Numbers or Letters?
Which game offers you the best chance of winning?
Age
14 to 16
Challenge level
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
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Given 3 numbers or 3 letters which are all different, the probability that they are in order is the same.
Call this probability $\text p$.
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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 $\dfrac{26\times25\times4}{26\times26\times26}\gt \dfrac3{25}$