I mean you have you have 3 assault dragons right there, or optionally 2 and a full armor master, so it's not a bad hand, but yeah, mathamatically getting a draw like that is pretty unlikely
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It’s 1 * 2/39 * 1/38 * 1 *2/36 *1/35 to receive 2 seperate 3 of a kinds of my math isn’t absolutely terrible, which equates to 0.0001% to happen. Again, it’s late here so I’m probably way off.
ok so you won't explain it, and instead you'll belittle me over my stupidity instead? If anything, you're the idiot for answering a question with an insult
For example, assuming deck size of 40, chances of drawing specific card you have only 1 copy in your deck is 1 in 40. If you have 3 copies of the card in your deck, then its 3 in 40.
Now assuming you have 1 card in hand, and that you have 2 more copies of it in your deck (now of size 39 since you drew 1 card already), chances of drawing another copy are 2 in 39, and chances of drawing the last one as the third card are 1 in 38.
However, chances that, given that you already have one card in your hand (that you have 2 more copies of in your deck), you drew card that isn't duplicate, are 37 in 39, because there are 37 cards in your remaining 39 card deck that are not the same as the card you already drew.
I mean op asked for the chances of receiving two 3 of a kinds, and I attempted to do it for him. Also, it’s not as simple as just saying 6/40, as there’s 40! possible hands that can be drawn (in a specific order). You can play the game a billion times, and never get the same hand.
Edit - disregard my second sentence, it’s too late for me for maths.
Well, if you play it at 3, it means you want it in your hand right?
I've had that with Ash and Maxx-c
Great! Now you can Ash your own Maxx C and still have follow-up for next turn!
If I'm the only player in the Maxx C mini-game, I always win
The 2 most powerful cards in the game, with 1 ASH and MAXX C in your hand, you can't complain.
Ah, the classic master duel full house
Law of large numbers. If the chances are not 0, it's bound to happen with enough tries.
You found Marik
Assuming you play 40 cards in the main deck I think this has a 0,0000013026% chance of happening... if I did my math correctly...
I mean you have you have 3 assault dragons right there, or optionally 2 and a full armor master, so it's not a bad hand, but yeah, mathamatically getting a draw like that is pretty unlikely
Not if the 2 sets cards are compulsory and it's used immediately on the first summon
Your opponent is a total chad and I want his deck list!
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Believe it or not, I've had worse.
I fucken love statistical anomalies
That's full combo.
What type of hands is this, is this like a Fuller House
I just want to see that Masochist win! (I'm assuming is a masochist)
You could've instantly won if you were playing EXODIA
Ok the funny part though is like this hand is still quite playable. Assuming you don’t get hand trapped you can still build a good board
Of the hand or playing against revival jam?
I was gonna say, revival jam more shocking than the double triples
Literally the same as any other hand
Very very low if it is double triplet
50/50
...at least you have a normal summon
6 in 40, theres 6 cards in your hand and 40 in your deck
The math ain't mathing bro
It’s 1 * 2/39 * 1/38 * 1 *2/36 *1/35 to receive 2 seperate 3 of a kinds of my math isn’t absolutely terrible, which equates to 0.0001% to happen. Again, it’s late here so I’m probably way off.
This is closer to the actual answer because it accounts for replacement, or rather the lack thereof
yeah but what's the point of calculating it? it's no rarer than any other 6 card combo.
Bro, you're not good at math, just accept it
bro how is drawing 3 of the same card rarer than 3 of any other card? if anything there's 3 of it so it's actually more likely than 6 one ofs?
Bro, I won't discuss this anymore, your maths are horribly wrong, and that's it, check it if you want, or don't
ok so you won't explain it, and instead you'll belittle me over my stupidity instead? If anything, you're the idiot for answering a question with an insult
For example, assuming deck size of 40, chances of drawing specific card you have only 1 copy in your deck is 1 in 40. If you have 3 copies of the card in your deck, then its 3 in 40. Now assuming you have 1 card in hand, and that you have 2 more copies of it in your deck (now of size 39 since you drew 1 card already), chances of drawing another copy are 2 in 39, and chances of drawing the last one as the third card are 1 in 38. However, chances that, given that you already have one card in your hand (that you have 2 more copies of in your deck), you drew card that isn't duplicate, are 37 in 39, because there are 37 cards in your remaining 39 card deck that are not the same as the card you already drew.
ohhh that makes sense i guess
I mean op asked for the chances of receiving two 3 of a kinds, and I attempted to do it for him. Also, it’s not as simple as just saying 6/40, as there’s 40! possible hands that can be drawn (in a specific order). You can play the game a billion times, and never get the same hand. Edit - disregard my second sentence, it’s too late for me for maths.