Q: 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.

Let&#39;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 <a href="https://en.wikipedia.org/wiki/Fibonacci_sequence" target=_blank>Fibonacci sequence</a>. The following list shows there are 89 ways to land on the 10th pad. <ul> <li>1 pad = 1 ways. </li> <li>2 pads = 2 ways. </li> <li>3 pads = 3 ways. </li> <li>4 pads = 5 ways. </li> <li>5 pads = 8 ways. </li> <li>6 pads = 13 ways. </li> <li>7 pads = 21 ways. </li> <li>8 pads = 34 ways. </li> <li>9 pads = 55 ways. </li> <li>10 pads = 89 ways. </li> </ul>

Question 1

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

Accepted Answer

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

The game starts with two unequal piles of stones.
The players will take turns.
Each turn will consist of removing any number of stones from either pile or the same number from both.
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.

Note the difference in stones between the two piles.
For difference of ten or less, consult the table above on what to play.
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.
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.
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.

Question 2

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?

Accepted Answer

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.

Question 3

There are ten lilllipads in a row on a lake.&nbsp; There is a fly on the tenth lillipad, which a frog on the shore wants to eat.&nbsp; The frog may jump in one direction only and may advance one or two lillipads per jump.&nbsp; 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.&nbsp;

Accepted Answer

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.

Wizard Recommends

Ask The Wizard #404

Jackpot Probability in Multi-Hand Video Poker gives insight into how frequently a player can expect to achieve each total win in a comprehensive simulation.

Wizard Recommends

Ask The Wizard #404

Jackpot Probability in Multi-Hand Video Poker gives insight into how frequently a player can expect to achieve each total win in a comprehensive simulation.

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