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NonCasino Games  FAQ
John from New York
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)*0.1^{n}*0.9^{8n}/10^{8}. Here is a table showing the probability for 0 to 8 of a specific number.
Specific Number Oddsin Liar's Poker
Number  Probability 

8  0.00000001 
7  0.00000072 
6  0.00002268 
5  0.00040824 
4  0.00459270 
3  0.03306744 
2  0.14880348 
1  0.38263752 
0  0.43046721 
Total  1.00000000 
The next table shows the probability of every possible type of bill, categorized by the number of each nofakind. For example, the serial number 66847680 would have one three of a kind, one pair, and three singletons, for a probability of 0.1693440.
General Probabilities in Liar's Poker
8 o.a.k.  7 o.a.k.  6 o.a.k.  5 o.a.k.  4 o.a.k.  3 o.a.k.  2 o.a.k.  1 o.a.k.  Probability 

1  0.0000001  
1  1  0.0000072  
1  1  0.0000252  
1  2  0.0002016  
1  1  0.0000504  
1  1  1  0.0012096  
1  3  0.0028224  
2  0.0000315  
1  1  1  0.0020160  
1  2  0.0015120  
1  1  2  0.0211680  
1  4  0.0211680  
2  1  0.0020160  
2  2  0.0141120  
1  2  1  0.0423360  
1  1  3  0.1693440  
1  5  0.0846720  
4  0.0052920  
3  2  0.1270080  
2  4  0.3175200  
1  6  0.1693440  
8  0.0181440  
Total  1.0000000 
o.a.k. = "of a kind"
For more information, see my page on liars poker.
Daniel from Las Vegas, USA
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 the same money over and over as many times as possible. However, your odds may be so bad that it isn't worth the bother.
Jansen from Toronto, Canada
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.
These bets can be made in both sic bo and chuck a luck.
James from Greeley, USA
That is probably the most frequently asked question I get which I don't have an answer to. An exhaustive Klondike Solitaire has never been done. Maybe, when computers are a million times faster, somebody will eventually do it. However, it is rumored that Vegas casinos used to offer the game at least back in the fifties. I've asked a number of Vegas oldtimers to verify that, but none have been able to, so far.
David from Sunland, USA
With every new roll the probability the next four rolls will be all double sixes is (1/36)^{4} = 1 in 1679616.
Atle from Porsgrunn, Norway
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.
"Anonymous" .
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.
"Anonymous" .
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 onethird. This is because I believe amateurs repeat less than onethird 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.
"Anonymous" .
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%
"Anonymous" .
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%.
"Anonymous" .
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.
"Anonymous" .
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.
Utpal from Lucknow
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%.
"Anonymous" .
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.
Jack from Troy
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.
Mike from Olympia
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 (1p)/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 (ta)/(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 (6360)/(63+1) = 3/64 = 4.69%. On the green bet the house edge is (1514)/(15+1) = 1/16 = 6.25%. On individual numbers the house edge is (6360)/(63+1) = 3/64 = 4.69%.
Tom from Buffalo, NY
VLT’s are glorified pulltab 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 pulltabs 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.
Ian F. from Provo
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 
Tony
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 hardcover 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 nearperfect 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 boneheaded 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.
Mike P.
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 6×20^{6} possible.
Player 1: 118,331,250Player 2: 126,566,457
Player 3: 139,102,293
DelRayVA from Fairfax, VA
The way you play, where a thirdcard match is a push, the odds swing in your favor when there are at least six ranks between the first two cards (a sixcard spread). The way I played in Orange County, a thirdcard match resulted in a double loss. Under that rule, the odds are breakeven with an eightcard spread. If a thirdcard match results in a 1x loss, then you need a sevencard spread for the odds to be in your favor.
Victor M. from Genoa City, WI
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.
Dorothy from Miami, FL
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.
John from Pointe Claire, Quebec, Canada
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.
"Anonymous" . from Mesa, AZ
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. 91,42940=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.
Jim from Boring, OR
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 
mkl654321
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.
Andrew from Queens, NY
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.
 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.
 The player starts paying a specified fee per play, such as $1.
 The player will spill 8 marbles onto an 11 by 13 grid. Each marble will fall into one of the 143 holes.
 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 BoardProbability 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  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 DazzleConversion Chart
Points Yards
Gained8 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  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 × 10^{168}, 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
WizardofEngland
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.
ahiromu
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 cashpoor 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 cashpoor game, favor the sets that are cheaper to develop, like the light blues. In a cashrich 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.
Anon E. Mouse
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 3^{8}=6,561 corner orientations. Likewise there are two ways each edge piece can be oriented, for a total of 2^{12}=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! × 3^{8} × 2^{12} = 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! × 3^{12} × 2^{12} / 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
"Anonymous" .
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.
AxiomOfChoice
I hope you're happy. To answer this question, I bought a big roll of tickets at the Office Depot. Then I put 500 of them in a paper bag, half folded in half, at about a 90degree angle, and the other half unfolded. Then I had six volunteers each draw 40 to 60 of them one at a time, with replacement, as I recorded the results. Here are the results.
Drawing Ticket Experiment
Subject  Folded  Unfolded  Total 

1  25  25  50 
2  38  22  60 
3  25  15  40 
4  34  16  50 
5  27  23  50 
6  26  24  50 
Total  175  125  300 
So, 58.3% of the tickets drawn were folded!
If it's assumed that folding had no effect, then these results would be 2.89 standard deviations away from expectations. The probability of getting this many folded tickets, or more, assuming folding didn't affect the odds, is 0.19%, or 1 in 514.
I might add the subjects who drew tickets hastily were much more likely to draw folded ones. Those who carefully took their time with each draw were at or near a 50/50 split.
So, my conclusion is definitely to fold them.
For discussion about this question, please visit my forum at Wizard of Vegas.