superprismatic

With each possible way of discarding, you have some expected value for the hand that result when you replace these cards from the 47 cards remaining in the deck. I wish to find the discard which gives the highest expected value for the resulting hand. Perhaps this will help: How I determine the best draw There are 2.6 million different poker hands but there are only 7,462 different values for those hands. I wrote a program that computes the value of a hand -- using the values 1 to 7,462. That isn't very hard to do. I could have just made a big (2.6 million-long) lookup table, but I wanted to be able to use it on machines with small memories. So, I optimized the thing to do on the order of a million hands a second on one core of a 6-core AMD processor. I tested it a lot, but I could have some small bugs. For this problem, I exhaused on all possible ways to discard (32) and repopulate from 47 cards (about 1.7 million) for each of the hands. I computed the average value of a hand resulting from each of the discard repopulation possibilities. The discard which gives the best average value wins.

What I mean is to make the expected value of the resulting hand as large as possible. Since there are only 7462 different values for poker hands, we can give the worst possible hand a value of 1,the best possible a value of 7462, and all the others appropriate values in between. I hope that's clear.

There is no opponent. Just you, a dealer, and a pack of cards.

Listed below are 20 poker hands. For each of the hands below, assume it was dealt to you from a full shuffled poker deck of 52 cards. You may stay with the hand you are given or discard any number of its cards. Any discarded cards will be replaced by a random draw from the remaining 47 cards. Your goal is to improve the value of your hand as much as possible. List which cards, if any, you should discard from each of the starting hands below: 1: 10♦ A♠ 10♥ 7♥ 8♣ 2: K♠ 8♣ 9♣ 10♠ J♠ 3: 9♣ 8♦ 8♥ 4♠ J♣ 4: 6♥ K♦ A♦ K♥ 4♦ 5: Q♠ 8♣ A♥ 6♦ 7♠ 6: 3♣ K♠ 5♦ 7♦ 3♥ 7: K♠ 7♥ 2♠ 5♠ J♣ 8: J♦ 10♣ 10♥ 8♣ 9♠ 9: 3♦ 6♠ 5♠ 4♠ 5♦ 10: 5♠ 7♠ 5♦ A♠ Q♥ 11: 4♠ 5♠ 6♠ 7♠ 7♣ 12: 2♠ 3♣ 4♣ 5♣ 6♣ 13: 3♠ 2♣ 3♣ 4♣ 5♣ 14: 2♠ 2♣ 3♣ 4♣ 5♣ 15: J♠ K♠ 2♥ 8♦ Q♠ 16: 2♠ 3♠ 4♠ 5♠ 6♣ 17: 3♦ 4♦ 6♦ K♦ 5♠ 18: 2♦ 8♥ 5♥ 9♣ 6♦ 19: 8♠ Q♦ A♠ K♠ 6♠ 20: 3♣ 4♦ J♥ 2♣ 10♥ [/code] Usual poker hand rankings apply. So, for example, "round the corner" straights are not allowed, although an ace may be either high or low card in a straight.

So did curr3nt in post #8. He did it in a different way, but he was first.

Looks good to me! You've got it!

Thanks for correcting my "series" instead of "sequence" error. It's a common mistake with me. Yes, you got the correct answer. However, the meat of this problem lies in the proof. Can you come up with one?

Suppose we allow digits in our base ten system to be any integer. Let's write the "unusual" digits inside brackets, {}. Then 123 would have its usual meaning, but 1{-2}3 would be interpreted as 1×100 + (-2)×10 + 3 = 83, and 7{12}9 would be interpreted as 7×100 + 12×10 + 9 = 829. Allowing these unusual digits, what would be the limit of this series: 9/1, 98/12, 987/123, 9876/1234, 98765/12345, 987654/123456, 9876543/1234567, 98765432/12345678, 987654321/123456789, 9876543210/123456789{10}, 9876543210{-1}/123456789{10}{11}, 9876543210{-1}{-2}/123456789{10}{11}{12}, ... ? Please provide a proof of your answer.

5 is not true because 0 and 20 could be married.

Thank You! I thought that an explanation of this problem was important because it embodies how Mathematics can lead us to surprising results. I'm happy that you appreciated my efforts.

It's not the area of the faces that count, it's the length of the edges. The snub cube (with vertices on the unit circle) has the closest neighbour of every point at a distance of 0.7442.

Thanks, smith, I misread the post. So, my reply was wrong.

According to the OP, each person was introduced to a different number of people. So, for every number 0,1,2,3,4,5,6,7,8,9 there must have been a person who met that many people. However 9 is impossible because that would mean that one of the couples had to be introduced to each other. That's impossible!

And here's a question for Superprismatic...

curr3nt's solution is indeed general. And he did it quickly! Nice work!