Simple Proablitiy Q: getting at least 40 pts from 3 coin-drops in pachinko

johndrummer

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Apr 27, 2016
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Background: I have built a "pachinko" game where a coin is dropped and makes a random path through evenly-spaced nails.
The coin has exactly a 1/5 chance of ending up (at the bottom) in a 30 point zone, a 2/5 chance of ending up in a 10 point zone and a 2/5 chance of scoring no points at all in a single drop. (the zones where the coin ends up at the bottom look like: [0 ][10][30][10][ 0])
My Question:
If the coin is dropped through the game 3 separate times, what are the chances that a total of at least 40 points will be scored? I am clueless as to how you would calculate this and would greatly appreciate any help!
 
Background: I have built a "pachinko" game where a coin is dropped and makes a random path through evenl
y-spaced nails.
The coin has exactly a 1/5 chance of ending up (at the bottom) in a 30 point zone, a 2/5 chance of ending up in a 10 point zone and a 2/5 chance of scoring no points at all in a single drop. (the zones where the coin ends up at the bottom look like: [0 ][10][30][10][ 0])
My Question:
If the coin is dropped through the game 3 separate times, what are the chances that a total of at least 40 points will be scored? I am clueless as to how you would calculate this and would greatly appreciate any help!

How many ways you can score 40 in 3 drops

[10][30][0]
[10][10][30]
[30][30][0]

and so on .... (not so much so forth)

and continue....
 
This is similar to a sum of dice

Two dice with the weights given.
zero
zero
ten
ten
forty
zero
0
0
10
10
40
zero
0
0
10
10
40
ten
10
10
20
20
50
ten
10
10
20
20
50
forty
40
40
50
50
80

Another fascinating point about dice problems:

The number of chances of a given sum coming up can be directly read
off the polynomial expansion of Sum(x^i)^numDice as the coefficients,
where Sum goes from 1 to sidesPerDie. For example, with three six-sided die:

( x^1 + x^2 + x^3 + x^4 + x^5 + x^6 )^3 expands to:

1x^3 + 3x^4 + 6x^5 + 10x^6 + 15x^7 + 21x^8 + 25x^9 + 27x^10 + 27x^11 + 25x^12
+ 21x^13 + 15x^14 + 10x^15 + 6x^16 + 3x^17 + 1x^18

The first term says there is 1 way to get a 3, the second says there are
3 ways to get a 4, the third says there are 6 ways to get a 5, etc.
 
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