Test your knowledge of Probability

[Kid R]

Re. ace from 2 cards: 2256/2652 does not include the probability that you are bluffing

 

[Balrog36]

Simple, take a sample size of 10,000 women aged 40-50 (to make the fractions easier). 80 of these women will have breast cancer, 9920 will not have it. (obviously "will" is being used as "mean probability").

Of the 80 women with breast cancer, 72 will test positive.
Of the 9920 women without, 694 (rounded down) will test false positives.
So out of 10,000 women, a total of 766 will have positive results.

Since the patient tests positive she falls into this category. There is a 72/766 chance that she is one of the true breast cancer patients, or 9.3%.

 

[ruff_hi]

Re. ace from 2 cards: 2256/2652 does not include the probability that you are bluffing

Pretty much all probability questions set as puzzles assume honesty and also true packs or dice. I have seen some questions the explicitly try to incorporate the issue of something 'dishonest' in the system.

 

[ParadigmShifter]

That would be game theory then.

 

[Kid R]

I think it's a bluff. That dog looks nervous, his nose is all shiny.

 

[Dhoomstriker]

Ok, someone pm'd for an answer ... let's see if I can answer part a) ...

1] I draw two cards. My outcomes are:

A A
A xA
xA A
xA xA

The chance of getting A A is 4/52 and 3/51 or 12 / 2652
The chance of getting A xA is 4/52 and 48/51 or 192 / 2652
The chance of getting xA A is 4/52 and 48/51 or 192 / 2652
The chance of getting xA xA is 48/52 and 47/51 or 2256 / 2652

The the chance of getting at least 1 Ace is 396 / 2652 (14.9%)
As it turns out, I have given enough information above to work out this part too.

In the bolded line, the numbers 4 and 48 should reverse locations, right?

I look at both cards and I tell you that I have at least 1 Ace.
b) What is the probability that I have another Ace?

Well, as you said:
The the chance of getting at least 1 Ace is 396 / 2652 (14.9%)

In a sense, though, we only really care about those 396 cases, right?

Out of those cases, 12 of them had us picking 2 Aces.

So, would we say that the probability of having a second Ace would be 12 / 396?

You can also use this method ...

P(at least 1 Ace) = 1 - P(No Aces) = 1 - 2256 / 2652.

This is particularly useful if I change the question to:

What is the probability of having an A if I draw 20 cards from a 52 card deck? No one wants to use the 'brute force and stupidity' method I outlined above when there are 20 cards involved.

Okay, it's a bit confusing here, since the P(No Aces) value that you substituted seems to apply for a different case than when 20 cards are drawn, if I understand things correctly.

Presumaly, if I follow what you are doing, then we'd want to calculate P(No Aces) for drawing 20 cards, which should be something like:
(48/52 * 47/51 * 46/50 * 45/49 * ... * 30/34 * 29/33)

Then you'd subtract that value from "1" in order to get the probability of finding at least one Ace.

Presumably, there is a formulaic way of displaying this series of fractions?

Also, is there a simple method of entering the calculation into a calculator or a program like MatLab without simply typing out all of the fractions?


Extending that question, can you answer how you would solve it if you were asked: What would you do if you were asked for the probability of having exactly 3A if you draw 20 cards from a 52 card deck?

 

[ruff_hi]

In the bolded line, the numbers 4 and 48 should reverse locations, right?

right - oops.

In a sense, though, we only really care about those 396 cases, right?

Out of those cases, 12 of them had us picking 2 Aces.

So, would we say that the probability of having a second Ace would be 12 / 396?

Right. I have explicitly ruled out the xA xA option, leaving the other three. So you take the A A option over the total of the other three and you get the answer - about 3%.

Presumably, there is a formulaic way of displaying this series of fractions?

yeah - its all about combinations and permutations (read the link about the factorial (!) function).

Extending that question, can you answer how you would solve it if you were asked: What would you do if you were asked for the probability of having exactly 3A if you draw 20 cards from a 52 card deck?

Yes. Basically, divide the 20 cards into 2 groups ... 3 cards and 17 cards. You need to calculate the ways you can get 3As from 4As and 17 xAs from 48 xAs. Then divide that by the total number of ways of selecting 20 cards from a deck of 52.

Getting 3As from 4As = ways of selecting 3 cards from 4 / ways of arranging 3 cards
= 4 . 3 . 2 / 3!
= 4

Getting AH, AD & AS is exactly the same as AD, AH, AS so the order is not important. If the order was important, you wouldn't divide by 3!.

 

[thadian]

great thread overall!