The Wizard’s Fruit Slot probability

I'm digging into The Wizard’s Fruit Slot working internals found atIt has the following reel distribution.

Quote:

Symbol Reel 1 Reel 2 Reel 3
Cherry 5 2 3
Orange 4 4 4
Bell 3 4 4
Globe 1 1 1
Plum 3 3 1
Lemon 3 5 6
Bar 1 1 1
Total 20 20 20

The author calculates the odds of hitting 3 symbols:

Quote:

Probability of three globes =(1/20)*(1/20)*(1/20)=1/8000=0.000125
Probability of three bars =(1/20)*(1/20)*(1/20)=1/8000=0.000125
Probability of three plums =(3/20)*(3/20)*(1/20)=9/8000=0.001125
Probability of three bells =(3/20)*(4/20)*(4/20)=48/8000=0.006000
Probability of three oranges =(4/20)*(4/20)*(4/20)=64/8000=0.008000
Probability of three cherries =(5/20)*(2/20)*(3/20)=30/8000=0.003750
Probability of two cherries =(5/20)*(2/20)*(17/20)=170/8000=0.021250

Everything's good and fine, but there's one thing I don't understand. Below in the article, the author states that the probability of hitting only one cherry(left aligned) is:

Quote:

Probability of one cherry =(5/20)*(18/20)*(20/20)=1800/8000=0.225000

I'm confused here, shouldn't the formula be (5/20)*(18/20)*(17/20)? because there's 3 cherries in third reel.

Also I'd be very greatful if someone could explain how to calculate the following combinations:
Reel1 Reel2 Reel3
Cherry Any Cherry

OR

Reel1 Reel2 Reel3
Any Cherry Cherry


I guess, it should be (5/20) * (18/20) * (3/20) + (15/20) * (2/20) * (3/20), but I want to be sure.

Thank you very much :)

Because of the left-aligned requirement, CHERRY-OTHER-CHERRY counts as one cherry because of the gap. CHERRY-CHERRY-OTHER would count as two cherries because both are left-aligned with no gap.

I see. Much clearer now. Thank you very much JB.

And what about my second question. Is my deduction correct?

Thanks again :)

If by "Any" you mean "Anything except a Cherry" then your math looks correct.

That's what I mean. Thanks again JB :)