That moment you casually get the same [useless] minion 3 times in a single turn from Shifter Zeruses. Seriously, what are the odds? 1/*number of minions in the game* x 1/*nomitg* x 1/*nomitg*. Someone please calculate that.
Somehow I managed to win this game only dropping a minion when I had access to 6 mana, btw :P tried to add the opponent afterwards to tell him about what happened but he never accepted. That's on you, stupid BM'ers.
I'm pretty sure it's not [1/nomitg]^3, it's actually just [1/nomitg]^2. It doesn't matter what minion the first Zerus is, just that the second and third are the same one.
The question is kinda flawed. You picked a random event and said "Wow, what were the odds of that?". It was in fact a certainty because you're asking for the odds of an event that had already occurred and picked the event that happened.
You could pick any event and ask the same thing. This is why I have no respect for Kripp's moaning in his chat when he says things like "damn that's no fireball in 10 arena runs, what are the chances? I'm so unlucky!!!", give us a break.
For future reference the probability of getting least 3 of the same minion out of 5 Shifter Zerus's is ((n-1)² * 10 + (n-1) * 5 + 1)/n⁴ where n is the size of the cardpool.
The question is kinda flawed. You picked a random event and said "Wow, what were the odds of that?". It was in fact a certainty because you're asking for the odds of an event that had already occurred and picked the event that happened.
You could pick any event and ask the same thing. This is why I have no respect for Kripp's moaning in his chat when he says things like "damn that's no fireball in 10 arena runs, what are the chances? I'm so unlucky!!!", give us a break.
I don't understand this logic. Yes, the events have occurred so now it is known that they happened, but that's not the question being asked. The question is 'what were the odds of that happening?' so any knowledge from the future is irrelevant. The question could as well be 'what are the odds of the exact same thing occurring again?'
If you design a program that, completely at random (ignoring anything about generated randomness etc.), picks one of two options - let's say right or left - and then were asked what the odds were of it picking right the first time it is run, you would answer 50%.
If the program is then run once, and it picks right, and you were asked what the odds were, would you answer 100% because right was picked? No, you would answer 50% because the odds never changed. Knowing the outcome doesn't change what the odds were before the event occurred.
I assume there are 598 minions in HS at the moment. This is why, please feel free to correct me:
According to the Innkeeper, there are 1083 collectable minion cards, and 113 of them are legendary.
This leaves (1083-113)/2 = 970/2 = 485 different non-legendary minions.
That gives the total of 485+113=598 minions you could get from Shifter Zerus in a Tavern Brawl.
You had 5 copies of Shifter Zerus that turn. There are (5 choose 3) = 10 choices of exactly 3 of them to become Wilfred Fizzlebang that turn. Hence the probability of that happening is
Bemooran is right at pointing out that if you only ask for any card to be repeated three times among your five Shifter Zeruses then the probability goes up to
The question is kinda flawed. You picked a random event and said "Wow, what were the odds of that?". It was in fact a certainty because you're asking for the odds of an event that had already occurred and picked the event that happened.
You could pick any event and ask the same thing. This is why I have no respect for Kripp's moaning in his chat when he says things like "damn that's no fireball in 10 arena runs, what are the chances? I'm so unlucky!!!", give us a break.
I don't understand this logic. Yes, the events have occurred so now it is known that they happened, but that's not the question being asked. The question is 'what were the odds of that happening?' so any knowledge from the future is irrelevant. The question could as well be 'what are the odds of the exact same thing occurring again?'
If you design a program that, completely at random (ignoring anything about generated randomness etc.), picks one of two options - let's say right or left - and then were asked what the odds were of it picking right the first time it is run, you would answer 50%.
If the program is then run once, and it picks right, and you were asked what the odds were, would you answer 100% because right was picked? No, you would answer 50% because the odds never changed. Knowing the outcome doesn't change what the odds were before the event occurred.
Yeah but the question is asked after the event. Before the event you don't even know what it is you want to calculate. In your scenario you've setup a situation where you know the possible outcomes beforehand (left or right) so it makes sense to calculate the odds. The OP posted a completely random event that could have been anything so it was a certainty.
The question could as well be 'what are the odds of the exact same thing occurring again?'
This is correct and that should have been the question asked but human nature stops people from doing so because they want to go "Wow look at that? What were the odds!!!".
That moment you casually get the same [useless] minion 3 times in a single turn from Shifter Zeruses. Seriously, what are the odds? 1/*number of minions in the game* x 1/*nomitg* x 1/*nomitg*. Someone please calculate that.
Somehow I managed to win this game only dropping a minion when I had access to 6 mana, btw :P tried to add the opponent afterwards to tell him about what happened but he never accepted. That's on you, stupid BM'ers.
I'm pretty sure it's not [1/nomitg]^3, it's actually just [1/nomitg]^2. It doesn't matter what minion the first Zerus is, just that the second and third are the same one.
The question is kinda flawed. You picked a random event and said "Wow, what were the odds of that?". It was in fact a certainty because you're asking for the odds of an event that had already occurred and picked the event that happened.
You could pick any event and ask the same thing. This is why I have no respect for Kripp's moaning in his chat when he says things like "damn that's no fireball in 10 arena runs, what are the chances? I'm so unlucky!!!", give us a break.
For future reference the probability of getting least 3 of the same minion out of 5 Shifter Zerus's is ((n-1)² * 10 + (n-1) * 5 + 1)/n⁴ where n is the size of the cardpool.
LOL are you serious?
If the program is then run once, and it picks right, and you were asked what the odds were, would you answer 100% because right was picked? No, you would answer 50% because the odds never changed. Knowing the outcome doesn't change what the odds were before the event occurred.
You can find me here! Good luck everyone!
I assume there are 598 minions in HS at the moment. This is why, please feel free to correct me:
You had 5 copies of Shifter Zerus that turn. There are (5 choose 3) = 10 choices of exactly 3 of them to become Wilfred Fizzlebang that turn. Hence the probability of that happening is
(5 choose 3) * (1/598)^3 (597/598)^2 = 0.0000000466.
Bemooran is right at pointing out that if you only ask for any card to be repeated three times among your five Shifter Zeruses then the probability goes up to
(5 choose 3) * (1/598)^2 (597/598)^2 = 0.00002787.
Auto squelch. That's all I'm asking for.
This is correct and that should have been the question asked but human nature stops people from doing so because they want to go "Wow look at that? What were the odds!!!".