33**★
I've just used my 10k 5* and 6* shards to draw my new heros. I thought after getting two times 5* Spiderman Classic, a 5* Rhino and a 5* Iron Man over the last 8 weeks (I am not a super frequent player anymore), what are the odds to draw another NULL. And here we go:

5* Cable and 6* Cable in a row!!

Since this game turned me down so much over the last year, I need at least half a year to get all 6* shards together. So it is effort, in time, in energy and emotional. And then KABAM's terrible RNG machine ruins you another night. The same thing occurs with 5* awakening gems (I've got 5 or 6 and solely 2x mutant, 3x skill, and 1x tech although I am waiting for mystic and cosmic for a year now). If you say, this is just randomness and probability, I'd say finally get a repaying RNG algorithm KABAM!!

For me this is the final killer. I am done with this game and ready for another one, done by better team that actually acknowledges the time and effort people invest in their games.

5* Cable and 6* Cable in a row!!

Since this game turned me down so much over the last year, I need at least half a year to get all 6* shards together. So it is effort, in time, in energy and emotional. And then KABAM's terrible RNG machine ruins you another night. The same thing occurs with 5* awakening gems (I've got 5 or 6 and solely 2x mutant, 3x skill, and 1x tech although I am waiting for mystic and cosmic for a year now). If you say, this is just randomness and probability, I'd say finally get a repaying RNG algorithm KABAM!!

For me this is the final killer. I am done with this game and ready for another one, done by better team that actually acknowledges the time and effort people invest in their games.

13

## Comments

16,000GuardianIt sounds like this game, or anything like it, is just not the kind of game you should be playing. You expect to hit the jackpot even after acknowledging you aren't opening as many crystals anymore. Someone who needs great openings specifically *because* they don't open many is precisely the kind of person that should not play games with random lootboxes, because they need the exact opposite to happen that will actually happen.

372★★★16,000GuardianThe odds of pulling a champion from the 5* basic and then the same champ again from the 6* basic is also about one in 150. Since order doesn't matter the odds of pulling a 5* champ and then an identical 6* champ are the same as pulling a 6* champ and then an identical 5* champ, which means the odds of pulling identical champs in a row are the same as above: one in 150. It doesn't even matter how many champs are in the 6* crystal: if there was only one champ in the 6* crystal then the odds of pulling the same champ twice in a row would still be one in 150. The 6* champ would be fixed (it could only be one thing) and then the odds of pulling that champ out of the 5* basic would be one in 150.

173★★1,276★★★★when its nick fury, AEgon,corvus etc its 0.008%.

572★★The easy way to explain this is... Roll a dice, you chance of rolling a 2 are 6:1

roll that same dice again and your chance of rolling a 2 are 6:1

The odds do not change and are not influenced in any way by your previous or subsequent rolls

(taking human influence, roll speed and force etc out of the equation that have no bearing in a RNG based calculation)

504★★132★Really really good

I opened my first 6*

Emma

I opened my second 6*

Emma again

Now she makes life miserable for people on defence

111★2,759★★★★★7,253★★★★★Not sure how many times this has been pointed out on the forums, but it amazes me that it has to be pointed out over and over again.

To @Plinko Think of it this way—you won the lottery with those Twin Janes. Granted, it was a Shirley Jackson kind of lottery, but you’re still a lottery winner.

Dr. Zola

117★So I believe the percentage of even pulling a 5* was like 3-5%. What would be the odds of pulling back to back 5* of the sames champ considering the odds??

Just curious..

143★3,374★★★★★725★★★94★I've had like 40-45 bad ones in a row, and then I got dipped stark spidey. As for 6*, I have a roaster of useless things

4,378★★★★★Yes, the draws are completely independent; and in a way itdoesn'tmake sense to calculate the odds; but the thing is, people like the OPkeep asking these questions.People want to know the odds, whether that desire is rational or not. Don't get angry with other community members for trying to help them.

564★★Would you you bet $100 that you could flip a coin on heads 10 times in row if I gave you 10:1 odds.

3,598Administrator ›3,091★★★★★16,000GuardianWe presume each crystal opening is an independent event that has nothing to do with each other, and from this assumption we calculate the odds of two crystal openings happening in a set. This is not only a fundamental calculation of statistics that is specifically intended to determine the odds of real world sequences of events, it isn't even a particularly interesting or controversial calculation. This is a completely trivial calculation.

I honestly have no idea where this misconception about "cumulative events" comes from. I can only assume it is a mental glitch where if you believe the first crystal and the second crystal are independent and thus do not affect each other, it is somehow "invalid" to treat the two rolls as a single event whose odds can be calculated. But that is precisely what the counting principle of statistics mathematically describes.

In a throw of two dice, the dice are independent, so the first throw can be any one of six results, each having the same chance to happen. Regardless of the first roll there are six possibilities for the second roll, all equally likely, because the second roll is independent of the first. That means there are thirty-six possible net outcomes: 1/1, 1/2, 1/3, 1/4, 1/5, 1/6, 2/1, 2/2, 2/3, 2/4, 2/5, 2/6, 3/1, 3/2, 3/3, 3/4, 3/5, 3/6, 4/1, 4/2, 4/3, 4/4, 4/5, 4/6, 5/1, 5/2, /53, /54, 5/5, 5/6, 6/1, 6/2, 6/3, 6/4, 6/5, 6/6. Out of the thirty six possible outcomes, six are identical throws: 1/1, 2/2, 3/3, 4/4, 5/5, 6/6. So there are six out of thirty six possibilities in which the dice are identical throws, which means the odds are six in thirty six or one in six.

All of statistical calculations are math that simplifies this basic fundamental counting of possible outcomes. We aren't somehow claiming that the two throws are a "cumulative event" we count all possibilities of two throws as independent events to determine the odds of a sequence of independent events to occur.

This is something you have to accept, or fail statistics on day one. In fact, I would like to see the face of a statistics teacher when you tell them "The odds for events like 'rolling [X] [Y] times in a row' are merely a concept." Yeah: that concept is called "statistics."

7,253★★★★★To @Kabam Vydious — ouch. But at least we have hope you’re on the inside putting in the good word for an IP buff.

Dr. Zola

16,000GuardianThe flaw here is that while each crystal opening is independent, the desired sequence is not an independent sequence. So the analysis has to look at the requirements of each crystal. The requirement on the second crystal is that it match the first crystal, and whatever the first one was there's only a one in 150 chance of matching. But there's no requirement on the first crystal, it can be anything. So the odds of the first crystal meeting the sequence requirements is 100%.

Alternatively, you can choose to use the fundamental counting principle, which is always correct, just maddeningly tedious. There are 150 possible drops for crystal 1. For each of those possibilities there's 150 possible drops for crystal 2. There are thus 22500 possibilities for the two crystals. However, there are 150 possibilities in which the two drops match. So the odds are 150 out of 22500, or one in 150.

Let's try this with the simplest possible case to make life easy. What are the odds of flipping a coin twice and getting the same result. By your calculation, each flip is independent and has two possibilities, so you multiply and get one in four. But that's clearly wrong. The fundamental counting principle says there are four possibilities: Heads/Heads, Heads/Tails, Tails/Heads, Tails/Tails, and two of them are matching: Heads/Heads, and Tails/Tails; so the odds are two out of four, or one in two. You count all possibilities, count all desired ones, and the odds are the latter out of the former. That's actually what saying "X out of Y" means. There are Y possibilities all of which have equal odds, and X of them are the desired ones.

So the two valid ways of calculating this are: calculate the number of possibilities and the number of desired ones and divide. Or use constraint logic, which is how Markov Chains fundamentally work. There's no constraint on the first flip, but a 50% constraint on the second flip (it must match the first one), for a net odds of 1/1 * 1/2 = 1/2 = 50% on the net result.

There's a simple double check on logic here that I used to use when teaching statistics. When someone says, for example, that the odds of pulling the same champ twice are one in 22500, because they just multiplied the numbers, I ask what are the odds of pulling Winter Soldier twice in two crystals. Of course that's one in 22500. But if that's the case, how can the odds of pulling *any* champion twice in two crystals be the same? Obviously, it has to be more likely to pull two identical champs than to pull WS specifically twice in a row. You have to realize that asking what the odds of pulling two identical champs in a row is really asking what are the odds of pulling Winter Soldier twice in a row, plus the odds of pulling Storm twice in a row, plus the odds of pulling Aegon twice in a row, etc. There's 150 ways to pull the same champ in a row, whereas there's only one way to pull Winter Soldier twice in a row.

16,000GuardianWhich is weird because I always thought probability and statistics was a required math subject for anyone who has graduated high school much less undergraduate college. Either you had to take it, or you had to take more sophisticated math that presumes it. I don't expect people to have command of the more sophisticated stuff, but I would think most people would at least be able to recognize it again when it is presented.