Probability

Coins, dice, cards, conditional probability and expected value, from the first rules to classic puzzles.

1. 01At least one high rollEasyAbout 2 min

   You roll three fair six-sided dice. What is the probability that at least one of them shows a 5 or a 6?

   Hint

   Count the opposite case first: no die shows a 5 or a 6.

   Answer

   19/27, about 0.704.

   Solution

   One die avoids 5 and 6 with probability 4/6 = 2/3.

   The dice are independent, so all three avoid them with probability (2/3)^3 = 8/27.

   At least one shows a 5 or a 6 with probability 1 - 8/27 = 19/27, about 0.704.

   Mark as done
2. 02Two chips of different colorEasyAbout 2 min

   A bag holds 5 red chips and 3 blue chips. You draw two chips without putting the first one back. What is the probability that the two chips have different colors?

   Hint

   There are two orders that give one chip of each color. Add them up.

   Answer

   15/28, about 0.536.

   Solution

   Red first and then blue has probability 5/8 × 3/7 = 15/56.

   Blue first and then red has probability 3/8 × 5/7 = 15/56.

   The two orders exclude each other, so the total is 30/56 = 15/28, about 0.536.

   Mark as done
3. 03Flagged tradesMediumAbout 5 min

   On a desk, 4% of all trades contain a booking error. An automatic check flags 90% of the trades with an error, and it also flags 5% of the trades without one. A trade gets flagged. What is the probability that it really contains an error?

   Hint

   Picture 1,000 trades and count how many of the flagged ones are real errors.

   Answer

   3/7, about 43%.

   Solution

   Out of 1,000 trades, 40 have an error and 960 do not.

   The check flags 0.9 × 40 = 36 trades with an error and 0.05 × 960 = 48 trades without one.

   So 84 trades are flagged, and 36 of them are real errors: 36/84 = 3/7, about 43%.

   A flag is far from proof here, because correct trades are so much more common than errors.

   Mark as done
4. 04Idle exchangesMediumAbout 5 min

   Six orders are sent out, and each one goes to one of four exchanges, chosen uniformly at random and independently of the other orders. What is the expected number of exchanges that receive no order at all?

   Hint

   Give each exchange an indicator that equals 1 if it stays empty. The expected value of a sum is the sum of the expected values.

   Answer

   729/1024, about 0.712.

   Solution

   Let I_k be 1 if exchange k receives no order and 0 otherwise. The number of empty exchanges is I_1 + I_2 + I_3 + I_4.

   A single order misses exchange k with probability 3/4, so all six miss it with probability (3/4)^6 = 729/4096.

   Linearity of expectation works even though the indicators are dependent, so the answer is 4 × 729/4096 = 729/1024, about 0.712.

   Mark as done
5. 05Up or down until a boundaryHardAbout 12 min

   You start with 3 dollars. Each round you win 1 dollar with probability 0.6 and lose 1 dollar with probability 0.4. You stop as soon as you have 6 dollars or 0 dollars. What is the probability that you reach 6 dollars?

   Hint

   Let P(k) be the chance of reaching 6 from k dollars. Condition on the first round and use P(0) = 0 and P(6) = 1.

   Answer

   27/35, about 0.771.

   Solution

   Conditioning on the first round gives P(k) = 0.6 P(k+1) + 0.4 P(k-1) for k from 1 to 5.

   The solutions have the form P(k) = A + B r^k with r = 0.4 / 0.6 = 2/3. The two boundaries give P(k) = (1 - r^k) / (1 - r^6).

   For k = 3: (1 - 8/27) / (1 - 64/729) = (19/27) / (665/729) = 27/35, about 0.771.

   A short simulation agrees:

   import random
   random.seed(1)
   wins = 0
   for _ in range(200_000):
       x = 3
       while 0 < x < 6:
           x += 1 if random.random() < 0.6 else -1
       wins += x == 6
   print(wins / 200_000)   # 0.77139

   Mark as done
6. 06Three short piecesHardAbout 12 min

   A rod of length 1 meter is cut at two points, chosen independently and uniformly along its length. What is the probability that all three pieces are shorter than 40 cm?

   Hint

   The three lengths (x, y, z) add up to 1 and are spread uniformly over the triangle of all such triples. Draw the region where every length stays below 0.4.

   Answer

   The probability is 1/25 = 0.04.

   Solution

   For two uniform cuts, the triple of lengths (x, y, z) with x + y + z = 1 is uniform over the triangle x, y, z ≥ 0.

   Write u = 0.4 - x, v = 0.4 - y and w = 0.4 - z. The condition becomes u, v, w > 0, and u + v + w = 1.2 - 1 = 0.2.

   Each of u, v, w is at most 0.4 automatically, so the region is a full triangle of the same shape, scaled by 0.2.

   Areas scale with the square of the length, so the probability is 0.2^2 = 1/25.

   In general the answer is (3a - 1)^2 for a limit a between 1/3 and 1/2. A simulation confirms the value:

   import random
   random.seed(7)
   n, hit = 400_000, 0
   for _ in range(n):
       a, b = sorted((random.random(), random.random()))
       hit += max(a, b - a, 1 - b) < 0.4
   print(hit / n)   # about 0.040

   Mark as done

Compare your own answer with the worked solution.