### Win or Lose?

A gambler bets half the money in his pocket on the toss of a coin, winning an equal amount for a head and losing his money if the result is a tail. After 2n plays he has won exactly n times. Has he more money than he started with?

### Fixing the Odds

You have two bags, four red balls and four white balls. You must put all the balls in the bags although you are allowed to have one bag empty. How should you distribute the balls between the two bags so as to make the probability of choosing a red ball as small as possible and what will the probability be in that case?

### Scratch Cards

To win on a scratch card you have to uncover three numbers that add up to more than fifteen. What is the probability of winning a prize?

# Introducing Distributions

##### Stage: 4 Challenge Level:

Situations like this in probability are all about clever counting. If we roll one die, we expect one of six possible outcomes... our first step is to ask how many possible rolls there are when we roll five dice. We label the dice 'one' to 'five', and denote the roll 'die one lands on $1$, die two lands on $3$, die three lands on $4$, die four lands on $6$, die five lands on $4$' as $(1,3,4,6,4)$, for example. This roll is not the same as $(1,4,3,6,4)$, although if we do not distinguish between dice they both look the same. Each die will land on one of six possible outcomes, and we can have any combination of these outcomes among the five dice, so there are in total $6\times 6\times 6\times 6\times 6 = 6^5 = 7776$ different rolls. Each of these rolls is equally likely.

If we are asked to find the probability of some event occuring, such as 'there are no sixes' or 'the sum of the dice adds up to 15', what we must do is count the number of rolls for which the event occurs. For example, the roll $(1, 2, 3, 4, 5)$ causes the event 'the sum of the dice adds up to $15$' to occur. There are many more rolls for which the sum of the dice is also $15$. The overall chance of this event occuring is the proportion of rolls for which it does occur. In other words, we count all rolls which add up to $15$ and divide by the total number of possible rolls.

We were asked which of the following events is more likely to occur: 'there are no sixes' or 'they are all sixes'. The only roll which gives us all sixes, unsuprisingly, is $(6, 6, 6, 6, 6)$! However, there are many rolls which don't contain a six: $(1, 2, 3, 4, 5)$, ... $(2, 3, 5, 4, 4)$, ... $(3, 2, 2, 5, 1)$ and so on, so we can safely say that the chance of getting all sixes is less than the chance of not getting any, or written formally:

$\mathbb{P}($'no sixes'$)> \mathbb{P}($'all sixes'$)$

Lets calculate some actual probabilities. Since only one roll gives rise to the event 'all sixes',

$\mathbb{P}($'all sixes'$) = \frac{1}{7776} \approx 0.01$%

Now, the number of rolls that don't conatin any sixes is the same as the number of rolls possible when we roll five $5$-sided dice, which is $5\times 5\times 5\times 5\times 5 = 5^5 = 3125$ So

$\mathbb{P}($'no sixes'$) = \frac{3125}{7776} \approx 40$%

which is indeed much greater than $\mathbb{P}($'all sixes'$)$.

Now for the event 'there is only one six'. How many rolls contain exactly one six? If die onerolls a six, then there are $5^4$ possible outcomes for the other four dice which ensure that dice $1$ is the only six. In other words there are $5^4$ rolls in which die onerolls a $6$ and the remaining four dice are not sixes. Similary there are $5^4$ rolls in which die twois a $6$ and the remaining four dice are not sixes and so on. The number of rolls containing exactly one six is then $5^4 + 5^4 + 5^4 + 5^4 + 5^4 = 5\times 5^4 = 5^5 =3125$ Therefore

$\mathbb{P}($'exactly one six'$) = \frac{3125}{7776} =\mathbb{P}($'no sixes'$)$

Now what about the number of rolls containing exactly two sixes? If we pick any two dice then the number of rolls for which these are both $6$'s and none of the remaining other dice roll $6$ are $5^3$, since there are three remaining dice which can take any value from $1$ to $5$. There are $10$ different ways of picking two dice (a quick count will convince you). Therefore the total number of rolls which contain exactly two $6$'s is:

(number of ways of picking two dice) $\times$ (number of rolls such that a given pair of dice are the only $6$'s)

$= 10\times 5^3 = 1250$

Therefore

$\mathbb{P}($'exactly two sixes'$) = \frac{1250}{7776} \approx 16$%

In the same way we can show that the total number of rolls which contain exactly three sixes is

(number of ways of picking three dice) $\times$ (number of rolls such that three given dice are the only $6$'s)

$= 10\times 5^2 = 250$

and the number of rolls which contain exactly four sixes is

(number of ways of picking four dice) $\times$ (number of rolls such that four given dice are the only $6$'s)

$= 5\times 5 = 25$

Therefore

$\mathbb{P}($'exactly three sixes'$) = \frac{250}{7776} \approx 3$%

$\mathbb{P}($'exactly four sixes'$) = \frac{25}{7776} \approx 0.3$%

To check that our calculations are correct we calculate

$\mathbb{P}($'no sixes'$) +\mathbb{P}($'exactly one six'$) +\mathbb{P}($'exactly two sixes'$) + \mathbb{P}($'exactly three sixes'$) + \mathbb{P}($'exactly four sixes'$) + \mathbb{P}($'all sixes'$)$

$=\frac{3125}{7776} +\frac{3125}{7776} +\frac{1250}{7776} +\frac{250}{7776} +\frac{25}{7776} +\frac{1}{7776} =\frac{7776}{7776} = 1$

Why does this make sense?

We plot our probability distribution below.

How does this distribution change if we change the number of dice in our roll?