Gaming nerds since '99

- I can't remember my stats.

- I don't know what to google for.

- I am feeling very tired today.

- I love you guys for your help.

How do I solve the following sort of problem?

- A user has a 74% chance to answer a question correctly.

- A user is presented with three questions.

- A user passes if: They answer any two questions correctly.

- They fail if they answer one or less questions correctly.

Thanks for any help!

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## Comments

http://www.mathwords.com/b/binomial_probability_formula.htm

Awesome.

The chance of answering 0 questions correctly is 1.7%

The chance of answering only 1 question correctly is 15%

(Yes, I edited this after reading Xocet's response and going back through my work . . . I punched something in wrong with my 1 question correct. This is why we should /think/ about things rather than just hammer them into a calculator, right?)

All 3 correct: You have a 74% chance of getting the first correct. Of that 74% of cases, you have a 74% chance of getting the second correct. And of that 74% of that 74% you have a 74% chance of getting the last one correct. So, (0.74)(0.74)(0.74) = (0.74)^3 = 40.5% chance

Looking at only getting two answers, there are three arrangements. You get the first wrong, the second wrong or the third wrong, and we can add up the probability of each of these three cases. We can write it out like this: (0.26)(0.74)(0.74) + (0.74)(0.26)(0.74) + (0.74)(0.74)(0.26) = 3 * 0.26 * (0.74)^2 = 42.7%

For one correct answer, there are three arrangements, as above, except in this case your three are first right, second right and third right. Add them up: (0.74)(0.26)(0.26) + (0.26)(0.74)(0.26) + (0.26)(0.26)(0.74) = (0.26)(0.26)(0.74) * 3 = 3 * (0.26)^2 * 0.74 = 15%

For no correct answers, it's just like the first case, except with the probability of getting the answer wrong cubed: (0.26)^3 = 1.76%

Note that these will all add to 100%. So in this case the probability of failing is 16.76% (adding the 1 correct and 0 correct numbers).

Thinking it out like this works well as long as you don't have too many trials. If things get crazy, you'll want to use the formulas. You'll notice that all this is the same as the formulas in Etter's link though. In our case p = 0.74, q = 0.26, n = 3, k = 0, 1, 2 or 3. To calculate the coefficient out front you need to use the following formula:

In the 3 correct and 0 correct cases you have a coefficient out front of 1:

n!/k!(n-k)! = 3!/(3!)(3-3)! = 3!/3!*1 = 1 or

n!/k!(n-k)! = 3!/(0!)(3-0)! = 3!/1*3! = 1

In the 1 correct and 2 correct cases, both coefficients are 3:

n!/k!(n-k)! = 3!/(1!)(3-1)! = 3!/1!*2! = 3*2*1/1*2*1 = 3 or

n!/k!(n-k)! = 3!/(2!)(3-2)! = 3!/2!*1! = 3*2*1/2*1*1 = 3

So, I am working on requirements for a personal security question system and trying to figure out the likelyhood that a) someone will remember enough questions to reset their password, b) their spouses won't be able to guess the answers too easily. I am looking at various scenarios and want to put some hard numbers to things (which is far more than anyone else involved seems to be willing to do, grumble).

Obviously there are a lot of potential factors for each case, but I have fairly reliable stats that:

-People remember any particular answer correctly 74% of the time

-Spouses can guess any particular answer correctly 20% of the time (I need to confirm that #, it is from memory).

Anyway, this helps a lot. I have one more question if Xocet or anyone else can spare some time to help:

How do I calculate the probability for the following scenario?

A user has 3 questions setup.

They are prompted for 2 answers randomly. If they get one or more wrong, they are then prompted with 2 answers (of the 3) randomly again. This could be the same 2 as before, it is completely random each time. They get 3 attempts overall.

At first I thought I might solve it as:

It is roughly 54% likely that they will get 2/2 correct, for any two questions. So, that is easy enough to plugin to the formula as n=3, k=1, p=0.54. That's a 90% chance of success but intuitively I know that is wrong. The problem is they will consistently get the same questions right and the same questions wrong, so if the question they get wrong keeps coming up it lowers the chances. Hmmm, I'll try thinking about this one logically, to try and make Xocet proud... but I am in somewhat of a rush to get this done so I would appreciate help. :)

I found this was a nice calculator btw:

http://faculty.vassar.edu/lowry/ch5apx.html

My initial, late night woke up and couldn't sleep guess is the chances go down by something like:

54% success round 1

36% success round 2

18% success round 3 (or maybe lower)

and so overall somewhere in the mide to low 70's...

thinkyou mean.Without worrying about which questions come up, we can consider the first option, which is that the know all three answers. In this case, they pass no matter what questions come up.

This happens (0.74)^3 = 40.5% of the time.

In they only know one answer (15% of the time) or they don't know any answers (1.76% of the time) then they'll fail no matter what questions come up.

So what about when they know two of the three answers? This happens in 42.7% of cases. However, in this 42.7% of cases, how likely is it that they get the question that they don't know three times in a row?

Well, they don't know one of the answers. So, there's a 66% chance that the question won't come up as the first question. But since we've now used up a question that they DO know, there's only a 50% chance of them getting a question that they know for the second question.

So the chances of them passing round one if they know two of three answers is (0.66)(0.5) = 33%.

Since probability has no memory, we know that their chances of passing rounds two and three are the same, at 33%.

Now, we only get to round two if they fail round one, which will happen 66% of the time. And they'll only get to round three if they fail round two, which happens 66% of the 66% of the time. And they'll only fail if they get round three wrong, which is 66% of 66% of 66% of the time.

So overall, what are the chances of success? Well, that'll be all the 3 answer people (40.5%) plus all of the two answer people (42.7%) minus the two answer people that failed all three rounds ((0.66)(0.66)(0.66)(0.427) = 12.3%):

0.405 + 0.427 - 0.66(0.66)(0.66)(0.427) = 70.9%

Let's look at this round by round.

Round 1: 40.5% chance of success from knowing all three answers. 33% chance of passing round 1 if they only know two answers, which happens 42.7% of the time. So chances of passing round 1 are 0.405 + 0.33(.427) = 54.6%

Round 2: If they got to round two, they don't know all three answers. So they either know 0, 1 or 2 of the three answers. We know that people that know 0 or 1 answers will fail here no matter what. So what proportion of people know two answers?

Well, 42.7% of the people who started know two, 15% know one and 1.76% know zero. But 33% of the 42.7% percent were successful in the first round, so they're not in this round. So the proportion of people who know two answers in this round is

(0.66)(0.427) / (0.66(0.427) + 0.15 + 0.0176) = 62.7%

So 62.7% of the people in round two know two answers. And they have a 33% chance of success. So (0.33)(0.627) = 20.7% of people pass round two.

Round 3: This is similar to round two. The people who passed round two are gone though, which is 33% of the people who were still in that knew two answers. So we work out the proportion again.

(0.66)(0.66)(0.427) / (0.66(0.66)(0.427) + 0.15 + 0.0176) = 52.6%

So 52.6% of the people in round three know two of the answers. And again, 33% of them will pass. So the number of people who pass round three is (0.33)(0.526) = 18.5%.

I'm just working this out as I type, so there may be mathematical or logical errors in here. This is how I'd reason through it though.

Summary

Overall: 70.9% chance of success

Round 1: 54.6% chance of success

Round 2: 20.7% chance of success

Round 3: 18.5% chance of success

I did a bad job explaining what I was after, sorry!

Here is a better summary I hope:

- Users have only three PVQs setup to draw on

- There will be up to three rounds for them to pass.

- Each round ONLY TWO (of three setup) questions will be randomly presented to the user.

- They must get both correct to proceed or they go on to the next round (up to three).

- Each round two questions are randomly picked again. They are always randomly picked. In theory (though unlikely) it could be the same two questions over and over again.

You said "probability has no memory" but I think in this case it does, right? I think the key is to calculate the probability in round 2 and 3 that one of the two questions will be a question that the user previously failed on. Each round that probability goes up.

Of course, a better system would be to include some smarts and not just randomly pick two each round, but I need to prove that this is a bad design for us.

In any case, what I explained up there exactly matches what you just outlined.

I mean, if they get to round 2, we now though that AT LEAST one of the three questions is guaranteed to fail, it's not like the user still has a 74% chance of passing each question.

Know what I mean? If I follow the logic above I end up with a 90% success rate which is better than just getting 2/3 correct at first, which makes no sense to me.

You'll notice I'm not using the 74% past grouping the people up into 0, 1, 2 and 3 groups. After that it's all about which question comes up (i.e. if they get a question they know or a question they don't know). In the round calculations the have a 100% or 0% chance of getting any given question right, but they have a 33% chance of getting the question they have a 0% chance of getting right.

What you posted makes complete sense.

Awesome! :)

I really have nothing useful to add to this thread, nor did I read all the posts, I just wanted to point that out.

The only other thing I have to add is that the way that works for ME is to imagine branching trees based on the percentages. This is no different in result than exactly what phro did, but for things that don't branch widely it's extremely intuitive (and even then you can apply it to parts of the "tree"). That's one of the many ways I remember being taught in stats classes and it's the only one that has really stuck with me -- I certainly wouldn't remember any of the more complex formulas even though they are really useful (combinations and Bernoulli Trials, etc.)

Get a baseball bat and ensure the user never gets any questions wrong... or else.