Probability and MtG

Factual Questions
1

So I was playing a little Magic the Gathering with my friends, and an improbable situation came up, and we were wondering just how improbable it was.

First, the scenario: I have a deck built around making my opponent discard all their cards, so they have nothing to play. This can get frustrating for them, especially when I play Distress. The card is, well, distressing. As you may or may not know, it requires two black mana to play Distress, which means the soonest I can play it is my second turn. Which I did–five games in a row–which opened me up to facetious accusations of cheating.

We tried to figure out the chances… but failed pretty miserably. But, aha! My genius internet-friends on the Dope can solve this conundrum! So here’s the data:
To play Distress on my second turn, I need two Swamps and one Distress. I have four Distress and 22 Swamps in my deck of 60 cards. By the second turn, I will have drawn eight cards. So what are the chances of playing it. How about 5 games in a row?

Thanks guys.

(And, yes, I know that after I do it the 4th time the chances of a 5th are the same as the first… you know what I mean)

2

You really should cut down to a 45 card deck, assuming that’s still the minimum for most tournament play. Whoever figures out the odds for you should also figure out those odds for a 45 card deck so you can see the difference. Myself, I’ve never been one for probability problems.

3

It seems a little easier to calculate the odds of not drawing what you need than to calculate the odds of drawing those cards.

There are 36 (60 - 22 - 4) cards in your deck that are not swamps or Distress. There are 36 Choose 8 ways of drawing those cards, out of a total of 60 C 8 possible hands, giving you a .012 chance of drawing no swamps and no Distress. The probability of drawing 1 swamp is (22 C 1) * (36 C 7) / (60 C 8) = .072. Probability of drawing 2 swamps and no distress is (22 C 2) * (36 C 6) / (60 C 8) = .18. Every other possibility involves drawing the cards you need, so the probability is 1 - .012 - .072 - .18 = .736

This isn’t perfect because these conditions can overlap (Distress with no swamps), but it’s pretty close and I don’t have time to fix it.

There’s roughly a 3/4 chance you can play your combo by the second turn. Five games in a row would be (.736)^5 = .215. One in five. Not too bad.

4

There are (60 C 3) different ways to pick 3 cards from 60 possible = 34220
There are (22 C 2) different ways to pick 2 swamps out of 22 possible = 231
There are (4 C 1) different ways to pick 1 distress out of 4 possible = 4

4 * 231 / 34220 = 0.027 = 2.7%

Doing that 5 times in a row is 0.027[sup]5[/sup] = 0.00000001435 = 0.000001435%

5

Don’t you also need to include the probability of 3 swamps & no distress, 4 swamps & no distress, etc.? In particular, the probability of 3 swamps & no distress, by your logic, is about 22.7% — not small.

6

I think your memory is faulty. There has never been an official tournament format with a 45-card minimum deck. The original game rules said 40 cards, but the limit for constructed tournaments has always been 60.

7

Isn’t that nine cards? Or have they changed the rules?

Punoqllads, you’re assuming that he only has three cards. He has more than that.

8

If you play first you skip your first draw step, so you’d only have 8. (And I always choose to play first, and my friends always choose second, so we don’t even bother to flip for the choice anymore ^^*)

9

Each card has a 1/15 chance of being distress. With 8 cards in your hand, it doesn’t seem to be too insanely unlikely that you draw one, even 4 games in a row.

10

D’oh, missed that part.

Minor note – with 4 distress and 22 swamp cards, there are 34 cards remaining, not 36.

The number of combinations of 8 cards that have at least 1 distress card and 2 swamp cards are (where C(a,b) is a choose b):

C(4,1) * C(22,2) * C(34,5) = 257108544
C(4,2) * C(22,2) * C(34,4) = 64277136
C(4,3) * C(22,2) * C(34,3) = 8293824
C(4,4) * C(22,2) * C(34,2) = 129591
C(4,1) * C(22,3) * C(34,4) = 285676160
C(4,2) * C(22,3) * C(34,3) = 55292160
C(4,3) * C(22,3) * C(34,2) = 3455760
C(4,4) * C(22,3) * C(34,1) = 52360
C(4,1) * C(22,4) * C(34,3) = 175091840
C(4,2) * C(22,4) * C(34,2) = 24622290
C(4,3) * C(22,4) * C(34,1) = 994840
C(4,4) * C(22,4) * C(34,0) = 7315
C(4,1) * C(22,5) * C(34,2) = 59093496
C(4,2) * C(22,5) * C(34,1) = 5372136
C(4,3) * C(22,5) * C(34,0) = 105336
C(4,1) * C(22,6) * C(34,1) = 10147368
C(4,2) * C(22,6) * C(34,0) = 447678
C(4,1) * C(22,7) * C(34,0) = 682176

Which sum to 948085402. The number of ways to choose 8 cards from 60 is C(60,8) = 2558620845. 948085402/2558620845 is about 0.37, i.e., 37%. The chances of doing that 5 times in a row is about 0.7%

11

MTGO’s online stat calculator verifies Punoqllads’ calculation (it has it at 36% and .6% respectively). A handy tool for quick and dirty probability calculations.

By the way, that doesn’t mean you cheated… improbable things are bound to happen and you tend to notice them more due to selective memory and confirmation bias.

12

original tourney rules were for 40 card decks and even played for ante. and DAMN were they brutal, no limits on the number of cards by type. the absolute ultimate deck ever was dark rituals, black lotuses, contract from belows and drain life’s. it was an almost certain first turn win.

early magic rules were weird, iffy to get a good call on from the ref’s, and definitely in need of some changes. (trust me, I convinced them of several of those changes personally.)

13

Right, the .7% chance is the chance of casting Distress on the second turn in five specific games, chosen a priori. But you’ve probably played hundreds or thousands of games. And even if they weren’t all with your forced-discard deck, your other decks have probably had second-turn cards which would be just as notable. Heck, your discard deck might even have other second-turn cards which are nearly as notable. So the fact that at sometime in your playing history, you managed five times in a row to get a good card out on the second turn, isn’t too significant.