As every Hearthstone player knows, it seems that every game against a Hunter starts with a turn 1 Undertaker. But really how likely is this to happen? It’s time to do some science.
If you don’t feel like reading, here is the short version: at most 41.25% going first, and 50.98% going second.
Probability with only one Undertaker in the deck and no mulligan
Let’s start simple, and assume that we are going first (i.e., we draw 3 cards) and have only 1 copy of Undertaker. Of course, no one plays only one copy of Undertaker but just bear with me. Using binomial coefficients, we know that there are C(30, 3) ways to draw 3 cards out of 30, which is 4060. Out of these possible 4060 starting hands, how many have the only Undertaker? The answer is C(29, 2) = 406 (i.e., Undertaker plus all the 2 card combinations of the remaining 29 cards), so the probability with one Undertaker is 406 / 4060 = 10%.
Probability with 2 Undertakers in the deck and no mulligan
Now, our beloved Hunters more likely have 2 copies of Undertaker. This complicates the analysis a bit, and it’s actually simpler to look at it this way:
P(having at least one Undertaker in the initial draw) = 1 – P(not having any Undertaker in the initial draw)
There are C(28, 3) ways to not draw Undertaker at all (all the 3 card combinations in the remaining 28 cards of the deck), so the probability with 2 undertakers is 1 – 3276 / 4060 = 19.31%.
Probability with 2 Undertakers and a hard mulligan
If we assume that we toss everything that is not an Undertaker in the opening hand, we get to draw 3 cards again. So the probability to have at least one Undertaker after mulligan is 1 – P(no Undertaker before mulligan) * P(no Undertaker after mulligan). We already know the probability to get no Undertaker in 3 cards from above (that is: 1 – 19.31% = 80.69%). For the probability after mulligan, we apply the same logic but in the remaining 27 cards, so the probability after mulligan is 1 – 0.8069 * C(25, 3) / C(27, 3) = 36.55%.
Probability with 2 Undertakers, mulligan and the turn 1 draw
Now, what matters is to play an Undertaker by turn 1, which means there is an additional card draw to take into account. The probability to have at least one is P(not having any after mulligan) + P(not having any after mulligan) * P(drawing one on turn 1) = 36.55% + (1 – 36.55%) * 2 / 27 = 41.25%
Probability with 2 Undertakers, mulligan and the turn 1 draw, going second
When going second, the initial card draw is 4, so we just need to apply the same logic as in the steps above. In the end, we find that the probability Undertaker will show up on turn 1 for the second player is 50.98%.
Of course, there is nothing special about the Undertaker, this gives you an estimate of how likely you are to mulligan what you want. I have double-checked my numbers by randomly sampling 1 million openings, the code is available in this gist.
EDIT: I updated the numbers to reflect that after the mulligan phase, you can’t draw the cards that have been tossed (as pointed out by masharpe)
One thought on “In Hearthstone, how likely is a turn 1 Undertaker?”
Still a huge geek danydan. :)