Uncanny Coincidences

Many years ago I dreamt the a reptile, a komodo dragon choose creature. The following day when walking with a shopping mall I quit dead in mine tracks once I saw a sculpture the that very animal in mine dream. Earlier then the experience blew mine mind.

Eerie coincidences like this room usually memmorable and because us ascribe mystical properties to desires we often tend to invest castle with definition (hidden or otherwise). The fact is, however, our innumeracy that probability concept has gained the better of us. Coincidences, yet uncanny, room to it is in expected and how often or how much we must expect them can be obtained with precision.

Take the simplest example: coin tosses. We know that periodically we gain a operation of 2, 3, 5, and even more, of top or tails. Nothing particularly extraordinary about that. However what about getting 100 tails or heads in a row? presume the coin is fair/unbiased wouldn"t that take a miracle? Let"s find out.


Let"s start at the beginning. How countless times carry out we have to flip a coin to attain a 50% opportunity of gaining at the very least one heads (yes, Virginia, that"s spelled with an s)? The probability of obtaining heads on a coin toss is 50% therefore we require only toss the once.

Let"s relocate up one notch. How countless times perform we have to flip a coin to attain a 50% probability of obtaining two top in a row at least once? If us flip a coin twice, the probability of gaining heads top top both tries is (0.5)(0.5) = 0.25. Therefore, the probability of not getting two heads = 1 - 0.25 = 0.75. If we flip the coin three times, the probability of not obtaining two heads on the an initial and 2nd flips is together we"ve checked out 0.75. An in similar way the probability the not obtaining two heads on the 2nd and third flips is also 0.75. Therefore, the probability of not obtaining two heads in a row in 3 coin flips is (0.75)(0.75) = 0.5625. With four tosses the probability of not getting any type of pair of heads in a heat is (0.75)(0.75)(0.75) = 0.421875. And also so the probability of having at least one pair of top in a heat in four flips = 1 - 0.421875 = 0.578125 or around 58%. Thus we should toss a coin 4 times to attain a 50% opportunity of acquiring two top in a row at least once.

What around a operation of 10 heads? We begin again through the minimum number of coin flips. In this instance it"s ten. The probability of gaining 10 heads in a heat = 0.510 = 0.0009765625. And so the probability the not obtaining 10 top in a row is its complement or 1 - 0.0009765625 = 0.9990234375. Keep in mental we"ve flipped the coin ten times. On the 11th flip the probability that 10 heads in a heat is the probability of acquiring ten top on the 2nd to 11th flip. ~ above the 12th upper and lower reversal we"re trying to find the probability of getting ten top on the 3rd to 12th flip. And so on. Therefore after having actually tossed the coin 11 time the probability the not getting ten top in a row is the probability the not getting ten heads in a heat on the first to 10th upper and lower reversal multiplied by the probability that not getting ten heads on the 2nd to 11th flip. That"s 0.99902343752. And on the 12th upper and lower reversal the probability = 0.99902343753. Take note and remember the exponent in the equation vis-a-vis the variety of coin flips actually made. It"s critical distinction.

If, after originally flipping the coin nine times, we toss the a hundreds times an ext the probability the NOT getting 10 top in a heat = 0.9990234375100. As such the probability of getting tens top in a heat at the very least once = 1 - 0.9990234375100 = .0923 approximately. That"s much less than 10%. What if us flip the coin 1000 times rather of 100? The probability climate becomes 1 - 0.99902343751000, i beg your pardon comes out to approximately 62%. Therefore to accomplish a 50% chance of getting 10 heads in a heat at least once we"d have to flip a coin somewhere in between 100 come 1000 times. But what is the exactly? Is the 500 times? 972? Is there a means to nail that number other than searching for it through trial and error? Well, I"m happy you asked.

Let:x = number of heads in a row we"re interested in ns = chances/probability of getting x top in a heat at the very least oncef = coin flips necessary to accomplish p

Actually f isn"t yes, really the exact variety of flips we need to attain p, together we saw over in the ten heads example. There"s one initial setup wherein we should flip the coin x - 1 times. Therefore in the situation of a operation of ten top we said we had actually to flip the coin 9 times. And then ~ above the tenth upper and lower reversal (that"s once f = 1) we can start computing for the probability of having actually x top in a row.

LetF = exact variety of coin flips crucial to achieve p

Because that the important "initial setup" of x - 1 flips we obtainF = f + x - 1

As we"ve watched above:

Probability of gaining x top in a heat = 0.5xProbability that not getting x heads in a row = 1 - 0.5xp = 1 - (1 - 0.5x)f

Solving for f us obtain:

(1 - 0.5x)f = 1 - ns log <(1 - 0.5x)f> = log(1 - p) (f)log (1 - 0.5x) = log(1 - p) f = log(1 - p) / log in <(1 - 0.5x)

Let"s take the formula out for a check run and also see what we get.

You are watching: Probability of flipping two heads in a row

p = 50%xf (rounded up)F (rounded up)
100≈ 8.79 x 1029≈ 8.79 x 1029

Instead of p = 50%, let"s view how many times we should toss a coin when p = 99.9% --very nearby to near certainty of acquiring x heads in a heat at least once. Applying the exact same equations we achieve the adhering to values.

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p = 99.9%xf (rounded up)F (rounded up)
100≈ 8.76 x 1030≈ 8.76 x 1030
1000≈ 7.4 x 10301≈ 7.4 x 10301

So getting a operation of 100 heads is fairly possible. And getting 1000 in a heat is theoretically achievable--if we deserve to flip coins quick enough before the cosmos runs the end of gas. Probably someone"s currently doing a simulation on a computer.


Instead of one experiment with an early stage x-1 flips and also then including one new toss and getting the probability that the last x tosses, we deserve to do one collection of x variety of coin flips, and also then perform another collection of x flips, and then an additional set, and also so on. In together a instance f is the variety of sets the x flips and therefore the total variety of actual coin flips is (f)(x). Example: because that x = 10 and also f = 7000, F = 7000 (10) = 70,000. The equation for f stays the same. Us just finish up with a larger number of coin tosses.

If you take it a look in ~ the values of x and also F in the tables, you"ll an alert a trend. Anytime x is incremented through 10 (i.e., x + 10), F increases by three orders of magnitude. In truth F boosts by 1024 or 210. As you can see that exponent is same to the difference in the values of x. Thus, if we boost x indigenous 20 come 50 F will boost by a variable of 230 because 50 - 20 = 30. Let"s write out the formula because that F in terms of the difference in values of x.

Letxi >= 10xn > xifxn = f offered x = xnfxi = f offered x = xi

for a details p, fxn = (2(xn - xi))(fxi)

Example: xi = 100xn = 500From the table whereby p = 99.9% we discover that for x = 100, f = 8.76 x 1030Therefore, for x = 500 f = (2(500 - 100))(8.76 x 1030) = 2.26 x 10151