I mulliganed 4 cards to start the game (was going second) and free the exact same cards. Has this happened to anyone else before or did it just glitch for me?
4/26 * 3/25 * 2/24 * 1/23 = 24/358800 ~= 0.000067 = 0.0067% so pretty unlikely but is bound to happen to someone eventually (EDIT: So once in every 14950 games where you had 4 cards you run 2 of in your opening hand pre-mulligan - which does not happen all the time if you run single copies of some cards, and the 4 cards shown have to be different of course).
EDIT2: However, given that the OP said he got the same cards back, then the above conditions must have applied to his situation, i.e. he saw 4 different cards that he had 2 copies of each in his deck. Therefore the probability of that occurring is as stated.
4/26 * 3/25 * 2/24 * 1/23 = 24/358800 ~= 0.000067 = 0.0067% so pretty unlikely but is bound to happen to someone eventually.
So if you do the math that's a 1 in 15000 chance. That's 1 or 2 people a day or so.
I think your math might be off though, it should be 4/30, 3/29, 2/28 and 1/27. So if my math is right that would be about a 1/330 mil chance, .00006% If you have two copies of each card it would drop to .0005% though, 1/200k.
It's 4/26 * etc. because the cards you mulligan do not get put into the pool of cards that you can get after the mulligan, you are picking from a pool of 26 cards not 30.
EDIT: It is less than 1 in 14950 though for the reasons explained in my edit, namely:
1) All cards seen pre-mulligan have to be 2 ofs in the deck
2) And each card must be different
You can work out the probability of that occurring too but it depends upon how many 1-ofs you are running.
However, given that the prior assumptions have occurred (you see 4 different cards you run 2 of in your deck and you mulligan them all), then the probability is indeed 1/14950 for that to happen.
It's 4/26 * etc. because the cards you mulligan do not get put into the pool of cards that you can get after the mulligan, you are picking from a pool of 26 cards not 30.
EDIT: It is less than 1 in 14950 though for the reasons explained in my edit, namely:
1) All cards seen pre-mulligan have to be 2 ofs in the deck
2) And each card must be different
You can work out the probability of that occurring too but it depends upon how many 1-ofs you are running.
However, given that the prior assumptions have occurred (you see 4 different cards you run 2 of in your deck and you mulligan them all), then the probability is indeed 1/14950 for that to happen.
Your first assumption is incorrect, I'm afraid. I have shuffled legendaries into the deck and redrew them before, I've also gotten the same card back in Razakus. The cards you mulligan absolutely do get put back into the deck before you draw the new cards
4/26 * 3/25 * 2/24 * 1/23 = 24/358800 ~= 0.000067 = 0.0067% so pretty unlikely but is bound to happen to someone eventually.
So if you do the math that's a 1 in 15000 chance. That's 1 or 2 people a day or so.
I think your math might be off though, it should be 4/30, 3/29, 2/28 and 1/27. So if my math is right that would be about a 1/330 mil chance, .00006% If you have two copies of each card it would drop to .0005% though, 1/200k.
Tldr it's rare but possible.
If the cards mulliganned could be returned immediately in the mulligan (which they can't), then the answer would not be 4/30 * 3/29 * 2/28 * 1/27. That's because you would have to consider any extra copies of each card in the pool of 30. (This would then also depend on the number of copies of each card in the deck). It would only be 4/30 * 3/29 * 2/28 * 1/27 if the cards you mulliganned were all 1-ofs [EDIT: If they were all 2-ofs, the answer would be 8/30 * 6/29 * 4/28 * 2/27]. Anyway, that's hypothetical since it does not apply to Hearthstone mulligans, where you can't immediately get a card back until you actually draw one at the beginning of the first turn.
It's 4/26 * etc. because the cards you mulligan do not get put into the pool of cards that you can get after the mulligan, you are picking from a pool of 26 cards not 30.
EDIT: It is less than 1 in 14950 though for the reasons explained in my edit, namely:
1) All cards seen pre-mulligan have to be 2 ofs in the deck
2) And each card must be different
You can work out the probability of that occurring too but it depends upon how many 1-ofs you are running.
However, given that the prior assumptions have occurred (you see 4 different cards you run 2 of in your deck and you mulligan them all), then the probability is indeed 1/14950 for that to happen.
Your first assumption is incorrect, I'm afraid. I have shuffled legendaries into the deck and redrew them before, I've also gotten the same card back in Razakus. The cards you mulligan absolutely do get put back into the deck before you draw the new cards
That is not correct (and even if it was, your calculation was not right either, since you did't consider the consequences of having up to 2 copies of a card in the card pool, see my post above). [EDIT: I see you did consider the possibility of different answers for having 1 or 2 copies, but I missed that since you didn't show your working. I only did 2 cases in my calculation for that case, all 1-ofs and all 2-ofs, The other cases are simple enough. But, it's hypothetical anyway]
The mulligan, or card selection stage, occurs at the start of each match. Each player is shown their randomly selected starting hand and is given the option to redraw as many of those cards as they like. Only one redraw of the selected cards is allowed; there is no chance to redraw again after seeing the results. Note that cards mulliganed away are not added back into the pool for possible replacements.[1]
As Ben points out, you can immediately draw a card you mulligan away on turn 1, or you can be given back a copy of a card you mulliganned but have 2 of in your deck.
This happened to me more than once and the last time it was with Patches so a pretty memorable event but I don't ever recall getting cards in the same order as before. When it occurs to you go and play some Lotto or whatever your national lottery is called where you live.
No it didn't :P
What happened is you mulliganned Patches and drew him on turn 1. [Probability this happens is 1/27 if going first, 1/26 if going 2nd]. It is not possible to get Patches in your opening hand [EDIT: before you draw a card at the beginning of your first turn] if you mulligan him. You can only get a card back if you have 2 copies in your deck, you get the other copy.
EDIT: Order the cards appear is irrelevant to probability calculations, just whether they appear or not.
This happened to me more than once and the last time it was with Patches so a pretty memorable event but I don't ever recall getting cards in the same order as before. When it occurs to you go and play some Lotto or whatever your national lottery is called where you live.
No it didn't :P
What happened is you mulliganned Patches and drew him on turn 1. [Probability this happens is 1/27 if going first, 1/26 if going 2nd]. It is not possible to get Patches in your opening hand [EDIT: before you draw a card at the beginning of your first turn] if you mulligan him. You can only get a card back if you have 2 copies in your deck, you get the other copy.
EDIT: Order the cards appear is irrelevant to probability calculations, just whether they appear or not.
@edit: the order is surely not irrelevant if you make it a condition. I.e. what is the probability to draw the same cards in the exact same order? Should be 1/27 *1/26 *1/25 going first.
Well yeah, it would be relevant if that was stated as a condition, and your calculation would be correct in that case, although the OP was going 2nd. Maybe that's what the OP meant when he said "exact same cards", since exact seems unnecessary there.
EDIT: The factor involved [difference between the 2 cases] is r! where r is number of cards, since it's the factor that is the difference between (n choose r) = n!/(r! * (n-r)!) [number of ways to choose r objects from a set of n] and (n permute r) = n!/(n-r)! [number of ways to arrange r objects in order from a set of n]. Reasons: Pascal's Triangle, in which the r-th entry in the n-th row is (n choose r). There are many interesting things about Pascal's triangle, try colouring in the odd numbers in the triangle and removing the even ones and see what pattern emerges ;)
It's 4/26 * etc. because the cards you mulligan do not get put into the pool of cards that you can get after the mulligan, you are picking from a pool of 26 cards not 30.
This is demonstrably incorrect (unless I've misunderstood your argument), since often when you mulligan away a legendary, that same legendary comes back to your hand. Which implies to me that all cards that you mulligan go back into the pool to draw from.
It does not come back to your hand in the mulligan phase, it comes back the first turn when you draw the first card. The cards you mulligan are added to the deck once the replacements are received.
In terms of what would happen in a real life card game, it works like this:
Look at your opening hand. Place cards you don't want to one side and then draw the same number from the deck.
Then, shuffle the cards you set aside into the deck.
Begin turn 1 and immediately draw a card.
If as you say it is often the case you throw back a legendary and get it back, it should be simple to find a video of this occurring.
EDIT: MtG does not work like that. In MtG you shuffle your entire hand back into the deck and draw a new one (with one less card unless you had 0 lands or all lands I think).
It's 4/26 * etc. because the cards you mulligan do not get put into the pool of cards that you can get after the mulligan, you are picking from a pool of 26 cards not 30.
This is demonstrably incorrect (unless I've misunderstood your argument), since often when you mulligan away a legendary, that same legendary comes back to your hand. Which implies to me that all cards that you mulligan go back into the pool to draw from.
Nope you're wrong.
When you mulligan away a card, you cannot draw the exact same card again because it doesn't go in to the pile of cards you can draw. This is very important to know because as soon as you know this, then you also know that you can mulligan a legendary card away and not draw it.
People who claim to have mulliganed a legendary card only to draw it again during the mulligan are either lying or didn't pay attention to the game. They most likely drew the legendary during their first draw on turn 1. Something that is very, very likely to happen to everyone on a monthly basis.
EDIT: MtG does not work like that. In MtG you shuffle your entire hand back into the deck and draw a new one (with one less card unless you had 0 lands or all lands I think).
An old myth that were never true. Some casual players made up their own rules but it was never part of the actual game.
In MTG you could NEVER reveal your hand to get to draw an entire new hand of the same number of cards simply because it had 100 % lands or 0 % lands. It's simply not true and never were. Of course people can make their own rules in their own houses but that has nothing to do with Magic: The Gathering.
It does not come back to your hand in the mulligan phase, it comes back the first turn when you draw the first card. The cards you mulligan are added to the deck once the replacements are received.
In terms of what would happen in a real life card game, it works like this:
Look at your opening hand. Place cards you don't want to one side and then draw the same number from the deck.
Then, shuffle the cards you set aside into the deck.
Begin turn 1 and immediately draw a card.
If as you say it is often the case you throw back a legendary and get it back, it should be simple to find a video of this occurring.
EDIT: MtG does not work like that. In MtG you shuffle your entire hand back into the deck and draw a new one (with one less card unless you had 0 lands or all lands I think).
For everyone other than mungo2613: I am certain this is how it works. Having played since beta I have never once mulligan-ed a unique card (be it a legendary or a card of which I have 1 golden and 1 non-golden copy in the deck) and got it back in my opening hand. Sure I have drawn it on turn 1 but that is an entirely separate issue.
It does not come back to your hand in the mulligan phase, it comes back the first turn when you draw the first card. The cards you mulligan are added to the deck once the replacements are received.
In terms of what would happen in a real life card game, it works like this:
Look at your opening hand. Place cards you don't want to one side and then draw the same number from the deck.
Then, shuffle the cards you set aside into the deck.
Begin turn 1 and immediately draw a card.
If as you say it is often the case you throw back a legendary and get it back, it should be simple to find a video of this occurring.
EDIT: MtG does not work like that. In MtG you shuffle your entire hand back into the deck and draw a new one (with one less card unless you had 0 lands or all lands I think).
For everyone other than mungo2613: I am certain this is how it works. Having played since beta I have never once mulligan-ed a unique card (be it a legendary or a card of which I have 1 golden and 1 non-golden copy in the deck) and got it back in my opening hand. Sure I have drawn it on turn 1 but that is an entirely separate issue.
If I'm wrong then I am wrong, but I am pretty sure I have often received the card I have mulliganed away. Obviously it's not the sort of thing I normally record, though, so I will do some science....
Speed up the scientific process by making a highlander deck (all 1-ofs). Always mulligan everything. Play the innkeeper and concede after the mulligan.
If the cards were available in the mulligan pool, then if you go 1st the chance to get one of them back would then be 3/30 or 10%. Going 2nd and mulligan everything the chance is 4/30 ~= 13.33%. Those odds are pretty good.
I suspect you'll be waiting a long time for a positive result though ;)
What will happen though is you draw one of the cards you threw away turn 1. Since you threw them all away the probability of getting one back immediately is 3/27 = 1/9 ~= 11.11% when going first or 4/26 ~= 15.38% when going 2nd.
I mulliganed 4 cards to start the game (was going second) and free the exact same cards. Has this happened to anyone else before or did it just glitch for me?
DREW the same 4 cards
Probability of that occurring is
4/26 * 3/25 * 2/24 * 1/23 = 24/358800 ~= 0.000067 = 0.0067% so pretty unlikely but is bound to happen to someone eventually (EDIT: So once in every 14950 games where you had 4 cards you run 2 of in your opening hand pre-mulligan - which does not happen all the time if you run single copies of some cards, and the 4 cards shown have to be different of course).
EDIT2: However, given that the OP said he got the same cards back, then the above conditions must have applied to his situation, i.e. he saw 4 different cards that he had 2 copies of each in his deck. Therefore the probability of that occurring is as stated.
It's 4/26 * etc. because the cards you mulligan do not get put into the pool of cards that you can get after the mulligan, you are picking from a pool of 26 cards not 30.
EDIT: It is less than 1 in 14950 though for the reasons explained in my edit, namely:
1) All cards seen pre-mulligan have to be 2 ofs in the deck
2) And each card must be different
You can work out the probability of that occurring too but it depends upon how many 1-ofs you are running.
However, given that the prior assumptions have occurred (you see 4 different cards you run 2 of in your deck and you mulligan them all), then the probability is indeed 1/14950 for that to happen.
It's about 5 times more likely to happen going first than going second,
3/27 * 2/26 * 1/25 = 6/17550 = 1/2995
Again, this assumes you see 3 different cards you run as 2-ofs in the mulligan.
(EDIT: Said 4.8 not 5 by mistake originally was using 14500 not 14950 for some reason. 14950/2995 ~= 4.99)
Well yeah, it would be relevant if that was stated as a condition, and your calculation would be correct in that case, although the OP was going 2nd. Maybe that's what the OP meant when he said "exact same cards", since exact seems unnecessary there.
EDIT: The factor involved [difference between the 2 cases] is r! where r is number of cards, since it's the factor that is the difference between (n choose r) = n!/(r! * (n-r)!) [number of ways to choose r objects from a set of n] and (n permute r) = n!/(n-r)! [number of ways to arrange r objects in order from a set of n]. Reasons: Pascal's Triangle, in which the r-th entry in the n-th row is (n choose r). There are many interesting things about Pascal's triangle, try colouring in the odd numbers in the triangle and removing the even ones and see what pattern emerges ;)
Which implies to me that all cards that you mulligan go back into the pool to draw from.
It does not come back to your hand in the mulligan phase, it comes back the first turn when you draw the first card. The cards you mulligan are added to the deck once the replacements are received.
In terms of what would happen in a real life card game, it works like this:
Look at your opening hand. Place cards you don't want to one side and then draw the same number from the deck.
Then, shuffle the cards you set aside into the deck.
Begin turn 1 and immediately draw a card.
If as you say it is often the case you throw back a legendary and get it back, it should be simple to find a video of this occurring.
EDIT: MtG does not work like that. In MtG you shuffle your entire hand back into the deck and draw a new one (with one less card unless you had 0 lands or all lands I think).
Speed up the scientific process by making a highlander deck (all 1-ofs). Always mulligan everything. Play the innkeeper and concede after the mulligan.
If the cards were available in the mulligan pool, then if you go 1st the chance to get one of them back would then be 3/30 or 10%. Going 2nd and mulligan everything the chance is 4/30 ~= 13.33%. Those odds are pretty good.
I suspect you'll be waiting a long time for a positive result though ;)
What will happen though is you draw one of the cards you threw away turn 1. Since you threw them all away the probability of getting one back immediately is 3/27 = 1/9 ~= 11.11% when going first or 4/26 ~= 15.38% when going 2nd.
EDIT Never mind. I am wrong on this one I think.