Conditional Probability with Dice

Conditional Probabilities in Practice

Focusing the Sample Space

Conditional probability, P(A|B), asks for the probability of event A happening given that event B has already happened. This "given" information is powerful because it allows us to narrow down our focus. Instead of considering all possible outcomes, we only look at the outcomes where B is true. This new, smaller set of possibilities becomes our reduced sample space.

When all outcomes in this reduced sample space are equally likely, the conditional probability can often be found by simple counting:
P(A|B) = (Number of outcomes where A and B both occur) / (Number of outcomes where B occurs)

EASY

You roll two standard fair six-sided dice. Given that their sum is 6, what is the probability that at least one of the dice shows a 4?

Explanation: Conditional Probability with Dice

Imagine you rolled two dice, and your friend (who can see them) tells you, "Hey, the sum of the two dice is 6!"

Now, given this information, what's the chance that at least one of those dice is showing a 4?

Possible ways to get a sum of 6: Let's list them (Die 1, Die 2):
- (1, 5)
- (2, 4)
- (3, 3)
- (4, 2)
- (5, 1)
There are 5 possible ways the dice could have landed to give a sum of 6. This is our new world of possibilities.
Which of these have at least one 4?
- (2, 4) - Yes, has a 4.
- (4, 2) - Yes, has a 4.
There are 2 ways to have at least one 4, given the sum is 6.
The Chance: Out of the 5 ways to get a sum of 6, 2 of them include a 4. So the probability is 2/5.

Let A be the event that at least one die shows a 4.
Let B be the event that the sum of the two dice is 6.

We want to find the conditional probability P(A | B), which is the probability of A occurring given that B has occurred.

1. Identify the Reduced Sample Space (Given Sum = 6)

We are given that the sum of the two dice is 6. We list all possible outcomes (Die 1, Die 2) for this event B:

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

There are 5 equally likely outcomes in this reduced sample space where the sum is 6.

2. Identify Favorable Outcomes within the Reduced Sample Space (At least one 4 AND Sum = 6)

Now, from the outcomes listed above (where the sum is 6), we identify those where at least one die shows a 4. This corresponds to the event (A ∩ B).

(2, 4) (Contains a 4)
(4, 2) (Contains a 4)

There are 2 favorable outcomes in the reduced sample space.

3. Calculate the Conditional Probability

The probability of A given B is the number of outcomes in (A ∩ B) divided by the number of outcomes in B (our reduced sample space):

P(At least one 4 | Sum = 6) = (Number of outcomes with sum=6 AND at least one 4) / (Number of outcomes with sum=6)

P(At least one 4 | Sum = 6) = 2 / 5

As a decimal, this is:

2/5 = 0.4 or 40%

Using the Formal Conditional Probability Formula:
The full sample space for two dice has 36 outcomes.

Event B (sum=6): {(1,5), (2,4), (3,3), (4,2), (5,1)}. So, P(B) = 5/36.
Event A (at least one 4): {(1,4), (2,4), (3,4), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,4), (6,4)}. There are 11 outcomes (P(A) = 11/36).
Event (A ∩ B) (sum=6 AND at least one 4): {(2,4), (4,2)}. So, P(A ∩ B) = 2/36.

Using P(A|B) = P(A ∩ B) / P(B):
P(A|B) = (2/36) / (5/36) = 2/5.
This confirms our result from directly using the reduced sample space.

Explanation: Dice Game Expected Value

Imagine playing this dice game many, many times. What would be your average winnings (or losses) per game? That's what "expected profit" means.

Cost to Play: $5 each time.
Winning: You win if the sum is 7. You get $21. Since you paid $5, your actual profit from winning is $21 - $5 = $16.
The chance of rolling a 7 is 6 out of 36 (or 1/6).
Losing: If the sum is not 7, you lose the $5 you paid. So your profit is -$5.
The chance of not rolling a 7 is 30 out of 36 (or 5/6).
Average it out:
(Chance of Winning × Profit if you Win) + (Chance of Losing × Profit if you Lose)
(1/6 × $16) + (5/6 × -$5)

If this average is positive, it's a good game to play long-term. If negative, you'd expect to lose money on average.

To calculate the expected profit, we need to identify all possible outcomes of the game, their associated profits (or losses), and their probabilities.

1. Identify Possible Outcomes and Their Profits

The cost to play is $5.

Outcome 1: Win (Sum of dice is 7)
You win $21.
Net Profit = Winnings - Cost to Play = $21 - $5 = $16
Outcome 2: Lose (Sum of dice is NOT 7)
You win $0.
Net Profit = Winnings - Cost to Play = $0 - $5 = -$5 (a loss of $5)

2. Determine the Probabilities of These Outcomes

Probability of Winning (P(Sum = 7)):
When rolling two fair dice, there are 36 total possible outcomes (6 outcomes for the first die × 6 outcomes for the second die).
The combinations that sum to 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 such combinations.
P(Win) = P(Sum = 7) = 6/36 = 1/6
Probability of Losing (P(Sum ≠ 7)):
This is the complement of winning.
P(Lose) = 1 - P(Win) = 1 - 1/6 = 5/6

3. Calculate the Expected Profit (E[Profit])

The expected value is the sum of each outcome's profit multiplied by its probability:

E[Profit] = (Profit of Win × P(Win)) + (Profit of Lose × P(Lose))

E[Profit] = ($16 × 1/6) + (-$5 × 5/6)
E[Profit] = $16/6 - $25/6
E[Profit] = -$9/6

Simplifying the fraction:

E[Profit] = -$3/2 = -$1.50

Final Result

The expected profit from playing this game once is -$1.50.

This means that, on average, you would expect to lose $1.50 each time you play this game in the long run.

Interpretation: An expected profit of -$1.50 indicates that the game is unfavorable to the player. While you might win in some individual plays (gaining $16), over many plays, the losses from the more frequent losing outcome will outweigh the wins, leading to an average loss of $1.50 per game

Fair Game? What are your thoughts on this? Try answering the questions yourself and share your insights or alternative approaches in the comments section below!

Back to Probability

Conditional Probability with Dice

Conditional Probabilities in Practice

Focusing the Sample Space