Ask the Wizard: Non-Casino Games - FAQ

First let me answer the unasked question on the probability that a specific number will show up n times on a random bill. There are 8 digits on a bill so the probability of n of a specific number is combin(8,n)*9^8-n/10⁸. The following are these probabilities for n=0 to 8.

 

0

0.43046721

1

0.38263752

2

0.14880348

3

0.03306744

4

0.00459270

5

0.00040824

6

0.00002268

7

0.00000072

8

0.00000001

Next is the probability that any number will n times on a dollar bill, where no other number appears more than n times.

 

0

0.0181440

1

0.6191640

2

0.3124800

3

0.0458955

4

0.0045927

5

0.0040824

6

0.0002268

7

0.0000072

8

0.0000001

I interpret your last question to be what is the probability that a specific number will show up at least once on at least one of 3 bills. The answer is 1-(9/10)²⁴=.920234.

Your expected win is the same regardless of how many times you divide your total deposits. A good strategy would be to deposit and withdraw and same money over and over as many times as possible. However your odds may be so bad that it isn’t worth the bother.

The probability of three matching is 1/216. The probability of two matching is 3*5/216. The probability of one matching is 25*5/216. The probability of 0 matching is 5*5*5/216. So the expected return is 3*(1/216)+2*(15/216)+1*(75/216)-1*(125/216)=-17/216=-7.87%. There is no optimal number of bets, you will give up an expected 7.87% of total money bet no matter what you do.

I’m afraid I don’t know. In the early days of Vegas the used to offer solitaire, I think Klondike, as a form of gambling but I don’t know the exact betting rules. The odds on this would be very hard to figure out.

With every new roll the probability the next four rolls will be all double sixes is (1/36)⁴ = 1 in 1679616.

Start by removing 2 pearls from the row with 3, leaving 1+4+5. Regardless of what your opponent does on your next turn leave him with any of the following: 1+1+1, 1+2+3, or 4+4. From any of these force the opponent to a situation of two piles of 2 or more each, or an odd number of piles of 1 each.

I like the orange set the best. It offers the best return on investment. For example a hotel costs $500 on the orange set and the average rent is $966.67, for a rent to expense ratio of 1.93. The only set with a higher ratio is the light blue set, at 2.27. However the maximum rent on the light blues is only $600. The rents with three houses on the oranges are the same as the hotels on the light blues, but cost 20% less, with room to build more. Also, the oranges are ripe for landing on just coming out of jail. So take my advice and when trading try to get the orange set.

The best piece of advice anywhere on this site may be this: The first round, ALWAYS PICK PAPER. That is because amateur players tend to pick rock the first time. Just hold out your hand in each position, one at a time, and you’ll see that rock is the most comfortable and natural choice. If you play repeated rounds you should pick whatever would beat your opponent the last round with probability less than one-third. This is because I believe amateurs repeat less than one-third of the time. If playing a pro who you fear can get into your head then randomize by looking at the second hand of your watch, divide the number of seconds by three and take the remainder, then map the remaider as follows 0=rock, 1=scissors, 2=paper (or any other mapping as long as determined in advance). So the next time you go to a restaurant Dutch style I suggest playing a single round for the check and then pick paper. You can thank me later.

For those who aren’t familiar with the game, Risk is the greatest board game ever made. Those who haven’t played haven’t lived yet. To answer your question in the common 3 on 2 battle the following are the possible outcomes:

• Defender loses both: 37.17%
• Each loses one: 33.58%
• Attacker loses both: 29.26%

The following table shows the probability of success on the last roll according to the number of additional dice you need to make a Yahtzee.

Last Roll Yahtzee Probabilities
Needed	Probability of Success
0	1
1	0.166667
2	0.027778
3	0.00463
4	0.000772

The next table shows the probabilities of improvement. The left column shows how many dice you need before any given roll and the top column shows how many you need after the roll. The body shows the probability of the given degree of improvement.

Probabilities of Improvement
Need Before Roll	0	1	2	3	4	Total
0	1	0	0	0	0	1
1	0.166667	0.833333	0	0	0	1
2	0.027778	0.277778	0.694444	0	0	1
3	0.00463	0.069444	0.37037	0.555556	0	1
4	0.000772	0.01929	0.192901	0.694444	0.092593	1

The next table shows the probability on the initial roll of needing 0 to 4 more dice to make a Yahtzee.

First Roll Yahtzee Probabilities
Needed	Probability
0	0.000772
1	0.019290
2	0.192901
3	0.694444
4	0.092593

The next table shows the probability of improvement and then eventual success according to the number needed after the first roll. For example, if the player needs 3 more dice to make a Yahtzee the probability of improving to needing 2 more after the second roll and making the Yahtzee on the third roll is 0.010288066.

Probabilities of Yahtzee after first roll according to number needed before and after second roll
Need Before Roll	0	1	2	3	4	Total
0	1	0	0	0	0	1
1	0.166667	0.138889	0	0	0	0.305556
2	0.027778	0.046296	0.01929	0	0	0.093364
3	0.00463	0.011574	0.010288	0.002572	0	0.029064
4	0.000772	0.003215	0.005358	0.003215	0.000071	0.012631

To get the final answer take the dot product of the number needed after the first roll two tables up and the probability of eventual success in the final column one table up. This is 0.092593*0.012631+ 0.694444*0.029064 + 0.192901*0.093364 + 0.019290*0.305556 + 0.000772*1 = 4.6028643%. To confirm this I did a 100,000,000 game simulation and the simulated probability was 4.60562%.

First, we can rule out ever playing paper. Regardless of what the other person throws you will make out equal or better by throwing dynamite over paper. Once paper is eliminated, dynamite essentially becomes the new paper, beating rock and losing to scissors. So the perfect strategy is to pick randomly, and with equal probability, between rock, scissors, and dynamite.

I asked this question to Randy Hill of Fun Industries Inc.. He said you should hold your arms straight out, palms down, and let the money blow up against the bottom of your hands and arms. When enough has accumulated you stuff it through the slot.

Lets call x the expected number of flips from the starting point.
Lets call y the expected number of remaining flips if one side is one flip in the majority.
Lets call z the expected number of remaining flips if one side is two flips in the majority.

E(x) = 1 + E(y)
E(y) = 1 + 0.5*E(x) + 0.5*E(z)
E(z) = 1 + 0.5*E(y)

It is then easy matrix algebra to see that E(x) = 9, E(y) = 8, and E(z) = 5. So on average it will take 9 flips for the disparity between heads and tails to be 3. So at 8 rupees it is a good bet for the person collecting the one rupee per flip, because he will receive on average 9 rupees, but pay back only 8. The house edge for the gambler is 11.11%. At 9 rupees it is a fair bet, at 7 the house advantage is 22.22%.

I explain in the 11/28/02 column how to play once there are only three rows left. Here is my strategy for four rows. When it is your turn look up the configuration along the left column and play what is on the right column. For example the starting position of 3456 is listed last and shows you should remove 4 pearls from the row with 5, leaving 1346. If the left column says "Lose" there is no way to win if the opponent plays optimal strategy, which the game at Transcience always seems to do.

A pattern to this table seems to be that you should force the opponent to a situation where the sum of the pearls in the smallest and greatest rows equals the sum of the two in the middle. This would include leaving zero in the row with the least number of pearls.

Pearls Before Swine II Strategy
You Have	Leave
1111	111
1112	111
1113	111
1114	111
1115	111
1116	111
1122	Lose
1123	1122
1124	1122
1125	1122
1126	1122
1133	Lose
1134	1133
1135	1133
1136	1133
1144	Lose
1145	1144
1146	1144
1155	Lose
1156	1155
1222	1122
1223	1122
1224	1122
1225	1122
1226	1122
1233	123
1234	123
1235	123
1236	123
1244	1144
1245	145
1246	246
1255	1155
1256	Lose
1333	1133
1334	1133
1335	1133
1336	1133
1344	1144
1345	145
1346	Lose
1355	1155
1356	1256
1444	1144
1445	1144
1446	1144
1455	1155
1456	1346
2222	Lose
2223	2222
2224	2222
2225	2222
2226	2222
2233	Lose
2234	2233
2235	2233
2236	2233
2244	Lose
2245	2244
2246	2244
2255	Lose
2256	2255
2333	2233
2334	2233
2335	2233
2336	2233
2344	2244
2345	Lose
2346	1346
2355	2255
2356	2345
2444	2244
2445	2244
2446	2244
2455	2255
2456	2345
3333	Lose
3334	3333
3335	3333
3335	3333
3336	3333
3344	Lose
3345	3344
3346	3344
3355	Lose
3356	3355
3444	3344
3445	3344
3446	3344
3455	3355
3456	1346

Brad S. wrote in to add a general strategy for any number of pearls and rows. First you break down each row into its binary components. For example the starting position of the Transcience game would be as follows.

• 3 = 2+1
• 4 = 4
• 5 = 4+1
• 6 = 4+2

Then you endeavor to leave an even number of each power of 2. For example in the above there are two 1’s, two 2’s, and three 4’s. So there is an extra 4. You then remove 4 from any of the rows with a 4 term. Keep doing this until you can get your opponent down to 2,2 or an odd number of 1’s.

Try this strategy on the Pearl 3 game, you’ll win every time. If you start with a losing scenario as I did on game 10 (4+7+8+11) you can click on "go" to make him go first.

You are on the right track with the binary numbers but that is not quite the winning strategy. First, if you can leave your opponent with an odd number of rows of one each then do so. Otherwise break down each row into its binary components. For example, 99 would be 64+32+2+1. Then add up the number of each component over all the rows. Then look for a play that will leave your opponent with an even number of all binary components over all the rows.

Let’s look at an example. Suppose it is your turn with the following scenario.

The following table breaks down each row into its binary components.

Player’s Turn 1
Row	1	2	4	8	16
6	0	1	1	0	0
9	1	0	0	1	0
4	0	0	1	0	0
5	1	0	1	0	0
25	1	0	0	1	1
Total	3	1	3	2	1

You can see that there is an odd number of ones, twos, fours, and sixteens. Clearly we need to get the row of 25 under 16 to eliminate the 16 unit. To keep the total of the binary components even we need to remove the 1, add a 2, add a 4, keep the 8, and remove the 16. That means the best play is 2+4+8=14 in the last row. Leaving 14 in the bottom row we have the following.

Computer’s Turn 1
Row	1	2	4	8	16
6	0	1	1	0	0
9	1	0	0	1	0
4	0	0	1	0	0
5	1	0	1	0	0
14	0	1	1	1	0
Total	2	2	4	2	0

The computer takes its turn, leaving us with this.

Here is the binary breakdown of that.

Player’s Turn 2
Row	1	2	4	8	16
6	0	1	1	0	0
9	1	0	0	1	0
2	0	1	0	0	0
5	1	0	1	0	0
14	0	1	1	1	0
Total	2	3	3	2	0

Here we need to remove a 2 and a 4, to get those totals even. There is only one row, the 14, which has both components. So remove 6 from that, leaving 8.

Computer’s Turn 2
Row	1	2	4	8	16
6	0	1	1	0	0
9	1	0	0	1	0
2	0	1	0	0	0
5	1	0	1	0	0
8	0	0	0	1	0
Total	2	2	2	2	0

The computer takes its turn, leaving us with this.

Now we need to change the 1, 4, and 8 columns.

Player’s Turn 3
Row	1	2	4	8	16
6	0	1	1	0	0
4	0	0	1	0	0
2	0	1	0	0	0
5	1	0	1	0	0
8	0	0	0	1	0
Total	1	2	3	1	0

That can be done by changing the row of 8 to 5 as follows.

Computer’s Turn 3
Row	1	2	4	8	16
6	0	1	1	0	0
4	0	0	1	0	0
2	0	1	0	0	0
5	1	0	1	0	0
5	1	0	1	0	0
Total	2	2	4	0	0

The computer takes its turn, leaving us with this.

Now we need to change the 2 and 4 totals.

Player’s Turn 4
Row	1	2	4	8	16
6	0	1	1	0	0
4	0	0	1	0	0
2	0	1	0	0	0
5	1	0	1	0	0
3	1	1	0	0	0
Total	2	3	3	0	0

This can be done by changing the 6 to a 0.

Computer’s Turn 4
Row	1	2	4	8	16
0	0	0	0	0	0
4	0	0	1	0	0
2	0	1	0	0	0
5	1	0	1	0	0
3	1	1	0	0	0
Total	2	2	2	0	0

The computer takes its turn, leaving us with this.

Now we need to change the 2s and 4s.

Player’s Turn 5
Row	1	2	4	8	16
0	0	0	0	0	0
2	0	1	0	0	0
2	0	1	0	0	0
5	1	0	1	0	0
3	1	1	0	0	0
Total	2	3	1	0	0

This can be accomplished by changing the row of 5 to 3. If you can ever get your opponent to an x,x,y,y situation you can’t help but win, if you can maintain the same situation until the end.

Computer’s Turn 5
Row	1	2	4	8	16
0	0	0	0	0	0
2	0	1	0	0	0
2	0	1	0	0	0
3	1	1	0	0	0
3	1	1	0	0	0
Total	2	4	0	0	0

The next few moves I keep the computer on x,x,y,y patterns. Here the computer leaves me with 2,2,3,2; so I leave it with 2,2,2,2.

The computer then gives me 2,2,1,2. I leave it with 2,2,1,1.

The computer then leaves me with 2,2,1. I leave it with 2,2. If you can ever get your opponent to two equal rows you can’t help but win, just keep the rows equal.

The computer then leaves me with a single pile of 2, and I remove 1.

Here is the end of the game.

So there are 30 black, 30 red, and 4 green numbers. That would make the probability of black 30/64, red 30/64 and green 4/64. If the probability of an event is p then the fair odds are (1-p)/p to 1. So fair odds for any red would be (34/64)/(30/64) = 34 to 30 = 17 to 15. Same for black. The fair odds on green are (60/64)/(4/64) = 60 to 4 = 15 to 1. For a specific number the fair odds are (63/64)/(1/64) to 63 to 1.

I suggest paying 1 to 1 on red and black, 14 to 1 on green, and 60 to 1 on any individual number. One formula for the house edge is (t-a)/(t+1), where t is the true odds, and a is the actual odds. In this case the house edge on the red or black bet is (63-60)/(63+1) = 3/64 = 4.69%. On the green bet the house edge is (15-14)/(15+1) = 1/16 = 6.25%. On individual numbers the house edge is (63-60)/(63+1) = 3/64 = 4.69%.

VLT’s are glorified pull-tab games. There is a predetermined pool of outcomes. When you play, the game picks an outcome from the pool at random, and displays the win to the player in the form of a slot machine or video poker game. Since the outcome is predestined, any element of skill is imaginary. For example, if you are dealt a royal flush and throw it away, you’ll get another one on the draw. Usually I say that in gambling the past doesn’t matter, but in this case there is an effect of removal. If you play one time and lose, then it will marginally improve the odds of the remaining game outcomes, until the supply of virtual pull-tabs is exhausted, and I presume the virtual drum is refilled. I believe that your hot and cold swings are just normal luck, and any predestination is imagined.

A reader later added the following to this topic.

I have a comment on your February 14 "Ask the Wizard" column (No. 183). It’s doesn’t really have anything to do with the question you answered. It’s just something you might find interesting.

Prior to the passing of Proposition 1A, that allowed to have full class 3 gaming, we had a small installation of VLT style for a couple of years. In our system, which was run by SDG (now part of Bally), the prize pool started with 4 million draws. When the pool was reduced and 2 million remained, the next pool of 4 million was added for a total pool of 6 million draws. When the pool was reduced to 2 million again, the process repeated.

Assuming the player always holds the most represented number, the average is 11.09. Here is a table showing the distribution of the number of rolls over a random simulation of 82.6 million trials.

Yahtzee Experiment
Rolls	Occurences	Probability
1	63908	0.00077371
2	977954	0.0118396
3	2758635	0.0333975
4	4504806	0.0545376
5	5776444	0.0699327
6	6491538	0.0785901
7	6727992	0.0814527
8	6601612	0.0799227
9	6246388	0.0756221
10	5741778	0.0695131
11	5174553	0.0626459
12	4591986	0.0555931
13	4022755	0.0487016
14	3492745	0.042285
15	3008766	0.0364257
16	2577969	0.0312103
17	2193272	0.0265529
18	1864107	0.0225679
19	1575763	0.019077
20	1329971	0.0161013
21	1118788	0.0135446
22	940519	0.0113864
23	791107	0.00957757
24	661672	0.00801056
25	554937	0.00671837
26	463901	0.00561624
27	387339	0.00468933
28	324079	0.00392347
29	271321	0.00328476
30	225978	0.00273581
31	189012	0.00228828
32	157709	0.00190931
33	131845	0.00159619
34	109592	0.00132678
35	91327	0.00110565
36	76216	0.00092271
37	63433	0.00076795
38	52786	0.00063906
39	44122	0.00053417
40	36785	0.00044534
41	30834	0.00037329
42	25494	0.00030864
43	21170	0.0002563
44	17767	0.0002151
45	14657	0.00017745
46	12410	0.00015024
47	10299	0.00012469
48	8666	0.00010492
49	7355	0.00008904
50	5901	0.00007144
51	5017	0.00006074
52	4227	0.00005117
53	3452	0.00004179
54	2888	0.00003496
55	2470	0.0000299
56	2012	0.00002436
57	1626	0.00001969
58	1391	0.00001684
59	1135	0.00001374
60	924	0.00001119
61	840	0.00001017
62	694	0.0000084
63	534	0.00000646
64	498	0.00000603
65	372	0.0000045
66	316	0.00000383
67	286	0.00000346
68	224	0.00000271
69	197	0.00000238
70	160	0.00000194
71	125	0.00000151
72	86	0.00000104
73	79	0.00000096
74	94	0.00000114
75	70	0.00000085
76	64	0.00000077
77	38	0.00000046
78	42	0.00000051
79	27	0.00000033
80	33	0.0000004
81	16	0.00000019
82	18	0.00000022
83	19	0.00000023
84	14	0.00000017
85	6	0.00000007
86	4	0.00000005
87	9	0.00000011
88	4	0.00000005
89	5	0.00000006
90	5	0.00000006
91	1	0.00000001
92	6	0.00000007
93	1	0.00000001
94	3	0.00000004
95	1	0.00000001
96	1	0.00000001
97	2	0.00000002
102	1	0.00000001
Total	82600000	1

Backgammon is one of my favorite gambling games. I don’t write about it because player vs. player games are extremely hard to analyze. I also can’t seem to find any new ground to break in the game. So, I’ll leave the advice to others. Here are my suggested resources:

Backgammon by Paul Magriel: If there were a Bible to backgammon, this would be it. I’m a proud owner of an old hard-cover edition. This book would be a great place to start. Although it was written in 1976, the advice still holds up well.

501 Essential Backgammon Problems by Bill Robertie: I’ve been trying to get through this book for years, and I’m still only half way there. It is disheartening to get half the problems wrong, enough to make me think I’m as bad at backgammon as I am at golf. However, with every problem missed, there is a valuable lesson to be learned. For the intermediate to advanced player, this book is a valuable, and humbling, learning tool.

Snowie backgammon software : I play about 1000 games a year against this game. Snowie not only plays a near-perfect game, but tells you exactly how costly your errors are, when you make them. There are lots of other features that I have never explored. If there is one thing I’ve learned from Snowie, it’s that the biggest problem with my game is bone-headed mistakes of not seeing perfectly obvious plays sometimes. Much like chess, one bad move can wipe out 100 good ones.

Motif website : Before I purchased Snowie, I played countless games against Motif. The strategy employed by Motif is very solid, in my opinion. There is nothing like playing against a stronger opponent to improve your own game.

The following table shows the probability of each player winning, according to the first player’s first spin, where player 1 goes first, followed by player 2, and player 3 last. The bottom row shows the overall probabilities of winning, before the first spin.

Probabilities in the Price is Right Showcase Showdown
Spin 1	Strategy	Player 1	Player 2	Player 3
0.05	spin	20.59%	37.55%	41.85%
0.10	spin	20.59%	37.55%	41.86%
0.15	spin	20.57%	37.55%	41.87%
0.20	spin	20.55%	37.55%	41.9%
0.25	spin	20.5%	37.56%	41.94%
0.30	spin	20.43%	37.56%	42.01%
0.35	spin	20.33%	37.58%	42.10%
0.40	spin	20.18%	37.60%	42.22%
0.45	spin	19.97%	37.64%	42.39%
0.50	spin	19.68%	37.71%	42.61%
0.55	spin	19.26%	37.81%	42.93%
0.60	spin	18.67%	37.96%	43.36%
0.65	spin	17.86%	38.21%	43.93%
0.70	stay	21.56%	38.28%	40.16%
0.75	stay	28.42%	35.21%	36.38%
0.80	stay	36.82%	31.26%	31.92%
0.85	stay	46.99%	26.35%	26.66%
0.90	stay	59.17%	20.36%	20.47%
0.95	stay	73.61%	13.19%	13.21%
1.00	stay	90.57%	4.72%	4.72%
Average		30.82%	32.96%	36.22%

Here are the winning number of combinations out of the 6×20⁶ possible.

Player 1: 118,331,250
Player 2: 126,566,457
Player 3: 139,102,293

The way you play, where a third-card match is a push, the odds swing in your favor when there are at least six ranks between the first two cards (a six-card spread). The way I played in Orange County, a third-card match resulted in a double loss. Under that rule, the odds are break-even with an eight-card spread. If a third-card match results in a 1x loss, then you need a seven-card spread for the odds to be in your favor.

I hope you’re happy; I spent all day on this. The answer and solution can be found on my other site mathproblems.info , problem 203, or the academic paper Game Theory and Poker by Jason Swanson.

For the benefit of other readers, a point is a commission charged for the loan. For example, on a $250,000 loan one point would be $2,500. I’m going to assume that the borrower would add the point to the principal balance, and never pay down the principle early.

The following table shows the equivalent interest rate without the point, according to the interest rate with one point and the term.

Equivalent Interest Rate with No Points
Interest Rate with One Point	10 years	15 years	20 years	30 years	40 years
4.00%	4.212%	4.147%	4.115%	4.083%	4.067%
4.25%	4.463%	4.398%	4.366%	4.334%	4.318%
4.50%	4.714%	4.649%	4.617%	4.585%	4.570%
4.75%	4.965%	4.900%	4.868%	4.836%	4.821%
5.00%	5.216%	5.151%	5.119%	5.088%	5.073%
5.25%	5.467%	5.402%	5.370%	5.339%	5.324%
5.50%	5.718%	5.654%	5.621%	5.590%	5.576%
5.75%	5.969%	5.905%	5.873%	5.842%	5.827%
6.00%	6.220%	6.156%	6.124%	6.093%	6.079%
6.25%	6.471%	6.407%	6.375%	6.344%	6.330%
6.50%	6.723%	6.658%	6.626%	6.596%	6.582%
6.75%	6.974%	6.909%	6.878%	6.847%	6.834%
7.00%	7.225%	7.160%	7.129%	7.099%	7.085%
7.25%	7.476%	7.412%	7.380%	7.350%	7.337%
7.50%	7.727%	7.663%	7.631%	7.602%	7.589%
7.75%	7.978%	7.914%	7.883%	7.853%	7.841%
8.00%	8.229%	8.165%	8.134%	8.105%	8.093%
8.25%	8.480%	8.416%	8.385%	8.357%	8.344%
8.50%	8.731%	8.668%	8.637%	8.608%	8.596%
8.75%	8.982%	8.919%	8.888%	8.860%	8.848%
9.00%	9.233%	9.170%	9.140%	9.112%	9.100%
9.25%	9.485%	9.421%	9.391%	9.363%	9.352%
9.50%	9.736%	9.673%	9.642%	9.615%	9.604%
9.75%	9.987%	9.924%	9.894%	9.867%	9.856%
10.00%	10.238%	10.175%	10.145%	10.119%	10.108%

This shows that a 5.75% interest rate with one point is equivalent to a 5.842% with no points. In other words the payment would be the same both ways, assuming the point charged is added to the principal balance. Your other offer was 5.875% with no points, which is higher than 5.842%, so I would take the 5.75% with the point.

P.S. For those of you wondering how I solved for i, I used the rate function in Excel.

According to Life: the Odds (and How to Improve Them) by Gregory Baer, the odds of a hole in one on a par 3 hole in the PGA tour is 1 in 2491. I believe those distances fall in the par 3 range.

A 1 handicap is darn good, so I’m not going to give much of a discount compared to PGA Tour players. Let’s say your son’s probability per par 3 hole is 1 in 3,000. A typical gold course will have about four par 3 holes. Let’s say your son plays every day. That would be 28 par 3 holes a week. The probability of making exactly two hole in ones would be combin(28,2)×(1/3000)²×(2999/3000)²⁶ = 1 in 24,017.

For the sake of simplicity, let’s ignore the fact that the more tickets you buy the lower the value of each ticket becomes because you compete with yourself. That said, the probability of losing all five tickets is (12/13)⁵ = 67.02%. So the probability of winning at least one prize is 32.98%. There are 7033×13=91,429 total tickets in the drum before you buy any. 91,429-40=91,389 are not big prizes. The probability of not winning any big prizes with five tickets is (91,389/91429)⁵ = 99.78%. So the probability of winning at least one big prize is 0.22%, or 1 in 458.

Probabilities for Long Suit in Hearts
Cards	Combinations	Probability
4	222766089260	0.35080524800183
5	281562853572	0.44339660045899
6	105080049360	0.16547685914958
7	22394644272	0.03526640326564
8	2963997036	0.00466761219692
9	235237860	0.00037044541245
10	10455016	0.00001646424055
11	231192	0.00000036407412
12	2028	0.00000000319363
13	4	0.00000000000630
Total	635013559600	1

The average number of cards in the long suit is 4.9.

First, the "rule of 72" is an approximation of the time needed to double your money, not an exact answer. The following table shows the "rule of 72" values and the exact number of years, for various annual interest rates.

Rule of 72 — Years to Double Money
Interest Rate	Rule of 72	Exact	Difference
0.01	72.00	69.66	2.34
0.02	36.00	35.00	1.00
0.03	24.00	23.45	0.55
0.04	18.00	17.67	0.33
0.05	14.40	14.21	0.19
0.06	12.00	11.90	0.10
0.07	10.29	10.24	0.04
0.08	9.00	9.01	-0.01
0.09	8.00	8.04	-0.04
0.10	7.20	7.27	-0.07
0.11	6.55	6.64	-0.10
0.12	6.00	6.12	-0.12
0.13	5.54	5.67	-0.13
0.14	5.14	5.29	-0.15
0.15	4.80	4.96	-0.16
0.16	4.50	4.67	-0.17
0.17	4.24	4.41	-0.18
0.18	4.00	4.19	-0.19
0.19	3.79	3.98	-0.20
0.20	3.60	3.80	-0.20

Why 72? It doesn’t have to be exactly 72. That is just the number that works out well for realistic interest rates you’re likely to see on an investment. It works out almost exactly for an interest rate of 7.8469%. There is nothing special about 72, like there is about π or e. Why does any number work? If the interest rate is i, then let’s solve for the number of years (y) it takes to double an investment.

2 = (1+i)^y
ln(2)= ln(1+i)^y
ln(2)= y×ln(1+i)
y = ln(2)/ln(1+i)

This may not be my best answer ever, but try to follow this logic: let y=ln(x).
dy/dx=1/x.
1/x =~ x at values of x close to 1.
So the dy/dx =~ 1 for values of x close to 1.
So the slope of ln(x) is going to be close to 1 for values of x close 1.
So the slope of ln(1+x) is going to be close to 1 for values of x close 0.
The "rule of 72" is saying that .72/i =~ .6931/ln(1+i).
We’ve established that i and ln(1+i) are similar for values of i close to 0.
So 1/i and 1/ln(1+i) are similar for values of i close to 0.
Using 72 instead of 69.31 adjusts for differences between i and ln(1+i) for values of i around 8%.

I hope that makes some sense. My calculus is rather rusty; it took hours to explain this to myself.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

The industry term for that game is Razzle Dazzle. I remember seeing it in southern California as a kid, and just last year in San Felipe, Mexico. It is usually skinned to look like a football game. This game is the worst of the carnival game scams, in my opinion. The state of New York should be ashamed for permitting it. Based on some research, the rules vary from place to place, but the gist of the con is always the same.

It is based on the same illusion as the field bet in craps. For those readers not familiar with the field bet, the player wins if the sum of the roll of two dice is 2, 3, 4, 9, 10, 11, or 12. Losing numbers are 5, 6, 7, and 8. Wins pay even money, except the 2 pays 2 to 1 and the 12 pays 3 to 1 (except at stingy Harrah’s casinos, where they pay 2 to 1 only on the 12). The mathematically challenged gambler may falsely reason it is a good bet because there are 7 totals that win and only 4 that lose. The reason the odds favor the house is that the losing numbers have the greatest chance to be rolled.

Here are the specific rules of Razzle Dazzle, as taken from the article Probabilities of Winning a Certain Carnival Game by Donald A. Berry and Ronald R. Regal, which appeared in the November 1978 issue of the The American Statistician.

1. The object of the game is to advance across the football field 100 yards. The player will be awarded some kind of nice prize when he does.
2. The player starts paying a specified fee per play, such as $1.
3. The player will spill 8 marbles onto an 11 by 13 grid. Each marble will fall into one of the 143 holes.
4. Each hole has a number of points from 1 to 6. The following table shows the frequency of each number of points.
   
   

   Razzle Dazzle Points Distribution
   Points	Number on Board	Probability
   1	11	0.076923
   2	19	0.132867
   3	39	0.272727
   4	44	0.307692
   5	19	0.132867
   6	11	0.076923
   Total	143	1.000000

5. The total number of points will be added. The carnie will look up the point total on a conversion chart to see how many yards the player advances. The conversion chart is shown below.
   
   

   Razzle Dazzle Conversion Chart
   Points	Yards Gained
   8	100
   9	100
   10	50
   11	30
   12	50
   13	50
   14	20
   15	15
   16	10
   17	5
   18 to 38	0
   39	5
   40	5
   41	15
   42	20
   43	50
   44	50
   45	30
   46	50
   47	100
   48	100

6. If the player rolls a total of 29, then the fee for all subsequent rolls will be doubled, and the player be awarded one extra prize if and when he reaches the other end of the football field.

The average points per marble is 3.52, and the standard deviation is 1.31. Note how 3 and 4 points have the highest probability. That keeps the standard deviation low, and the sum of many marbles close to expectations. The standard deviation of the roll of a single die is 1.71, by comparison.

Next, notice how there are 20 winning totals and 21 losing totals on the yardage conversion chart. The kind of sucker who gambles on carnival games might incorrectly reason his probability of advance is 20/41 or 48.8%. It wouldn’t surprise me if the carnies falsely claimed these were the odds of advancing. However, much like the field bet, the most likely outcomes don’t win anything.

The next table show the probability of each number of points per turn, yards gained, and expected yards gained. The lower right cell shows the average yards gained per turn is 0.0196.

Expected Yards Gained per Turn
Points	Probability	Yards Gained	Expected Yards Gained
8	0.00000000005	100	0.00000000464
9	0.00000000176	100	0.00000017647
10	0.00000002586	50	0.00000129285
11	0.00000022643	30	0.00000679305
12	0.00000143397	50	0.00007169849
13	0.00000713000	50	0.00035650022
14	0.00002926510	20	0.00058530196
15	0.00010234709	15	0.00153520642
16	0.00031168305	10	0.00311683054
17	0.00083981462	5	0.00419907311
18	0.00202563214	0	0.00000000000
19	0.00441368617	0	0.00000000000
20	0.00874847408	0	0.00000000000
21	0.01586193216	0	0.00000000000
22	0.02642117465	0	0.00000000000
23	0.04056887936	0	0.00000000000
24	0.05757346716	0	0.00000000000
25	0.07566411880	0	0.00000000000
26	0.09221675088	0	0.00000000000
27	0.10431970222	0	0.00000000000
28	0.10958441738	0	0.00000000000
29	0.10689316272	0	0.00000000000
30	0.09677806051	0	0.00000000000
31	0.08125426057	0	0.00000000000
32	0.06317871335	0	0.00000000000
33	0.04540984887	0	0.00000000000
34	0.03009743061	0	0.00000000000
35	0.01833921711	0	0.00000000000
36	0.01023355162	0	0.00000000000
37	0.00520465303	0	0.00000000000
38	0.00239815734	0	0.00000000000
39	0.00099365741	5	0.00496828705
40	0.00036673565	5	0.00183367827
41	0.00011909673	15	0.00178645089
42	0.00003349036	20	0.00066980729
43	0.00000797528	50	0.00039876403
44	0.00000155945	50	0.00007797235
45	0.00000023832	30	0.00000714969
46	0.00000002632	50	0.00000131607
47	0.00000000176	100	0.00000017647
48	0.00000000005	100	0.00000000464
Totals	1.00000000000	0	0.01961648451

Here are some results of a random simulation of 17.5 million games.

Razzle Dazzle Simulation Results
Question	Answer
Probability of advancement per turn	0.0028
Expected yards gained per turn	0.0196
Expected yards gained per advancement	6.9698
Expected turns per game	5238.7950
Average doubles per game	559.9874
Averages prizes per game	560.9874

I would have liked to indicate the average total bet per game, but my computer can not handle numbers so large. The average game had the player doubling his bet 560 times over the average of 5,239 turns per game. One game in the simulation had the player doubling his bet 1,800 times. Even at the average of 560 doubles, the bet per roll would be $3.77 × 10¹⁶⁸, assuming a starting bet of $1. That is many orders of magnitutude greater than the number of atoms in the known universe (source ).

Even the most naive player will not play for long if he is advancing once every 355 plays only. What the carnies will do is cheat in the player’s favor at first. He may spot the player free rolls, or lie in adding up the points, giving the player winning totals to boost his confidence. I’ve never played the game, but I imagine that when the player gets close to the red zone (20 yards or less from a touchdown), then the carnie will start playing fairly. The player may wonder why he is suddenly getting nowhere, but with money already invested, and being so close to the goal line, he would hesitate to walk away and give up the yardage he already paid for.

Links

• Razzle Dazzle , excerpt from the book On the Midway.
• Razzle Dazzle Carny Board Game Arcade Scam .
• Probabilities of Winning a Certain Carnival Game by Donald A. Berry and Ronald R. Regal

There are eight ways two win: three rows, three columns, and two diagonals. There are combin(9,3)=84 ways to pick 3 squares out of 9. So the probability of winning is 8/84 = 9.52%. This question was raised and discussed in the forum of my companion site Wizard of Vegas .

Here is my Wizard’s basic strategy for Monopoly:

• Buy everything. Advanced players may make exceptions if the property won’t help you make a monopoly, block someone else, and has little value as a bargaining chip. Utilities can also be declined in a cash-poor situation.
• Trade as well as you can. This is where the skill comes in. Try to trade for the best set you can. Here is how I rank them, in general: Orange, Yellow, Light Blue, Dark Blue, Light Purple, Red, Green, Dark Purple. This will vary depending on circumstances. In a cash-poor game, favor the sets that are cheaper to develop, like the light blues. In a cash-rich game, go for the ones where there is more potential to spend money on, like the yellows or dark blues.
• Once you get a set, whether naturally or by trade, build up quickly. Try to get to three houses on each property as quickly as possible. The marginal return per house drops after three. Mortgage most of your other properties and spend your cash. You want to leave a little equity for small expenses. Not spending your money is like a soldier in battle not using his bullets.
• Oppose all the silly house rules. This especially goes for the money pot on Free Parking (I can’t stand that one!). If you are more skilled than your opponents, you want to minimize the randomness of the game.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

The six center faces of the cube are fixed. By turning the faces all you can do is rearrange the corners and edges. If you took the cube apart, then there would be 8!=40,320 ways to arrange the eight corners, without respect to the orientation of each piece. Likewise, there are 12!=479,001,600 ways to arrange the 12 edges without regard to orientation.

There are 3 ways each corner can be oriented, for a total of 38=6,561 corner orientations. Likewise there are two ways each edge piece can be oriented, for a total of 212=4,096 edge orientations.

So if we could take the cube apart, and rearrange the edge and corner groups, then there would be 8! × 12! × 38 × 212 = 519,024,039,293,878,000,000 possible permutations. However, not all of these permutations can be arrived at from the starting position by rotating the faces.

First, it is impossible to rotate just one corner and leave everything else the same. No combination of turns will achieve that. Basically, every action has to have a reaction. If you wish to rotate one corner, it would disturb the other pieces somehow. Likewise, it is impossible to flop just one edge piece. For these reasons, we have to divide the number of permutations by 3 × 2 = 6.

Second, it is impossible to switch two edge pieces without disturbing the rest of the cube. This is the hardest part of this answer to explain. All you can do with a Rubik’s Cube is rotate one face at a time. Each movement rotates four edge pieces and four corner pieces for a total of eight pieces moved. A sequence of rotations can be represented by a number of piece movements divisible by 8. Often a sequence of moves will result in two movements canceling each other out. However, there will always be an even number of pieces moved with any sequence of rotations. To swap two edge pieces would be one movement, an odd number, which can not be achieved with the sum of any set of even numbers. Mathematicians would call this a parity problem. So we have to divide by another 2 because two edge pieces cannot be swapped without other pieces being disturbed.

So there are 3 × 2 × 2 = 12 possible groups of Rubik’s Cube permutations. If you disassembled a Rubik’s Cube and put it back together randomly, there is a 1 in 12 chance that it would be solvable. So the total number of permutations in a Rubik’s Cube is 8! × 12! × 38 × 212 / 12 = 43,252,003,274,489,900,000. If you had seven billion monkeys, about the human world population, playing randomly with the Rubik’s cube, at a rate of one rotation per second, a cube will pass through the solved position on average once every 196 years.

Links

• Group Theory via Rubik’s Cube by Tom Davis (PDF)
• Rubik’s Cube Group at Wikipedia

For those unfamiliar with the rules of Hearts, play starts with dealing 13 cards each to four players. The hearts suit is significant to the game, so how many you get is important. The following table shows the odds of being dealt 0 to 13 hearts.

Probability of 0 to 13 Hearts out of 13 Cards
Hearts	Combinations	Probability	Inverse
13	1	0.0000000000016	1 in 635,013,559,600.0
12	507	0.0000000007984	1 in 1,252,492,228.0
11	57,798	0.0000000910185	1 in 10,986,773.9
10	2,613,754	0.0000041160601	1 in 242,950.8
9	58,809,465	0.0000926113531	1 in 10,797.8
8	740,999,259	0.0011669030492	1 in 857.0
7	5,598,661,068	0.0088166008164	1 in 113.4
6	26,393,687,892	0.0415639752774	1 in 24.1
5	79,181,063,676	0.1246919258321	1 in 8.0
4	151,519,319,380	0.2386080062219	1 in 4.2
3	181,823,183,256	0.2863296074662	1 in 3.5
2	130,732,371,432	0.2058733541286	1 in 4.9
1	50,840,366,668	0.0800618599389	1 in 12.5
0	8,122,425,444	0.0127909480376	1 in 78.2
Total	635,013,559,600	1.0000000000000

This question was raised and discussed in the forum of my companion site Wizard of Vegas .