10 Rolls of a Fair 16-sided Die

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If I roll a presumably fair 16-sided die 10 times, what are the odds that I will roll at least three sevens? In no specific order, just three sevens or more total during the 10 rolls.

In attempting to calculate the problem I tried to use the formula of 1 minus the sum of the probability of getting A) getting exactly 0 sevens, B) getting exactly 1 seven, and C) getting exactly 2 sevens.

This is what that looked like:

A = (1) * (15/16)^10
B = (10) * (1/16)^1 * (15/16)^9
C = [(10*9)/(1*2)] * (1/16)^2 * (15/16)^13

So I added A, B, and C together to get 0.95. Leaving me with a 5% chance to get at least 3 sevens? Am I doing that correctly? Instinctively it seems like 5% is too high to me.

Would greatly appreciate any advice that can be spared for verifying this problem.
 
If I roll a presumably fair 16-sided die 10 times, what are the odds that I will roll at least three sevens? In no specific order, just three sevens or more total during the 10 rolls.
In attempting to calculate the problem I tried to use the formula of 1 minus the sum of the probability of getting A) getting exactly 0 sevens, B) getting exactly 1 seven, and C) getting exactly 2 sevens.
This is what that looked like:
A = (1) * (15/16)^10
B = (10) * (1/16)^1 * (15/16)^9
C = [(10*9)/(1*2)] * (1/16)^2 * (15/16)^13
So I added A, B, and C together to get 0.95. Leaving me with a 5% chance to get at least 3 sevens? Am I doing that correctly? Instinctively it seems like 5% is too high to me.

Although I have not done the actual calculations, your ideas are correct.
You have a minor error: the ^13 should be ^8.
 
If I roll a presumably fair 16-sided die 10 times,
what are the odds that I will roll at least three sevens? In no specific order,
just three sevens or more total during the 10 rolls.

It's not a complete question. There is no automatic notion that each side has
a unique number or that there are consecutively-numbered sides, or that there
are any sevens on any sides at all.

In contrast, if you had stated that you had a fair 6-sided die, then I would not have
questioned the numbers being 1, 2, 3, 4, 5, 6 on the faces of the die, one per face.
 
Last edited:
If I roll a presumably fair 16-sided die 10 times, what are the odds that I will roll at least three sevens? In no specific order, just three sevens or more total during the 10 rolls.

In attempting to calculate the problem I tried to use the formula of 1 minus the sum of the probability of getting A) getting exactly 0 sevens, B) getting exactly 1 seven, and C) getting exactly 2 sevens.

This is what that looked like:

A = (1) * (15/16)^10
B = (10) * (1/16)^1 * (15/16)^9
C = [(10*9)/(1*2)] * (1/16)^2 * (15/16)^13

So I added A, B, and C together to get 0.95. Leaving me with a 5% chance to get at least 3 sevens? Am I doing that correctly? Instinctively it seems like 5% is too high to me.

Would greatly appreciate any advice that can be spared for verifying this problem.
Perhaps the error of ^13 where you should have ^8 made the difference .. I get 2.1%
 
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