# Drawing three cards and pick away the lowest, what is the expected value of this cards and the other cards?

Assume a 52 deck of cards. Values A = 1, 2=2 ...... 10 = 10, J = 11, Q = 12 , K = 13.

So the expected value of a randomly picked card is 364/52=7

Draw three cards and remove the lowest value of these three cards. Now without replacement pick a new card. What will then be the expected value of these three cards?

What is the expected value of the lowest card you picked up?

What is the expected value of the card you pick to replace the lowest card with?

I am having a hard time with these questions. They are not for an exam or anything, just some brainteasers I found online.

Thanks!

• Hint: to get started, let $p_n$ denote the probability that the lowest value is $≥n$. Compute $p_n$. Then observe that the probability that the lowest value is exactly $n$ is $p_n - p_{n+1}$ – lulu Jan 16 '17 at 20:54
• Thank you. So for lowest value A (1) or higher is: 100%. Lowest value 2 or higher is (48/52*47/51*46/50) = 78%. Therefore probability of A as lowest value are 100%- 78% = 22%? – Rosy Jan 16 '17 at 21:17
• Yes, that looks correct. For the second part of your question I think you need to clarify the intent. When you go to draw the replacement card, are you conditioning the calculation on the cards you have seen or not? If not...well, the expected value of one card is the same as any other. – lulu Jan 16 '17 at 21:50