How do I create a probability model?

Question

You draw a card from a deck If you get a red card ou win nothing. If you get a spade you win $5. If you draw a club you win $10 plus an extra $20 for the ace of clubs. Create a probablity model for the amount you win in this game.

Answer 1

You don't state if you put the card back into the deck after each attampt or not, and if there are Jokers in the deck or not (or only 52 cards of 13 in each of 4 suits). If yes (cards are reinserted into the deck and it's a standard deck of 52 with no jokers), then you have a 1/4 chance of winning $5, a 1/4 chance of winning $10, and a 1/52 chance of winning $20. But if you DON'T replace the card after any attempt, then the odds depend on what you have already pulled, and can quickly get complicated... :-)

Answer 2

I don't fully agree with the given answer. When we say probability model (at least in the context of the problem), in essence is asking for the probability distribution, in which case the probabilities should add up to 1.

card Winnings Probability

Red 0 1/2

Spade 5 1/4

Clubs (except Ace) 10 12/52 or 3/13

Ace of Clubs 30 (10+20) 1/52

Answer 3

🎯 Game Rules Recap:

1. Red cards (hearts or diamonds) → Win $0
2. Spades → Win $5
3. Clubs (except Ace) → Win $10
4. Ace of clubs → Win $30 ($10 + $20 bonus)

🃏 Total cards: 52

1. Red cards = 26 (13 hearts + 13 diamonds)
2. Spades = 13
3. Clubs = 13, but one is the Ace of clubs, so:
4. 12 regular clubs → $10
5. 1 Ace of clubs → $30

🎲 Step 1: List all possible outcomes and their probabilities

We list distinct dollar amounts you can win and calculate the probability of each.

Amount WonDescriptionNumber of CardsProbability
$0	Red cards (hearts/diamonds)	26	26 / 52 = 0.50
$5	Spades	13	13 / 52 = 0.25
$10	Clubs (excluding Ace)	12	12 / 52 ≈ 0.2308
$30	Ace of clubs only	1	1 / 52 ≈ 0.0192

✅ Final Probability Model:

P($0)=0.50P($5)=0.25P($10)=1252≈0.2308P($30)=152≈0.0192\begin{align*} P(\$0) &= 0.50 \\ P(\$5) &= 0.25 \\ P(\$10) &= \frac{12}{52} \approx 0.2308 \\ P(\$30) &= \frac{1}{52} \approx 0.0192 \end{align*}P($0)P($5)P($10)P($30)​=0.50=0.25=5212​≈0.2308=521​≈0.0192​

Answer 4

Hi Ej,

Creating a probability model for this card game involves listing all the possible outcomes and their associated probabilities. In a standard deck of 52 cards, there are 26 red cards (hearts and diamonds), 13 spades, and 13 clubs, including the ace of clubs.

Winning Nothing (Red Card):

1. Number of outcomes: 26 (hearts and diamonds)
2. Probability: 26/52 = 0.5

Winning $5 (Spade):

1. Number of outcomes: 13 (spades)
2. Probability: 13/52 = 0.25

Winning $10 (Club, but not Ace):

1. Number of outcomes: 12 (clubs, excluding the ace)
2. Probability: 12/52 = 0.2308

Winning $30 (Ace of Clubs):

1. Number of outcomes: 1 (ace of clubs)
2. Probability: 1/52 = 0.0192

Putting it all together, your probability model would look like this:

1. Win $0: Probability = 0.5
2. Win $5: Probability = 0.25
3. Win $10: Probability = 0.2308
4. Win $30: Probability = 0.0192

This model gives you a complete picture of the possible winnings and their associated probabilities. Now you can make informed decisions about whether or not to play the game, or how to adjust the game to make it more or less favorable.

I hope this helps!

| Outcome | Number of Outcomes | Probability |

|-------------------|--------------------|-------- |

| Winning Nothing (Red) | 26 | 0.5 |

| Winning $5 (Spade) | 13 | 0.25 |

| Winning $10 (Club) | 12 | 0.2308 |

| Winning $30 (Ace Club) | 1 | 0.0192