Probability Flashcards

How many straight flushes are there (including and not including royal)?

Not royal:
Pretend all suits are spades, then multiply by 4. With 5 cards, K=11, Q=12, J=13, A=1 (ace can only be 14 in a royal flush). Biggest possible number is 9,10,11,12,13. That's 9 possibilities × 4 suits = 36 total.

Royal:
Add the royal flush: 10, J, Q, K, A × 4 suits = 40 total.

Area of red = ¼ area of black (π vs 4π), so about ¼ of the 100 dots, or 25, would be expected in the red circle.

$12 if you pull out with replacement. RGG, RBP, and YYB. Find the probability.

3! = 6

RBP → 6 (ways to order) × 1/10 (red) × 1/5 (blue) × 3/10 (purple) = 3/1000 × 6 = 18/500

YYB → ways to order = 3!/2 = 3
3 × (3/20) × (3/20) × (1/5) = 27/2000 × 3 = 81/2000

RGG → 3 × (1/10) × 1/4 × 1/4 = 3/160

These sum up to 273/4000

Full house with 5 dice. Find the probability.

Denom = 6⁵ = 7776

Num = 6C1 (set with 3) × 5C1 (set with 2) × 5C3 (to add order)

This gives 300/7776 = 25/648

Three of a kind dice odds. Denom = 7776, num = ?

6C1 for the set with 3.
5C3 to choose which dice show that value = 10.
5P2 for the other 2 (order matters) = 20.

6C1 = 6, 5C3 = 10, 5P2 = 20
10 × 6 × 20 = 1200

1200/7776 = 50/324 = 25/162

Large straight with 5 dice. Find the probability.

There are 2 large straights: 12345 and 23456.
Each has 5! ways to arrange it.
Total = 5! × 2 = 240

Denom = 7776
240/7776 = 5/162

Odds 2 dice are rolled, 2nd is ≥ first.

6 first gives 6 options, 5 gives 5, etc.
6 + 5 + 4 + 3 + 2 + 1 = 21

21/36 = 7/12

10 girls and 5 boys, 2 randomly leave room. Most likely ratio = ?

All will have denom 15C2.
Odds 2 girls leave: 10C2 = 45
Odds 2 boys leave: 5C2 = 10
15C2 = 105. 105 - 55 = 50

9 girls to 4 boys

1, 2, 3, 4, 5, 6, 7, 8, 9
You are choosing 3 of them, what is the expected value.

All are equally likely, each balances with another. 9 and 1, 8 and 2, 7 and 3, 6 and 4, and then there is the average 5. However, you are picking 3, so multiply by 3, and get 15

Let y = number of $1 tokens. Expected value is 2.25, equal $1 and $2, with 18 $5. How many tokens?

The expected value will be total value divided by total number of tokens. n + 2n + 90 = 3n + 90. 3n + 90 is the total value, 2n + 18 is the number of tokens. (3n + 90)/(2n + 18) = $2.25. 3n + 90 = 4.5n + 40.5. 1.5n = 49.5, n = 33. 2n + 18 = number of tokens, 66 + 18 = 84

Expected loss is 50 cents, red gains one, k loses 1.

The expected value is 5/(5+k) - k/(5+k). This is just (5-k)/(5+k). That is the expected value, and it equals -0.5. 5 - k = -2.5 - 0.5k. 7.5 = 0.5k, k = 15

Expected value = 1.20. How many 1 dollars are needed to offset a $5 to 1.20.

If the average is 1.2, $5 is 3.8 extra, $1 is 0.2 under. 3.8/0.2 = 19, meaning you need 19 $1s per 1 $5. 19 + 1 = 20

Game costs $5. Fair game with 6 dice.

1/2 × 1/6 = 1/12 chance at some number. 5 × 12 = 60

4/38(8) + 34/38(-1) = -2/38 = -1/19

8C2 = 28 ways to pick 2 points. Possible distances are 1 (8 ways) 2 (6 ways) √5 (8 ways) 2√2 (2 ways) and √2 (4 ways).

8 + 12 + 8√5 + 4√2 + 4√2 = (20 + 8√5 + 8√2)/28 = (5 + 2√5 + 2√2)/7

This is just sum up all the terms of Pascal's that are above choosing 5, since 10 - 5 = 5, and we need more heads than tails.

210 + 120 + 45 + 10 + 1 = 386. 2¹⁰ = 1024. 386/1024 = 193/512

Roll 7 dice.

Sum = 6 → 0
Sum = 7 → 1/279936
Sum = 23 → 66641, 66632, 66551, 66542, 66533, 66416, 66443

>40 sum, max is 42. Law of symmetry can be used here. 42 is the same as 7, 41 is the same as 8. 8 has 7 ways. 7 + 1 = 8. 8/279936 = 1/34992

Sum is greater than 29

52C13 = denom. num = 48C9

52C13 / 48C9 = 11/4165

Roll 12 6 sided dice, odds you get a prime with product?

Simple: In order to have a prime, it must be all ones and then a single prime. The odds you get all ones are (1/6)¹¹, and then the odds you get a prime is 3/6 = 1/2. You then must multiply by the 12 ways to order, since the order you roll the numbers in matters. Doing this, you get (1/6)¹¹ × (12/2) = (1/6)¹¹ × 6 = (1/6)¹⁰

1296 = denom. All we are doing here is choosing dice in a specific order, but allowing repeats. Basically, all you have to do is stars and bars, since repeats are allowed.

Let's do it based on rolls. There are 4 rolls, and 6 numbers. Let's say these 4 rolls are stars, and these 6 numbers are 5 bars. 4 + 5 = 9, of those choosing 4 to be stars, 9C4 = 126. 126/1296 = 7/72

(5C2 × 15C3) / 20C5

14 total marbles. 2 red, 4 green, 8 white. Removing two at random, no replacement. Odds 2 not the same.

Different bad cases. Red is 1/7 × 1/13, green is 2/7 × 3/14, and white is 4/7 × 7/13. Subtract from total of 1. 1 - (1/7 × 1/3 + 2/7 × 3/14 + 4/7 × 7/13) = 8/13