Five Probability Questions 1. A deck of 52 cards are well-shuffled and split int

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Five Probability Questions

1. A deck of 52 cards are well-shuffled and split into 13 piles of 4 cards each. Label these piles as A,2,...,10, J, Q, K. Take the first card from pile ‘K’, and next card from pile ‘X’ where X is the value of the previous card taken. What is the probability that all cards are removed?

2. A hat contains a number of cards, with 30% white on both sides, 50% red/white, and 20% red on both sides. The cards are mixed up, and a single card is drawn at random and placed on the table. If the top is red, what is the probability that the bottom is white?

3.   Mark is about to visit London. He wants to know if he should bring an umbrella. According to his experience, it rains 1 day out of 4 in London. He calls 3 random friends who live there and ask each independently if it is raining. Each of his friends has a 2/3 chance of telling him the truth and a 1/3 chance of messing with him by lying. All 3 friends tell him that Yes, it is raining. What is the probability that it is actually raining in London?

4..At a party, 100 guests check their hats at the entrance. Mark is the first person to leave. He is drunk and takes away a hat at random. Each succeeding person takes his own hat if it is not taken. If it is taken, he picks a hat at random from the remaining hats. What is the probability that the last person gets his own hat?

5..There is a deck of 52 cards, and you pick one by one without replacement. You observe the past sequence and have one chance to stop and bet that the next card is red. Can you find a strategy that guarantees a probability of winning greater than 50%?

Explanation / Answer

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1)

Remark 1. The last card we take before the game ends (either winning or loosing) is a “K”. Proof: Let us assume that at iteration j we draw card X but pile X is empty (thus the game terminates). Let X = K (i.e. we lose). Because pile X is empty and X = K, we must have already drawn (prior to draw j) 4 X cards. But then we can not draw an X card at the jth iteration, a contradiction. Note: There is no contradiction if the last card is a “K” and all other cards have been already removed (in that case the game terminates with win)

Remark 2. We win if the fourth “K” card is drawn at the 52 iteration. Proof: whenever we draw for the 1st, 2nd or 3rd time a“K” card, the game does not terminate because the K pile is not empty so we can continue (see remark 1). when the fourth K is drawn at the 52nd iteration then all cards are removed and the game’s result is “win”.

Because of remark 2, it is: Pr{win} = Pr{4th “K” at the 52nd iteration} = = #game evolutions: 52nd card = 4th “K” /#all game evolutions

- we actually have 13x4=52 distinct positions (13 piles, 4 positions each) where 52 distinct cards are placed. This gives a total of 52! different placements. -

each game evolution actually corresponds to an ordered permutation of the 52 cards.

- The winning permutations are those where the 52nd card is a “K” (4 ways) and the 51 preceding cards are arbitrarily chosen (51!).

Thus: Pr{win} = 4·51! /52! = 4 /52 = 1 /13

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