Lottery probability: Matching first 6 of 7 is x6 harder then any 6?

Question

In the UK lottery, you choose 6/59. The machine produces 7 balls in order (same machine all 7, same number pool all 7). The First 6 are main and number 7 is a bonus.

You win

1. jackpot if you 6 = main 6. 6 is few millions
2. second prize if 5 + 1 = 5 main + bonus ball. Only one million

In Wikipedia 5+1 is much better odds then main 6 (link in bottom). What I do not understand, if the same machine produces 7 numbers from the same number pool why 5+1 is less? If something the order for bonus important and the order for others not so others should be better odds?

https://en.wikipedia.org/wiki/National_Lottery_(United_Kingdom)#2018_changes

Answer 1

In scenario 1 the probability all 6 numbers is computed as follows:

$$ \frac{{6 \choose 6} \times {53 \choose 0}}{59 \choose 6} \approx \frac{1}{45 \text{ million}} $$

i.e. We want to match 6 out of the 6 we chose on our lottery ticket and 0 from the numbers left in the machine (this makes up our numerator). The number of ways we can draw the main 6 numbers from the pool of 59 numbers is given in the denominator.

For scenario 2, we do a similar computation but now we want to also match the bonus ball. Now we still draw 6 numbers from the machine but now we want to match 5 on our ticket with the 6 drawn and 1 from the numbers which remain in the machine. The bonus ball is drawn from the remaining pool of 53 numbers so we have to make sure we include the probability of matching this, i.e. 1/53.

$$ \frac{{6 \choose 5} \times {53 \choose 1}}{59 \choose 6} \times \frac{1}{53} \approx \frac{1}{7.5 \text{ million}} $$

Answer 2

Imagine you bought one ticket for each possible subset of $6$ of the $59$ numbers. (This is just hypothetical; you would need to buy more than $45$ million tickets to do this.)

When the six main balls and one bonus ball have been drawn, you will have one ticket that wins the jackpot and six that win second prize.