Probability of a match

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SteveBozell

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I'm prototyping a matching game. I'm generating shapes below with the following characteristics.

One of five colors (red, green, blue, yellow, cyan)
One of five shapes (circle, square, star, triangle, diamond)
One of five fill patterns (e.g. solid, outline, crosshatched, checkerboard, dithered)

If I generate two shapes, randomly selecting a color/shape/fill for each, what are the chances that at least one attribute matches. So, a solid green circle would match a solid red square (solid), and a dithered yellow star would match a crosshatched yellow diamond (yellow).
 
SteveBozell said:
I'm prototyping a matching game. I'm generating shapes below with the following characteristics.

One of five colors (red, green, blue, yellow, cyan)
One of five shapes (circle, square, star, triangle, diamond)
One of five fill patterns (e.g. solid, outline, crosshatched, checkerboard, dithered)

If I generate two shapes, randomly selecting a color/shape/fill for each, what are the chances that at least one attribute matches. So, a solid green circle would match a solid red square (solid), and a dithered yellow star would match a crosshatched yellow diamond (yellow).
Click to expand...
We know that p(0 attribute match)+ p(1 attribute match)+ p(2 attribute match) + p(3 attribute match)= 1.
Then p(1 attribute match)+ p(2 attribute match) + p(3 attribute match)= 1 - p(0 attribute match).

Can you figure out what p(0 attribute match) equals??
 
Jomo said:
We know that p(0 attribute match)+ p(1 attribute match)+ p(2 attribute match) + p(3 attribute match)= 1.
Then p(1 attribute match)+ p(2 attribute match) + p(3 attribute match)= 1 - p(0 attribute match).

Can you figure out what p(0 attribute match) equals??
Click to expand...

1 - p(1) - p(2) - p(3). Correct? And since each of those is 20%, the p(0) is 1 - 60%, or 40%.

Or did I butcher that up real good?
 
Oh, I think I did that wrong - I was thinking "attribute 1 match" instead of "1 attribute match".
 
Let me try again. The probability of all 3 matching is 1/5 * 1/5 * 1/5, or 1/125. The probability of 2 matching is 1/5 * 1/5, or 1/25. The probability of 1 matching is 1/5.

So the probability of 0 matching is 1 - 1/5 - 1/25 - 1/125, or roughly 75%.
 
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SteveBozell said:
1 - p(1) - p(2) - p(3). Correct? And since each of those is 20%, the p(0) is 1 - 60%, or 40%.

Or did I butcher that up real good?
Click to expand...
You want the probability of at least 1. What does at least one mean in this case? It means 1, 2 and 3.

So p(at least 1) = P(exactly 1) + p( exactly 2) + p( exactly 3). Instead of computing 3 probabilities and 2 additions why not do use the complement which is 1 - p( no matches)? This only requires us to compute 1 probability and 1 subtraction.

What is p(no matches)??
 
Jomo said:
You want the probability of at least 1. What does at least one mean in this case? It means 1, 2 and 3.

So p(at least 1) = P(exactly 1) + p( exactly 2) + p( exactly 3). Instead of computing 3 probabilities and 2 additions why not do use the complement which is 1 - p( no matches)? This only requires us to compute 1 probability and 1 subtraction.

What is p(no matches)??
Click to expand...

p(no matches) is p(no shape match) * p(no color match) * p(no fill match). I think. Which is 4/5 for each. So p(no match) is 4/5 * 3, or .512.

I feel like a cat chasing a laser pointer :)
 
SteveBozell said:
Let me try again. The probability of all 3 matching is 1/5 * 1/5 * 1/5, or 1/125. The probability of 2 matching is 1/5 * 1/5, or 1/25. The probability of 1 matching is 1/5.

So the probability of 0 matching is 1 - 1/5 - 1/25 - 1/125, or roughly 75%.
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The whole idea behind computing 1- p(0) is so you do not have to compute p(1) + p(2) + p(3)
 
Please entertain me and find p(0) directly. You claimed to have found p(1), p(2), and p(3) so please find p(0).
 
Jomo said:
Better, but please write (4/4)^3 but NOT 4/5 * 3.

So p(at least one) 1 - .512= .488.

Note that what you got for p(1)+p(2)+p(3) was not correct!
Click to expand...

Yes, 4/5 cubed, not times 3. Thanks for the help and quick replies!
 
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