Pick a Number

Here's an easy game. Pick a random number between 1 and 10. No, not the rating of your favorite TV show. Just a random number. Run the numbers through your head and pick one. Got it? Don't read further until you settle on a number.

Your number is most likely 3 or 7.

We don't want to pick 1 or 10 since those we feel are outliers. And 5 is definitely not random since it is right in the middle. So some number greater than 1 and less than 5 or a number greater than 5 or less than 10 sounds like a reasonably random number. 2, 4, 6, or 9 may not be random enough since they are so close to the middel and extreme values. So we pick 3 or 7.

Of course in reality ALL numbers are equally random.

Us humans generally do not do a very good job picking random numbers nor ascertaining probabilities or likely events for any situation beyond the most simple scenarios using our intuition or common sense alone.

Thus many probability problems that apply to very common scenarios become paradoxical and fun to explore!

We will explore more of these paradoxes in this blog.

Pick a Card Part 2

Continuation of our puzzle from yesterday.

Let's solve this problem using brute force. We do this by looking at all of the possibilities. Let's label the cards as follows:
R1 - the first red card
R2 - the second red card
B1 - the first black card
B2 - the second black card

Next, list all of the possibilities of the cards from left to right:

1. R1 R2 B1 B2
2. R1 R2 B2 B1
3. R2 R1 B1 B2
4. R2 R1 B2 B1
5. R1 B1 R2 B2
6. R1 B1 B2 R2
7. B1 R1 R2 B2
8. B1 R1 B2 R2
9. B1 B2 R1 R2
10. B1 B2 R2 R1
11. B2 B1 R1 R2
12. B2 B1 R2 R1
13. B2 R2 R1 B1
14. B2 R2 B1 R1
15. R2 B2 R1 B1
16. R2 B2 B1 R1

We have exactly 16 different ways these cards can be dealt. Check it for yourself.

Now since these cards are dealt face down, we don't know the card values. Let's say we pick the left 2 cards. How many of the 16 possibilities gives us either 2 red cards or 2 black cards? Just check the list above and count them. We get 8.

Thus our probability of getting a pair of cards of the same color is 8/24 or 1/3!

Is that our final answer? Yes it is! :) Not 2/3, nor 1/2 as we might have expected intuitively.

Next time we will see if we can devise some shortcuts and see if we can extend this to more general problems.

Pick a card, any card

Pick a card, any card. Better yet, pick 2! 

Let's say I have a small deck of 4 cards. 2  black and 2 red.  I shuffle this deck, place them face down, and ask you to pick any 2 cards. What is the probability that these card are the same color?

One person may say there are 3 possibilities. Both are black, both are red, or both are different colors. Therefore, the probability is 2/3.

Hold on! Says another person. That isn't right. They can  both be black, both be red, the first card you picked is black and the other is red, or the first card you picked is black and the other is red.  Thus, the cards match in 2 out 4 cases, giving us a probability of 1/2.

In fact, both are wrong!

What is the true probability and why?

Dice Paradox and Conditional Probability

Let's look at another example.

The probability of rolling a 7 with 2 dice is 1/6. We can verify this by first listing all possible combinations of the 2 dice:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).

Now, we scan through this list of 36 possible events and count all events in which the sum of the 2 dice is 7. The number of events in which this is true is 6. Check for yourself. THus the probability of obtaining a 7 is 6/36 or 1/6. This is true regardless of whether you throw one die followed by the second die or throw them both at the same time.

If I bet on 7 for any random throw of the dice, I would have a chance of winning 1/6 of the time. Now, being an astute gambler, I would like to improve the odds of me winning. Let's have an impartial observer take a look at the dice after they are rolled, but not allow me to see the result. Let's say our favorite number is '4', so anytime the observer sees a '4' on at least one of the die thrown, he shouts out "I see a four!" With this extra bit of information provided before we make our bet, we have improved our odds of winning from 1 in 6 to 2 in 11! We have improved our odds in spite of the fact that we have not changed the probability of the resulting roll of the dice.

How can that be? Well, we just need to go back to our list of all possible events and count the number of events in which a '4' appears on at least one of the die. 11 of the 36 combinations have at least one '4' appear. Now of those 11, how many have a sum equal to 7? Exactly 2. Since we consider only consider betting on those dice rolls in which the number '4' is called out, we have only 11 possibilities and of those only 2 can sum up to 7.

This is an example of conditional probability. This extra bit of information we receive can affect the original probability. The conditional probability theorem gives us a shortcut method to obtain the probability without having to count all combinations and subset of combinations as we did in this example. We will explore the formulation of this conditional probability equation next time.

To be continued...