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Throwing dice is more complicated than tossing coins, as there are more than 2 values. If you throw a single dice, then it can fall six ways, each of which is equally likely if the dice is true. So the probability of getting one particular value is 1/6. If you want either of two values it is 2/6 or 1/3, and so on.

It gets more interesting when you have two dice. One thing that you can do is work out what the total of the dice is. The dice experiment allows you to simulate throwing pairs of dice and see what the result is. This is a good introduction to probability, since you can see which combinations are more likely. But the real world, or even a simulated real world, never matches completely with calculated probability. So how do we calculate it? The first thing is to work out what the range is. You can't have a total less than 2 (both dice being 1) and you can't have a total more than 12 (both dice being 6). The easiest way to see what the probabilities is to write out the possible totals. There are 36 of them in all (6 x 6).

The percentages above are rounded. Compare these with the dice experiment. It won't match exactly, although the more throws you do, the closer it will get.

There are plenty of other conditions that you can have with two dice. Often, the easiest way to get the probability of a condition is to list all possible throws, and count the ones that fit the condition. The best way to make sure you have remembered all throws is to list them in a 6 x 6 table, like below.

Imagine two dice, red and green. This table shows all possible throws of these dice, with red numbers for the red dice, and green for the green dice. The background is black if the condition is satisfied. You can change the numbers and the condition to see what happens. You find the probability of this condition by counting the black squares, and dividing by 36 (all possible throws).

The probability of one dice being a particular number is 1/6. You would assume that it would be twice as likely that either of two dice being a particular number, or 1/3, but this would be wrong. If you select the first condition above, you will see why. While there are six conditions where one dice is a number, and six conditions where the other dice is that number, there is one condition in both. So the probability is (12-1)/36 or11/36.

There is a better way to calculate this. If you click on the second condition, then back on the first, you will find that they are opposites of each other. That means that their probabilities add up to one. The second condition is easier to calculate directly (see below), as 25/36. So this probability is (36-25)/36 = 11/36.

The probability of one dice not being a particular number is 5/6. The probability of two dice not being a that number is 5/6 x 5/6 = 25/36. You can do this as they don't overlap as all (the correct term is that they are independent events).

The probability of one dice being a particular number is 1/6. The probability of two dice being the same particular number is 1/6 x 1/6 =1/36. This is not the same as saying that both dice are the same number. There are six different possible numbers, so that would be 6/36or 1/6.

This is the opposite of both dice being the same particular number, so the probabilities will add up to one. So the probability of at least one dice isn't a particular number = 1 - (1/36) = 35/36.

This is a fun conditions, but a little strange to work out! Try experimenting with different numbers, and you will get different collections of throws fitting the conditions, and so different probabilities. To take an example, considering one dice, if the particular number is 5 then there are 5 throws which fit the condition. There are another 5 for the other dice, and there is one overlap, so the result is (5 + 5 - 1)/36 = 9/36 or 1/4. But this varies for other numbers.

The probabilities vary considering what the number is. But the probability of the lowest dice being a particular number is the same as the probability of the highest dice being six minus that particular number.

Like the first condition, there are two ways to work this out. The first is doing it directly. Each dice has 6 throws for each number, but there are four overlaps. So the probability is (4 x 6 - 4)/36 = 20/36.

An easier way is to realise that this condition is the opposite of the next one, dice are not either number. That probability is 16/36 so this one is 1 - 16/36 = 20/36 or 5/9.

The probability of one dice not being either of two numbers is 4/6, so the probability of both dice fitting this condition is 4/6 x 4/6 = 16/36. We can do this as the conditions don't overlap, so they are independent events.

There are only two ways that two dice can have two values (try the exercise above to see why), so the probability is 2/36 or 1/18.

As the number of dice increase, then the different conditions can become more complicated. You can work out some simply.

The easiest is the probability of a number of dice being a particular number. For n dice, this is 1/(6ⁿ). For example, the probability for throwing 5 dice and getting them all sixes is 1/(6⁵) = 1/(6x6x6x6x6) =1/7776 = 0.000128 or 7775 to 1.

You can work out what is the chance of getting various numbers when throwing several dice (or one dice several times), but you do it in a back-to-front way, as above. For example, if you throw three dice, what is the chance of one of the dice getting less than five? This means that one of the dice (at least) will be a one, two, three or four. Now this is quite hard to work out directly. What we can do is work out the opposite. The chance of one dice not getting less than five means the chance of it getting a five or six, and this is 2/6, or 1/3. But we need all dice to get this. This is 1/3x1/3x1/3 or 1/27. This was the opposite of what we want, so we must take it away from one. 1 - 1/27 or 26/27 which is about 96%.

People occasionally point out to me that the singular of 'dice' is 'die'. Yes, I know that. However, a lot of people don't, and I prefer to use 'dice' since then my meaning is clear. It stops them wondering why I'm talking about dying! However, if you think that I should be accurate rather than comprehensible, I'm afraid that the authorities are not on your side. The Oxford English Dictionary says "The form dice (used as pl. and sing.) is of much more frequent occurrence in gaming and related senses than the singular die." I think that probability is related to gaming! It gives a quote: 1474 CAXTON Chesse 132 "He caste thre dyse and on eche dyse was a sise." (Translation - "He casts three dice and on each dice was a six.")

Probabilities are written as numbers between zero and one. A probability of one means that the event is certain. If you toss a coin, it will come up a head or a tail. So there is a probability of one that either of these will happen. A probability of zero means that an event is impossible. If you toss a coin, you cannot get both a head and a tail at the same time, so this has zero probability. Anything that can happen but is not certain is written as a number less than one. It could be a decimal, a fraction, a percentage, or described as "one in a thousand", which is another way of writing a fraction. All these are ways of describing probabilities. The coins experiment and dice experiment pages use percentages. Much of the theory is easier in fractions.

There are a couple of important points. Firstly, probabilities do not tell you what is going to happen, they merely tell you what is likely to happen. It is unlikely that you will toss twenty coins and that they will all come up heads. But if enough people toss enough coins for long enough, then this may well happen. It will startle the person it happens to, but think of all the people it didn't happen to! Secondly, if you toss a coin nineteen times and it comes up heads each time, then it is not more likely that the next toss will be a tail. The odds stay the same, at 50%. The tosses are called 'independent events' which means that the coin can't remember what has happened to it. While twenty heads in a row is unlikely, once you have nineteen heads in a row, the unlikely event has already happened. The potential twentieth head has the same probability as the first head. Another way of looking at it is that any sequence of twenty tosses is unlikely as twenty heads in a row, even if it looks random. But you have to write down the sequence before you start tossing to see if you get it!

When we toss a coin, there are two possibly outcomes. It can be a head or a tail, which are both equally likely. If we toss two coins, there can be two heads, two tails, or a head and a tail. It is tempting to say that there are three equally possible outcomes. But this would be wrong. You must think of the coins separately. It might be easier to imagine tossing one coin first and the other after (or even tossing the same coin twice, which has exactly the same effect). Or you could imagine two different values of coins, so they can be told apart. Now you can see that there are four possibilities: both heads, both tails, first coin a head and the second a tail, and first coin a tail and the second a head.

You can see that the number of possible outcomes gets bigger and bigger. Click on One more coin to see how the number of possible outcomes increases:

It is quite easy to find the different outcomes, since they are represented by the binary numbers with that amount of digits, with H representing the digit one and T representing zero. This also shows us how many outcomes there are, since there are 2ⁿ possible binary numbers with ndigits.

Once you have listed all possible outcomes, then you can work out the probabilities quite easily. Say that you are going to toss three coins, and you want to work out the probability of only one head (and so two tails). The possible outcomes are:

TTT, TTH, THT, THH, HTT, HTH, HHT, HHH

All these outcomes are different, and they are all equally likely. There are8 of them. There are 3 tosses with only one head:

TTH, THT, HTT

So the probability is 3/8. You can convert this into a decimal 0.375 or a percentage 37.5%, which you can round to 38% if you wish. Or you can describe it as a three in eight chance. All these mean the same.

You can list the possible outcomes above for any dice up to 6 and count the tosses which match the probability that you want. Then divide it by the total number of throws. Or you can use this calculator, where the computer counts the number of tosses for you!

Now you can work out the probabilities for various combinations for heads and coins, then you can experiment to see if reality matches the probability! You will find it won't necessarily match (although the more numbers of throw, the closer it will get), but the experiments should give figures fairly close to the probability.

The counting method works, and is very good for getting the right answer with a small number of coins. However, for larger numbers, we need a more mathematical approach. We know that the probability will be a fraction, and we know that the denominator (the number underneath) is2ⁿ for n coins. The problem is working out the numerator (the number on top). We don't want to count all the cases where it happens. What we can do is start with one coin, then add a coin at a time, and see what difference it makes to the probability. This starts to build up a pattern.

The pattern that emerges from the above example is something called Pascal's triangle (after Blaise Pascal a French mathematician). You work this out like this. Every line is made by adding the two numbers in the line above. You assume that there are zeros at the start and end of each line, and you start with a one in the top row. Pascal's triangle crops up a lot in probability. In particular, take the nth row of Pascal's triangle, and the mth number in it, and you have the numerator (top bit) of the probability of finding m heads when tossing n coins.