You might have noticed that before the commencement of a cricket match, a decision is to be made, which team would bat or bowl first. How is this done? You see that the captains of the two teams participate in a coin toss wherein they pick one side of a coin each, that is head or tail. The umpire tosses the coin in the air. The team which wins the toss gets to make the decision of batting or bowling first. This is one of the most common applications of the coin toss experiment.

Why do you think this method is used? This is because the possibility of obtaining a Head in a coin toss is as likely as obtaining a tail, that is, 50%. So when you toss one coin, there are only two possibilities – a head (H) or a tail (L). However, what if you want to toss 2 coins simultaneously? Or say 3, 4 or 5 coins? The outcomes of these coin tosses will differ. Let us learn more about the coin toss probability formula.

**Coin Toss Probability**

Probability is the measurement of chances – the likelihood that an event will occur. If the probability of an event is high, it is more likely that the event will happen. It is measured between 0 and 1, inclusive. So if an event is unlikely to occur, its probability is 0. And 1 indicates the certainty for the occurrence.

Now if I ask you what is the probability of getting a Head when you toss a coin? Assuming the coin to be fair, you straight away answer 50% or ½. This is because you know that the outcome will either be head or tail, and both are equally likely. So we can conclude here:

Number of possible outcomes = 2

Number of outcomes to get head = 1

Probability of getting a head = ½

Hence,

### Solved Examples

**Question**: Two fair coins are tossed simultaneously. What is the probability of getting only one head?

**Solution**:

When 2 coins are tossed, the possible outcomes can be {HH, TT, HT, TH}.

Thus, the total number of possible outcomes = 4

Getting only one head includes {HT, TH} outcomes.

So number of desired outcomes = 2

Therefore, probability of getting only one head

**Question**: Three fair coins are tossed simultaneously. What is the probability of getting at least 2 tails?

**Solution**:

When 3 coins are tossed, the possible outcomes can be {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

Thus, total number of possible outcomes = 8

Getting at least 2 tails includes {HTT, THT, TTH, TTT} outcomes.

So number of desired outcomes = 4

Therefore, probability of getting at least 2 tails =

## What Are Coin Toss Probability Formulas?

The probability of an event is defined as the ratio of the number of favorable outcomes to the total number of outcomes. The same applies to the coin toss probability formula as well.

Probability = Number of favorable outcomes/Total number of outcomes

When a coin is tossed, there are only two possible outcomes. Therefore, using the probability formula:

- On tossing a coin, the probability of getting head is:

P(Head) = P(H) = 1/2 - Similarly, on tossing a coin, the probability of getting a tail is:

P(Tail) = P(T) = 1/2

## Solved Examples Using Coin Toss Probability Formulas

**Example 1:** Coin-A is tossed 100 times, and the relative occurrence of Tails is 0.5. Coin-B is tossed an unknown number of times, but it is known that the relative occurrence of Heads is 0.48. Which coin is fairer?

**Solution: **

To find: Fairer coin

Given: For coin A:

Total number of outcomes for coin = 100

Relative Occurrence of tails = 0.5

For coin B:

Total number of outcomes for coin = unknown

Relative Occurrence of tails = 0.48

It is not possible to comment on the fairness of Coin-B, because the number of times it was tossed is not known.

On the other hand, Coin-A seems to be fair, as the relative occurrence of tails over a large number of tosses is almost 1/2.

**Answer: **It cannot be commented on which is a fairer coin based on given information.

**Example 2:** On tossing a coin twice, what is the probability of getting only one tail?

**Solution:**

To find: the probability of getting only one tail

On tossing a coin twice, the possible outcomes are {HH, TT, HT, TH}

Therefore, the total number of outcomes is 4

Getting only one tail includes {HT, TH}

Therefore, the number of favorable outcomes is 2

Hence, using the coin toss probability formula,

the probability of getting exactly one tail is 2/4 = 1/2

**Answer: **The** **probability of getting exactly one tail = 1/2.

**Binomial Distribution**

*Learning Objectives*

- Define binomial outcomes
- Compute the probability of getting X successes in N trials
- Compute cumulative binomial probabilities
- Find the mean and standard deviation of a binomial distribution

When you flip a coin, there are two possible outcomes: heads and tails. Each outcome has a fixed probability, the same from trial to trial. In the case of coins, heads and tails each have the same probability of 1/2. More generally, there are situations in which the coin is biased, so that heads and tails have different probabilities. In the present section, we consider probability distributions for which there are just two possible outcomes with fixed probabilities summing to one. These distributions are called binomial distributions.

**A Simple Example**

The four possible outcomes that could occur if you flipped a coin twice are listed below in Table 1. Note that the four outcomes are equally likely: each has probability 1/4. To see this, note that the tosses of the coin are independent (neither affects the other). Hence, the probability of a head on Flip 1 and a head on Flip 2 is the product of P(H) and P(H), which is 1/2 x 1/2 = 1/4. The same calculation applies to the probability of a head on Flip 1 and a tail on Flip 2. Each is 1/2 x 1/2 = 1/4.

Table 1. Four Possible Outcomes.

Outcome | First Flip | Second Flip |
---|---|---|

1 | Heads | Heads |

2 | Heads | Tails |

3 | Tails | Heads |

4 | Tails | Tails |

The four possible outcomes can be classified in terms of the number of heads that come up. The number could be two (Outcome 1), one (Outcomes 2 and 3) or 0 (Outcome 4). The probabilities of these possibilities are shown in Table 2 and in Figure 1. Since two of the outcomes represent the case in which just one head appears in the two tosses, the probability of this event is equal to 1/4 + 1/4 = 1/2. Table 2 summarizes the situation.

Table 2. Probabilities of Getting 0, 1, or 2 Heads.

Number of Heads | Probability |
---|---|

0 | 1/4 |

1 | 1/2 |

2 | 1/4 |

Figure 1. Probabilities of 0, 1, and 2 heads.

Figure 1 is a discrete probability distribution: It shows the probability for each of the values on the X-axis. Defining a head as a “success,” Figure 1 shows the probability of 0, 1, and 2 successes for two trials (flips) for an event that has a probability of 0.5 of being a success on each trial. This makes Figure 1 an example of a *binomial distribution*.

The Formula for Binomial Probabilities

The binomial distribution consists of the probabilities of each of the possible numbers of successes on N trials for independent events that each have a probability of π (the Greek letter pi) of occurring. For the coin flip example, N = 2 and π = 0.5. The formula for the binomial distribution is shown below:

where P(x) is the probability of x successes out of N trials, N is the number of trials, and π is the probability of success on a given trial. Applying this to the coin flip example,

If you flip a coin twice, what is the probability of getting one or more heads? Since the probability of getting exactly one head is 0.50 and the probability of getting exactly two heads is 0.25, the probability of getting one or more heads is 0.50 + 0.25 = 0.75.

Now suppose that the coin is biased. The probability of heads is only 0.4. What is the probability of getting heads at least once in two tosses? Substituting into the general formula above, you should obtain the answer .64.

**Cumulative Probabilities**

We toss a coin 12 times. What is the probability that we get from 0 to 3 heads? The answer is found by computing the probability of exactly 0 heads, exactly 1 head, exactly 2 heads, and exactly 3 heads. The probability of getting from 0 to 3 heads is then the sum of these probabilities. The probabilities are: 0.0002, 0.0029, 0.0161, and 0.0537. The sum of the probabilities is 0.073. The calculation of cumulative binomial probabilities can be quite tedious. Therefore we have provided a binomial calculator to make it easy to calculate these probabilities.

Mean and Standard Deviation of Binomial Distributions

Consider a coin-tossing experiment in which you tossed a coin 12 times and recorded the number of heads. If you performed this experiment over and over again, what would the mean number of heads be? On average, you would expect half the coin tosses to come up heads. Therefore the mean number of heads would be 6. In general, the mean of a binomial distribution with parameters N (the number of trials) and π (the probability of success on each trial) is:

μ = Nπ

where μ is the mean of the binomial distribution. The variance of the binomial distribution is:

σ^{2} = Nπ(1-π)

where σ^{2} is the variance of the binomial distribution.

Let’s return to the coin-tossing experiment. The coin was tossed 12 times, so N = 12. A coin has a probability of 0.5 of coming up heads. Therefore, π = 0.5. The mean and variance can therefore be computed as follows:

μ = Nπ = (12)(0.5) = 6

σ^{2} = Nπ(1-π) = (12)(0.5)(1.0 – 0.5) = 3.0.

Naturally, the standard deviation (σ) is the square root of the variance (σ^{2}).

## Coin Flip Probability Calculator

Welcome to the **coin flip probability calculator**, where you’ll have the opportunity to learn how to calculate the probability of obtaining a set number of heads (or tails) from a set number of tosses. This is one of the fundamental classical probability problems, which later developed into quite a big topic of interest in mathematics. For example, maybe you like Batman, and know of one of his many villains, Two-Face? You’d think that his name comes from the fact that half of his face is burnt, but no! (Okay, maybe a little bit.) He has a lucky coin that he always flips before doing anything. As this coin has two faces on it, his coin toss probability of getting a head is 1. Better not get on the wrong side (or face) of him!

We here at Omni Calculator wonder what the odds are that you’ll toss a head to your Witcher…

## Classical probability

**The probability** of some event happening is a mathematical (numerical) representation of how likely it is to happen, where a probability of 1 means that an event will always happen, while a probability of 0 means that it will never happen. **Classical probability problems** often need to you find how often one outcome occurs versus another, and how one event happening affects the probability of future events happening. When you look at all the things that may occur, the formula (just as our **coin flip probability formula**) states that

`probability = (no. of successful results) / (no. of all possible results)`

.

**Take a die roll as an example**. If you have a standard, 6-face die, then there are six possible outcomes, namely the numbers from 1 to 6. If it is a fair die, then **the likelihood of each of these results is the same**, i.e., 1 in 6 or `1 / 6`

. Therefore, the probability of obtaining 6 when you roll the die is `1 / 6`

. The probability is the same for 3. Or 2. You get the drill. If you don’t believe me, take a dice and roll it a few times and note the results. Remember that **the more times you repeat an experiment, the more trustworthy the results**. So go on, roll it, say, a thousand times. We’ll be waiting here until you get back to tell us we’ve been right all along.

**But what if you repeat an experiment a hundred times and want to find the odds that you’ll obtain a fixed result at least 20 times?**

Let’s look at another example. Say that you’re a teenager straight out of middle school and decide that you want to meet the love of your life this year. More specifically, you want to ask ten girls out and go on a date with only four of them. One of those has got to be the one, right? The first thing you have to do in this situation is look in the mirror and **rate how likely a girl is to agree to go out with you** when you start talking to her. If you have problems with assessing your looks fairly, go downstairs and let your grandma tell you **what a handsome, young gentleman you are**. So a solid `9 / 10`

then.

As you only want to go on four dates, that means you only want four of your romance attempts to succeed. This has an outcome of `9 / 10`

. This means that you want the other six girls to reject you, which, based on your good looks, has only a `1 / 10`

change of happening (The sum of all events happening is always equal to 1, so we get this number by subtracting `9 / 10`

from 1). If you multiply the probability of each event by itself the number of times you want it to occur, you get the chance that your scenario will come true. In this case, your odds are `210 * (9 / 10)`

, where the ^{4} * (1 / 10)^{6} = 0.000137781`210`

comes from the number of possible fours of girls among the ten that would agree. Not very likely to happen, is it? Maybe you should try being less beautiful!

## How to calculate probability?

“*Hey man, but girls and coins are two different things! I should know, I’ve seen at least one of each.*” Well, let me explain that **these two problems are basically the same**, that is, from the point of view of mathematics. Whether you want to toss a coin or ask a girl out, there are only two possibilities that can occur. In other words, if you assign the success of your experiment, be it getting tails or the girl agreeing to your proposal, to one side of the coin and the other option to the back of the coin, **the coin toss probability will determine the answer**. It all boils down to getting your hands on a coin that is weighted appropriately.

Let’s look at a step-by-step example to see how to calculate the probability of an event using the coin toss probability calculator:

**Determine your experiment**. What are the two possibilities that can happen? Assign*heads*to one of them and*tails*to the other.- How many times are you going to repeat the experiment? Put that number as the
*number of flips*in the calculator. - What do you want to achieve? An exact number of successful tries? At least a set number of successful tries? Or no more than a certain number of successful tries? Choose the correct option from the list.
- How many successful (exact, at least, or at most) attempts you want to have? Put that number before
*heads*. - (Optional) If your
*heads*and*tails*don’t have the same probability of happening, go into advanced mode, and set the right number in the new field. Remember that in classical probability, the likelihood cannot be smaller than 0 or larger than 1. - The coin flip probability calculator will automatically calculate the chance for your event to happen.

## What is the Coin Toss Probability Formula?

Let’s look at a few things about flipping a coin before studying the coin toss probability formula. There are two potential consequences when flipping a coin: heads or tails. We don’t know which way the coin will land on a given toss, but we do know it will either be Head or Tail. Tossing a coin, on the other hand, is a random experiment since you know the set of outcomes but not the exact outcome for each random experiment execution. The probability formula for a coin flip can be used to calculate the probability of some experiment.

**Coin Toss Probability**

Each result has a predetermined likelihood, which remains constant from trial to trial. Heads and tails share the same chance of 1/2 when it comes to coins. In addition, there are cases where the coin is skewed, resulting in varying odds for heads and tails. In this part, we’ll look at probability distributions with just two potential outcomes and fixed probabilities that add up to one. Binomial distributions are the name for certain types of distributions.

**Here are Some Examples of Problems Involving Coin Toss Chance.**

If a coin is flipped, there are two potential outcomes: a ‘head’ (H) or a ‘tail’ (T), and it is difficult to determine whether the toss will end in a ‘head’ or a ‘tail.’

Assuming the coin is equal, then the coin probability is 50% or 12

This is because you know the result would be either head or tail, and both are equally probable.

**The Probability for Equally Likely Outcomes is:**

Total number of favourable outcomes Total number of possible outcomes

Where,

Total number of possible outcomes = 2

(i) Coin toss probability formula for heads

If the favourable outcome is head (H).

A number of favourable outcomes = 1.

Therefore, P(getting head)

**Coin Probability Problems**

**1. A Coin is Tossed Thrice at Random. What is the Probability of Getting**

- At Least One Head
- The Same Face?

**Solution:**

The possible outcomes of a given event are {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

So, a total number of outcomes = 8.

(i) Number of favourable outcomes for event E

= Number of outcomes appears at least one head

= 4 (as HHH, HHT, HTH, HTT are having at least one head).

= 4/8

= 1/2

So, by definition, P(F) = 1/4

(ii) Number of favourable outcomes for event E

= Number of outcomes having the same face

= 2 (as HHH, TTT has the same face).

= 2/8

= 1/4

So, by definition, P(F) = ¼

**2. What is the Probability of Getting a Head When Tossing a Coin?**

**Solution:**

When a single coin is tossed, the possible outcomes can be {H, T}.

Thus, the total number of possible outcomes = 2

Getting head includes {H} outcomes.

So the number of favourable outcomes = 1

Therefore,

the probability of getting head is,

**3. Two Coins are Tossed Randomly 150 Times and it is Found That Two Tails Appeared 60 Times, One Tail Appeared 74 Times and No Tail Appeared 16 Times.**

If two coins are tossed at random, then what is the probability of,

- Getting 2 Tails
- Getting 1 Tail
- Getting 0 Tail

**Solution:**

Total number of trials = 150

Number of times 2 tails appear = 60

Number of times 1 tail appears = 74

Number of times 0 tail appears = 16

Let T_{1}, T_{2}, and T_{3} be the events of getting 2 tails, 1 tail and 0 tail respectively.

(i) P(getting 2 tails)

The number of times two tails appear is 60.

The probability of getting one tail is 0.4933

(iii) P(getting 0 tail)

The number of times zero tail appear is 16

### FAQs (Frequently Asked Questions)

**Q.1) What is the Sample Space When Four Coins are Tossed?**

Answer: Four coins are tossed simultaneously, then the sample space is,

Tossed coins = 4

Hence, The number of faces = 2^{4}= 16

{HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

**Q.2) Where Did the Coin Toss Originate From?**

Answer: The view of a random result as an act of divine will is the historical basis of coin-flipping.

The Romans named coin flipping Navia Aut Caput (“ship or head”) since certain coins had a ship on one hand and the emperor’s head on the other. This was known as cross and pile in England.

**Q.3) Who Won the Super Bowl 2020 Coin Toss?**

Answer: In the 2020 year, the 49ers won the Super Bowl LIV coin toss after picking tails.

## Introduction to Probability and Statistics

## Calculation and Chance

Most experimental searches for paranormal phenomena are statistical in nature. A subject repeatedly attempts a task with a known probability of success due to chance, then the number of actual successes is compared to the chance expectation. If a subject scores consistently higher or lower than the chance expectation after a large number of attempts, one can calculate the probability of such a score due purely to chance, and then argue, if the chance probability is sufficiently small, that the results are evidence for the existence of some mechanism (precognition, telepathy, psychokinesis, cheating, etc.) which allowed the subject to perform better than chance would seem to permit.

Suppose you ask a subject to guess, before it is flipped, whether a coin will land with heads or tails up. Assuming the coin is fair (has the same probability of heads and tails), the chance of guessing correctly is 50%, so you’d expect half the guesses to be correct and half to be wrong. So, if we ask the subject to guess heads or tails for each of 100 coin flips, we’d expect about 50 of the guesses to be correct. Suppose a new subject walks into the lab and manages to guess heads or tails correctly for 60 out of 100 tosses. Evidence of precognition, or perhaps the subject’s possessing a telekinetic power which causes the coin to land with the guessed face up? Well,…no. In all likelihood, we’ve observed nothing more than good luck. The probability of 60 correct guesses out of 100 is about 2.8%, which means that if we do a large number of experiments flipping 100 coins, about every 35 experiments we can expect a score of 60 or better, purely due to chance.

But suppose this subject continues to guess about 60 right out of a hundred, so that after ten runs of 100 tosses—1000 tosses in all, the subject has made 600 correct guesses. The probability of *that* happening purely by chance is less than one in seven billion, so it’s time to start thinking about explanations other than luck. Still, improbable things happen all the time: if you hit a golf ball, the odds it will land on a given blade of grass are millions to one, yet (unless it ends up in the lake or a sand trap) it is certain to land on *some* blades of grass.

Finally, suppose this “dream subject” continues to guess 60% of the flips correctly, observed by multiple video cameras, under conditions prescribed by skeptics and debunkers, using a coin provided and flipped by The Amazing Randi himself, with a final tally of 1200 correct guesses in 2000 flips. You’d have to try the 2000 flips more than 5×10^{18} times before you’d expect that result to occur by chance. If it takes a day to do 2000 guesses and coin flips, it would take more than 1.3×10^{16} years of 2000 flips per day before you’d expect to see 1200 correct guesses due to chance. That’s more than a million times the age of the universe, so you’d better get started soon!

Claims of evidence for the paranormal are usually based upon statistics which diverge so far from the expectation due to chance that some other mechanism seems necessary to explain the experimental results. To interpret the results of our RetroPsychoKinesis experiments, we’ll be using the mathematics of probability and statistics, so it’s worth spending some time explaining how we go about quantifying the consequences of chance.

Note to mathematicians:The following discussion of probability is deliberately simplified to consider only binomial and normal distributions with a probability of 0.5, the presumed probability of success in the experiments in question. I decided that presenting and discussing the equations for arbitrary probability would only decrease the probability that readers would persevere and arrive at an understanding of the fundamentals of probability theory.

## Twelve and a half cents: *one bit!*

In slang harking back to the days of gold doubloons and pieces of eight, the United States quarter-dollar coin is nicknamed “two bits”. The Fourmilab radioactive random number generator produces a stream of binary ones and zeroes, or *bits*. Since we expect the generator to produce ones and zeroes with equal probability, each bit from the generator is equivalent to a coin flip: heads for one and tails for zero. When we run experiments with the generator, in effect, we’re flipping a *binary coin*, **one bit**—twelve and a half cents!

(We could, of course, have called zero heads and one tails; since both occur with equal probability, the choice is arbitrary.) Each bit produced by the random number generator is a flip of our one-bit coin. Now the key thing to keep in mind about a genuine random number generator or flip of a fair coin is that it has *no memory* or, as mathematicians say, each bit from the generator or flip is *independent*. Even if, by chance, the coin has come up heads ten times in a row, the probability of getting heads or tails on the next flip is precisely equal. Gamblers who’ve seen a coin come up heads ten times in a row may believe “tails is way overdue”, but the coin doesn’t know and couldn’t care less about the last ten flips; the next flip is just as likely to be the eleventh head in a row as the tail that breaks the streak.

Even though there is no way whatsoever to predict the outcome of the next flip, if we flip a coin a number of times, the laws of probability allow us to predict, with greater accuracy as the number of flips increases, the probability of obtaining various results. In the discussion that follows, we’ll ignore the order of the flips and only count how many times the coin came up heads. Since heads is one and tails is zero, we can just add up the results from the flips, or the bits from the random generator.

### Four Flips

Suppose we flip a coin four times. Since each flip can come up heads or tails, there are 16 possible outcomes, tabulated below, grouped by the number of heads in the four flips.

**Number of Ways** summarises how many different ways the results of the four flips could end up with a given number of heads. Since the only way to get zero heads is for all four flips to be tails, there’s only one way that can occur. One head out of four flips can happen four different ways since each of the four flips could have been the head. Two heads out of four flips can happen six different ways, as tabulated. And since what’s true of heads applies equally to tails, there are four ways to get three heads and one way to get four.

Mathematically, the number of ways to get *x* heads (or tails) in *n* flips is spoken of as the “number of combinations of *n* things taken *x* at a time”, which is written as:

This, it turns out, can be calculated for any positive integers *n* and *x* whatsoever, as follows.

For example, if we plug in 4 for *n* and 2 for *x*, we get4! / (2! (4 − 2)!) = 4! / (2! 2!) = 24 / (2 2) = 24 / 4 = 6

as expected. Plotting the number of ways we can get different numbers of heads yields the following graph.

## Probability

Since the coin is fair, each flip has an equal chance of coming up heads or tails, so all 16 possible outcomes tabulated above are equally probable. But since there are 6 ways to get 2 heads, in four flips the *probability* of two heads is greater than that of any other result. We express probability as a number between 0 and 1. A probability of zero is a result which cannot ever occur: the probability of getting five heads in four flips is zero. A probability of one represents certainty: if you flip a coin, the probability you’ll get heads *or* tails is one (assuming it can’t land on the rim, fall into a black hole, or some such).

The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of total results of four flips, 16. We can then tabulate the probabilities as follows.

Number of Heads | Number of Ways | Probability |
---|---|---|

0 | 1 | 1/16 = 0.0625 |

1 | 4 | 4/16 = 0.25 |

2 | 6 | 6/16 = 0.375 |

3 | 4 | 4/16 = 0.25 |

4 | 1 | 1/16 = 0.0625 |

Since we are absolutely certain the number of heads we get in four flips is going to be between zero and four, the probabilities of the different numbers of heads should add up to 1. Summing the probabilities in the table confirms this. Further, we can calculate the probability of any collection of results by adding the individual probabilities of each. Suppose we’d like to know the probability of getting fewer than three heads from four flips. There are three ways this can happen: zero, one, or two heads. The probability of fewer than three, then, is the sum of the probabilities of these results, 1/16 + 4/16 + 6/16 = 11/16 = 0.6875, or a little more than two out of three. So to calculate the probability of one outcome **or** another, **sum** the probabilities.

To get probability of one result **and** another from two separate experiments, **multiply** the individual probabilities. The probability of getting one head in four flips is 4/16 = 1/4 = 0.25. What’s the probability of getting one head in each of two successive sets of four flips? Well, it’s just 1/4 × 1/4 = 1/16 = 0.0625.

The probability for any number of heads *x* in any number of flips *n* is thus:

the number of ways in which *x* heads can occur in *n* flips, divided by the number of different possible results of the series of flips, measured by number of heads. But there’s no need to sum the combinations in the denominator, since the number of possible results is simply two raised to the power of the number of flips. So, we can simplify the expression for the probability to:

### More Flips

Let’s see how the probability behaves as we make more and more flips. Since we have a general formula for calculating the probability for any number of heads in any number of flips, we can graph of the probability for various numbers of flips.

In every case, the peak probability is at half the number of flips and declines on both sides, more steeply as the number of flips increases. This is the simple consequence of there being many more possible ways for results close to half heads and tails to occur than ways that result in a substantial majority of heads or tails. The RPKP experiments involve a sequence of 1024 random bits, in which the most probable results form a narrow curve centred at 512. A document giving probabilities for results of 1024 bit experiments with chance expectations greater than one in 100 thousand million runs is available, as is a much larger table listing probabilities for all possible results. (The latter document is more than 150K bytes and will take a while to download, and contains a very large table which some Web browsers, particularly on machines with limited memory, may not display properly.)

### Perfectly Normal

As we make more and more flips, the graph of the probability of a given number of heads becomes smoother and approaches the “bell curve”, or *normal distribution*, as a limit. The normal distribution gives the probability for *x* heads in *n* flips as:

where μ=*n*/2 and σ is the *standard deviation*, a measure of the breadth of the curve which, for equal probability coin flipping, is:

We keep the standard deviation separate, as opposed to merging it into the normal distribution probability equation, because it will play an important rôle in interpreting the results of our experiments. To show how closely the probability chart approaches the normal distribution even for a relatively small number of flips, here’s the normal distribution plotted in red, with the actual probabilities for number of heads in 128 flips shown as blue bars.

The probability the outcome of an experiment with a sufficiently large number of trials is due to chance can be calculated directly from the result, and the mean and standard deviation for the number of trials in the experiment. For additional details, including an interactive probability calculator, please visit the **z** Score Probability Calculator.

## Calculation and Reality

This is all very persuasive, you might say, and the formulas are suitably intimidating, but does the real world actually behave this way? Well, as a matter of fact, it does, as we can see from a simple experiment. Get a coin, flip it 32 times, and write down the number of times heads came up. Now repeat the experiment fifty thousand times. When you’re done, make a graph of the number of 32-flip sets which resulted in a given number of heads. Hmmmm…32 times 50,000 is 1.6 million, so if you flip the coin once a second, twenty-four hours per day, it’ll take eighteen and a half days to complete the experiment….

Instead of marathon coin-flipping, let’s use the same HotBits hardware random number generator our experiments employ. It’s a simple matter of programming to withdraw 1.6 million bits from the generator, divide them up into 50,000 sets of 32 bits each, then compute a histogram of the number of sets containing each possible number of one bits (heads). The results from this experiment are presented in the following graph.

The red curve is the number of runs expected to result in each value of heads, which is simply the probability of that number of heads multiplied by the total number of experimental runs, 50,000. The blue diamonds are the actual number of 32 bit sets observed to contain each number of one bits. It is evident that the experimental results closely match the expectation from probability. Just as the probability curve approaches the normal distribution for large numbers of runs, experimental results from a truly random source will inexorably converge on the predictions of probability as the number of runs increases.

## Ensembles of Experiments: the *X*² (Chi-Square) Test

So far, we’ve seen how the laws of probability predict the outcome of large numbers of experiments involving random data, how to calculate the probability of a given experimental result being due to chance, and how one goes about framing a hypothesis, then designing and running a series of experiments to test it. Now it’s time to examine how to analyse the results from the experiments to determine whether they provide evidence for the hypothesis and, if so, how much.

Since its introduction in 1900 by Karl Pearson, the chi-square (*X*²) test has become the most widely used measure of the significance to which experimental results support or refute a hypothesis. Applicable to any experiment where discrete results can be measured, it is used in almost every field of science. The chi-square test is the final step in a process which usually proceeds as follows.

- Based either on theory or examination of empirical data, a phenomenon is suggested to exist.
- A hypothesis is framed incorporating the supposed phenomenon.
- An experiment is designed to test the hypothesis. The experiment must produce results which differ based on the validity of the hypothesis. The results of the experiment for the hypothesis-false (
*null hypothesis*) case are calculated and, if appropriate, verified by a control experiment in which the hypothesised effect is excluded. - A series of experiments are conducted, and the results tabulated.
- The chi-square test is applied to determine the significance of the discrepancy between the calculated null hypothesis case and the experimental results, and the probability that the discrepancy is the result of chance. If that probability is very low, the experiment provides evidence for the hypothesis.

No experiment or series of experiments can ever *prove* a hypothesis; one can only rule out other hypotheses and provide evidence that assuming the truth of the hypothesis better explains the experimental results than discarding it. In many fields of science, the task of estimating the null-hypothesis results can be formidable, and can lead to prolonged and intricate arguments about the assumptions involved. Experiments must be carefully designed to exclude selection effects which might bias the data.

Fortunately, retropsychokinesis experiments have an easily stated and readily calculated null hypothesis: “The results will obey the statistics for a sequence of random bits.” The fact that the data are prerecorded guarantees (assuming the experiment software is properly implemented and the results are presented without selection or modification) that run selection or other forms of cheating cannot occur. (Anybody can score better than chance at coin flipping if they’re allowed to throw away experiments that come out poorly!) Finally, the availability of all the programs in source code form and the ability of others to repeat the experiments on their own premises will allow independent confirmation of the results obtained here.

So, as the final step in analysing the results of a collection of *n* experiments, each with *k* possible outcomes, we apply the chi-square test to compare the actual results with the results expected by chance, which are just, for each outcome, its probability times the number of experiments *n*.

Mathematically, the *chi-square* statistic for an experiment with *k* possible outcomes, performed *n* times, in which *Y _{1}, Y_{2},… Y_{k}* are the number of experiments which resulted in each possible outcome, where the probabilities of each outcome are

*p*is:

_{1}, p_{2},… p_{k}It’s evident from examining this equation that the closer the measured values are to those expected, the lower the chi-square sum will be. Further, from a chi-square sum, the probability *Q* that the *X*² sum for an experiment with *d* degrees of freedom (where *d*=*k*−1, one less the number of possible outcomes) is consistent with the null hypothesis can be calculated as:

Where Γ is the generalisation of the factorial function to real and complex arguments:

Unfortunately, there is no closed form solution for *Q*, so it must be evaluated numerically. If your Web browser supports JavaScript, you can use the Chi-Square Calculator to calculate the probability from a chi-square value and number of possible outcomes, or calculate the chi-square from the probability and the number of possible outcomes.

In applying the chi-square test, it’s essential to understand that only very small probabilities of the null hypothesis are significant. If the probability that the null hypothesis can explain the experimental results is above 1%, an experiment is generally not considered evidence of a different hypothesis. The chi-square test takes into account neither the number of experiments performed nor the probability distribution of the expected outcomes; it is valid only as the number of experiments becomes large, resulting in substantial numbers for the most probable results. If a hypothesis is valid, the chi-square probability should converge on a small value as more and more experiments are run.

Now let’s examine an example of how the chi-square test identifies experimental results which support or refute a hypothesis. Our simulated experiment consists of 50,000 runs of 32 random bits each. The subject attempts to influence the random number generator to emit an excess of one or zero bits compared to the chance expectation of equal numbers of zeroes and ones. The following table gives the result of a control run using the random number generator without the subject’s attempting to influence it. Even if the probability of various outcomes is easily calculated, it’s important to run control experiments to make sure there are no errors in the experimental protocol or apparatus which might bias the results away from those expected. The table below gives, for each possible number of one bits, the number of runs which resulted in that count, the expectation from probability, and the corresponding term in the chi-square sum. The chi-square sum for the experiment is given at the bottom.

Entering the *X*² sum of 23.8872 and the degrees of freedom (32, one less than the 33 possible outcomes of the experiment) into the Chi-Square Calculator gives a probability of 0.85. This falls within the “fat region” of the probability curve, and thus supports the null hypothesis, just as we expected.

Next, we invite our subject to attempt to influence the random output of our generator. How? *Hypotheses non fingo.* Let’s just presume that by some means: telekinesis, voodoo, tampering with the apparatus when we weren’t looking—whatever, our subject is able to bias the generator so that out of every 200 bits there’s an average of 101 one bits and 99 zeroes. This seemingly subtle bias would result in an experiment like the following.

The graph of the result versus the expectation doesn’t show any terribly obvious divergence from the expectation, yet the chi-square test unambiguously fingers the bias. A *X*² sum of 89.6775 in an experiment with 33 possible outcomes (32 degrees of freedom) has a probability of occurring by chance of 0.00000022—about two in ten million—a highly significant result, worthy of follow-up experiments and investigation of possible mechanisms which might explain the deviation from chance.

## Coin Flip Probability – Explanation & Examples

The image of a flipping coin is invariably connected with the concept of “chance.” So it is no wonder that coin flip probabilities play a central role in understanding the basics of probability theory.

*Coin flip probabilities deal with events related to a single or multiple flips of a fair coin. A fair coin has an equally likely chance of coming up Heads or Tails.*

It might be advisable to refresh the following concepts to understand the material discussed in this article.

- Set theory.
- Basic Probability theory.
- Independent Events.
- Tree Diagrams.

After reading this article, you should understand:

- What is meant by coin flip probabilities.
- How to calculate probabilities associated with multiple coin flips using sample spaces.
- How to calculate probabilities associated with multiple flips using tree diagrams.
- How to calculate probabilities associated with multiple flips using the formula for probabilities of independent events.

## How to calculate the probability of coin flips

To understand how to calculate the probability of coin flips, we first need to discuss the concept of sample spaces.

### Sample Spaces:

A sample space is a set (i.e., collection) of all possible events in a probabilistic experiment.

### Probabilities of multiple coins flip using tree diagrams

It is more convenient to rely on tree-diagrams to find multiple coin flip probabilities than to use the sample space method in many cases. We illustrate the concept using examples

* Example 3*:

A coin is flipped three times. Draw a tree diagram that represents all possible outcomes. Also, calculate the probabilities of the following events:

- Getting three Heads.
- Getting two Tails.
- Getting no Heads.
- Getting at least one Tails.

Solution:

1) **Getting three Heads**

From the tree diagram, we can see that only one outcome corresponds to the event of getting all three heads. To get probabilities out of a tree diagram, we multiply the probabilities along the branches. So, the probability of getting three heads is

### Multiple flips and independent events

When the number of flips is large, both the tree diagrams and the sample space methods might become too cumbersome. In such cases, we can rely on the fact that multiple flips are independent events. **Two events are said to be independent if one event does not affect the probabilities of the other**. When we flip a coin multiple times, the outcome of any one flip does not affect the other flips’ outcomes, so the

We can use the above expression to solve multiple coin flips’ problems, as shown in the examples below.

* Example 4:* A fair coin is flipped twice. What is the probability of the following events:

- Getting at least one Heads.
- Getting at most one Heads.
- Getting Tails twice.
- Getting no Tails.

Solution:

We have already solved this example using the sample space method. Now, we solve it using the concept of independent probabilities.

1) **Getting at least one Heads**

We first find the probability of finding no Heads, i.e., the probability that both flips are Tails.

### Practice Questions:

- A coin is flipped 4 times. Draw a tree diagram to show the probability that three heads and one tail appear?
- Three fair coins are tossed simultaneously. What is the probability of the following:
- The first is the head, and the second is the tail.
- Three heads in a row.
- Two tails and one head.

- Three fair coins are tossed simultaneously. Use a tree diagram to determine the probability of getting:
- At least 2 Tails.
- At most two Heads.
- No Tails at all.

- A fair coin is tossed 5 times. What is the probability of the following events?
- At least one Heads.
- No Tails.
- The coin comes up Heads for the first time after 3 attempts.
- First Heads in the first three attempts.

### Answers:

1)

**How to Determine Probable Outcomes with Coins and Dice**

Although the basic probability formula isn’t difficult, sometimes finding the numbers to plug into it can be tricky. One source of confusion is in counting the number of outcomes, both favorable and possible, such as when tossing coins and rolling dice.

*Tossing coins*

*Tossing coins*

When you flip a coin, you can generally get two possible outcomes: heads or tails. When you flip two coins at the same time — say, a penny and a nickel — you can get four possible outcomes:

When you flip three coins at the same time — say, a penny, a nickel, and a dime — eight outcomes are possible:

Notice the pattern: Every time you add an additional coin, the number of possible outcomes doubles. So if you flip six coins, here’s how many possible outcomes you have: 2 2 2 2 2 2 = 64 The number of possible outcomes equals the number of outcomes per coin (2) raised to the number of coins (6): Mathematically, you have 2^{6} = 64. Here’s a handy formula for calculating the number of outcomes when you’re flipping, shaking, or rolling multiple coins, dice, or other objects at the same time: Number of outcomes per object^{Number of objects} Suppose you want to find the probability that six tossed coins will all fall heads up. To do this, you want to build a fraction, and you already know that the denominator — the number of possible outcomes — is 64. Only one outcome is favorable, so the numerator is 1:

So the probability that six tossed coins will all fall heads up is 1/64. Here’s a more subtle question: What’s the probability that exactly five out of six tossed coins will all fall heads up? Again, you’re building a fraction, and you already know that the denominator is 64. To find the numerator (favorable outcomes), think about it this way: If the first coin falls tails up, then all the rest must fall heads up. If the second coin falls tails up, then again all the rest must fall heads up. This is true of all six coins, so you have six favorable outcomes:

Therefore, the probability that exactly five out of six coins will fall heads up is 6/64, which reduces to 3/32.

*Rolling dice*

*Rolling dice*

When you roll a single die, you can get six possible outcomes: 1, 2, 3, 4, 5, or 6. However, when you roll two dice, this number jumps to 36, as shown in the following figure.

Every time you add an additional die, the number of possible outcomes is multiplied by 6. So if you roll four dice, here’s the number of possible outcomes: 6^{4} = 6 6 6 6 = 1,296 Suppose you want to calculate the possibility of rolling four 6s. The probability is a fraction, and you already know that the denominator of this fraction is 1,296. In this case, only one outcome — all four dice coming up 6 — is favorable, so here’s how you build your fraction:

So the probability that you’ll roll four 6s is 1/1,296 — a very small probability, indeed. Here’s a more interesting question: What’s the probability that all four dice will come up 4, 5, or 6? Again, you’re building a fraction whose denominator is 1,296. To find the numerator, think about it this way: For the first die, there are three favorable outcomes (4, 5, or 6). For the first two dice, there are 3 3 = 9 favorable outcomes as shown here:

For three dice, there are 3 3 3 = 27 favorable outcomes. So for all four dice, there are 3 3 3 3 = 81 favorable outcomes. So

Thus, the probability that all four dice will come up 4, 5, or 6 is 81/1,296. This fraction reduces to 1/16.

✅ Math Formulas ⭐️⭐️⭐️⭐️⭐

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