# Numbers or Letters?

Which game offers you the best chance of winning?

## Problem

Here are two games that you might play:**Game one**

You have a bag of 26 tiles, each with a different letter of the alphabet.

You choose a tile, write down the letter, and put it back in the bag.

Then you choose a second tile, write down the letter, and put it back in the bag.

Then you choose a third tile.

If the three letters came out in strictly ascending alphabetical order, you win. Otherwise, you lose.**Game two**

You roll a 10-sided dice three times, writing down the number that is shown each time. If the three numbers are in strictly ascending numerical order, you win. Otherwise, you lose.**Which game offers you the best chance of winning?**

## Student Solutions

**Answer**: you have a better chance of winning the letters game

**Using a tree diagram beginning by checking for duplicated numbers/letters**

Image

Image

Given 3 numbers or 3 letters which are all different, the probability that they are in order is the same.

Call this probability $\text p$.

Image

Image

Which is greater out of $\text a$ and $\text b$?

More likely to get repeats using 10 tiles than 26 tiles, so more likely to get 3 different with 26 tiles (letters).

So $\text a \gt \text b$ - you are more likely to win the letters game.

**Creating all of the winning combinations**

Probability of winning numbers game is $$\frac{\text{number of number combinations in the right order}}{\text{total number of possible number combinations}}$$

Total number of possible number combinations is $10\times10\times10 = 1000$

Number of winning combinations:

First tile |
Second tile |
Third tile |
Number of options |

0 | 1 | 2, 3, 4, ..., 9 | 8 |

2 | 3, 4, ..., 9 | 7 | |

3 | 4, ..., 9 | 6 | |

4 | 5 | ||

5 | 4 | ||

6 | 3 | ||

7 | 8, 9 | 2 | |

8 | 9 | 1 | |

1 | 2 | 3, 4, 5, ..., 9 | 7 |

3 | 6 | ||

4 | 5 | ||

... | ... | ||

8 | 9 | 1 | |

2 | 3 | 4, 5, ..., 9 | 6 |

4 | 5 | ||

5 | 4 | ||

... | ... | ||

8 | 1 | ||

3 | 5 + 4 + 3 + 2 + 1 | ||

4 | 4 + 3 + 2 + 1 | ||

5 | 3 + 2 + 1 | ||

6 | 2 + 1 | ||

7 | 8 | 9 | 1 |

Total number of options: $$1\times8 + 2\times 7 +3\times6+4\times5+5\times4+6\times3+7\times2+8\times1\\

= (8+14+18+20)\times 2\\

=120$$

$\therefore$ probability of winning numbers game is $\dfrac{120}{1000}$

Total number of possible letter combinations is $26\times26\times26$

Total number of options will be: $$(1\times24 + 2\times23 + 3\times22 + ... + 12\times13)\times2\\

= (24+46+66+84+100+114+126+136+144+150+154+156)\times2\\

= 2600$$

$\therefore$ probability of winning letters game is $\dfrac{2600}{26\times26\times26}$

Which is bigger?

$\dfrac{120}{1000}=\dfrac{12}{100}=\dfrac{24}{200}$

$\dfrac{2600}{26\times26\times26} = \dfrac {100}{26\times26} = \dfrac{25}{13\times13} = \dfrac{25}{169}>\dfrac{24}{200}$

More likely to win the letters game

**Counting winning possibilities by considering combinations**

Probability of winning numbers game is $$\frac{\text{number of number combinations in the right order}}{\text{total number of possible number combinations}}$$

Total number of possible number combinations is $10\times10\times10 = 1000$

Now we need the number of number combinations in the right order.

Must have 3 different numbers, which then can be put in order:

Tile A: $10$ options

Tile B: $9$ options

Tile C: $8$ options

So $10\times9\times8 = 720$ ways get three different numbers.

But some are counted twice: $4, 6, 3$ and $6, 3, 4$ and so on - but both score a point as $3, 4, 6.$

How many times is $3, 4, 6$ counted?

Tile A: $3$ options ($3, 4$ or $6$)

Tile B: $2$ options

Tile C: $1$ option

$3\times2\times1 = 6$ times.

So there are $720\div6 = 120$ winning combinations.

Probability of winning numbers game is $\dfrac{120}{1000}=\dfrac{3}{25}$

Similarly, for letters:

$26\times26\times26$ possible combinations.

$(26\times25\times24)\div(3\times2\times1) = 26\times25\times4$ winning combinations.

Probability of winning letters game is $\dfrac{26\times25\times4}{26\times26\times26}\gt \dfrac3{25}$