Question

Professor Meyer has a deck of 52 randomly shuffled playing cards, 26 red, 26 black. He proposes the following game: he will repeatedly draw a card off the top of the deck and turn it face up so that you can see it. At any point while there are still cards left in the deck, you may choose to stop, and he will turn over the next card. If the turned up card is black you win, and otherwise you lose. Either way, the game ends. Suppose that after drawing off some top cards without stopping, the deck is left with r red cards and b black cards. (a) Show that if you choose to stop at this point, the probability of winning is b/(r + b). (b) Prove if you choose not to stop at this point, the probability of winning is still b/(r + b), regardless of your stopping strategy for the rest of the game. Hint: Induction on r + b.​

Answer 1

Step-by-step explanation:

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Answer 2

Given:

Number of black and red cards = 26 each

Number of red cards left = r

Number of black cards left = b

To find:

Probability to win

Solution:

Therefore to win the game, we need a black card to appear at the top.

Hence Probability of a black card getting picked up

$= \frac{Number of black cards}{Total number of cards} = \frac{b}{r+b}$

If we choose not to stop at this point, the probability of winning in the next round

$=\frac{b-1}{r+b-1} \frac{b}{r+b} +\frac{b}{r+b-1} \frac{r}{r+b} \\\\=\frac{b}{r+b}$

Hence like this, for all the rounds from now, the total probability to win is still $\frac{b}{r+b}$.

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