5
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Both players are dealt $26$ cards from a regular $52$ card deck. He discards $1$ and makes $5$ hands (from the remaining $25$) in any order, each with $5$ cards. His score is given by the sum of points scored by each of his individual hands as per the following table:

  • Royal flush - 10
  • Straight flush - 9
  • 4 of a kind - 8
  • Full house - 7
  • Flush - 6
  • Straight - 5
  • 3 of a kind - 4
  • 2 pairs - 3
  • Pair - 2
  • High card - 1

The actual ranks used do not matter. You are playing against an omniscient opponent (who works directly against your goal). You, however, have hacked into the computer that shuffles the deck. Hence, you can decide exactly which cards to deal to whom.

Question 1: Deal $26$ cards to your opponent to minimize his score.

Question 2: Deal $26$ cards (from the same deck) each to yourself and your opponent to maximize the difference between your and your opponent's score.

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5
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Answer 1, showing his 26 cards:

X H C S D
A * *
K * *
Q * *
J * *
0 * *
9 * *
8 * *
7 * *
6 * *
5 * *
4 * *
3 * *
2 * *

He has

2 flush (6x2)
2 straights (5x2)
1 2-pair (3)
25 points

The secret here is to

Avoid all 3 of a kinds and as many flushes as possible. This can be done with a spread of 9 of 2 suits and 4 of the other 2 suits. Stagger the cards to avoid straight flushes, and you're done.

Answer 2, showing my 26 cards:

X H C S D
A * *
K * *
Q * *
J * *
0 * * * *
9 * *
8 * *
7 * *
6 * *
5 * * * *
4
3 *
2 *

I have

2 Royal flushes (10x2)
2 Straight flushes (9x2)
1 4 of a kind (8)
46 points total

He has

1 Full house (7)
4 Flush (6x4)
31 points

For a difference of

15 points

My goal was to frustrate his choices. Give him enough to get a 4 of a kind or a flush (in one suit), then he can have 1 full house, but a second full house costs him 2 flushes. I also took away the ability to get a straight, because that seemed like easy points from leftover cards.

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  • 1
    $\begingroup$ I was about to post this answer. The opponent can make four flushes and a full house for 31 points. $\endgroup$ – f'' Oct 6 '15 at 19:37
  • $\begingroup$ Nice -- but for answer 1, doesn't the opponent have 2 flushes and 3 straights? I think you can avoid this if you stagger the cards a bit differently ... $\endgroup$ – Tyler Seacrest Oct 6 '15 at 20:28
  • $\begingroup$ @TylerSeacrest OK. Staggered to prevent 3 spades or diamonds in a straight. Should solve the extra straight issue. Thanks. $\endgroup$ – Joel Rondeau Oct 6 '15 at 20:38
  • $\begingroup$ Now there are four flushes and a two-pair in #1. $\endgroup$ – f'' Oct 6 '15 at 21:00
  • $\begingroup$ @f'' Yeah, accidentally had one set of H/C in the D/S area. Hopefully that's the last edit (barring a completely different mistake) $\endgroup$ – Joel Rondeau Oct 6 '15 at 21:04
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Initial response for question 1:

A reasonable goal is to avoid straights and full houses and flushes.
For straights, you can do this by clustering the number values you give them into chunks of 4. 2,3,4,5,7,8,9,10,Q,K,A (if ace is high). Sadly this is only 11 values. To avoid full houses we cannot have sets of 3. So this only allows us to allocate 22 cards of 26.
Looking at the flushes restriction we have 26 cards over 4 suits and all we want to do is minimize the number of times they can have 5 of a suit. So this recommends a 9,9,4,4 distribution of suits.

This is where it gets slightly interesting, what if you have it so that the cards necessary for a flush, take away all the three of a kind? We can allocate 4 numbers to be 3 of a kind. Sadly, the opponent could just use the other 5 of the suit.
Oh well another interesting problem I don't have the time at the moment to rigorify. I hope someone can use some thought I had to lead to something.

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  • $\begingroup$ Your approach looks good so far except for one thing (A,2,3,4,5 make a straight). I only managed to get as far as you are, so you might just get the answer. $\endgroup$ – ghosts_in_the_code Oct 4 '15 at 16:37
  • $\begingroup$ Fair point, I was unsure of the Aces high and what not for real poker. I suppose A, 2, 3, 4, 6, 7, 8, 9, J,Q,K works then. $\endgroup$ – Going hamateur Oct 4 '15 at 17:47
  • $\begingroup$ Yes, the new order works fine. And I think A2345 and 10JQKA are both valid straights (hence ace is both the lowest and the highest card). $\endgroup$ – ghosts_in_the_code Oct 5 '15 at 9:41
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The lowest thing you can attempt to avoid is the straight and 3 of a kinds. With both, you're unable to do it, since you can only leave 2 gaps, filling 22/26 cards. Straights are worth more than 3 of a kind, but 3 of a kind leads to full houses.

So, how about something like this:

     S D H C
A      x
K    x x
Q       
J    x x
10   x x
9    x x x x 
8        x  
7    x     x
6    x x x x 
5        x
4      x   x
3      x x x
2        x x

So, either 2x four of a kind + 1 full house, 2x 2 pairs = 29, or
2 straights, 2 full house, 2 pair = 31

On our end, we get 2 royal flushes, 1 straight flush, 1 full house , 2 pairs= 39

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  • $\begingroup$ @JoelRondeau has done better. $\endgroup$ – ghosts_in_the_code Oct 7 '15 at 13:39

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