# What are the odds of drawing a given card after discarding my initial hand?

Let's say that I have 4 of a card in my 60 card deck.

What are my odds of drawing at least 1 of that card in my initial 7 card hand?

If I discard those 7, what are the odds of drawing 1 of the 4 in another 7 card draw? (I guess you could think of a Wheel of Fortune type situation)

• I amnot sure I understand the question. Are you asking if you don't draw it in your initial 7, what are the odds if you draw it in the next 7 when mulliganing (HGD(same answer)), without reshuffling (1-HGD(53,4,7,0) i.e. just remove 7 cards from the "deck", or something different. – user1873 Jan 14 '13 at 17:47
• No, there is no way to send private messages, welcome to the site :) – Stephen Jan 14 '13 at 18:11
• Thanks for the info Stephen. @user1873, I'm not sure how a mulligan works... are those 7 cards removed altogether? Or do you reshuffle and draw a fresh 7 cards? – Osidan Jan 14 '13 at 18:44
• @Osidan For a mulligan, you actually shuffle your original hand back into your deck and then draw the new hand (usually a hand with one card fewer, per tournament rules, but in friendly play and certain odd formats there are exceptions). This means that the contents of your original hand have no influence whatsoever on the contents of your mulligan hand. – Steven Stadnicki Jan 14 '13 at 21:17
• corsiKa and Stephen have the right odds if it truly is a "Wheel of Fortune" where you discard the first 7 and draw another 7. user1873 has the right odds if it's the more typical Magic multiplayer first mulligan: put the first 7 back into the deck and reshuffle, draw another 7. – ghoppe Jan 17 '13 at 15:51

Since no one has provided the correct answer, at least not they current way you have it worded, I will take a stab at it.

What are my odds of drawing at least 1 of that card in my initial 7 card hand?

Answer: 1-Hypergeometric Distribution(population=60,successes in population=4,sample size=7,successes in sample=0) = 1 - .600500 = 39.95%

Everyone got this right. I personally like corsiKa's answer, because if your aren't interested in drawing at least 1 success then it is easier to do the calculation (especially by hand) by figuring the odds of not drawing any cards and subtracting from 1.

If I discard those 7, what are the odds of drawing 1 of the 4 in another 7 card draw? (I guess you could think of a Wheel of Fortune type situation)

This is where everyone mucks it up. First realize that this question is fundamentally different than any of the following

• What are the odds of drawing at least 1 of the 4 in 14 cards? 66.54%
• What are the odds of drawing at exactly 1 of the 4 in 14 cards? 43.58%
• After discarding 7 cards that aren't successes, What are the odds of drawing exactly 1 of the 4 in 7 more cards?
• What are the odds of drawing at least 1 of the 4 in 7 cards, with two chances? (This is what I thought you saying originally. You wanted to know how unlikely you were to miss, then mulligan shuffling the cards back into the deck, and then missing again) 1 - (.6005 * .6005) = 63.94%
• etc.

What you actually asked for was the bolded question above. So, what that means is that at that moment, you have removed 7 cards from the deck that weren't successes, leaving you 53 cards. The odds of drawing exactly 1 card can easily be found, HGD(53,4,7,1) = 36.29% or of getting at least 1 is (1 - HGD(53,4,7,0) = 44.27%).

Now that you have finally clarified what it is that you want. The answer that you seek is above. It is the last answer, which is = all possibilities - (chance of missing first 7 cards * chance of missing again after replacement) = 1 - (.6005 * .6005) = 63.94%

In general, you either multiply probabilities (when determining how likely two independent events will occur in a row), or add them together (when determining if either of two mutually exclusive events will occur). As for when to "add" these probabilities together and when to "multiply" them, it is very difficult to know when to do which.

In your above question, you might realize that there are 4 independent mutually exclusive events for drawing 7 cards, replacing those 7 cards, shuffling and drawing 7 cards again. From the answer from the first question, you know that you have about a 40% chance of drawing at least one copy, and a 60% chance of drawing zero copies. (0.60 + 0.40 = 1 these mutually exclusive events equal 100%, so that makes sense)

• Miss twice in a row. (0.6 * 0.6) = 0.36
• Miss the first time, succeed the second. (0.6 * 0.4) = 0.24
• Succeed the first time, fail the second. (0.4 * 0.6) = 0.24
• Succeed twice in a row. = . (0.4 * 0.4) = 0.16

Your first and second (draw 7 cards and see if you succeed) events are independent of each other, because you are replacing the cards you drew back into the deck. Neither event has any effect on the results of the other, so you can multiply the first and second events together to see how likely it is to get a particular result twice in a row. Finally, because you are interested in the chances of succeeding at least once, you can either add all the mutually exclusive results above that succeed one or more times, or just subtract 1 from the chance of missing twice. (realizing that those two values should be the same)

• Doh, sorry for the number of edits. I knew what I wanted to write as the correct answer, but I kept copying the wrong values. – user1873 Jan 15 '13 at 3:26
• Note that edits within a five minute grace period are collapsed together. So it's not that big of a deal. – mattdm Jan 15 '13 at 3:30
• Thanks for the answer. I didn't mean a discard, though, but a new draw from a reshuffled deck. Basically, you have twice a chance of 39.95% to get one... how do you add these odds to a total chance in two draws? – Osidan Jan 15 '13 at 8:33
• .3995 + (1 - .3995)* .3995 = 0.64 Would be your chances for getting a success or a failure followed by a success. – Affe Jan 17 '13 at 0:59
• @Affe, with or without replacement about 40%, since the first 7 didn't count. – user1873 Jan 17 '13 at 2:08

The chance of drawing the card will be `1 - (chance of not drawing)`. I know that sounds silly, but bear with me.

You get to draw 7 cards. The chance of avoiding it on your first draw is `56 / 60` because there's 56 cards that are not that card. The next card is a chance `55 / 59`. This carries on for 7 cards.

``````56 * 55 * 54 * 53 * 52 * 51 * 50
--------------------------------
60 * 59 * 58 * 57 * 56 * 55 * 54
``````

You can cancel out some of those values (56, 55, 54)

``````53 * 52 * 51 * 50    7,027,800
----------------- = ---------- = .6005
60 * 59 * 58 * 57   11,703,240
``````

So there's a 60.05% chance of NOT getting the card on the first go. That means there's a 39.95% chance of getting it.

You can carry this out to 14 cards to represent your wheel. As in the last one, we can keep the 60, 59, 58, and 57 on the bottom, and the last four on top (46, 45, 44, 43). The rest will cancel out

``````46 * 45 * 44 * 43 = 3,916,440
----------------- = ---------- = .3346
60 * 59 * 58 * 57   11,703,240
``````

So you have a 33.46% chance of avoiding your target card, which means a 66.54 chance of getting it.

Of course, the answer to the first part of your question is here, and describes to some degree what a hypergeometric distribution is.

For the first part of your question, your odds are about 39.95% of drawing a particular card in your opening 7 card hand given 4 in a 60 card pool.

If you are in a situation where you draw 7 then draw 7 more without replacement, you are effectively drawing 14 cards instead, then your probability changes to 66.53%

• What I meant is: what are the odds that I end up with one of those cards if I redraw another 7 if it wasn't in the first sample. Does it not increase the odds overall if you have two chances of getting it in those 7 cards drawn? – Osidan Jan 14 '13 at 18:45
• @Osidan I've assumed you have 4 chances to draw that card in your initial 7 out of 60 total. – Stephen Jan 14 '13 at 18:59
• Yes Stephen, 4 chances out of 60. What I'm wondering is: what are the combined odds of drawing two hands of 7 out of 60, where only one of them has to hit at least 1 of the 4 cards. Does it not give you higher odds if you can draw another 7 if the first 7 didn't leave you with one of the 4 cards you want? How do you "add" those precentages? – Osidan Jan 15 '13 at 8:29
• @Osidan You don't have to add the percentages. The odds are the same as if you had drawn 14 cards. If you draw 7 you get about 40%, if you draw 14 you get about 67%. – Stephen Jan 15 '13 at 17:56
• @Stephen, "the odds are the same as if you had drawn 14 cards" not so sure about this. What exactly are you saying it is equal to? (drawing 7 after having missed with the first 7, with/out replacement, etc. Your 24% seems to be wrong, or it is unclear what those odds are in reference to. – user1873 Jan 17 '13 at 2:19