What is the optimal strategy for the game of Tsyan Shizi (sometimes called "take stones" or "Chinese nim")?

anonymous

For the benefit of other readers, let me explain the rules of Tsyan Shizi.

1. The game starts with two unequal piles of stones.
2. The players will take turns.
3. Each turn will consist of removing any number of stones from either pile or the same number from both.
4. The player to remove the last stone WINS.

Consider the following table for my strategy.

 Difference Play 0 All 1 1,2 2 3,5 3 4,7 4 6,10 5 8,13 6 9,15 7 11,18 8 12,20 9 14,23 10 16,26

For a longer table of up to differences of 89, please click the spoiler box below.

 Difference Play 0 All 1 1,2 2 3,5 3 4,7 4 6,10 5 8,13 6 9,15 7 11,18 8 12,20 9 14,23 10 16,26 11 17,28 12 21,33 13 22,35 14 24,38 15 25,40 16 29,45 17 30,47 18 32,50 19 33,52 20 35,55 21 37,58 22 38,60 23 42,65 24 43,67 25 45,70 26 46,72 27 55,82 28 56,84 29 58,87 30 59,89 31 63,94 32 64,96 33 66,99 34 67,101 35 76,111 36 77,113 37 79,116 38 80,118 39 84,123 40 85,125 41 87,128 42 88,130 43 90,133 44 92,136 45 93,138 46 97,143 47 98,145 48 100,148 49 101,150 50 110,160 51 111,162 52 113,165 53 114,167 54 118,172 55 119,174 56 121,177 57 122,179 58 144,202 59 145,204 60 147,207 61 148,209 62 152,214 63 153,216 64 155,219 65 156,221 66 165,231 67 166,233 68 168,236 69 169,238 70 173,243 71 174,245 72 176,248 73 177,250 74 199,273 75 200,275 76 202,278 77 203,280 78 207,285 79 208,287 80 210,290 81 211,292 82 220,302 83 221,304 84 223,307 85 224,309 86 228,314 87 229,316 88 231,319 89 232,321

Here is my strategy, based on the table above.

1. Note the difference in stones between the two piles.
2. For difference of ten or less, consult the table above on what to play.
3. If both piles have enough stones to play according to the table above, then remove the same number of stones from both piles to achieve the state in the "play" column.
4. If both piles do NOT have enough stones to play according to the table above (for example with 6 and 11), then take stones from one pile to achieve any of the states in the table. For example, with 6 and 11, you would take 1 from the 11 pile to achieve a state of 10,6.
5. The only other possibility is you ARE in one of the states above. Then you are screwed if playing with a skilled player, as he will be able to force you onto another losing positions in the table no matter what you do. If not playing with a skilled player, I recommend taking just one stone from either pile, to give your opponent more opportunity to screw up.

What is the probability of getting a jackpot (win of \$1,200 or more) in 9/6 Jacks or Better in multi-play video poker?

anonymous

Of course, it depends on the denomination and number of plays. The following table shows those probabilities.

### Jackpot Probability in Multi-Play Video Poker

Denomination 3 Play 5 Play 10 Play 25 Play 50 Play 100 Play
\$0.01 0.000000 0.000000 0.000000 0.000000 0.000001 0.000001
\$0.05 0.000000 0.000000 0.000001 0.000001 0.000002 0.000022
\$0.10 0.000001 0.000001 0.000002 0.000006 0.000047 0.000378
\$0.25 0.000002 0.000003 0.000008 0.000053 0.000369 0.000556
\$0.50 0.000070 0.000115 0.000238 0.000782 0.001247 0.008527
\$1.00 0.000070 0.000128 0.000473 0.000786 0.009518 0.072671
\$2.00 0.000083 0.000363 0.000488 0.010002 0.070029 0.239753
\$5.00 0.000720 0.001290 0.012978 0.100374 0.318838 0.768839
\$25.00 0.041494 0.124818 0.348811 0.835708 0.995943 0.999983

This table was taken from my video poker appendix 2, where I show how often the player gets each total win in a large simulation.

There are ten lilllipads in a row on a lake.  There is a fly on the tenth lillipad, which a frog on the shore wants to eat.  The frog may jump in one direction only and may advance one or two lillipads per jump.  How many different sets of lillipads are there that the flog lands on? Note that the frog must land on the tenth lillipad to eat the fly.

anonymous

Let's reduce this problem to one lillipad and then add one more at a time, to try to find a pattern.

If there were just one lillipad, then the answer is obviously 1.

If there were two lillipads, then the frog could either jump on the first pad on the way or over it, for a total of 2 sets.

If there were three lillipads, then the first jump could advance the frog either one or two pads. That will put him either 1 or 2 pads way. We have seen there is one way to advance one pad and two ways to advance two. Adding the choice of the first move, there are 1+2 = 3 sets.

If there were four lillipads, then the first jump could advance the frog either one or two pads. That will put him either 2 or 3 pads way. We have seen there is are 2 ways to advance 2 and 3 ways to advance 3. Adding the choice of the first move, there are 2+3 = 5 sets.

If there were five lillipads, then the first jump could advance the frog either one or two pads. That will put him either 3 or 4 pads way. We have seen there is are 3 ways to advance 3 and 5 ways to advance 4. Adding the choice of the first move, there are 3+5 = 8 sets.

This is following the Fibonacci sequence. The following list shows there are 89 ways to land on the 10th pad.

• 1 pad = 1 ways.
• 2 pads = 2 ways.
• 3 pads = 3 ways.
• 4 pads = 5 ways.
• 5 pads = 8 ways.
• 6 pads = 13 ways.
• 7 pads = 21 ways.
• 8 pads = 34 ways.
• 9 pads = 55 ways.
• 10 pads = 89 ways.