Odds

**ibetan** · March 18th 2010, 08:09 AM

Aki is a participant
on a trivia-based game show. He has an
equal likelihood on any given trial of being
asked a question from one of six categories:
Hollywood, Strange Places, Number Fun,
Who?, Having a Ball, and Write On! Aki
feels that he has a 50/50 chance of getting
Having a Ball or Strange Places questions
correct, but thinks he has a 90% probability
of getting any of the other questions right. If
Aki has to get two of three questions
correct, what are his odds of winning?

**ibetan** · March 18th 2010, 10:24 PM

I tried finding all the different combinations and using the probabilities but cannot get the right answer

**Archie Meade** · March 19th 2010, 03:19 AM

Originally Posted by ibetan

Aki is a participant
on a trivia-based game show. He has an
equal likelihood on any given trial of being
asked a question from one of six categories:
Hollywood, Strange Places, Number Fun,
Who?, Having a Ball, and Write On! Aki
feels that he has a 50/50 chance of getting
Having a Ball or Strange Places questions
correct, but thinks he has a 90% probability
of getting any of the other questions right. If
Aki has to get two of three questions
correct, what are his odds of winning?

First, if he has an equal likelihood of being asked a question from any of the 6 categories,
and he has to answer at least 2 of 3 questions correctly,
the situation can be divided up according to the questions he must answer.

3 "Having a Ball" questions.
2 "Having a Ball" questions and 1 "Strange Places" question.
1 "Having a Ball" question" and 2 "Strange Places" questions.
3 "Strange Places" questions.
1 "Having a Ball" question, 1 "Strange Places" question and 1 "other four" question.
2 "Having a Ball" questions and 1 "other four" question.
1 "Having a Ball" question and 2 "other four" questions.
2 "Strange Places" questions and 1 "other four" question.
1 "Strange Places" and 2 "other four" questions.
3 "other four" questions.

then, beginning with the first category (3 Having a Ball questions), you can calculate the probability that
he is asked the question and he answers 2 correctly or 3 correctly.
If he answers 2 correctly, then he can get the first, second or third one wrong.

Hence that probability he'd get 2 correct would be

$\frac{1}{6}\ \color{red}0.5\color{black}\ \frac{1}{6}\ 0.5\ \frac{1}{6}\ 0.5+\frac{1}{6}\ 0.5\ \frac{1}{6}\ \color{red}0.5\color{black}\ \frac{1}{6}\ 0.5+\frac{1}{6}\ 0.5\ \frac{1}{6}\ 0.5\ \frac{1}{6}\ \color{red}0.5$

$=\left(\frac{1}{6}\right)^3(0.5)^2\color{red}(0.5) \color{black}(3)=\left(\frac{1}{6}\right)^3(0.5)^3 (3)$

and the probability of getting all 3 correct is

$\frac{1}{6}\ 0.5\ \frac{1}{6}\ 0.5\ \frac{1}{6}\ 0.5=\left(\frac{1}{6}\right)^3(0.5)^3$

Continuing on with the other categories, the remaining probabilities can be calculated and then summed.

**Grandad** · March 19th 2010, 07:12 AM

Hello ibetan

Originally Posted by ibetan

Aki is a participant
on a trivia-based game show. He has an
equal likelihood on any given trial of being
asked a question from one of six categories:
Hollywood, Strange Places, Number Fun,
Who?, Having a Ball, and Write On! Aki
feels that he has a 50/50 chance of getting
Having a Ball or Strange Places questions
correct, but thinks he has a 90% probability
of getting any of the other questions right. If
Aki has to get two of three questions
correct, what are his odds of winning?

If we group the questions into 'hard' and 'easy', then there are $2$ categories in the 'hard' group (Having a Ball and Strange Places) and $4$ categories in the 'easy' group (the remainder of the $6$ categories).

Assuming that the categories are chosen at random, the probability that a 'hard' question is chosen is therefore $\tfrac13$ , and the probability that an 'easy' question is chosen is $\tfrac23$ .

If a 'hard' question is chosen, the probability that Aki gets it right is $\tfrac12$ ; and if an 'easy' question is chosen, the probability he gets it right is $\tfrac{9}{10}$ .

So the probability that Aki gets a question chosen at random right is:

$\frac13\times\frac12 + \frac23\times\frac{9}{10}$

$= 0.7667$

Another assumption now: Aki is asked all three questions, even if he gets the first two right. (The working will be a bit different if a third question isn't asked if this happens.)

So the probability that he wins is the probability that, out of three questions, he gets all three right, or two out of three right

$=(0.7667)^3 + 3(0.7667)^2(1-0.7667)^1$

$= 0.862$ (to 2 d.p.)

Grandad

**ibetan** · March 19th 2010, 08:18 AM

Originally Posted by Grandad

Hello ibetanIf we group the questions into 'hard' and 'easy', then there are $2$ categories in the 'hard' group (Having a Ball and Strange Places) and $4$ categories in the 'easy' group (the remainder of the $6$ categories).

Assuming that the categories are chosen at random, the probability that a 'hard' question is chosen is therefore $\tfrac13$ , and the probability that an 'easy' question is chosen is $\tfrac23$ .

If a 'hard' question is chosen, the probability that Aki gets it right is $\tfrac12$ ; and if an 'easy' question is chosen, the probability he gets it right is $\tfrac{9}{10}$ .

So the probability that Aki gets a question chosen at random right is:

$\frac13\times\frac12 + \frac23\times\frac{9}{10}$

$= 0.7667$

Another assumption now: Aki is asked all three questions, even if he gets the first two right. (The working will be a bit different if a third question isn't asked if this happens.)

So the probability that he wins is the probability that, out of three questions, he gets all three right, or two out of three right

$=(0.7667)^3 + 3(0.7667)^2(1-0.7667)^1$

$= 0.862$ (to 2 d.p.)

Grandad

Thanks. I understand everything you did except the part in $3(0.7667)^2$ , why did you multiply by 3?

**Grandad** · March 19th 2010, 08:58 AM

Hello ibetan

Originally Posted by ibetan

Thanks. I understand everything you did except the part in $3(0.7667)^2$ , why did you multiply by 3?

This is an example of the Binomial Distribution Formula. The probability of getting $r$ successes from $n$ trials is:

$\binom nrp^r(1-p)^{n-r}$

Here, $p = 0.7667, \;n = 3,\; r = 2$ . The probability of getting 2 questions right out of 3 is:

$\binom 32(0.7667)^2(1-0.7667)^1$

and $\binom32 = 3$ . This is where the 3 comes from.

It is simply the number of ways of choosing 2 questions out of 3, these being the questions that Aki gets right. He could get:

right, right, wrong

or

right, wrong, right

or

wrong, right, right

3 ways. Do you see now?

Grandad

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