Numbers or Letters?

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

Using a tree diagram beginning by checking for duplicated numbers/letters
    
Given 3 numbers or 3 letters which are all different, the probability that they are in order is the same. 

Call this probability $\text p$.
  
  
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 $\dfrac{26\times25\times4}{26\times26\times26}\gt \dfrac3{25}$