On a game show, the - contestants each day can win $1, 000, 000 by correctly guessing an integer between 1 and 100 inclusive (which is chosen randomly each day). Before guessing the contestant can ask one yes/no question of his or her choice. Monday's contestant asked Is the number 57? and Tuesday's contestant asked ìs the number greater than 50?. Let P (M) be the probability of Monday's contestant winning and let P(T) be the probability of Tuesday's contestants winning (assume each contestant properly uses the information gained from the question). Which of the following is true? A. P (M)/ P (T) <.1 B. 0.1 < P (M)/ P (T) < 0.9 C. 0. 9< P (M)/ P (T) < 1.1 D. 1.1 < P (M)/ P (T) < 2 E. P (M)/ P (T) < 2

  • 1
    $\begingroup$ Just so you know, your Erdös number is definitely not zero... unless you are somehow actually Erdös. $\endgroup$
    – Prince M
    Aug 18, 2018 at 7:06

1 Answer 1


If $W$ denotes the winning number then:$$P(M)=P(M\mid W=57)P(W=57)+P(M\mid W\neq57)P(W\neq57)=$$$$1\cdot\frac1{100}+\frac1{99}\cdot\frac{99}{100}=\frac1{50}$$

More directly you could say that Monday's contestant uses the possibility to ask two distinct numbers. If the first happens to be correct already then the asking of the second is virtual.

Further we also find:$$P(T)=\frac1{50}$$This because after answer "yes" as well as answer "no" there is a choice between $50$ equiprobable numbers.

edit (to give some clarity) concerning monday contestant.

The event $M$ (i.e. the monday contestant wins) can be split up in two mutually exclusive events:

the answer on his question is "yes" and (of course) his guess will be $57$ or the answer on his question is "no" and (of course) he guesses a number that differs from $57$ which appears to be the correct number.

In mathematical language this is: $$M=(M\cap\{W=57\})\cup(M\cap\{W\neq57\})$$

Note that $\cap$ corresponds with "and" and $\cup$ corresponds with "or".

If two events $A,B$ are mutually exclusive (i.e. they cannot happen both, or mathematically $A\cap B=\varnothing$ then we have the rule $P(A\cup B)=P(A)+P(B)$ and applying that here we find:$$P(M)=P(M\cap\{W=57\})+P(M\cap\{W\neq57\})$$Now it remains to find the two terms.

If $A,B$ are independent events (i.e. the probability that $A$ happens will not affect the probability that $B$ happens, and vice versa) then we have the rule:$$P(A\cap B)=P(A)P(B)$$

However we cannot apply that here because we are definitely not dealing with events $M$ and $\{W=57\}$ that are independent here!

Then how to find $P(A\cap B)$ if $A,B$ are events that are not independent? For that we use another rule that also results in a multiplication:$$P(A\cap B)=P(A\mid B)P(B)$$

Here $P(A\mid B)$ stands for the probability that event $A$ happens under the condition that event $B$ happens.

Applying that on the first term we find: $$P(M\mid \{W=57\})P(\{W=57\})$$or a bit more shortly written:$$P(M\mid W=57)P(W=57)$$

Now what is the probability that monday wins if $57$ is the winning number? $1$ of course, because the answer on his question has been answered with "yes" and then he simply knows that $57$ is the winning number. He will surely guess that number and win. This tells us that $P(M\mid W=57)=1$. Further what is the probability that $57$ is indeed the winning number? $\frac1{100}$ of course because there are $100$ equiprobable candidates. This tells us that $P(W=57)=\frac1{100}$ and we can calculate the first terms as:$$P(M\mid W=57)P(W=57)=1\cdot\frac1{100}=\frac1{100}$$

Now it remains to find the second term which will have the looks:$$P(M\mid W\neq57)P(W\neq57)$$ What is the probability that monday wins if the winning number is not $57$? His question has been answered with "no" and this information tells him that the winning number must be one of the elements of $\{1,2,\dots,56,58,\dots,99,100\}$. They have equal chances so the conclusion is evidently that $P(M\mid W\neq57)=\frac1{99}$. But we must not forgot the other factor: $P(W\neq 57)$, i.e. the probability that the winning card is not $57$ or equivalently the probability that his question is answered by "no". For that we find quite easily $P(W\neq57)=1-P(W=57)=1-\frac1{100}=\frac{99}{100}$. Now we are equipped to find the second term: $$P(M\mid W\neq57)P(W\neq57)=\frac1{99}\frac{99}{100}=\frac1{100}$$This results in:$$P(M)=P(M\cap\{W=57\})+P(M\cap\{W\neq57\})=\frac1{100}+\frac1{100}=\frac1{50}$$

  • 1
    $\begingroup$ for tuesday player if the answer is yes then there is only 49 numbers greater then 50 ( 50 is not included) . if No then the numbers less then 50 then there are 49 to chose from including the number 50 would be 50 choices so I think it would be 1/49 + 1/50 $\endgroup$
    – Randin
    Aug 18, 2018 at 20:56
  • $\begingroup$ For tuesday player if the answer is "yes" then the numbers $51,52,\dots,100$ are still candidates. These are the numbers $>50$ and there are $50$ such numbers! $\endgroup$
    – drhab
    Aug 19, 2018 at 8:09
  • $\begingroup$ My former comment only concerns a mistake in counting. Much more severe is the fact that you are completely wrong in adding up both probabilities. What is your reason to do that? Try it out on the same situation where $100$ is replaced by $4$ and $50$ is replaced by $2$. According to your reasoning and without repairing the discount mistake mentioned in my former comment your probability would be $1/1+1/2>1$. According to your reasoning with repairing the former discount it would be $1/2+1/2=1$ (so certainty that "tuesday" will guess well). Absurdity in both cases. $\endgroup$
    – drhab
    Aug 19, 2018 at 8:11
  • $\begingroup$ You are free to react or not, and I respect that freedom. But I use the freedom to speak out that it disappoints me that you do not react on my comments. This answer is here for a week now with a downvote on it that is not justified (I am not saying that you are the downvoter). Sad. $\endgroup$
    – drhab
    Aug 24, 2018 at 10:43
  • $\begingroup$ This is why universities like the uw madison suck ! The professors are never around to teach such advanced math ..questions like this ..a contest problem never taught. I appreciate your answer and wish i could understand it . I didnt downvote and will try to upvote $\endgroup$
    – Randin
    Aug 26, 2018 at 6:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .