Thread: How long to wait

1. How long to wait

HI;
The problem in my book says "A basket ball player has a 68% success rate, How long is it before he /she misses?

I get success = 0.68%, failure = 0.32%.

Expected wait time is 0.68 / 0.32 = 2.1throws.

is this the correct method?

Thanks.

2. Re: How long to wait

Yes, that is correct.

3. Re: How long to wait

Ok thanks now I have another question.

A coin is tossed until a head is obtaines, On average how many trials would it take
I get success =1/2 failure =1/2. therefore (1/2) / (1/2) = 1.
is this correct.

4. Re: How long to wait

For the first one, your answer is correct only if you are looking for shots made, not shots missed. So, the player will make on average 2.1 shots in a row, but miss the next shot. For this example, you are tossing a coin until you get a heads. That is $\dfrac{1}{\text{Pr}(\text{tails})} = \dfrac{1}{\tfrac{1}{2}} = 2$. So, on average, you have to flip the coin twice before the coin winds up showing heads. (Here, you are requiring the end state to be heads while in the previous example, you did not require the end state to be a missed shot).

5. Re: How long to wait

In the first problem I'm looking for how long before a missed shot oh I see its the
same thing we worded it differently.

In the second one can you explane the formula, I get the denominator part
but why is the numerator 1?

6. Re: How long to wait

I can prove that the numerator is one, but the proof likely makes use of more advanced mathematics than you are ready for (an infinite geometric sum).

7. Re: How long to wait

You can prove it like this:To calculate the expected value of the first "failure" (i.e. missing the basketball shot, or tossing a head instead of a tails in the above examples) is the infinite sum of the number of turns times the probabilty of that number of turns being the first failure. For the basketball problem this yields:

For N=1: P(miss first shot) = 1-.68 = 0.32
For N=2: P(make first and miss 2nd) = 0.68 x 0.32
For N=3: P((make first 2 and miss third) = 0.68^2 x 0.32
For N= 4: P(make first 3 qnd miss 4th) = 0.68^3 x 0.32
etc.

The expected value for N is then:
$1 \times 0.32 + 2 \times 0.86 \times 0.32 + 3 \times 0.68^2 \times 0.32 + 4 \times 0.68^3 \times 0.32 + ....$

This turns out to be equal to 1/(1-0.62) = 3.1. Thus the expected value of the first missed shot is 3.1.

To prove this consider that if p = probability of success for a given turn, then (1-p) = probability of failure, and the expected number of the first failure is:

$1(1-p) + 2p(1-p) + 3p^2(1-p) + 4p^3(1-p) + 5p^4(1-p) + ....$
$= \ 1 - p + 2p - 2p^2 + 3p^2 - 3p^3 +4p^3 - 4p^4 + 5 p^4 - ...$
$= \ 1+ p + p^2 + p^3 }+ p^4 + ...$

But note if you perform long divisdion on the fraction $\frac 1 {1-p}$ you get the same result: $1 + p + p^2 + p^3 + ...$. Thus given p = probability of success for a given turn, the expected number of turns for the first failure is: 1/(1-p). And the expected number of successful turns before that first failure is one less than this: $\frac 1 {1-p} - 1 = \frac p {1-p}$. Thus if you want to determine the number of successes before failure use p/(1-p), and if you want to know the expected number of turns for the first failure to occur use 1/(1-p).

The wording of the basketball problem is a bit ambiguous: "how long is it before he misses" can be interpreted to mean "what's the expected shot number for the first miss," and if so then the correct answer is shot number 3.1, not 2.1.