# Silly probability question

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Okay, this is a slightly embarrassing question, it's something I've learned back in high school but I can't remember it now for the life of me. If an archer has 30% chance to hit the target in a single shot, what is the chance to hit that target at least once in 3 consecutive shots? And of course, the actual question is: how do you work that result out? :)

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Better to ask, what is the probability that he won't ever hit the target in 3 shots? That's a lot easier: He never hits the target if: he doesn't hit the target the first time, AND he doesn't hit the target the second time, AND he doesn't hit the target the third time. Three independent random variables, each with P=0.7. So you have 0.7*0.7*0.7=0.343 probability he doesn't make any shot, or P=0.657 he does.

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Yeah, that was it, thanks.

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If you're feeling masochistic enough to calculate it the other way:
P(at least once) = P(3 hits) + P(2 hits) + P(1 hit)                 = (3 nCr 3) * .3 * .3 * .3 +                    (3 nCr 2) * .3 * .3 * .7 +                   (3 nCr 1) * .3 * .7 * .7                 = 0.657

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You can use generating functions for this type of problem.

(.7 + .3*x)^3 = .343 + .441*x + .189*x^2 + .027*x^3

That gives you that
P(0 hits) = .343
P(1 hit) = .441
P(2 hits) = .189
P(3 hits) = .027

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deleted in order to satisfy "don't talk to much - it can hurt people" rule.

[Edited by - PlayStationX on January 11, 2009 12:41:15 AM]

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Quote:
 Original post by PlayStationXSuppose you're on a game show, and you're given the choice of three doors with one car and two goats behind them. Say, you pick DOOR-1. The host then, who knows what's behind the doors opens another, say DOOR-2, which of course has a goat. He then says to you: "For the final choice, do you want to switch to DOOR-3?" Assuming you want a car rather than a goat,- is it to your advantage to switch your final choice to DOOR-3, or to stay with DOOR-1? ...or, it does not matter?

If you'd like, you can start a new thread to ask about that; hijacking Talin's thread isn't really appropriate. Before you do that, though, you might want to look over the previous discussions here, not to mention on other forums.

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Quote:
 Original post by TalinIf an archer has 30% chance to hit the target in a single shot, what is the chance to hit that target at least once in 3 consecutive shots?
The original question is adequately answered upthread. An interesting variant of this question is "How many times will I have to shoot in order to be reasonably certain of hitting the target at least once?"

In order to answer this version, we have to define reasonably certain. A willingness to be wrong -- perhaps in this case it is being right 19 times out of 20 (95% certainty).

By experiment, I've found nine trials produce the correct answer using the polynomial method above:

0.0403536 (no hits)
0.15565 x (one hit)
0.266828 x^2 (two hits)
0.266828 x^3 (three hits)
0.171532 x^4 (etc)
0.0735138 x^5
0.0210039 x^6
0.00385787 x^7
0.000413343 x^8
0.000019683 x^9

But the process of finding out that it takes nine?

This version of the question has some other interesting applications as well. For instance, let's say a game not written by us as MagicDrops, and we have two current model guesses as to how they drop. The models differ by maybe 5%. How many drops will it take in order clearly select one model over the other? (Same question, just rephrased).

[Edited by - AngleWyrm on January 13, 2009 2:34:10 AM]

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You want to find what is the first power of 0.7 that falls below 0.05.

0.7^n < 0.05 <=> log(0.7^n) < log(0.05) <=> n*log(0.7) < log(0.05) <=> n > log(0.05)/log(0.7) = 8.39905...

Notice the change from <' to >' in the last step, since I divided both sides by a negative number

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Nice.

Restating as formula for trials need to get at least one success:

NumTrials = Ceiling( Log(1-Confidence) / Log(1-Success) )

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