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Medium featured0907 card analysis fiery war axe

This guide first appeared here.

Summary

Without Coin: 44.35%

With Coin: 52.60%

If the deck is running only 1 copy of Fiery War Axe, the odds will be 25% and 30.67% for without coin and with coin respectively.

Information

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I asked myself frequently how likely is it for me to secure that 1 card I needed for my starting hand. Given the deck size, mulligan and all, what's the probability?

We know having a second copy will increase the consistency/probability; but by how much?

This exercise is not just about Fiery War Axe, but it is an example that many of us can easily relate to. The same theory can be applied to other cards.

The motivation behind this exercise was also triggered by the revival of Resurrect Priest. Both Resurrect and Injured Blademaster were existing cards, but the archetype was not popular. With the introduction of Onyx Bishop (additional 2 resurrect cards), the deck becomes slightly viable. The consistency of being able to resurrect increased tremendously due to having 4 resurrect cards instead of 2; but by how much? (I understand Onyx and Resurrect are not an apple-to-apple comparison, and the revival of rez priest is not just because of the higher rez probability, but other factors too; example Onyx being a body itself is a big factor contributing to the viability. But here, I am just isolating on the probability, just to provide perspective on this one aspect.)

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I hope to put these 'gut feels' into actual numbers, so that deck builders can use them as a guide when deciding between "should I have 3 or 4 enrage/patron activator", and having some numbers as a guide. I understand it is definitely not the only factor, but having this perspective may help.

The goal of the full exercise goes beyond probability of having 1 card in mulligan/starting hand, but a more generalized "probability of N cards by turn X".

Hopefully, this can help answer questions like "what are the odds of getting my N-card-combo by turn X", and "probability of getting at least 2 activators by turn X", "Should I include a 2nd copy of Doomhammer to increase consistency by Y%?" etc.

Will post the generated tables when I get some time on hand to work on them.

Detailed Breakdown

For this part 1 of the exercise, I am posting the details of the manual workings.

Part 2 will be generated tables, I may only post the end results; and possibly the pseudo code/algorithm if I have time to clean them up.

Without Coin, only 1 copy

[A1] - Number of combinations = 30 choose 3 = 4060

[A2] - Number of combinations with desired card = 29 choose 2 = 406

[A3] - Probability of getting desired card before mulligan = [A2]/[A1] = 10%

In general, if probability of an event is n, the probability of event happening in an independent 2nd try is also n. The probability of n happening at least once in 2 tries is: n + (1-n)*n = 2n - n2 Hence,

[A4] - Probability of getting desired card after hard mulligan = 2*[A3] - [A3]2 = 19%

[A5] - If mulligan failed, probability of drawing by turn 2 = 1/27 + (26/27)*(1/26) = 2/27 = 7.41%

[A6] - Probability of getting desired card by turn 2 = [A4] + (1 - [A4])*[A5] = 25%

Without Coin, 2 copies

[B2] - Number of combinations with at least 1 copy of desired card = 2*(29choose2) - 28choose1 = 784

[B3] - Probability of getting at least 1 copy of desired card before mulligan = [B2] / [A1] = 19.31%

[B4] - Probability of getting desired card after hard mulligan = 2*[B3] - [B3]2 = 34.89%

[B5] - If mulligan failed, probability of drawing by turn 2 = 2/27 + (25/27)*(2/26) = 14.53%

[B6] - Probability of getting desired card by turn 2 = [B4] + (1 - [B4])*[B5] = 44.35%

With Coin, only 1 copy

[C1] - Number of combinations = 30 choose 4 = 27405

[C2] - Number of combinations with desired card = 29 choose 3 = 3654

[C3] - Probability of getting desired card before mulligan = [C2]/[C1] = 1/7.5 = 13.33%

[C4] - Probability of getting desired card after hard mulligan = 2*[C3] - [C3]2 = 24.89%

[C5] - If mulligan failed, probability of drawing by turn 2 = 1/26 + (25/26)*(1/25) = 2/26 = 7.69%

[C6] - Probability of getting desired card by turn 2 = [C4] + (1 - [C4])*[C5] = 30.67%

With Coin, 2 copies

[D2] - Number of combinations with at least 1 copy of desired card = 2*(29choose3) - 28choose2 = 6930

[D3] - Probability of getting at least 1 copy of desired card before mulligan = [D2] / [C1] = 25.29%

[D4] - Probability of getting desired card after hard mulligan = 2*[D3] - [D3]2 = 44.18%

[D5] - If mulligan failed, probability of drawing by turn 2 = 2/26 + (24/26)*(2/25) = 15.08%

[D6] - Probability of getting desired card by turn 2 = [D4] + (1 - [D4])*[D5] = 52.60%

Comments (2)

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GeneralAcorn

There's two answers to your question:

1.) When I'm playing against a warrior - 100% he'll have it.
2.) When I'm playing warrior - 0% chance of having it.

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Ragnaroky

I think you 've swapped the probabilities of coin and not-coin. Also same typo on # of copies. Just need to reverse titles, right?