Episode 111: Use Probability to Cheat Your Friends! [Discussion]

[jackierulesall]

Win free beers by taking advantage of their bad judgment!

Watch or download now! (http://revision3.com/scamschool/statistics)

[masherscf]

Win free beers by taking advantage of their bad judgment!

Watch or download now! (http://revision3.com/scamschool/statistics)


Not just bad judgment. Human beings are incapable of assesing risk. This why they play the lottery and buy extended warranties.

Don't think so? Look up "the birthday problem" on wikipedia.

[cephas]

Ok, Brian says somebody's going to do this, so I'll take a crack at it, First the face card one.
Consider what it takes to NOT get a face card three times in a row.
The first time, you have a 40/52 chance of not getting a face card.
The second time, there are only 51 cards left, and and the one you removed was not a face card, so you have a 39/51 chance of not getting a face card.
By the same logic, the third time you have a 38/50 chance of not getting a face card.
The trick is that all three of these things have to happen at once, which means you have to multiply the probabilities.

(40/52) * (39/51) * (38/50) = 38/85.

But remember, this is the odds of NOT getting a face card. 1 - 38/85 gives us your odds of getting a face card as 47/85 or 47 to 38.
The odds Brian stated (6 to 5) are close-ish (off by about .7%). A closer approximation would be 5 to 4 (off by about .3%)

Now onto the Monty. Consider what it takes to NOT get a queen both times.
The first time, you have a 4/6 chance of not getting a queen.
The second time, there are only 5 cards left, and the one you removed was not a queen, so you have a 3/5 chance of not getting a queen.
Again, the trick is that both of these things have to happen at once, wich means you have to multiply the probabilities.

(4/6) * (3/5) = 2/5

But remember, this is the odds of NOT getting a queen. 1 - 2/5 gives us your odds of getting a queen as 3/5 or 3 to 2.
Brian's stated odds (6 to 4) are mathematically equivalent. (i.e. exactly correct)

[timestreamer]

To elaborate a little more on Cephas' great explanation:

The two scams in this episode are both examples of phenomena described by the hypergeometric distribution. We are taking several samples without replacement (3 in the first scam, 2 in the second scam). What "without replacement" means in this context is simply that once a sample is taken, that particular card cannot be picked again as part of the same sequence. If the cards were sampled one at a time, reinserting & reshuffling between samples, then the binomial distribution would be appropriate.

I don't know how (or if it's possible) to post equations on this board, so please see the attached PDF.


One quick note: I've met more than a few people that use "odds of" and "probability of" interchangeably. This is wrong. The odds of event A occurring is given by:

Probability(A) : Probability(not A)

The x : y notation is deliberately used instead of x/y to avoid confusing odds with probability.

[darkknight512]

Very interesting problem. This reminds me of this question imagine if you were a cell phone producer "If you have a 1% chance of producing a broken cell phone, what is the chance of making 100 cell phones in a row without having one faulty one?"

I was actually unsure of the answer, I suspected it to be 0.99 to the power of 100 but was unsure, so I wrote a simulator in C++ that simulated this problem 1 million times (user settable), and sure enough it was 0.366 (Aprox 0.99^100)

People are easily fooled by probability because the math behind it can be quite complex in the end.

[timestreamer]

This reminds me of this question imagine if you were a cell phone producer "If you have a 1% chance of producing a broken cell phone, what is the chance of making 100 cell phones in a row without having one faulty one?"

I was actually unsure of the answer, I suspected it to be 0.99 to the power of 100 but was unsure, so I wrote a simulator in C++ that simulated this problem 1 million times (user settable), and sure enough it was 0.366 (Aprox 0.99^100)

Your example is also interesting. If we assume that each phone has the same 1% defect chance independent of the other phones (i.e. it's not a systemic problem like a malfunctioning device in the assembly line) then we can use the binomial distribution I mentioned in my previous post.

The probabilities for a 100 unit production run are (approximately):
0 defective: 0.366032341
1 defective: 0.369729638
2 defective: 0.184864819
3 defective: 0.060999166
4 defective: 0.014941715
5 defective: 0.002897787
6-100 defective: 0.000534534

[latrine]

Another simple way to think of the problems is with combinations, although it's harder to do for the first problem because of the large numbers involved. And again the easiest thing to do is consider the case where the combination contains the complement to the desired outcome.

For the face cards you're trying to find how many combinations of 3 cards can contain a face card. The total number of combinations is 52 choose 3 = 22100. The number of combinations without a face card is 40 choose 3 = 9880. The probability of getting at least one face card is (22100 - 9880) / 22100 which is 47/85 as mentioned above.

For the queens you're trying to find how many combinations of 2 cards can contain a queen. The total number of combinations is 6 choose 2 = 15. The number of combinations without a queen is 4 choose 2 = 6. The number of combinations with at least one queen is 15 - 6 = 9. So the odds are 9:6 or 3:2.