Event 1:
I want 1 of 12 possible cards out of the 99 in my opening hand.

Event 2:
I want 1 of 8 other cards (different from the first 12) by the next 3 draw steps.

What is the probability both of these events happen? Don't worry about mulls, cantrip cards (ponder, etc), land, other factors.

I'm no math wiz but here is what I did:

(Probably of Event 1) * (Probability of Event 2)

Probability of Event 1 = 1 - (Probability of it not happening) = 1 - [(87/99)^7] = .59

Probabilty of Event 2 = 1 - (Probability of it not happening) = 1 - [(90/98)^9] = .46

.59 * .46 = 27%

Is this correct? If not, what is the correct answer?

your computation is sampling WITH replacement. This is like rolling a 99-sided die seven times.

What you need to do is use sampling without replacement, since you draw a card, and it's no longer in your library.

When you wrote (87/99)^7, you are drawing a card, shuffling it back into your library, drawing a second card and shuffling it back, and so on.

What I suspect what you really want (from the calculations you've done and the setup) is the odds that you will draw
at least one of the 12 cards out of 99 in your opening hand.

You have a (1 - 87/99) chance to draw one of the twelve as the first card, but if you draw two cards, this isn't the same as flipping a coin, which gives you 50% every flip. If you draw one of the 87, now there's 86 left that aren't one of the 12, and only 98 cards left.

So, you get:

P(X) = (87/99)*(86/98)*(85/97)*(84/96)*(83/95)*(82/94)*(81/93).

I'm letting X = event that you draw seven cards which contain 0 of the 12 you want.

1 - P(X) gets you the odds that you have an opening hand of at least 1 copy of that card.

quote:

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Originally posted by yakusoku:
It looks like you're really close, much closer than most people get, but you're still making a common error:

your computation is sampling WITH replacement. This is like rolling a 99-sided die seven times.

What you need to do is use sampling without replacement, since you draw a card, and it's no longer in your library.

When you wrote (87/99)^7, you are drawing a card, shuffling it back into your library, drawing a second card and shuffling it back, and so on.

What I suspect what you really want (from the calculations you've done and the setup) is the odds that you will draw
at least one of the 12 cards out of 99 in your opening hand.

You have a (1 - 87/99) chance to draw one of the twelve as the first card, but if you draw two cards, this isn't the same as flipping a coin, which gives you 50% every flip. If you draw one of the 87, now there's 86 left that aren't one of the 12, and only 98 cards left.

So, you get:

P(X) = (87/99)*(86/98)*(85/97)*(84/96)*(83/95)*(82/94)*(81/93).

I'm letting X = event that you draw seven cards which contain 0 of the 12 you want.

1 - P(X) gets you the odds that you have an opening hand of at least 1 copy of that card.

Then, you need to apply similar math for your event 2.

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quote:

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Originally posted by simbayu:
This question is in regard to an EDH deck.

Event 1:
I want 1 of 12 possible cards out of the 99 in my opening hand.

Event 2:
I want 1 of 8 other cards (different from the first 12) by the next 3 draw steps.

Thanks in advance!
-Henry

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What you want is a hypergeometric distribution

So first case you have n=12 cards, which you want one in your opening hand. I am assuming opening hand size is 7, it's easy to adjust.

Do the following:
(12 choose 0)(87 choose 7)/(99 choose 7)=0.39251317092
which is prob(0 show up in 7 cards)

so 1-prob(0) = prob(1 or more show up) = .6074

This is the probability out of a group of 12 cards you desire in your opener, you will draw any number between 1 and 7 (if you really want in depth probability help feel free to pm me).

For the second one I'm unsure of the setup of the question. Can I assume you drew your opening hand and didn't find one of those 8 cards, or do you mean I need one of these 8 cards by turn 3?

Here's the math assuming you need to find exactly 1 card in a group of 8 by turn 3 (10 cards drawn).

quote:

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Originally posted by JoshSherman:
I wish I had paid more attention in my probability/ stats class. I kinda feel like the system cheated me by letting me cheat it.

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If I had been your teacher that wouldn't have happened. I NEVER use true-false or multiple choice questions. Everything is open-ended. Getting the right number is only the start. Your explanation supporting your answer is worth as much, or more, than the correct answer.

Back to the original math question. I need to confirm exactly what it is you are asking:

You want one of 12 cards (out of a deck of 99) to show up in your initial hand of 7 cards. More than 1 of these 12 cards is allowable, but at least one must show up in the original 7 card pull.

You want 1 of 8 cards (out of the same original deck of 99) that is either dealt in the original 7 card pull, or is available as one of the subsequent 3 cards drawn.

SIMPLIFIED PROBABILITY CALCULATION

Let “X” = cards in group of 12
Let “Y” = cards in group of 8

We use the complementary function to simplify. The odds of getting at least one “X” in the first 7 cards, as noted above, equal 1 minus the probability of getting no X’s. Calculating "no X's" is easier.

For the first card drawn the odds of not getting an X equals: (99 – 12) / 99 = 87 / 99

For the second card to not be an X, given that the first was not an X, equals (98 -12) / 98 = 86 / 98. We have subtracted the non-X chosen in the first pick from both the numerator and denominator.

These incremental probabilities need to be multiplied, so for a hand of 7 cards that has no X, the equation becomes:

P(no X out of 7 cards) = (87 x 86 x 85 x 84 x 83 x 82 x 81) / (99 x 98 x 97 x 96 x 95 x 94 x 93)

If you have a permutation function, this equals (87P7) / (99P7) = 0.392513

The probability of getting at least one X is the complementary of getting no X’s, which equals:
1 - 0.392513 = 0.607487 = 0.6075

Given the above result, we can then use the same logic to calculate the probability of picking at least 1 Y out of the first ten cards picked, as follows:

We will assume we got the desired X on the first card (we’ll substitute the probability result from above), and calculate the probability of getting at least one Y in the next 9 draws.

The probability of not getting at least one Y in the next nine draws equals:
P(no Y’s) = (90 x 89 x 88 x 87 x 86 x 85 x 84 x 83 x 82) / (98 x 97 x 96 x 95 x 94 x 93 x 92 x 91 x 90)

This equals (90P9)/(98P9) = 0.448794

Probability of getting at least one Y in 9 cards = 1 – 0.448794 = 0.551206

Overall probability of getting at least one X in initial draw of 7, and getting at least one Y in the first 10 cards, with simplification, equals:

SIMPLIFIED FINAL PRBOBILITY = 0.6075 x 0.5512 = 0.3348 = 0.335 or 33.5%

The simplifications required for this answer are as follows:

1. I have taken the overall probability for getting at least one X in the original draw of 7, and have substituted it in as the probability of success for the first card chosen. This simplification is mainly one of cleaning up otherwise messy math, by combining all the different permutations that get me at least one X in the initial draw of 7 cards.

DETAILED PROBABILITY CALCULATION

The actual probability of meeting the stated requirements is the sum of the following:

P(one X in first 7 cards) times P(at least one Y in first 10 cards) given one X in first 7 cards, plus
P(two X’s in first 7 cards) times P(at least one Y in first 10 cards) given two X’s in first 7 cards, plus
P(three X’s in first 7 cards) times P(at least one Y in first 10 cards) given three X’s in first 7 cards, plus
P(four X’s in first 7 cards) times P(at least one Y in first 10 cards) given four X’s in first 7 cards, plus
P(five X’s in first 7 cards) times P(at least one Y in first 10 cards) given five X’s in first 7 cards, plus
P(six X’s in first 7 cards) times P(at least one Y in first 10 cards) given six X’s in first 7 cards, plus
P(seven X’s in first 7 cards) times P(at least one Y in first 10 cards) given seven X’s in first 7 cards.

For the first parts of the above statements:

P(one X in first 7 cards) times P(at least one Y in first 10 cards) given one X in first 7 cards, plus

P(one X out of 7 cards): This is a variation of the original calculation shown above for no X’s. There are two components to the variation:

- The first is replacing the lowest non-X numerator variable (the 81) with the highest possible X variable. Since there are 12 possible X cards, the replacement number is 12.
- The above calculation gives us the probability of a single permutation, so we will need to multiply that result by the number of permutations which result in the same combination (of one X and six non-X cards). This turns out to be ((total cards in hand)!/((# of non-X)! (# of X)1!)) = 7!/(6! X 1!) = 7

After making these adjustments, the resulting probability, P(one X out of 7 cards), is:

P(one X) = [(87 x 86 x 85 x 84 x 83 x 82 x 12) / (99 x 98 x 97 x 96 x 95 x 94 x 93)] x 7 = 0.4070507

Continuing for all possible values of X:

P(two X) = [(87 x 86 x 85 x 84 x 83 x 12 x 11) / (99 x 98 x 97 x 96 x 95 x 94 x 93)] x 21 = 0.1638131
P(three X) = [(87 x 86 x 85 x 84 x 12 x 11 x 10) / (99 x 98 x 97 x 96 x 95 x 94 x 93)] x 35 = 0.0328942
P(four X) = [(87 x 86 x 85 x 12 x 11 x 10 x 9) / (99 x 98 x 97 x 96 x 95 x 94 x 93)] x 35 = 0.0035244
P(five X) = [(87 x 86 x 12 x 11 x 10 x 9 x 8) / (99 x 98 x 97 x 96 x 95 x 94 x 93)] x 21 = 0.0001990
P(six X) = [(87 x 12 x 11 x 10 x 9 x 8 x 7) / (99 x 98 x 97 x 96 x 95 x 94 x 93)] x 7 = 0.0000054
P(seven X) = [(12 x 11 x 10 x 9 x 8 x 7 x 6) / (99 x 98 x 97 x 96 x 95 x 94 x 93)] x 1 = 0.0000001

As a double check, the sum of these partial probabilities equals:

0.4070507
0.1638131
0.0328942
0.0035244
0.0001990
0.0000054
0.0000001
0.6074869 = 0.6075 (which was the same cumulative probability noted in the simplified calculation)

We now need to evaluate the second part of each probability statement. The first example is:

“P(at least one Y in first 10 cards) given one X in first 7 cards”

This probability is based upon how many “free” cards there are to chose. If we have one X, we have nine chances to draw at least one Y. If we have two X’s, we only have 8 chances to draw at least one Y. The more X’s, the less chances to draw at least one Y. We will start with the calculation from above.

If we have exactly one X, we have nine draws to get at least one Y. The probability of not getting at least Y in the next nine draws equals:

P(no Y’s in nine cards) = (90 x 89 x 88 x 87 x 86 x 85 x 84 x 83 x 82) / (98 x 97 x 96 x 95 x 94 x 93 x 92 x 91 x 90)

This equals (90P9) / (98P9) = 0.448794, so the probability of getting at least one Y in nine cards = (1 – 0.448794) = 0.551206

If we have exactly two X’s, we have eight draws to get at least one Y. The probability of not getting at least Y in the next eight draws equals:

P(no Y’s in eight cards) = (89 x 88 x 87 x 86 x 85 x 84 x 83 x 82) / (97 x 96 x 95 x 94 x 93 x 92 x 91 x 90)

This equals (89P8) / (97P8) = 0.488688, so the probability of getting at least one Y in 9 cards = (1 – 0.488688) = 0.511312

For three X’s, (88P7) / (96P7) = (1 - 0.532615) = 0.467385
For four X’s, (87P6) / (95P6) = (1 - 0.581034) = 0.418966
For five X’s, (86P5) / (94P5) = (1 - 0.634463) = 0.365537
For six X’s, (85P4) / (93P4) = (1 - 0.693482) = 0.306518
For seven X’s, (84P3) / (92P3) = (1 - 0.758751) = 0.241249

P(1 X, 1 or more Y) = 0.4070507 x 0.551206 = 0.224369
P(2 X, 1 or more Y) = 0.1638131 x 0.511312 = 0.083760
P(3 X, 1 or more Y) = 0.0328942 x 0.467385 = 0.015374
P(4 X, 1 or more Y) = 0.0035244 x 0.418966 = 0.001477
P(5 X, 1 or more Y) = 0.0001990 x 0.365537 = 0.000073
P(6 X, 1 or more Y) = 0.0000054 x 0.306518 = 0.000002
P(7 X, 1 or more Y) = 0.0000001 x 0.241249 = 0.000000

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Originally posted by oneofchaos:
I still think my method is more intuitive AND less work...

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Intuition, based upon fact, is a wonderful thing. But in statistics, and specifically when determining probabilities, accuracy has to be the prime goal.

If you exactly know the probability or statistics of your competitive position, then your intuition can decide how to proceed.

The opposite, where intuition decides the competitive position and then logic determines how to proceed, is never as successful.

And simplicity is in the eye of the beholder. In probability and statistics most time is usually consumed trying to determine exactly what a question is asking.

quote:

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Originally posted by hammr7:
Intuition, based upon fact, is a wonderful thing. But in statistics, and specifically when determining probabilities, accuracy has to be the prime goal.

If you exactly know the probability or statistics of your competitive position, then your intuition can decide how to proceed.

The opposite, where intuition decides the competitive position and then logic determines how to proceed, is never as successful.

And simplicity is in the eye of the beholder. In probability and statistics most time is usually consumed trying to determine exactly what a question is asking.

My method didn't take that long to calculate (took longer to type than to calculate). Of course it helps to use statistics regularly, which I do.

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My method gives an exact answer not quite sure what the point was. It's easier to divide cards into a couple groups and say you want some from this and some from that as opposed to multiplying a change of shrinking numerators over denominators.

It's really no big deal, whatever he prefers.