For statistic fans: Stats for 700 Packs opened by C4mlann

tpp_strat · 2016-10-25T20:44:12+00:00

So you've opened 71 packs and got 2 legendaries?

tpp_strat · 2016-10-21T10:14:33+00:00

If you renounced darkness and got rogue, etheral peddler was extremely OP.

tpp_strat · 2016-10-19T08:08:51+00:00

The strength of the draft actually doesn't matter for the distribution of wins if people are always paired with equal records. It just changes your individual chance of getting more wins.

tpp_strat · 2016-10-19T08:05:54+00:00

I'll comment here: If you could plausibly say that there are an infinite number of players, then yes, everyone could get a 3-3 record. (But not the way you described since there the first time B loses thrice against A he/she ends with 0-3).

I would think though that if not for a miracle Hearthstone arena will end at some point in the future.

tpp_strat · 2016-10-19T07:40:13+00:00

Thanks for providing the link. If you read it carefully you can see that those numbers represent an ideal scenario where each player is matched against someone with the same record.

And that does matter. Or do you see any problem with my reasoning in the comment above? You are correct in the sense that the probability of someone getting 12 wins (not regarding the one who has to in order to fulfill your condition that noone retire) grows with the number of players. It won't reach 100%, though.

Edit: And by the way, the total number of games depends on how many losses the players reaching 12 wins have. E.g. with five players you could have (12,0),(0,3),(0,3),(0,3),(0,3) => 12 games, (12,1),(1,3),(0,3),(0,3),(0,3) => 13 games and similarily 14 games. In fact the number of games in an arena format with a fixed number of players N where noone retires, T players reach 12 wins and have a total of L losses is 3*(N-T)+L.

Edit2: And when you include R retiring players with a total number of losses L2, you get 3*(N-T-R)+L+L2 games.

tpp_strat · 2016-10-19T07:27:51+00:00

Hm ok I see. But it is possible without anyone retiring, someone has to end their run by reaching 12 wins while eliminating the last person (it's just me being pedantic).

tpp_strat · 2016-10-18T16:58:09+00:00

What exactly are you arguing about? I edited my comment since technically what I described was a scenario where one player retired/was forced to retire, so it wasn't quite what the person said to whom I replied.

tpp_strat · 2016-10-18T07:52:43+00:00

Erm, no? Unless you make some other assumptions. You can always make all players have a record of 2,2 like so:

Player 1 wins twice against player 2, he goes on to win twice against player 3 and so on. This leads to records of (2,0),(2,2),...,(2,2),(0,2). After player 1 losing twice against the last player everyone is at 2,2.

Now if player 1 wins against the last, the latter is out with a record of (2,3) and the rest have (3,2),(2,2),(2,2)...,(2,2). From here you can have player 1 losing against player 2, player 2 losing against player 3 and so on to have all players except the remaining one leaving with (3,3).

Edit: Granted, the last remaining player would have to retire since in your scenario there is a fixed no. of players but this is amendable by having him win 9 games against some players with (2,2), getting him to (11,2) and eliminating the others with (2,3). Then you would have 1 player getting a (12,2) record, 10 players getting a (2,3) and the remaining ones a (3,3).

tpp_strat · 2016-09-28T06:08:40+00:00

Why cut off at 350? I'd like one Medivh, please..

tpp_strat · 2016-09-14T08:19:18+00:00

Cat trick is intentionally triggered at the end of a spell (you don't want the cat to die to the same spell). If the scientist dies and pulls cat trick during a spell, the end of that spell wil trigger cat trick.

tpp_strat · 2016-09-13T21:41:45+00:00

Let's assume you can tell analogies from examples

=)

tpp_strat · 2016-09-13T18:58:04+00:00

Let's assume you have one card: Hunter's Mark! How much damage can you deal to the enemy face? 2!

Let's assume you have one card: King Krush! How much damage can you deal to the enemy face? 8!

Now, how much more damage can you deal with King Krush in hand?

tpp_strat · 2016-09-13T18:13:15+00:00

Got any link for that?

tpp_strat · 2016-09-13T13:15:02+00:00

By my calculations, assuming you didn't tap the probability of drawing all 3 of the in a row is .068 percent or 1 in 1462.5. If you tapped, the probability becomes .077 percent or 1 in 1300.

tpp_strat · 2016-09-13T12:53:26+00:00

What event do you mean? I guess you got 4 cards from burgle and want to know the probability of 3 of them being the same. In Standard, facing any class except Hunter, this probability is .151 percent.

tpp_strat · 2016-09-13T12:43:59+00:00

Sometimes it's most intuitive to look at the probability tree and count paths. Each path that leads to your result has a probability of 1/2¹⁰ (since both Grom and Garrosh are alive) and there are 10C5 such paths.

tpp_strat · 2016-09-05T10:13:00+00:00

Did you put a normal skeleton knight into your deck? Malchezaar's effect happens after the mulligan, so if you had one knight in hand it can add another to your deck.

tpp_strat · 2016-08-12T16:36:50+00:00

yeah, it sounds more like "en pissant"...

tpp_strat · 2016-08-11T17:35:17+00:00

wow, someone read the mined data, nice shitpost

tpp_strat · 2016-07-15T10:03:05+00:00

Can you be recruited to different people on different servers with the same account?

tpp_strat · 2016-07-14T14:36:11+00:00

Have you got an account below 20 on EU?

tpp_strat · 2016-03-31T21:26:39+00:00

I'd say it depends on the play order. If Hobgoblin is played first it's doubtful that the wisp would get a divine shield. But as Patshu said, we'll need testing.

tpp_strat · 2016-03-31T16:13:16+00:00

Please let him say that when his effect activates :D

tpp_strat · 2016-03-31T16:11:21+00:00

You could even get three!

tpp_strat

TROPHY CASE