cryptography math homework questions

Please see attached for questions

1. Here are two (simplified) versions of Roulette. Compute the probabilities of losing in each.
   1. (a) Las Vegas roulette. There are 38 (uniformly random) outcomes: two green numbers 0 and 00, and 36 red and black numbers 1 to36 (half red, half black). You bet on a red or black color (say red). You then win if you get a red number, and lose if you get a non-red number. What’s the probability that you lose?
   2. (b) Monte-Carlo roulette. There are 37 (again, uniformly random) outcomes: 0 is green, and 36 red and black numbers 1 to 36(again half red, half black). You bet on a red or black color (say red). You then win if you get a red number, and lose if you get a black number. If you get the green 0, you get a special reroll, where if you get a red then you get your money back (so you don’t win but don’t lose either), but you lose if you get a green or black on the reroll. What is your probability of losing?

Then, comment on the values themselves.

1. Suppose you have 15 people coming together to form a committee of5 people. 6 out of 15 of the people are spies. What’s the probability that at least half of the committee are spies?
2. True or false? (Warning: as in all my problems, you need to explain your answers to get any credit.)(a) (Z/6Z)∗ has a primitive root. (b) (Z/13Z)∗ has a generator. (c) (Z/15Z)∗ has a generator.
3. Similar to Di?e-Hellman, here’s the ?Zhang key-exchange?:? We pick a big prime p and a generator g.? Alice has a secret a ∈ Z/(p − 1)Z. 1

• ? Bobhasasecretb∈Z/(p−1)Z.
• ? Alice sends ga to Bob. Bob sends gb to Alice.

1. (a) Show that a key is indeed exchanged; that is, Alice and Bob can both compute g−a−b (mod p).
2. (b) Showthatthiskey-exchangeisverybadcomparedtoDi?e-Hellman.

1. How fast (big O notation is enough) can you check if two lists of n integers each has at least one integer in common? (hint: sorting a list of n integers ? meaning putting them in order, takes O(n log(n)) time with the best algorithms we have).