Problem #1: In Cribbage, a collection of cards scores as follows:
Usually, Cribbage is played with 5 cards. It is well known that the highest scoring 5 card cribbage hand is J5555, worth 29 points. What is the highest scoring n card Cribbage hand, for other values of n?
Problem #2: In a Poker variant called Guts, each player is dealt cards, and then each player declares simultaneously whether he is in or out. Of all the players who are in, the best poker hand wins the pot and all the others match the pot.
If there are p players, and n cards dealt to each player, what is the worst hand that a player should stay in with?
Problem #3: My favorite magic trick involving cards is the following. Alice gives Bob any 5 cards without Chuck seeing any of them. Bob reveals 4 of those cards face up, one at a time. Chuck, who is working with Bob, names the fifth card without seeing it.
How do Bob and Chuck do this trick? Here's a more general question: If Alice gives Bob n cards, and Bob can always play k of them so that Chuck can predict the other cards, what is the maximum number of cards in the deck? If the cards can be labeled in any way, what is the "easiest" strategy for achieving this trick?
Problem #4: Pinochle is played with a 48 card Pinochle deck: the aces, kings, queens, jacks, tens and nines from a double deck. A hand scores as follows:
Unlike Cribbage, a card in Pinochle can only be used once within any given category. Usually, Pinochle is played with 12 cards. I think the maximal score with 12 cards is 190, for a double run in spades and two jacks of diamonds. What is the highest scoring n card Pinochle hand, for various values of n?
Gavin Theobald found the highest scoring hands for n≤20, including one improvement.
Jeremy Galvagni found the highest scoring hands for n≤6.
Joe DeVincentis found the highest scoring hands for n≤7, but proved his assertions for n≤4.
Philippe Fondanaiche found the highest scoring hands for n=6 and n=7.
Cards | Points | Hand |
---|---|---|
1 | 1 | J |
2 | 3 | JJ or 5J |
3 | 8 | 555 |
4 | 20 | 5555 |
5 | 29 | 5555J |
6 | 46 | 445566 |
7 | 72 | 4455566 |
8 | 112 | 33334455 |
9 | 188 | AA2233444 |
10 | 332 | AA22334455 |
11 | 576 | AAA22233344 or AAA22233444 or AAA22333444 or AA222333444 (GT) |
12 | 1,024 | AAA222333444 |
13 | 1,836 | (A2)33344 |
14 | 3,206 | (A23)44 |
15 | 5,086 | (A23)444 |
16 | 7,440 | (A234) |
17 | 9,601 | (A234)5 |
18 | 12,980 | (A234)55 |
19 | 16,555 | (A234)555 |
20 | 20,328 | (A2345) |
... | ... | ... |
52 | 872,449,969 | (A23456789TJQK) |
Problem #2:
Joe DeVincentis sent a well thought out analysis. You should stay in when your expected winning is larger staying in than going out. Ignoring that the same cards can not be dealt twice, he found out that you should stay in with probability .5 with 2 players, .293 with 3 players, and .207 with 4 players.
n \ p | 2 | 3 | 4 |
---|---|---|---|
1 | 8 | J | Q |
2 | J7 | K5 (JD) | KQ (JD) |
3 | K63 | A85 (TG) | AQT (JD) |
4 | A875 | ||
5 | AKQJ7 | ||
6 | 66A74 |
Trevor Green did not ignore that cards can not be dealt twice, and found that the critical hands changed slightly. His complete analysis for n=5 and p=2 convinced me that this was a VERY HARD problem.
n \ p | 2 | 3 | 4 |
---|---|---|---|
1 | 8 | J | Q |
2 | J8 (TG) | K4 (TG) | A2 (TG) |
3 | K42 (TG) | A82 (TG) | |
4 | A532 (TG) | ||
5 | AKQ98* (TG) |
Then in 2013, Trevor Green gave an even more detailed analysis of the 2-player game. His final results were:
n | hand |
---|---|
1 | T (TG) |
2 | K2 (TG) |
3 | A32 (TG) |
4 | AKJ4 (TG) |
5 | 77982 or 77A32 (TG) |
Problem #3:
Joe DeVincentis solved the problem with n=5 and k=4. He also found the upper bounds below.
The method for 124 cards for n=5 and k=4 is basically this: The cards are numbered 0 to 123, given 5 cards c_{0} < c_{1} < c_{2} < c_{3} < c_{4}, the assistant hides c_{i} where i = c_{0} + c_{1} + c_{2} + c_{3} + c_{4} (mod 5). Let the sum of the displayed cards be s (mod 5). If you renumber the 120 non-displayed cards consecutively 0 to 119, then the hidden card must be -s (mod 5), limiting it to precisely 24 of the non-displayed cards. The order of the four cards then determines which of the 24 it is. Philippe Fondanaiche also mentioned that it is possible to predict the last card when the deck has size d = n! + n – 1 cards.
Trevor Green found explicit strategies for k=2 and 3≤n≤5, and the lower bound d≥n+k.
Here are (rotationally symmetric) strategies for k=2.
n=4: 1234: play 12 / 1235: play 53 / 1236: play 63 / 1245: play 54 / 1246: play 46
n=5: 45678: play 54 / 35678: play 53 / 34678: play 63 / 34578: play 73 /
25678: play 56 / 24678: play 46 / 24578: play 47
n \ k | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|
2 | 3 | |||||
3 | 4 | 8 | ||||
4 | 5 | 7 | 27 | |||
5 | 6 | 8 | 8-14 | 124 | ||
6 | 7 | 8 | 9-13 | 10-31 | 725 | |
7 | 8 | 9 | 10-13 | 11-22 | 12-76 | 5046 |
Problem #4:
Cards | Points | Hand |
---|---|---|
1 | 1 | 9 |
2 | 4 | Q J or KQ |
3 | 8 | KQ J (JD) |
4 | 40 | QQ JJ |
5 | 44 | KQQ JJ |
6 | 48 | KKQQ JJ |
7 | 49 | KKQQ9 JJ |
8 | 100 | AA AA AA AA |
9 | 101 | AA9 AA AA AA |
10 | 150 | AAKKQQJJTT |
11 | 154 | AAKKQQJJTT J or AAKKQQJJTT Q |
12 | 190 | AAKKQQJJTT JJ or AAKKQQJJTT QQ |
13 | 192 | AAKKQQJJTT KQQ |
14 | 196 | AAKKQQJJTT QQ Q Q (JD) |
15 | 200 | AAKKQQJJTT AJJ A A or AAKKQQJJTT AQQ A A |
16 | 250 | AAKKQQJJTT AA AA AA or AAKKQQJJTT QQ QQ QQ |
17 | 254 | AAKKQQJJTT AAJ AA AA or AAKKQQJJTT AAQ AA AA |
18 | 290 | AAKKQQJJTT AAJJ AA AA or AAKKQQJJTT AAQQ AA AA |
19 | 292 | AAKKQQJJTT AAKQQ AA AA |
20 | 296 | AAKKQQJJTT AAQQ AAQ AAQ |
21 | 298 | AAKKQQJJTT AAQQ AAKQ AAQ (JD) |
22 | 350 | AAKKQQJJTT AAQQ AAQQ AAQQ (JD) |
23 | 352 | AAKKQQJJTT AAKQQ AAQQ AAQQ (GT) |
24 | 374 | AAKKQQJJTT AAKKQQ AAKK AAKK (GT) |
25 | 376 | AAKKQQJJTT AAKKQQ AAKKQ AAKK (GT) |
26 | 384 | AAKKQQJJTT AAKKQQ AAKKQ AAKKQ (GT) |
27 | 386 | AAKKQQJJTT AAKKQQ AAKKQQ AAKKQ (GT) |
28 | 442 | AAKKQQJJTT AAKKQQ AAKKQQ AAKKQQ (GT) |
29 | 443 | AAKKQQJJTT9 AAKKQQ AAKKQQ AAKKQQ (GT) |
30 | 444 | AAKKQQJJTT99 AAKKQQ AAKKQQ AAKKQQ (GT) |
31 | 446 | AAKKQQJJTT AAKKQQJ AAKKQQJ AAKKQQJ (GT) |
32 | 447 | AAKKQQJJTT9 AAKKQQJ AAKKQQJ AAKKQQJ (GT) |
33 | 448 | AAKKQQJJTT99 AAKKQQJ AAKKQQJ AAKKQQJ (GT) |
34 | 482 | AAKKQQJJTT AAKKQQJJ AAKKQQJJ AAKKQQJJ (GT) |
35 | 483 | AAKKQQJJTT9 AAKKQQJJ AAKKQQJJ AAKKQQJJ (GT) |
36-48 | 484 | AAKKQQJJTT99 AAKKQQJJ AAKKQQJJ AAKKQQJJ (GT) |
If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 7/27/13.