This week's Riddler Express

[Bob78164]

Here's a link to this week's Riddler column. The Riddler Express puzzle references some game show that used to be on the air.

My answer is

Spoiler

0.875

. --Bob

[silverscreenselect]

Spoiler

No, it's the same 70% as before. Before, the odds were A - 70%, Not A - 30%. You are removing two "not A's" which makes the odds for A still 70%. It's the odds for B that have gone up from 10 to 30%.

[ShamelessWeasel]

Spoiler

My thought is the answer is indeterminate given what you know. Bob is assuming the chance of the other 3 answers being right is equal so you have a 7 out of 8 chance. SSS is assuming that the choice is between B and not B.

But what if your thought process is that the chance of B is to be right 70%, C is 25%, D is 5% and A is 0%. Depending on which of the answers are left the odds could be 100% (A&B) or 74%(B&C are left which is probably what would happen).

[ShamelessWeasel]

Spoiler

And to be fair I kinda ran into this on my $64K question. I really did not know so I asked the audience. My ask the audience came back A23 B33 C 7 D 37. I used 50/50 which left A&D. My odds for D before the 50/50 was 37% It was not 37% after the 5050. D was my going to be my guess and since it was highest on the Ask the Audience and still there after 50 50 I went with it.

https://millionaire.fandom.com/wiki/Eric_Grosskurth

[Bob78164]

Spoiler

My thought is the answer is indeterminate given what you know. Bob is assuming the chance of the other 3 answers being right is equal so you have a 7 out of 8 chance. SSS is assuming that the choice is between B and not B.

But what if your thought process is that the chance of B is to be right 70%, C is 25%, D is 5% and A is 0%. Depending on which of the answers are left the odds could be 100% (A&B) or 74%(B&C are left which is probably what would happen).

That possibility is ruled out by the conditions of the question, which say that the a priori probabilities of each of the other three answers being correct are equal to each other. --Bob

[Bob78164]

Spoiler

No, it's the same 70% as before. Before, the odds were A - 70%, Not A - 30%. You are removing two "not A's" which makes the odds for A still 70%. It's the odds for B that have gone up from 10 to 30%.

You're mistaking this for the Monty Hall problem. This isn't the Monty Hall problem. Unlike in the Monty Hall problem, if B were one of the wrong answers, it could have gone away. So the fact that it remains gives you additional information that you didn't have before.

Spoiler

After all, by your reasoning the remaining spoiler also has to have the same chance of being right, and both things can't possibly be true because now you know with 100% certainty (not 80% certainty) that one or the other is correct.

Here is how to think about the problem:

Spoiler

Assume this question comes up 30 times. B will be the correct answer 21 of those times, and each of the other three responses will be the correct answer 3 times, for a total of 9. Each of these trials is equally likely.

In 3 of the 9 trials where B is not the correct answer, it will survive the 50-50. In 21 of the 21 trials where B is the correct answer, it will survive the 50-50.

You now know that you're in 1 of the 24 trials where B survives. Each of those 24 trials was constructed to be equally likely. Since B is the correct answer is 21 of those 24 trials, the probability that it's correct has increased from 70% to 87.5% (and the possibility that the remaining spoiler is correct has increased from 10% to 12.5%).

--Bob

[BackInTex]

I’m with SSS

Spoiler

its still 70%. You were 70% confident not because of the other answers but only because there were other answers. Had there been 100 choices with 99 of them being no better than the others then narrowing down to twenty would not make you more sure or less sure. This is not a pure statistical probability problem. This is a gut feeling and the number of other answers don’t change how you feel about your 1st choice.

[Bob78164]

I’m with SSS

Spoiler

its still 70%. You were 70% confident not because of the other answers but only because there were other answers. Had there been 100 choices with 99 of them being no better than the others then narrowing down to twenty would not make you more sure or less sure. This is not a pure statistical probability problem. This is a gut feeling and the number of other answers don’t change how you feel about your 1st choice.

Spoiler

You thought the remaining answer had a 10% probability of being right. Do you still think that?

--Bob

[BackInTex]

I’m with SSS

Spoiler

its still 70%. You were 70% confident not because of the other answers but only because there were other answers. Had there been 100 choices with 99 of them being no better than the others then narrowing down to twenty would not make you more sure or less sure. This is not a pure statistical probability problem. This is a gut feeling and the number of other answers don’t change how you feel about your 1st choice.

Spoiler

You thought the remaining answer had a 10% probability of being right. Do you still think that?

--Bob

Nothing is ever said or assumed about any answer other than the 70% one. This is not a random probability situation.

[littlebeast13]

I’m with SSS

Spoiler

its still 70%. You were 70% confident not because of the other answers but only because there were other answers. Had there been 100 choices with 99 of them being no better than the others then narrowing down to twenty would not make you more sure or less sure. This is not a pure statistical probability problem. This is a gut feeling and the number of other answers don’t change how you feel about your 1st choice.

Spoiler

You thought the remaining answer had a 10% probability of being right. Do you still think that?

--Bob

Nothing is ever said or assumed about any answer other than the 70% one. This is not a random probability situation.

Spoiler

I agree with the strange bedfellows. If you subjectively believe that a certain answer has a 70% probability of being correct, the only way that changes by removing any of the possible answers is if the 70% answer itself disappears. The remaining 30% probability isn't spread out among the other answers in any special way... that 30% just gets consolidated into whichever other choice remains after the 50/50...

lb13

[littlebeast13]

Spoiler

You thought the remaining answer had a 10% probability of being right. Do you still think that?

--Bob

Nothing is ever said or assumed about any answer other than the 70% one. This is not a random probability situation.

Spoiler

I agree with the strange bedfellows. If you subjectively believe that a certain answer has a 70% probability of being correct, the only way that changes by removing any of the possible answers is if the 70% answer itself disappears. The remaining 30% probability isn't spread out among the other answers in any special way... that 30% just gets consolidated into whichever other choice remains after the 50/50...

lb13

Spoiler

FWIW, I get what Bob is trying to do here since the very fact that B survives the 50/50 should theoretically increase the odds that it is correct by taking its base probability from 25% to 50%. My issue is that the 70% is entirely based on the player's gut, and in the real world that shouldn't change since it becomes exactly what SSS originally said, it's a A vs. Not A problem...

lb13

[Bob78164]

I’m with SSS

Spoiler

its still 70%. You were 70% confident not because of the other answers but only because there were other answers. Had there been 100 choices with 99 of them being no better than the others then narrowing down to twenty would not make you more sure or less sure. This is not a pure statistical probability problem. This is a gut feeling and the number of other answers don’t change how you feel about your 1st choice.

Spoiler

You thought the remaining answer had a 10% probability of being right. Do you still think that?

--Bob

Nothing is ever said or assumed about any answer other than the 70% one. This is not a random probability situation.

That's not correct. You are expressly told that "none of the remaining choices looks more plausible than another."

Lat's assume you think B probably is not the answer, so that it only has a 10% probability of being right, but that none of the three looks more plausible than another. Now the 50-50 leaves B behind. Do you still think the other choice has only a 30% probability of being right?

[BackInTex]

Spoiler

You thought the remaining answer had a 10% probability of being right. Do you still think that?

--Bob

Nothing is ever said or assumed about any answer other than the 70% one. This is not a random probability situation.

That's not correct. You are expressly told that "none of the remaining choices looks more plausible than another."

Lat's assume you think B probably is not the answer, so that it only has a 10% probability of being right, but that none of the three looks more plausible than another. Now the 50-50 leaves B behind. Do you still think the other choice has only a 30% probability of being right?

Again, this is not a problem of random events. Your initial confidence in being correct is based on non-random criteria thus the correct answer is not randomly distributed between the 4 initial answers. Your probabilities are grouped into two choices (A, B1, B2, B3).

Let's assume that that the question was about a famous person. You can't remember who it is but you are 70% sure that person is male. Your choices are: Bob, Judy, Kim, and Terri. Now you are 70% sure it's Bob (because he is male) and none of the other answers, should the correct answer be female, are better than the any of the other females. You do the 50/50 and are left with Bob and Judy.

You are still left with 70% male (Bob) and 30% female (Judy).

[BackInTex]

Here is a similar question:

I drew a random ball out of a bin of 10 consecutively numbered balls. Which of the below is correct?

I drew:

a) an even numbered ball
b) ball #3
c) ball #5
d) ball #9

One of the above IS correct.

Let's assume we do this question 100 times. The correct answer will always be given. Answer a will always be even or odd (randomly selected) with b, c, & d always being 3 numbers opposed to what ever answer a is (thus if a is incorrect, b,c, or d will be correct). Also, the 50/50 is not random and will always leave answer a (see below note).

The probability of answer a will always be 50% before and after the 50/50. This is because the distribution of possible answers is not random, nor is the 50/50.

The probability of the initial set of 4 answers is 50% a, 50% (bcd). After the 50/50 it is still 50% a, 50% not a.

Because the 50/50 is not random, the probabilities do not shift off of a, they get consolidated among b,c, and d.

With the Let's Make a Deal problem, it is assumed that the elimination of lower tier prizes is random. With WWTBAM we all (almost) know the 50/50 is not random, more often than not, the greatest distractor was left along with the correct answer.

[Bob78164]

Here is a similar question:

I drew a random ball out of a bin of 10 consecutively numbered balls. Which of the below is correct?

I drew:

a) an even numbered ball
b) ball #3
c) ball #5
d) ball #9

One of the above IS correct.

Let's assume we do this question 100 times. The correct answer will always be given. Answer a will always be even or odd (randomly selected) with b, c, & d always being 3 numbers opposed to what ever answer a is (thus if a is incorrect, b,c, or d will be correct). Also, the 50/50 is not random and will always leave answer a (see below note).

The probability of answer a will always be 50% before and after the 50/50. This is because the distribution of possible answers is not random, nor is the 50/50.

The probability of the initial set of 4 answers is 50% a, 50% (bcd). After the 50/50 it is still 50% a, 50% not a.

Because the 50/50 is not random, the probabilities do not shift off of a, they get consolidated among b,c, and d.

With the Let's Make a Deal problem, it is assumed that the elimination of lower tier prizes is random. With WWTBAM we all (almost) know the 50/50 is not random, more often than not, the greatest distractor was left along with the correct answer.

You have it exactly backward. In the Monty Hall problem, the elimination of distractors is not random. Monty knows which door conceals the grand prize and never reveals that door.

In this problem, the 50-50 is assumed to be random, so that fact that B survived the 50-50 does provide information. I promise you that when the answer is published next Friday, it will be 87.5%. --Bob

[BackInTex]

Here is a similar question:

I drew a random ball out of a bin of 10 consecutively numbered balls. Which of the below is correct?

I drew:

a) an even numbered ball
b) ball #3
c) ball #5
d) ball #9

One of the above IS correct.

Let's assume we do this question 100 times. The correct answer will always be given. Answer a will always be even or odd (randomly selected) with b, c, & d always being 3 numbers opposed to what ever answer a is (thus if a is incorrect, b,c, or d will be correct). Also, the 50/50 is not random and will always leave answer a (see below note).

The probability of answer a will always be 50% before and after the 50/50. This is because the distribution of possible answers is not random, nor is the 50/50.

The probability of the initial set of 4 answers is 50% a, 50% (bcd). After the 50/50 it is still 50% a, 50% not a.

Because the 50/50 is not random, the probabilities do not shift off of a, they get consolidated among b,c, and d.

With the Let's Make a Deal problem, it is assumed that the elimination of lower tier prizes is random. With WWTBAM we all (almost) know the 50/50 is not random, more often than not, the greatest distractor was left along with the correct answer.

You have it exactly backward. In the Monty Hall problem, the elimination of distractors is not random. Monty knows which door conceals the grand prize and never reveals that door.

In this problem, the 50-50 is assumed to be random, so that fact that B survived the 50-50 does provide information. I promise you that when the answer is published next Friday, it will be 87.5%. --Bob

Not giving Judy her due.

[Bob78164]

Here is a similar question:

I drew a random ball out of a bin of 10 consecutively numbered balls. Which of the below is correct?

I drew:

a) an even numbered ball
b) ball #3
c) ball #5
d) ball #9

One of the above IS correct.

Let's assume we do this question 100 times. The correct answer will always be given. Answer a will always be even or odd (randomly selected) with b, c, & d always being 3 numbers opposed to what ever answer a is (thus if a is incorrect, b,c, or d will be correct). Also, the 50/50 is not random and will always leave answer a (see below note).

The probability of answer a will always be 50% before and after the 50/50. This is because the distribution of possible answers is not random, nor is the 50/50.

The probability of the initial set of 4 answers is 50% a, 50% (bcd). After the 50/50 it is still 50% a, 50% not a.

Because the 50/50 is not random, the probabilities do not shift off of a, they get consolidated among b,c, and d.

With the Let's Make a Deal problem, it is assumed that the elimination of lower tier prizes is random. With WWTBAM we all (almost) know the 50/50 is not random, more often than not, the greatest distractor was left along with the correct answer.

You have it exactly backward. In the Monty Hall problem, the elimination of distractors is not random. Monty knows which door conceals the grand prize and never reveals that door.

In this problem, the 50-50 is assumed to be random, so that fact that B survived the 50-50 does provide information. I promise you that when the answer is published next Friday, it will be 87.5%. --Bob

Told you so. --Bob

[BackInTex]

Here is a similar question:

I drew a random ball out of a bin of 10 consecutively numbered balls. Which of the below is correct?

I drew:

a) an even numbered ball
b) ball #3
c) ball #5
d) ball #9

One of the above IS correct.

Let's assume we do this question 100 times. The correct answer will always be given. Answer a will always be even or odd (randomly selected) with b, c, & d always being 3 numbers opposed to what ever answer a is (thus if a is incorrect, b,c, or d will be correct). Also, the 50/50 is not random and will always leave answer a (see below note).

The probability of answer a will always be 50% before and after the 50/50. This is because the distribution of possible answers is not random, nor is the 50/50.

The probability of the initial set of 4 answers is 50% a, 50% (bcd). After the 50/50 it is still 50% a, 50% not a.

Because the 50/50 is not random, the probabilities do not shift off of a, they get consolidated among b,c, and d.

With the Let's Make a Deal problem, it is assumed that the elimination of lower tier prizes is random. With WWTBAM we all (almost) know the 50/50 is not random, more often than not, the greatest distractor was left along with the correct answer.

You have it exactly backward. In the Monty Hall problem, the elimination of distractors is not random. Monty knows which door conceals the grand prize and never reveals that door.

In this problem, the 50-50 is assumed to be random, so that fact that B survived the 50-50 does provide information. I promise you that when the answer is published next Friday, it will be 87.5%. --Bob

Told you so. --Bob

Not really. I should not have confused you by even bringing up the Monty Hall problem. It is not relevant. By bringing it up I think you assumed I was using it as an example but my intention was to show it was not relevant here. I did not do a good job at that.

Go back to the WHY you were 70% sure the answer is male (or 50% sure it's even) example. You could have 100 alternative all female (or odd). Eliminate 99 of the females (or odds) and you'd still be 70% sure of your answer (or 50% sure it's even).

Without knowing what the question is or the answers given were you can't give a correct answer other than the "at least 70%".