Ok this is a question for the math-heads out there. I am wondering what is the probability of a rogue getting Prince Keleseth - Shadowstep - Prince Keleseth on turn 1/2 and how high are they for prince - step - prince - step-prince in round 1 or 2? I tried calculating it but my stupid math-hating brain just laughed at me. The reason i want to know is because i played 5 rogues today, 3 of them had the one-time-shadowstep-prince-combo in round 1 or 2 and one even had the two shadowsteps plus prince aggro so i am wondering if am having a shitty day or a really shitty day. I know i am basically asking for someone to calculate something that i am just to lazy and dumb to do, but i would really apreciate it. Cheers.

It will depend on if you are going first or second, since you have one more card. I'll do the maths when I get home, but I think someone will have done it already :)

EDIT : It appears to be more complicated than I thought since, if you're going first with Swashburglar or SD on turn 1, you'll have more chance to have the turn 2 dream since Patches will be out of the deck, offering you more chances to draw your last piece next turn

Let's assume you mulligan very hard for Keleseth (by which I mean you throw back Shadowstep if you get it without Keleseth.) Most players don't keep a first-turn Shadowstep.

I'll start with chances to get Keleseth:

With the coin, you draw 4 cards, which have a cumulative 70.4% chance NOT to have Keleseth in them:

(29/30) * (28/30) * (27/30) * (26/30) = 0.704

If there's no keleseth, you do it again: 0.704 * 0.704 = 0.495, or 49.5% chance not to get Keleseth in either hand.

Then, on turn 2, you draw another card, which has a 25/26 chance not to be Keleseth:

0.495 * (25/26) = 0.476, or 47.6% chance not to get Keleseth at all the first 2 turns.

Turn that around, and you have a 52.4% chance to get Keleseth.

Without the coin, you draw 3 cards. If none are Keleseth you throw them all back.

Chances of not getting Keleseth to start are:

(29/30) * (28/30) * (27/30) = 0.812 or 81.2% no Keleseth.

Square that and you get 65.9% chance of no Keleseth in mulligan either. And multiply by 26/27 to find chances of no Keleseth first two turns, or 63.5%. Turn that around and you have a 36.5% chance of getting Keleseth with no coin.

You can do something similar to figure out the chances with Shadowstep, but it's a lot of work. I'll hand-wave it and guess that the odds of Keleseth and a shadowstep are somewhere around 10-20% with the coin and under 10% without, but I'll leave that step to someone else.

That math is pretty wrong With the coin, 29/30*28/29*27/28*26/27=26/30=.8666

If you do not get keleseth you hard mulligan for keleseth 25/26*24/25*23/24*22/23=.846 (You cannot get those 4 cards again)

Probability of not getting Keleseth on turn 1 = .8666*.846= .733

So you have 26.7% chance of getting Keleseth on turn 1 if you hard mulligan for it.

Probability to get it on turn 2 =25/26 .733*25/26=.704

So you have 30 % chance of having Keleseth till turn 2 with coin.

Keleseth in opening hand: hard mulligan for Shadowstep

First case (86,66%):

Keleseth not in opening hand, mulligan for 4.

Probability that this case occurs: 29/30*28/29*27/28*26/27=0.8(6).

I will have 5 random cards at the start of turn 1. There is a 1/30 chance to draw Keleseth in the first card and a 2/29 chance to draw a Shadowstep in the second. There are 3276 ways to choose the other 3 cards. 1/30*2/29*3276=7,53%.

Second case (13,33%):

Keleseth in opening hand, mulligan for 3.

Probability that this case occurs: 4/30=0.1(3).

I already have Keleseth and I am going to calculate what is the chance that I get a Shadowstep in the next 4 cards. Probability of failure (that I miss both Shadowsteps): 27/29*26/28*25/27*24/26=73,89% -> Probability of success: 26,1%.

Overall probability:

86,66% * 7,53% + 13,33% * 26,1% = 10,00% to have Keleseth and Shadowstep on turn 1 given you started second.

Just played against a Rogue two times in a row Casual. He conceded first game turn one then queued into him again.

Guess what he played turn 2 next game? That's right, Keleseth turn 2.

So if Rogues always seem to have Keleseth on turn 2, you're not crazy. They concede until they have Keleseth turn 2, at least in Casual...

Funny part about it is that I beat him with my Zoo Warlock that doesn't even have Keleseth.

And that is the true lesson regarding Rogue Keleseth. Just because they play it on turn 2 doesn't mean you will lose. Yes, there is increased likelyhood, but smart play and good deck techs can still earn you a win.

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Quit complaining and just have fun with this game!

Keleseth in opening hand: hard mulligan for Shadowstep

First case (86,66%):

Keleseth not in opening hand, mulligan for 4.

Probability that this case occurs: 29/30*28/29*27/28*26/27=0.8(6).

I will have 5 random cards at the start of turn 1. There is a 1/30 chance to draw Keleseth in the first card and a 2/29 chance to draw a Shadowstep in the second. There are 3276 ways to choose the other 3 cards. 1/30*2/29*3276=7,53%.

Second case (13,33%):

Keleseth in opening hand, mulligan for 3.

Probability that this case occurs: 4/30=0.1(3).

I already have Keleseth and I am going to calculate what is the chance that I get a Shadowstep in the next 4 cards. Probability of failure (that I miss both Shadowsteps): 27/29*26/28*25/27*24/26=73,89% -> Probability of success: 26,1%.

Overall probability:

86,66% * 7,53% + 13,33% * 26,1% = 10,00% to have Keleseth and Shadowstep on turn 1 given you started second.

This is pretty much in line with personal experience. I don't have a lot of tempo rogue games under my belt other than ones done for quests, but through that experience If found it's more common to get to turn 5-6 without him than it is to get him on turn 2 let alone with shadowstep.

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Ok this is a question for the math-heads out there. I am wondering what is the probability of a rogue getting Prince Keleseth - Shadowstep - Prince Keleseth on turn 1/2 and how high are they for prince - step - prince - step-prince in round 1 or 2? I tried calculating it but my stupid math-hating brain just laughed at me.

The reason i want to know is because i played 5 rogues today, 3 of them had the one-time-shadowstep-prince-combo in round 1 or 2 and one even had the two shadowsteps plus prince aggro so i am wondering if am having a shitty day or a really shitty day. I know i am basically asking for someone to calculate something that i am just to lazy and dumb to do, but i would really apreciate it. Cheers.

Edit: Thanks Kraken296 for correcting my math.

It will depend on if you are going first or second, since you have one more card. I'll do the maths when I get home, but I think someone will have done it already :)

EDIT : It appears to be more complicated than I thought since, if you're going first with Swashburglar or SD on turn 1, you'll have more chance to have the turn 2 dream since Patches will be out of the deck, offering you more chances to draw your last piece next turn

With the coin,29/30*28/29*27/28*26/27=26/30=.8666

If you do not get keleseth you hard mulligan for keleseth

25/26*24/25*23/24*22/23=.846

(You cannot get those 4 cards again)Probability of not getting Keleseth on turn 1 = .8666*.846= .733

So you have

26.7% chance of getting Keleseth on turn 1if you hard mulligan for it.Probability to get it on turn 2 =25/26

.733*25/26=.704

So you have 30 % chance of having Keleseth till turn 2 with coin.without coin it reduces to 23%Nothing to add. Thanks again.

Yeah, I was mistaken, but I will point out that the moment I posted that I did look it up and found the same sources you linked. :)

Posts edited.

Scenario:First case (86,66%):Keleseth not in opening hand, mulligan for 4.

Probability that this case occurs: 29/30*28/29*27/28*26/27=0.8(6).

I will have 5 random cards at the start of turn 1. There is a 1/30 chance to draw Keleseth in the first card and a 2/29 chance to draw a Shadowstep in the second. There are 3276 ways to choose the other 3 cards. 1/30*2/29*3276=

7,53%.Second case (13,33%):Keleseth in opening hand, mulligan for 3.

Probability that this case occurs: 4/30=0.1(3).

I already have Keleseth and I am going to calculate what is the chance that I get a Shadowstep in the next 4 cards. Probability of failure (that I miss both Shadowsteps): 27/29*26/28*25/27*24/26=73,89% -> Probability of success:

26,1%.Overall probability:86,66% * 7,53% + 13,33% * 26,1% =

10,00%to have Keleseth and Shadowstep on turn 1 given you started second.Thanks alot guys! So turns out my day was pretty shitty but not as shitty as i thought, yaye.

Just played against a Rogue two times in a row Casual. He conceded first game turn one then queued into him again.

Guess what he played turn 2 next game? That's right, Keleseth turn 2.

So if Rogues always seem to have Keleseth on turn 2, you're not crazy. They concede until they have Keleseth turn 2, at least in Casual...

Funny part about it is that I beat him with my Zoo Warlock that doesn't even have Keleseth.

Quit complaining and just have fun with this game!