probability/expected value exercise

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August 25th 2007, 12:58 AM
shirel

probability/expected value exercise

how many
August 25th 2007, 02:16 AM
CaptainBlack

Quote:

Originally Posted by shirel

how many times does one have to roll a die in order to have face 5 showing up with a 90 % certainty?
if face 5 did not show up among the estimated number of tries, would that legitimate you to suspect the die be unfair?
consider the same questions with an underlying 99 % certainty!

calculations are necessary, but important is the discussion of the questions..
please help me with all your knowledge and mathematical way of thinking;)

thanks!

In $N$ trials the number of times, $n$ that face $5$ shows on a fair dice has a binomial distribution $B(N, 1/6)$ .

So your question is asking for the smallest $N$ such that:

$<br /> p(n=0) \le 0.1<br />$

so:

$<br /> p(n=0) = b(0;N,1/6)=\left(\frac{1}{6}\right)^N \le 0.1<br />$

so now find N.

RonL
August 25th 2007, 02:20 AM
CaptainBlack

Quote:

Originally Posted by shirel

how many times does one have to roll a die in order to have face 5 showing up with a 90 % certainty?
if face 5 did not show up among the estimated number of tries, would that legitimate you to suspect the die be unfair?
consider the same questions with an underlying 99 % certainty!

Whether you suspect that a coin is unfair if it fails a test that a fair coin
passes 90% of the time depends on you, do you have some other reasom to
suspect the coin? Otherwise you may suspect it and possibly do somemore tests.

If it fails a 99% test you may have more reason to suspect the coin.

RonL
August 25th 2007, 03:57 PM
Soroban

Hello, shirel!

Another (slightly different) explanation . . .

Quote:

How many times does one have to roll a die in order
to have face 5 showing up with a 90% certainty?

We know that for one roll: . $P(5) = \frac{1}{6},\;\;P(\text{not 5}) = \frac{5}{6}$

Consider $n$ consecutive rolls without a 5.
. . This probability is: . $\left(\frac{5}{6}\right)^n$

And we want this probability to be less than 10%.

. . Hence, we have: . $\left(\frac{5}{6}\right)^n \:<\:0.1$

Take logs: . $\log\!\left(\frac{5}{6}\right)^n \:<\:\log(0.1) \quad\Rightarrow\quad n\cdot\log\!\left(\frac{5}{6}\right) \;<\;\log(0.1)$

. . Divide by $\log\!\left(\frac{5}{6}\right)$ which is negative: . $n \;> \;\frac{\log(0.1)}{\log\!(\frac{5}{6})} \;\approx\;12.629$

Therefore, one must roll a die at least 13 times.