Recreational Mathematics
Probability Puzzles
Many games are played with the element of chance, hence probability can be used to analyze them. In Game Theory, strategies are analyzed based on probabilities and logic. Many games are more than pure luck, there are some form of strategies that can be employed if you work through the logical steps. E.g. it's always possible to force a draw in the game of tic-tac-toe, it is not a game of pure chance.
(1) On the counter of 6 squares marked 1 to, players are to throw 3 dice. If the counter number you choose appears on 1 die only, you get your money back plus the same amount you placed on the square. If 2 dice show your number, you get back your money plus twice the amount you bet; and the same if you have all dice showing your number. Of course if the number is not on any of the dice, you lose. Is the game in favour of you?
(2) How many times on average must a die be thrown so as to get all 6 numbers on the die to appear at least once?
(3) 3 prisoners A, B and C are kept in separate cells. One day, the governor decides to pardon one of them. This news spread among the prisoners, but they have no idea who is pardoned. A asks the governor:" Can you tell me one of the 2 who's not pardoned? If B is pardoned, tell me C. If C is pardoned, tell me B. If I'm to be pardoned, toss a coin to decide who to tell me." The governor says:" If you see me toss the coin, you'll know that you're pardoned. And if you don't see me toss a coin, you'll know it's either you or the person I didn't name." A then says:" Then tell me tomorrow." The governor, who has no knowledge of probability thought this is not going to improve A's chance of surviving, so next day he told him B is one of those not pardoned. Upon hearing this, A, who has some knowledge of probability, reasoned that now B has been eliminated, both C and my chance of survival is 1/2. Is this correct?
(4) In a duel between 3 persons A, B and C, it's decided that they shoot in order. A goes last because he's a 100% marksman, B second as he's a 2/3 shooter (accurate 2 in 3 times on average), and C goes first as he's a 1/3 shooter. This order repeats until last man standing. Since C goes first, who should he shoot?
(5) A and B each writes down an integer from 1 to 100. The game is to guess each other's number first. The rule is they can ask each other questions with Yes or No answers and as long as the answer is Yes, they can carry on asking. Suppose A goes first, what should he ask?
(6) A die is rolled repeatedly until the running total exceeds 12. What is the likely total that'll be obtained?
Answers:
(1) If you reason that the probability of winning is 1/2 because the chance of your number showing on 1 die is 1/6, but since there are 3 dice, the chances must be 3×1/6=1/2, then your are wrong. First of all, the chance of winning is 75/216 for 1 six in 3 dice, 15/216 for 2 6's and 1/216 for 3 6's. Therefore, total chance of winning is 91/216. Moreover, the payoff is different for 1 six than more than 1 6's. The expected payoff is -1×125/216 + 1×75/216 + 2×16/216 = -18/216 = -1/18 i.e. a loss of $1/18 for every $1 bet.
(2) What do we mean when we say that the probability of getting any number on a die is 1/6? It means we need to throw on average 6 times (reciprocal of 1/6) to get the number. Now the probability of getting the first number is 1 (any number will do); the chance of getting the second number to be different from the first is 5/6; the chance for the third to be different from the first 2 is 4/6; and so on, till the chance for the last to be different from the rest is 1/6. Hence answer is 1+6/5+6/4+6/3+6/2+6/1 = 14.7 times.
(3) A's chance is still 1/3, C's chance is 2/3. There are 4 possibilities:
a) C is pardoned, governor names B (chance=1/3)
b) B is pardoned, he names C (1/3)
c) A is pardoned, he names B (1/6)
d) A is pardoned, he names C (1/6)
Since he named B, the sample space reduces to a and c. Therefore A's chance is (1/6)/(1/3+1/6)=1/3 and C's is (1/3)/(1/3+1/6)=2/3.
(4) Nobody. Suppose he aims at B and kills him, A will finish him off. If he aims at A and kills him, B's chance of killing him is 6/7 (2/3 + 1/3×2/3×2/3 + 1/3×2/3×1/3×2/3×2/3 + ...). By shooting at nobody, the next round, C will face either A or B only. With 2/3 chance, B will hit A and C will have a chance of 3/7. With 1/3 chance, B will miss A and A will finish the stronger B off; and C's chance against A is 1/3. Hence, by shooting at nobody, C's overall chance is 25/63 (approx. 40%); B's is 8/21 (approx. 38%); and A's 2/9 (approx. 22%).
(5) A just has to ask the questions "Is your number bigger than n" starting from n=1 till he gets a No, which will pinpoint the number B has. Then on the next round A can guess the correct number. The only way for B to win is for him to guess the number correctly in his first round of questions, probability = 1/100.
(6) The answers is obviously 13 and in general the most likely total that first exceeds the number n (n≥6) is n+1. To formally prove this consider the throw before the last. The total must be either 7, 8, 9, 10, 11 or 12. If it is 12, the final result will be 13, 14, 15, 16, 17 or 18 with equal chance. If it is 11, the result will be 12-17. And so on, with 13 occurring most frequently.