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Don't laugh, this is a dumb question, but my brain just doesn't work mathematically. A question in my math class says

A coin is tossed 4 times. Compute the probability of at least 2 tails occurring.

OK, so I know I figure out how many total events are in the sample, then figure out how many possible ways at least 2 tails are occurring, and divide. My problem is, I can NEVER seem to figure out how many total events there are! I start with HHHH, HHHT, HHTH, HTHH, and so on, but I always get lost somewhere along the way, miss an event, and never get them all. My book says there are 16 different possibilities. Is there a better way of figuring out how many different events could happen??

mvw
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SuzieQ
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    That's a very common problem people have when starting to learn probability, not a dumb question at all. – GFauxPas Nov 27 '16 at 02:20
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    To make sure you get all the events you can think of going "alphabetically" or "numerically." The first alphabetically is HHHH, then HHHT, then HHTH, then HHTT, ... (if you replace H by 0 and T by 1, this is the same as counting in binary: 0000, 0001, 0010, 0011, ...) – kccu Nov 27 '16 at 02:25
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    There are never dumb questions in math! – Brandon Nov 27 '16 at 02:29
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    You have exactly the correct idea. You just need a little practice! Describing how to ride a bike is easy. Getting good enough to do it correctly is hard. – The Count Nov 27 '16 at 02:43
  • In probability, you need to learn to count without actually counting! That's why they teach combinatorics before probability. When I was in high school, the book Mathematics of Choice: Or, How to Count Without Counting by Ivan Niven helped me a lot – polfosol Nov 27 '16 at 06:23
  • If 2n coins are flipped, the probability that exactly n will be heads is (2n C n)/2^n, or, equivalently, (2n)!/(n!n!2^n). – SuperJedi224 Nov 27 '16 at 12:43

4 Answers4

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For every toss you have two different outcomes, there are four tosses, so you have $2\cdot 2 \cdot 2 \cdot 2 = 2^4 = 16$ different outcomes in total.

You could draw a binary tree to visualize the different roads to an outcome.

binary tree For this problem it might be helpful to consider the not interesting events, here the ones if no tail ($1$ case) and exactly one tail ($4$ cases) is tossed, so there must be $16-5 = 11$ events with at least two tails.

Image source: Link

mvw
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  • That is very helpful, but how do I figure out how many tosses there are that have at least 2 tails, without writing them all out? Not sure what a binary tree is? – SuzieQ Nov 27 '16 at 02:24
  • @SuzieQ I added an image of such a tree. – mvw Nov 27 '16 at 02:53
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    oh my gosh that makes sense to me!! Thanks so much!! I do so much better when I can "see" how things work. Just as in formulas, I can memorize all I want, but until I understand why they work, they won't ever make sense. Thanks again!! – SuzieQ Nov 27 '16 at 02:54
  • @SuzieQ: When used in probability, such trees are very helpful when you first get your head around the sample space. Drawing them out will make it much easier to think about the problem, at least until you can see them mentally. – user21820 Nov 27 '16 at 07:14
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If you want to list all the events, the important thing is to do it systematically so you don't miss any.

For instance, you can count in binary:

$$0000 \\ 0001 \\ 0010 \\ 0011 \\ 0100 \\ 0101 \\ 0110 \\ 0111 \\ 1000 \\ 1001 \\ 1010 \\ 1011 \\ 1100 \\ 1101 \\ 1110 \\ 1111$$

and then replace each $0$ by an H and each $1$ by a T (or the other way around).

(By the way, at the end, it's a good idea to count your entries to double-check that you didn't miss one. In this case, the total number of entries should be $2^4=16.)$

Mitchell Spector
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  • great visual example. – Brandon Nov 27 '16 at 02:28
  • I have been doing this kind of nonsense as part of my research for awhile now, and it NEVER occurred to me write everything in binary to be sure to cover all cases. Brilliant idea. – The Count Nov 27 '16 at 02:37
  • @TheCount ya until you have thousands of counts lol – Brandon Nov 27 '16 at 02:40
  • @Brandon If you have thousands of events that you need to enumerate, you'll probably be writing a computer program for it, not doing it by hand. But you'll still need to do it systematically, likely using this idea or some variant of it. (And historically people did do things like that by hand, of course.) – Mitchell Spector Nov 27 '16 at 02:41
  • @Brandon that's what computer programs are for. – The Count Nov 27 '16 at 02:41
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    @MitchellSpector whoa, that was uncanny. – The Count Nov 27 '16 at 02:42
  • As a statistician I never knew of such programs ;) My original comment was quite sarcastic, but thanks! – Brandon Nov 27 '16 at 02:50
  • @MitchellSpector: I'd say that once you can even write a program to count such things, you would have known enough recursion to be well aware of binary counting. – user21820 Nov 27 '16 at 07:12
  • @user21820 That's why I didn't bring this up in the answer -- it didn't seem appropriate to the actual question, which I interpreted as being about how to do a small-scale homework problem intended to be worked out by hand. – Mitchell Spector Nov 27 '16 at 07:15
  • @MitchellSpector: Ah oops I didn't notice you were the poster of the answer haha.. Yeap I fully agree. Though at a later stage it is always a good idea to see how iteration and recursion are related, which is very concrete for a program. – user21820 Nov 27 '16 at 07:17
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As mvw stated, there are a total of 16 different outcomes, which you will use as your denominator.

Now you need to find the total outcomes for at least 2 tails occurring among 4 coin flips. Then divide this by the total amount of outcomes (16).

There are ${^4C_2} + {^4C_3} + {^4C_4} = 6+4+1 = 11$ ways to get at least 2 tails.

Thus $11/16$ or $68.75\%$ chance of at least 2 tails occurring among 4 coin flips

PS: $^nC_r$ is the binomial coefficient, also written as $\binom n r$ or $\frac{n!}{r!(n-r)!}$, and counts ways to select $r$ things from a set of $n$; in this case, the coin tosses which turn up heads.

Graham Kemp
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Brandon
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  • What does the C mean in your example? I'm in a beginning class, so struggling quite a bit.. – SuzieQ Nov 27 '16 at 02:31
  • That means choose, so 4 choose 2 and so on... It is used to compute combinations, have you seen the formula for computing combinations? If not no worries, don't want to confuse you but you will learn it eventually. – Brandon Nov 27 '16 at 02:32
  • Yes I have, but wasn't sure how that would apply to this situation... you're talking about C(n,k)? – SuzieQ Nov 27 '16 at 02:34
  • So we do 4C2 + 4C3 + 4C4, because that's saying "choose 2 tails from 4 throws, then choose 3 tails from 4 throws, then 4 tails from 4 throws", because the question asks "at least 2 tails".. at least implies 2 or more. in this case, 4 total coin flips so it could be 2 tails, 3 tails or 4 tails. – Brandon Nov 27 '16 at 02:35
  • @SuzieQ yes C(n,k) – Brandon Nov 27 '16 at 02:35
  • C(n,k) applies to this situation because you have to consider all the combinations of flipping a coin 4 times. We found that all the combinations are 16. then we used the same formula to find all the combinations of at least 2 tails occurring among 4 throws – Brandon Nov 27 '16 at 02:38
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Partition the possible results into three sets: $T_2$ (exactly two tails occur), $T_{>2}$ (more than two tails occur), and $T_{<2}$ (fewer than two tails occur). These sets are exclusive and exhaustive, so $P(T_2)+P(T_{>2})+P(T_{<2})=1$. Furthermore, $P(T_{>2})=P(H_{<2})$, and by symmetry, $P(T_2)+2P(T_{<2})=1$, so $P(T_{2})=1-2P(T_{<2})$ It’s not hard to find the probability P(T_2), which is the probability that the sequence of throws is an arrangement of $2$ $H$’s and $2$ $T$’s, of which there are $4\choose2$ many, giving a probability of ${4\choose2} / 2^4=\frac38$, so $P(T_{<2})=\frac{1-\frac38}2=\frac5{16}$, and therefore the probability of having not fewer than $2$ tails is $1-\frac5{16}=\frac{11}{16}$.

Steve Kass
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