Trump said that DK Rexxar was his "best card" of the set, if only it wasn't in Hunter. Can see that logic, as it's a card that can only get help the more control style cards are made.
Forgot for a few seconds that there was the streamer with that name. Was trying to remember news about a certain political figure tweeting re: Hearthstone...my brain shorted out. THANKS Obama MNC Dover.
You just know THAT Trump would play Jade Druid.
I'm pretty sure the Pres would play an all-Legendaries deck. Ineffective, not viable, somewhat nonsensical, no coherent strategy, no synergy... but flashy and expensive.
(Edit: For the record, playing an all-Legendary deck is a completely unviable strategy that everyone should try at least once. It's hilarious and I kinda want to do it this week. Haven't done it in a couple of years.)
And he would have a 100% win record. Any reports otherwise are false news.
In REAL Hearthstone discussion, I love playing completely horribly crap decks around rank 20 for the "Play X Class Cards" quests. Rarely win, but when you do pull off your 8 card combo it's hilarious.
Do you, in fact, have any builds in this shop at all?
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Nope, you've just been getting lucky. The only starting hand influence is the 100% Un'Goro quest card inclusion.
Trump said that DK Rexxar was his "best card" of the set, if only it wasn't in Hunter. Can see that logic, as it's a card that can only get help the more control style cards are made.
Forgot for a few seconds that there was the streamer with that name. Was trying to remember news about a certain political figure tweeting re: Hearthstone...my brain shorted out. THANKS Obama MNC Dover.
You just know THAT Trump would play Jade Druid.
I'm pretty sure the Pres would play an all-Legendaries deck. Ineffective, not viable, somewhat nonsensical, no coherent strategy, no synergy... but flashy and expensive.
(Edit: For the record, playing an all-Legendary deck is a completely unviable strategy that everyone should try at least once. It's hilarious and I kinda want to do it this week. Haven't done it in a couple of years.)
And he would have a 100% win record. Any reports otherwise are false news.
In REAL Hearthstone discussion, I love playing completely horribly crap decks around rank 20 for the "Play X Class Cards" quests. Rarely win, but when you do pull off your 8 card combo it's hilarious.
I had to pull out murlocs and priest cards this weekend so made up a crappy Shaman Murloc deck. Actually beat a Jade Druid since he had no idea what I was doing. Got my quest done in two games. Then played my Big Fatties Priest deck for the other quest. Lost to a Jade Druid, Mage and then pounded a poor baby seal playing a hodge podge Shaman deck. End of the month is really weird.
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
I'm really late to this party but I started playing last week. Lots of fun so far. Being able to play a game I'm really engaged with on my PC or phone interchangeably is pretty novel.
Right now I'm trying to grind up the mage class to level 20 and doing ranked at the same time (to get end of month rewards). I was going 50/50 in ranked until I looked up a starter net deck and I won 3 or 4 in a row last night.
Probably my highlight of this game so far was beating some gimmicky looking deck a rogue was running last night. They played a bunch of cards in one turn and then played a minion that powered up based on how many cards you had already played that turn. It was an impressive 13/13 unit after this trick.
I immediately polymorphed it.
+16
lwt1973King of ThievesSyndicationRegistered Userregular
I'm really late to this party but I started playing last week. Lots of fun so far. Being able to play a game I'm really engaged with on my PC or phone interchangeably is pretty novel.
Right now I'm trying to grind up the mage class to level 20 and doing ranked at the same time (to get end of month rewards). I was going 50/50 in ranked until I looked up a starter net deck and I won 3 or 4 in a row last night.
Probably my highlight of this game so far was beating some gimmicky looking deck a rogue was running last night. They played a bunch of cards in one turn and then played a minion that powered up based on how many cards you had already played that turn. It was an impressive 13/13 unit after this trick.
I immediately polymorphed it.
The next step in your growth is for him to friend you to rage about RNG. Be sure to screen shot it and post it here.
Today I had a Pirate Warrior get mad because I was playing Jade and I had no skill.
"He's sulking in his tent like Achilles! It's the Iliad?...from Homer?! READ A BOOK!!" -Handy
I'm really late to this party but I started playing last week. Lots of fun so far. Being able to play a game I'm really engaged with on my PC or phone interchangeably is pretty novel.
Right now I'm trying to grind up the mage class to level 20 and doing ranked at the same time (to get end of month rewards). I was going 50/50 in ranked until I looked up a starter net deck and I won 3 or 4 in a row last night.
Probably my highlight of this game so far was beating some gimmicky looking deck a rogue was running last night. They played a bunch of cards in one turn and then played a minion that powered up based on how many cards you had already played that turn. It was an impressive 13/13 unit after this trick.
I immediately polymorphed it.
The next step in your growth is for him to friend you to rage about RNG. Be sure to screen shot it and post it here.
Today I had a Pirate Warrior get mad because I was playing Jade and I had no skill.
You don't :P
So i was getting wrecked by exodia mage and murloc paladin as my hunter deck. Teched in flare, and it is so satisfying flaring away iceblock + icebarrier, or getting the getaway kodo for their Tirion. It has also randomly helped against rng minions people got from random cards with stealth. it is kind of dumb that if i don't have 3 minions out on the first turn that i literally have no board. When alley cat rotates hunter is going to be super screwed unless it gets something crazy.
I tried N'zoth pally today because I unlocked Arthas and had a win 5 quest. I beat TWO exodia mages and a jade druid with Eye for an eye from Hydrologist! That druid will never again emote Well played attack with his highest attack minion first instead of his lowest :P. Well played in deed...
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Isn't that an oversimplification tacking the first draw on to the sample size to make 7? Your first draw is 1 from the population of 27 cards that are not in your hand after the mulligan, which includes the 3 cards pitched (or not) from your mulligan.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
3 fucking deathgrips and only one combo piece drops out of this exodia mage. And its a friggin apprentice. And of course he gets a simulacrum from glyph.
+1
Tynnanseldom correct, never unsureRegistered Userregular
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Isn't that an oversimplification tacking the first draw on to the sample size to make 7? Your first draw is 1 from the population of 27 cards that are not in your hand after the mulligan, which includes the 3 cards pitched (or not) from your mulligan.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
If you'd like to figure out the precisely correct model for this and present your work, go for it.
It's a simplification because mulligans interact weirdly with the calculation, but doing incremental probabilities for each new deck size doesn't gain you a significant amount of accuracy.
My legends this expansion: Liliana Voss, Sindragosa, Archbishop Benedictus, Malfurion (from Lich king intro)
The pack I got from beating the Lich King, I rolled over and saw two legendaries got super hype thinking I'd finally get some DK heroes that I'd actually enjoy playing (Anduin, Jaina, Rexxar, even Gul'dan).
Rotface and Blood-Queen Lana'thel.
Womp Womp. I then impulse crafted Jaina and threw Sindragosa in the deck so at least I could play with one of the legends I pulled.
I'm sooooooooo tired of shadow essence summoning Barnes. I think I'm 3 for 3 with it today. Gonna add another minion or two to the deck to dilute the chances of it happening again.
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Isn't that an oversimplification tacking the first draw on to the sample size to make 7? Your first draw is 1 from the population of 27 cards that are not in your hand after the mulligan, which includes the 3 cards pitched (or not) from your mulligan.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
They find that going first, mulliganing for a specific 2-of card you'll have a 41.25% chance of seeing at least one copy on the first turn, which is indistinguishable from the hypergeometric calculator result of 41.8%.
My legends this expansion: Liliana Voss, Sindragosa, Archbishop Benedictus, Malfurion (from Lich king intro)
The pack I got from beating the Lich King, I rolled over and saw two legendaries got super hype thinking I'd finally get some DK heroes that I'd actually enjoy playing (Anduin, Jaina, Rexxar, even Gul'dan).
Rotface and Blood-Queen Lana'thel.
Womp Womp. I then impulse crafted Jaina and threw Sindragosa in the deck so at least I could play with one of the legends I pulled.
Oh hey, I also pulled Sindragosa and Voss on expansion day.
And I've heard Rotface is fun.
Currently Playing:
The Division, Warframe (XB1)
GT: Tanith 6227
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Isn't that an oversimplification tacking the first draw on to the sample size to make 7? Your first draw is 1 from the population of 27 cards that are not in your hand after the mulligan, which includes the 3 cards pitched (or not) from your mulligan.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
They find that going first, mulliganing for a specific 2-of card you'll have a 41.25% chance of seeing at least one copy on the first turn, which is indistinguishable from the hypergeometric calculator result of 41.8%.
Thanks for checking into that. It still basically tells us that 3 games out 5, that pirate warrior shouldn't have N'zoth's First Mate. I wish he had also computed the odds given that it's not in your opening mulligan, since as the person making the mulligan decision, you really want to know how bad or good of an idea it is to go all-in with your mulligan.
It's also probably worth noting that the original post this discussion spawned from mentioned only having one of the card that kept showing up, so all these 2-of calculations are the wrong ones to apply.
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Isn't that an oversimplification tacking the first draw on to the sample size to make 7? Your first draw is 1 from the population of 27 cards that are not in your hand after the mulligan, which includes the 3 cards pitched (or not) from your mulligan.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
They find that going first, mulliganing for a specific 2-of card you'll have a 41.25% chance of seeing at least one copy on the first turn, which is indistinguishable from the hypergeometric calculator result of 41.8%.
Thanks for checking into that. It still basically tells us that 3 games out 5, that pirate warrior shouldn't have N'zoth's First Mate. I wish he had also computed the odds given that it's not in your opening mulligan, since as the person making the mulligan decision, you really want to know how bad or good of an idea it is to go all-in with your mulligan.
It's also probably worth noting that the original post this discussion spawned from mentioned only having one of the card that kept showing up, so all these 2-of calculations are the wrong ones to apply.
Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Actually that term refers to the fact that the deck contains two copies of the card you're trying to find. The "success condition" output here is for the combined set of hands that have any number of copies of the target card, whether that's one or two. It's not worth calculating the chance of getting exactly one copy because that's not how mulligan decision-making works. When you're aggressively mulliganing for a specific card, you're accepting the risk that you get both copies or you don't consider that to be a risk.
And I'm not sure I follow your complaint about whether the card is in your initial mulligan. The calculations in that link properly account for that possibility. Essentially they take the total likelihood that you fail to find the card and work backwards, so part of the result is the set of hands that included the correct card in the initial draw.
Edit: sorry if I'm coming off as overly blunt here, it's not my intention to be rude. I am speaking a bit forcefully here because I think mulliganing is one of the most interesting phases of the game and it's where so many games are won or lost. And because it's all math based I like to make sure the correct information is getting conveyed.
Edit edit: just saw your addendum about the fact that only one macaw was in that deck. My bad. That observation is low sample size and confirmation bias (remembering the times it was there, forgetting the games it wasn't). I think discussing probabilities for two-ofs is generally more relevant for most decks.
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Isn't that an oversimplification tacking the first draw on to the sample size to make 7? Your first draw is 1 from the population of 27 cards that are not in your hand after the mulligan, which includes the 3 cards pitched (or not) from your mulligan.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
They find that going first, mulliganing for a specific 2-of card you'll have a 41.25% chance of seeing at least one copy on the first turn, which is indistinguishable from the hypergeometric calculator result of 41.8%.
Thanks for checking into that. It still basically tells us that 3 games out 5, that pirate warrior shouldn't have N'zoth's First Mate. I wish he had also computed the odds given that it's not in your opening mulligan, since as the person making the mulligan decision, you really want to know how bad or good of an idea it is to go all-in with your mulligan.
It's also probably worth noting that the original post this discussion spawned from mentioned only having one of the card that kept showing up, so all these 2-of calculations are the wrong ones to apply.
Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Actually that term refers to the fact that the deck contains two copies of the card you're trying to find. The "success condition" output here is for the combined set of hands that have any number of copies of the target card, whether that's one or two. It's not worth calculating the chance of getting exactly one copy because that's not how mulligan decision-making works. When you're aggressively mulliganing for a specific card, you're accepting the risk that you get both copies or you don't consider that to be a risk.
And I'm not sure I follow your complaint about whether the card is in your initial mulligan. The calculations in that link properly account for that possibility. Essentially they take the total likelihood that you fail to find the card and work backwards, so part of the result is the set of hands that included the correct card in the initial draw.
Ummm, read the post I quoted again. The poster literally said he has one Macaw in his deck. So computing draw probabilities for that should be done of with the 1-of charts.
Whether your card is in your initial mulligan is of critical importance. If it's in it, boom, you won. If it's not, you probably want to know what your odds of getting it are from mulliganing 1, 2, 3, 4 cards. It doesn't help you a lot to know what your odds were of getting it pre AND post-mulligan, because when it's not in your mulligan, you want to know what the consequences of your next actions will be.
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Isn't that an oversimplification tacking the first draw on to the sample size to make 7? Your first draw is 1 from the population of 27 cards that are not in your hand after the mulligan, which includes the 3 cards pitched (or not) from your mulligan.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
They find that going first, mulliganing for a specific 2-of card you'll have a 41.25% chance of seeing at least one copy on the first turn, which is indistinguishable from the hypergeometric calculator result of 41.8%.
Thanks for checking into that. It still basically tells us that 3 games out 5, that pirate warrior shouldn't have N'zoth's First Mate. I wish he had also computed the odds given that it's not in your opening mulligan, since as the person making the mulligan decision, you really want to know how bad or good of an idea it is to go all-in with your mulligan.
It's also probably worth noting that the original post this discussion spawned from mentioned only having one of the card that kept showing up, so all these 2-of calculations are the wrong ones to apply.
Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Actually that term refers to the fact that the deck contains two copies of the card you're trying to find. The "success condition" output here is for the combined set of hands that have any number of copies of the target card, whether that's one or two. It's not worth calculating the chance of getting exactly one copy because that's not how mulligan decision-making works. When you're aggressively mulliganing for a specific card, you're accepting the risk that you get both copies or you don't consider that to be a risk.
And I'm not sure I follow your complaint about whether the card is in your initial mulligan. The calculations in that link properly account for that possibility. Essentially they take the total likelihood that you fail to find the card and work backwards, so part of the result is the set of hands that included the correct card in the initial draw.
Ummm, read the post I quoted again. The poster literally said he has one Macaw in his deck. So computing draw probabilities for that should be done of with the 1-of charts.
Whether your card is in your initial mulligan is of critical importance. If it's in it, boom, you won. If it's not, you probably want to know what your odds of getting it are from mulliganing 1, 2, 3, 4 cards. It doesn't help you a lot to know what your odds were of getting it pre AND post-mulligan, because when it's not in your mulligan, you want to know what the consequences of your next actions will be.
I did miss the fact that there was only one macaw in that deck (see the edit I was making as you wrote your post), so that's my bad. But you're still missing that these values already include the possibility you're raising.
The math for singletons is easier than for two-ofs so I'll quote the r/competitvehs author for why this is accounted for:
Case 1 - The card is legendary, or you just have 1 copy of it (ahem, Reno Jackson and Acidic Swamp Ooze)
If you are OFF THE COIN you can just draw 3 cards in your initial hand; there's a 90% chance (27/30) that the card you are looking for is not there. If you mulligan for it again, returning all 3 cards to your deck, your odds of not drawing the card you are looking for are 88.9% (24/27).
Overall, there's an 80% chance you don't hit the card off of the mulligan.
With each successive card drawn, the odds of you not having drawn that card drop by 2.926% (80%/27). Overall, we have the following table
Skulking Geist actually managed to win me a game against Jade Druid. That's a first.
I only really put it in my Paladin deck to shore up my games against Evolve Shaman. (It doesn't really weaken my match-ups elsewhere).
Skulking Geist can help the decks that are prepared to handle Jades up to 9/9 in size or so. It can also help burn decks which might lose to Earthen Scales. It's definitely not a "beat Jade Druid for free card" if you don't fall into those categories. Although one slightly more subtle benefit in any matchup where it hits cards... it can remove cards from your opponent's hand, which is basically a discard effect which is much stronger than just removing from the deck.
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Isn't that an oversimplification tacking the first draw on to the sample size to make 7? Your first draw is 1 from the population of 27 cards that are not in your hand after the mulligan, which includes the 3 cards pitched (or not) from your mulligan.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
They find that going first, mulliganing for a specific 2-of card you'll have a 41.25% chance of seeing at least one copy on the first turn, which is indistinguishable from the hypergeometric calculator result of 41.8%.
I don't know what the hypergeometric calculator is doing, but actually calculating the probability isn't that hard. It's just a basic discrete math problem.
The odds of having at least one copy of a card in your opening hand before mulligan as player 1 is 1 - ((28 choose 3)/(30 choose 3)) = 1 - 3276/4060 = 19.31%
(Basically you are finding 1 - (the number of ways to draw 3 of the 28 other cards)/(the number of ways to draw 3 cards from 30))
You have a 80.69% to not get the card before the mulligan. If you mulligan all three you then have another 19.31% to get the card. This will happen .8069 * 19.31% = 15.58% of the time.
So you are at 15.58% + 19.31% = 34.89% chance to see it before drawing your first card.
In the 65.11% of times you still haven't seen it, you have a 2/27 chance of drawing it. 2/27 * 65.11% = 4.82%
34.89% + 4.82% = 39.71%
So you have a 39.71% to see a card on the first turn if you aggressively mulligan for it as player 1.
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Isn't that an oversimplification tacking the first draw on to the sample size to make 7? Your first draw is 1 from the population of 27 cards that are not in your hand after the mulligan, which includes the 3 cards pitched (or not) from your mulligan.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
They find that going first, mulliganing for a specific 2-of card you'll have a 41.25% chance of seeing at least one copy on the first turn, which is indistinguishable from the hypergeometric calculator result of 41.8%.
Thanks for checking into that. It still basically tells us that 3 games out 5, that pirate warrior shouldn't have N'zoth's First Mate. I wish he had also computed the odds given that it's not in your opening mulligan, since as the person making the mulligan decision, you really want to know how bad or good of an idea it is to go all-in with your mulligan.
It's also probably worth noting that the original post this discussion spawned from mentioned only having one of the card that kept showing up, so all these 2-of calculations are the wrong ones to apply.
Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Actually that term refers to the fact that the deck contains two copies of the card you're trying to find. The "success condition" output here is for the combined set of hands that have any number of copies of the target card, whether that's one or two. It's not worth calculating the chance of getting exactly one copy because that's not how mulligan decision-making works. When you're aggressively mulliganing for a specific card, you're accepting the risk that you get both copies or you don't consider that to be a risk.
And I'm not sure I follow your complaint about whether the card is in your initial mulligan. The calculations in that link properly account for that possibility. Essentially they take the total likelihood that you fail to find the card and work backwards, so part of the result is the set of hands that included the correct card in the initial draw.
Ummm, read the post I quoted again. The poster literally said he has one Macaw in his deck. So computing draw probabilities for that should be done of with the 1-of charts.
Whether your card is in your initial mulligan is of critical importance. If it's in it, boom, you won. If it's not, you probably want to know what your odds of getting it are from mulliganing 1, 2, 3, 4 cards. It doesn't help you a lot to know what your odds were of getting it pre AND post-mulligan, because when it's not in your mulligan, you want to know what the consequences of your next actions will be.
I did miss the fact that there was only one macaw in that deck (see the edit I was making as you wrote your post), so that's my bad. But you're still missing that these values already include the possibility you're raising.
The math for singletons is easier than for two-ofs so I'll quote the r/competitvehs author for why this is accounted for:
Case 1 - The card is legendary, or you just have 1 copy of it (ahem, Reno Jackson and Acidic Swamp Ooze)
If you are OFF THE COIN you can just draw 3 cards in your initial hand; there's a 90% chance (27/30) that the card you are looking for is not there. If you mulligan for it again, returning all 3 cards to your deck, your odds of not drawing the card you are looking for are 88.9% (24/27).
Overall, there's an 80% chance you don't hit the card off of the mulligan.
With each successive card drawn, the odds of you not having drawn that card drop by 2.926% (80%/27). Overall, we have the following table
Yes, at that one point in his post he did a single computation for post-mulligan odds (the 88.9%). However, that's not what all the charts are based on. The odds in the charts include the possibility of the first 3/4 cards having the desired card, which isn't very helpful for mulligan decision making.
Edit: As a comparison, someone flipping a coin twice in a row has a 75% chance to get heads at least once. It would be reasonable to bet even money that you will flip heads at least once with two flips. However, if your first flip ends up tails, there is now a 50% chance that your next flip will be heads. If (in this nonsensical and heavily biased in your favor game) you were allowed to withdraw your bet after this first flip or go on with it, the realization that at this point forward your chance to win is only 50% would probably influence your willingness to maintain vs. withdraw your bet, as the odds are no longer in your favor.
The math in the thread I just linked is correct. Most of the numbers I am seeing here are not correct. I would be happy to explain how to calculate the probability if anyone is interested. I did a quick search on hearthstone probabilities and I saw a lot of bad math out there.
You should be getting a 34.89% to chance to see a card if running 2 prior to your first draw when you hard mulligan as player 1. Then you have a 2/27, 2/26, 2/25 ... chance to draw the card after that.
As player 2 you 44.18% chance to see the card prior to your first draw if you hard mulligan and then a 2/26,2/25,2/24 chance to draw the card with each successive draw.
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Tynnanseldom correct, never unsureRegistered Userregular
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Isn't that an oversimplification tacking the first draw on to the sample size to make 7? Your first draw is 1 from the population of 27 cards that are not in your hand after the mulligan, which includes the 3 cards pitched (or not) from your mulligan.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
They find that going first, mulliganing for a specific 2-of card you'll have a 41.25% chance of seeing at least one copy on the first turn, which is indistinguishable from the hypergeometric calculator result of 41.8%.
I don't know what the hypergeometric calculator is doing, but actually calculating the probability isn't that hard. It's just a basic discrete math problem.
The odds of having at least one copy of a card in your opening hand before mulligan as player 1 is 1 - ((28 choose 3)/(30 choose 3)) = 1 - 3276/4060 = 19.31%
(Basically you are finding 1 - (the number of ways to draw 3 of the 28 other cards)/(the number of ways to draw 3 cards from 30))
You have a 80.69% to not get the card before the mulligan. If you mulligan all three you then have another 19.31% to get the card. This will happen .8069 * 19.31% = 15.58% of the time.
So you are at 15.58% + 19.31% = 34.89% chance to see it before drawing your first card.
In the 65.11% of times you still haven't seen it, you have a 2/27 chance of drawing it. 2/27 * 65.11% = 4.82%
34.89% + 4.82% = 39.71%
So you have a 39.71% to see a card on the first turn if you aggressively mulligan for it as player 1.
The bolded assumes independence between the initial hand and the next cards drawn off the mulligan. Cards rejected in the mulligan are not eligible to be drawn in the second hand (the earliest you'd see any of those specific cards again is on the first turn draw), so this calculation slightly underestimates the probability of seeing the target card.
now has anyone ever run the numbers on whether these probabilities are borne out by the in-game data? i'm not saying blizzard are fudging the numbers, but it's possible. it'd be very easy for them to calculate what represents a good or bad opening hand for any given deck and, er, encourage it to happen or not happen in certain circumstances. as we see with things like the pity timer - which i believe took a lot of painstaking research by the playerbase to discover - sometimes random is just too random for a tightly crafted and highly addictive player experience
on the other hand, the game's reputation would probably be destroyed if something like this was revealed. would such shenanigans ever be worth the risk? probably not. still, unless we know for sure...
I don't think it'd be very easy for blizzard to calculate what represents a good or bad opening hand for a deck at all, especially considering that what constitutes a good or bad hand is matchup-dependent
Speaking of tweaking the odds... does the game make it more likely that you'll have certain cards in your starting hand? Because I've got a Hunter deck with one Jeweled Macaw and I nearly always have it in my starting hand.
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Isn't that an oversimplification tacking the first draw on to the sample size to make 7? Your first draw is 1 from the population of 27 cards that are not in your hand after the mulligan, which includes the 3 cards pitched (or not) from your mulligan.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
They find that going first, mulliganing for a specific 2-of card you'll have a 41.25% chance of seeing at least one copy on the first turn, which is indistinguishable from the hypergeometric calculator result of 41.8%.
I don't know what the hypergeometric calculator is doing, but actually calculating the probability isn't that hard. It's just a basic discrete math problem.
The odds of having at least one copy of a card in your opening hand before mulligan as player 1 is 1 - ((28 choose 3)/(30 choose 3)) = 1 - 3276/4060 = 19.31%
(Basically you are finding 1 - (the number of ways to draw 3 of the 28 other cards)/(the number of ways to draw 3 cards from 30))
You have a 80.69% to not get the card before the mulligan. If you mulligan all three you then have another 19.31% to get the card. This will happen .8069 * 19.31% = 15.58% of the time.
So you are at 15.58% + 19.31% = 34.89% chance to see it before drawing your first card.
In the 65.11% of times you still haven't seen it, you have a 2/27 chance of drawing it. 2/27 * 65.11% = 4.82%
34.89% + 4.82% = 39.71%
So you have a 39.71% to see a card on the first turn if you aggressively mulligan for it as player 1.
The bolded assumes independence between the initial hand and the next cards drawn off the mulligan. Cards rejected in the mulligan are not eligible to be drawn in the second hand (the earliest you'd see any of those specific cards again is on the first turn draw), so this calculation slightly underestimates the probability of seeing the target card.
Oooooh good catch! That means you have a 1 - ((25 choose 3)/(27 choose 3) chance with the hard mulligan to draw a copy running 2.
That means you have a 36.55% + 4.82% chance on turn 1 as player 1. So 41.37%
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HADRONOX.
*womp womp*
Whoa, an epic in the same pack??
FATESPINNER
Oh, come on!
The Division, Warframe (XB1)
GT: Tanith 6227
And he would have a 100% win record. Any reports otherwise are false news.
In REAL Hearthstone discussion, I love playing completely horribly crap decks around rank 20 for the "Play X Class Cards" quests. Rarely win, but when you do pull off your 8 card combo it's hilarious.
I had to pull out murlocs and priest cards this weekend so made up a crappy Shaman Murloc deck. Actually beat a Jade Druid since he had no idea what I was doing. Got my quest done in two games. Then played my Big Fatties Priest deck for the other quest. Lost to a Jade Druid, Mage and then pounded a poor baby seal playing a hodge podge Shaman deck. End of the month is really weird.
Steam: betsuni7
Drawing cards from a random deck can be modeled by a hypergeometric distribution, so we can calculate the probability that you'll have any given card in your opening hand (plus mulligan). For seeing one copy of a two-of in your deck of 30 cards, the chance of seeing it in your opening hand (on the play, full mulligan) comes out to 42% (bottom output row).
Population size = deck size = 30;
number of successes = 2 since it's a two-of;
sample size = 7 (3 looks at first mulligan, 3 new cards from full mulligan, 1 card drawn for first turn);
success criterion = 1 or 2 macaws in opening hand (X >= 1 success).
So then your likelihood of having a string of successful Jeweled Macaw hands is pretty similar to your likelihood of a string of the same coin flip result. That's pretty common.
Right now I'm trying to grind up the mage class to level 20 and doing ranked at the same time (to get end of month rewards). I was going 50/50 in ranked until I looked up a starter net deck and I won 3 or 4 in a row last night.
Probably my highlight of this game so far was beating some gimmicky looking deck a rogue was running last night. They played a bunch of cards in one turn and then played a minion that powered up based on how many cards you had already played that turn. It was an impressive 13/13 unit after this trick.
I immediately polymorphed it.
The next step in your growth is for him to friend you to rage about RNG. Be sure to screen shot it and post it here.
Today I had a Pirate Warrior get mad because I was playing Jade and I had no skill.
You don't :P
So i was getting wrecked by exodia mage and murloc paladin as my hunter deck. Teched in flare, and it is so satisfying flaring away iceblock + icebarrier, or getting the getaway kodo for their Tirion. It has also randomly helped against rng minions people got from random cards with stealth. it is kind of dumb that if i don't have 3 minions out on the first turn that i literally have no board. When alley cat rotates hunter is going to be super screwed unless it gets something crazy.
Steam: https://steamcommunity.com/profiles/76561198004484595
I've seen two this week, one on my main account and one on my F2P.
My last like 4 arenas have had 2-3 spells, and I don't think I was skipping any.
And of course all that only applies if you mulligan every card that isn't the one you're hoping to get in your initial hand, which realistically doesn't happen all the time since you're often going to hang on to a playable 1 or 2 drop.
If you'd like to figure out the precisely correct model for this and present your work, go for it.
It's a simplification because mulligans interact weirdly with the calculation, but doing incremental probabilities for each new deck size doesn't gain you a significant amount of accuracy.
My legends this expansion: Liliana Voss, Sindragosa, Archbishop Benedictus, Malfurion (from Lich king intro)
The pack I got from beating the Lich King, I rolled over and saw two legendaries got super hype thinking I'd finally get some DK heroes that I'd actually enjoy playing (Anduin, Jaina, Rexxar, even Gul'dan).
Rotface and Blood-Queen Lana'thel.
Womp Womp. I then impulse crafted Jaina and threw Sindragosa in the deck so at least I could play with one of the legends I pulled.
PSN: ChemENGR
Witty signature comment goes here...
wra
I DID IT PA!
GOD DAMMIT!
YES! YES!
Legends of Runeterra: MNCdover #moc
Switch ID: MNC Dover SW-1154-3107-1051
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Here's someone who went though and did conditional probabilities, factoring in deck size after each draw, for trying to find specific 1-of or 2-of cards: https://www.reddit.com/r/CompetitiveHS/comments/5hkfx6/math_in_hearthstone_1_aggressively_mulliganing/
They find that going first, mulliganing for a specific 2-of card you'll have a 41.25% chance of seeing at least one copy on the first turn, which is indistinguishable from the hypergeometric calculator result of 41.8%.
Ahh the roller coaster of emotions that is hearthstone
And I've heard Rotface is fun.
The Division, Warframe (XB1)
GT: Tanith 6227
It's also probably worth noting that the original post this discussion spawned from mentioned only having one of the card that kept showing up, so all these 2-of calculations are the wrong ones to apply.
Actually that term refers to the fact that the deck contains two copies of the card you're trying to find. The "success condition" output here is for the combined set of hands that have any number of copies of the target card, whether that's one or two. It's not worth calculating the chance of getting exactly one copy because that's not how mulligan decision-making works. When you're aggressively mulliganing for a specific card, you're accepting the risk that you get both copies or you don't consider that to be a risk.
And I'm not sure I follow your complaint about whether the card is in your initial mulligan. The calculations in that link properly account for that possibility. Essentially they take the total likelihood that you fail to find the card and work backwards, so part of the result is the set of hands that included the correct card in the initial draw.
Edit: sorry if I'm coming off as overly blunt here, it's not my intention to be rude. I am speaking a bit forcefully here because I think mulliganing is one of the most interesting phases of the game and it's where so many games are won or lost. And because it's all math based I like to make sure the correct information is getting conveyed.
Edit edit: just saw your addendum about the fact that only one macaw was in that deck. My bad. That observation is low sample size and confirmation bias (remembering the times it was there, forgetting the games it wasn't). I think discussing probabilities for two-ofs is generally more relevant for most decks.
Whether your card is in your initial mulligan is of critical importance. If it's in it, boom, you won. If it's not, you probably want to know what your odds of getting it are from mulliganing 1, 2, 3, 4 cards. It doesn't help you a lot to know what your odds were of getting it pre AND post-mulligan, because when it's not in your mulligan, you want to know what the consequences of your next actions will be.
i sure wish the synergy change for arena had rhyme or reason.
I did miss the fact that there was only one macaw in that deck (see the edit I was making as you wrote your post), so that's my bad. But you're still missing that these values already include the possibility you're raising.
The math for singletons is easier than for two-ofs so I'll quote the r/competitvehs author for why this is accounted for:
I only really put it in my Paladin deck to shore up my games against Evolve Shaman. (It doesn't really weaken my match-ups elsewhere).
Skulking Geist can help the decks that are prepared to handle Jades up to 9/9 in size or so. It can also help burn decks which might lose to Earthen Scales. It's definitely not a "beat Jade Druid for free card" if you don't fall into those categories. Although one slightly more subtle benefit in any matchup where it hits cards... it can remove cards from your opponent's hand, which is basically a discard effect which is much stronger than just removing from the deck.
Its too bad the card is unplayable.
I don't know what the hypergeometric calculator is doing, but actually calculating the probability isn't that hard. It's just a basic discrete math problem.
The odds of having at least one copy of a card in your opening hand before mulligan as player 1 is 1 - ((28 choose 3)/(30 choose 3)) = 1 - 3276/4060 = 19.31%
(Basically you are finding 1 - (the number of ways to draw 3 of the 28 other cards)/(the number of ways to draw 3 cards from 30))
You have a 80.69% to not get the card before the mulligan. If you mulligan all three you then have another 19.31% to get the card. This will happen .8069 * 19.31% = 15.58% of the time.
So you are at 15.58% + 19.31% = 34.89% chance to see it before drawing your first card.
In the 65.11% of times you still haven't seen it, you have a 2/27 chance of drawing it. 2/27 * 65.11% = 4.82%
34.89% + 4.82% = 39.71%
So you have a 39.71% to see a card on the first turn if you aggressively mulligan for it as player 1.
Edit: As a comparison, someone flipping a coin twice in a row has a 75% chance to get heads at least once. It would be reasonable to bet even money that you will flip heads at least once with two flips. However, if your first flip ends up tails, there is now a 50% chance that your next flip will be heads. If (in this nonsensical and heavily biased in your favor game) you were allowed to withdraw your bet after this first flip or go on with it, the realization that at this point forward your chance to win is only 50% would probably influence your willingness to maintain vs. withdraw your bet, as the odds are no longer in your favor.
The math in the thread I just linked is correct. Most of the numbers I am seeing here are not correct. I would be happy to explain how to calculate the probability if anyone is interested. I did a quick search on hearthstone probabilities and I saw a lot of bad math out there.
You should be getting a 34.89% to chance to see a card if running 2 prior to your first draw when you hard mulligan as player 1. Then you have a 2/27, 2/26, 2/25 ... chance to draw the card after that.
As player 2 you 44.18% chance to see the card prior to your first draw if you hard mulligan and then a 2/26,2/25,2/24 chance to draw the card with each successive draw.
The bolded assumes independence between the initial hand and the next cards drawn off the mulligan. Cards rejected in the mulligan are not eligible to be drawn in the second hand (the earliest you'd see any of those specific cards again is on the first turn draw), so this calculation slightly underestimates the probability of seeing the target card.
on the other hand, the game's reputation would probably be destroyed if something like this was revealed. would such shenanigans ever be worth the risk? probably not. still, unless we know for sure...
Oooooh good catch! That means you have a 1 - ((25 choose 3)/(27 choose 3) chance with the hard mulligan to draw a copy running 2.
That means you have a 36.55% + 4.82% chance on turn 1 as player 1. So 41.37%
seriously i don't understand it.
Was offered a dragon on my first pick this arena run. took it.
did not get offered a single dragon the rest of the 29.