Ed answers probability and other math-related questions
 by  Edward D. Collins

 In 1999 I wrote a couple of articles that were published in the magazine European Backgammon News.  I posted one of them (Dice and the Laws of Probability) here on my website.   Almost from the moment I posted that article, hundreds of people who have read that article have written to me, asking for help with probability related questions on dice, playing cards, football betting, etc!  (Ok, maybe not hundreds of people, but dozens and dozens and dozens.)  What am I, an expert or something? I've also helped countless others in many of the forums and discussion boards I've participated in over the years. Below are just some of the actual questions I've answered.

 Hi Ed, When flipping a coin, the is a 50% chance that it will land on heads and 50% chance that it will land on tails.  (Assuming the coin is evenly balanced.)  How do you calculate the probability of getting x heads in a row, say, five heads in a row? Ah, this one is easy.   To determine the overall odds of all of the individual events occurring, you simply multiply the odds of each event together. For example, assume you wish to know what the odds are of flipping two coins and having a head appear on each.  The chances of a head appearing on the first coin is 1/2 or .50. The chances of a head appearing on the second coin is also 1/2.   Multiply 1/2 by 1/2 to arrive at our answer:  1/4.   There is one chance out of four (25%) that a head will appear on both coins when two coins are flipped. This is easy to confirm when you list the total possibilities.  There are only four of them:  Head-Head, Head-Tail, Tail-Head, or Tail-Tail.  Only one of these four meets our criteria.. a head on both.  Again, one chance out of four. Back to your question.  The odds of a head appearing each time, when the coin is flipped five times, is (1/2)^5 or (1/2 x 1/2 x 1/2 x 1/2 x 1/2) or 1/32.  There is just one chance out of 32 of a head appearing each time, when a coin is flipped five times.

 Hi Ed, I need to know the probability of getting five heads (or more) in a row, when I flip a coin 200 number of times?   To clarify, if I flip a coin 200 times, what are the chances of seeing at least one run of five consecutive heads? Apparently this solution is not trivial!  I just spent more time than I care to admit researching this seemingly simple equation.  And I STILL have not found the definitive answer!   So, just for fun, I wrote a computer program that simulates coin flips.  Of course, it's BETTER to know the EXACT formula, and the exact percentage, but in cases like this, when maybe you just don't know what that formula is in the first place, then a computer simulation that can roll the dice (or flip the coin) millions and millions of times will produce a percentage VERY VERY close to the exact percentage.  Below is the output of my program: Number of times the coin was flipped 200 times: 5,000,000 Probability of a single event occurring: .5 (a fair coin) Number of total times we will flip this coin: 200 Number of consecutive runs of heads we are looking for: 5 Number of times out of the total games played we saw our specified event occur: 4,829,647 Percentage: .9659294 I ran this simulation several times, and each time the results were different, but not enough to change the percentage.  The answer you are looking for is about 96.5%.

 What are the chances of this parlay hitting? Titans +4 Texans +9 Panthers +13.4 I don't see any value in any of those lines so my answer is about the same as any other 3-team parlay... about 1 out of 8, or 7 to 1 against you. (12.5%)

 Good morning Edward, I liked your dice probability work on the chances of getting one 6 when rolling different number of dice. What is the probabilities of getting at least a 1 OR a 5 with 1 die, 2 dice, 3 dice, etc. I think I know why you're asking this.  About 20 years ago or so I remember my brother teaching me a dice game in which you roll dice and rolling ones and fives are good.   Anyway...   The chances of getting at least one 1 or one 5 when rolling one die is 1/3...or 33.33%.   The chances of getting at least one 1 or one 5 when rolling two dice is 20/36...or 55.55%.   The chances of getting at least one 1 or one 5 when rolling three dice is 70.37%.   The chances of getting at least one 1 or one 5 when rolling four dice is 80.24%.   The chances of getting at least one 1 or one 5 when rolling five dice is 86.83%.   The chances of getting at least one 1 or one 5 when rolling six dice is 91.22%.     The formula for the above is easy.  For example, with three dice you do it this way:  Multiply the chances of NOT getting a one or a five (4/6) on the first die by the chances of NOT getting a one or a five on the second die (4/6) by the chances of NOT getting a one or a five on the third die, and the subtract the whole thing from 1.   example: 1 - (4/6 x 4/6 x 4/6) = .7037

 Hi Edward, My name is Thomas.  Can you please help me with these two questions?    1)  Seven dice are rolled.  What is the probability that the number that comes up on the fifth die equals the number on the seventh die?   2)  Four coins are flipped.  Given that at least one of the coins is a Head, what is the probability that the first coin is a Head?  1)  The probability the number on the 5th die equals the number on the 7th die is 1/6 or 16.666%   (The number on the 5th die has an equal chance of being 1 through 6... likewise with the 7th dice.) It's completely irrelevant that seven dice are rolled.  If only three dice were rolled, for example, the probability the number on the 1st die equals the number on the 3rd die is also 1/6. 2) There are 16 different "results," all equally likely, after four coins are flipped: H-H-H-H, H-H-H-T, H-H-T-H, H-H-T-T H-T-H-H, H-T-H-T, H-T-T-H, H-T-T-T T-H-H-H, T-H-H-T, T-H-T-H, T-H-T-T T-T-H-H, T-T-H-T, T-T-T-H, T-T-T-T Only ONE of these 16 results can we throw out... since we know for sure it didn't occur.  (We can throw out T-T-T-T, since we know at least one head appeared.) However, all of the remaining 15 possibilities are equally likely.  Thus, the chances that the first flip was a head is 8/15 or 53.333%.

 1)  If 2 people are picked from 5 men, 6 women, 7 children, find the probability that they are not both children. 2)  7 dice are rolled.  What is the probability that exactly four show an even number?   1)  There are 18 people in all.  11 of them are not children.  The chances of picking ONE person, and it not being a child, is very easy...  11/18... or 61.11% After you pick one person, assuming it's not a child, you have 17 people left.  10 of these are not children.  The odds of picking a person now, that is not a child, is 10/17. Thus, multiply the chances of NOT picking one child (11/18) by the chances of NOT picking another child (10/17) to get the chances of not picking BOTH children.  11/18 x 10/17 = 110/306 which is 35.94% The probability of picking two people, and both not being children, is 35.94%.  2)  Exactly four?  76,545 / 279,936  which reduces to 35/128 or 27.34375%.   7C4 is the expression, which is 7! / (4! * ( (7-4)! )) 7*6*5*4*3*2*1 / (4*3*2*1 * (3*2*1)) 5,040 / (24*6) 5,040 / 144 35 is the numerator and 128 (2 to the 7th power) is the denominator.   I shouldn't be doing your homework for you!

 I looked at the dice explanation on your web site it was very helpful in understanding the probability of dice.   http://www.edcollins.com/backgammon/diceprob.htm   However, I have an equally weighted dice with 9 sides.  What is the probability of rolling a 2 and a 4 on a single roll? Andrew Hi Andrew,   You wrote "...what is the probability of rolling a 2 and a 4 on a single roll?"   Before I answer, let me clarify what I believe is the question.  With two dice, each with nine sides, (the numbers 1 through 9) you wish to know the probability of rolling a 2 on one of them AND a 4 on the other.   Note that I capitalized 'AND'.  (You used the word 'and' in your question.  The answer is different, of course, if you wish to say OR instead of AND.) If this is all correct, the answer is easy.  Throw the dice one at a time.  In order to satisfy the criteria, your first dice-roll must be either a 2 OR a 4.   The chances of this happening are 2/9.   Once you do that, throw the other die.  NOW you must roll a 4 (if you rolled a 2 earlier) OR you must roll a 2 (if you rolled a 4 earlier).  Either way, just one number helps.  Thus, the probability of that, obviously, is 1/9.   The probability of both occurring is 2/9 x 1/9 or 2/81... a little more than 40 to 1 against.   You can verify this by hand by listing all the possible rolls of two dice.  There are only 81 of them.   1-1  1-2  1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 1-9 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 2-7, 2-8, 2-9 3-1, 3-2, 3-3, 3-4, 3-5, 3-6, 3-7, 3-8, 3-9 4-1, 4-2, 4-3, 4-4, 4-5, 4-6, 4-7, 4-8, 4-9 etc.   After listing the dice rolls, you'll see that just two of them (2-4 and 4-2) satisfy the criteria of needing to roll a 2 AND a 4.   However, it's possible I'm misreading your question.  If you want to know the probability of rolling a 2 OR a 4 using two, nine-sided dice, you take the chances of NOT rolling a 2 or a 4 on the first die (7/9) and multiply that by the chances of NOT rolling a 2 or a 4 on the sceond die (7/9).   The result is 49/81.   49 times out of 81 you will not roll a 2 or a 4.  81 - 49 = 32.  This means that 32 times you WILL roll either a 2 or a 4.   This can also be verified by listing the 81 possible dice rolls, and noting the number of rolls where a 2 OR a 4 occurred.  You'll see that it is indeed 32. 1-1, 1-2, 1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 1-9 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 2-7, 2-8, 2-9 3-1, 3-2, 3-3, 3-4, 3-5, 3-6, 3-7, 3-8, 3-9 4-1, 4-2, 4-3, 4-4, 4-5, 4-6, 4-7, 4-8, 4-9 etc.

 I liked your "Dice and the Laws of Probability" article but have a question about a different scenario.  What is the chance of rolling a 4 or under when you roll three different dice? That's another easy one.  Multiply the chances of NOT rolling a 4 or under (2/6) with the first die, by the chances of NOT rolling a 4 or under with the second die (2/6 by the chances of NOT rolling a 4 or under with the third die.  (Also 2/6) 2/6 x 2/6 x 2/6 = 8/216.   Thus, the chances OF rolling a 4 or under with three dice is a whopping 208/216 or 96.29% This is confirmed by listing all of the 216 ways three dice can be rolled and then circling the ones in which a 4 or under isn't rolled.  If you do so you will get these 8 rolls: 5 5 5   5 5 6   5 6 5   5 6 6  6 5 5   6 5 6   6 6 5   6 6 6

 What is the probability of throwing 3-of-a-kind with 3, 4, 5 or 6 dice?
 Number of Dice Thrown A SPECIFIC set of threes (like three 6s) ANY set of threes (like three 2s or three 6s) 3 1 / 216 6 / 216 4 20 / 1,296 120 / 1,296 5 250 / 7,776 1,500 / 7,776 6 2,500 / 46,656 14,700 / 46,656

 Sniper2008:  I was wondering if playing the Oklahoma moneyline being -1000 would be to risky of a play or is Oklahoma winning a sure thing. I need some suggestions. Dougie2138:  I don't care how much of a "sure thing" you think it is. Laying -1000 is never a smart play in ANY case.   That's the most incorrect statement I've read in this forum in awhile.  It's all about the odds and your chances of winning.  Assume Team A wins 91% of the time, and \$100 is bet each time.  Do the math: Team A wins 91 games out of 100:  \$10 win x 91 games = +\$910 Team A loses 9 games out of 100  -\$100 loss x 9 games = -\$900 Net win of \$10 each 100 games. As you can see, as long as Team A wins 91% of the time, or more, laying -1000 is +EV.  I'd lay -1000 all day long that a person can't roll a single die three times and have it land on a six each time.  Why will I do this?  Because the chances of it happening are less than 1/2 of 1%.  (1 chance out of 216) Laying "only' -1000 would be a very smart play.   To say that is is never a smart play to lay -1000 is completely incorrect.

 In five-card draw poker, what are the chances of being dealt a Royal Flush, before the draw? With your first card dealt, you can receive any of 20 cards and your chances of a Royal are still alive.  But no matter what card you receive, you then have just 4 'live' cards left... and then 3 and then 2 and then 1.  Thus, your chances are 20/52 x 4/51 x 3/50 x 2/49 x 1/48 or about 0.00000153907.

 A fast-food chain includes one of five different toy cars with every kid's meal.  Each toy is equally likely to be included with the meal.  How many meals must be purchased, on the average, before all five different toy cars are collected? This is another type of question that is simply easier for me to answer by letting my computer 'tell' me the answer.   I wrote a quick and dirty program to simulate this, and ran it 5,000,000 times. (And then re-ran it several times to double check the results.) My program reports that, on average, you have to purchase 11.41 meals before you have all five toys. As you probably know, this is the average amount. If you are very LUCKY, you only have to purchase five meals. If you are very UNLUCKY, you may have to purchase a lot more.  In fact, out of my simulated 5 million people who purchased meals, the unluckiest person had to purchase 77 meals before he/she had all five toys! From a probability standpoint, you have a slightly greater than 50-50 chance of getting all five toys if you purchase 12 meals. You have a slightly less than 50-50 chance of you purchase 11 meals.

 From a well shuffled deck, what is the probability the last three cards to be dealt are all diamonds?   .012941 or 1.29%   (13/52 * 12/51 * 11/50)

 Help!  A box has 24 bulbs.  Four of these are known to be defective.  Four bulbs are selected at random. What is the probability that none of the four selected are defective? 20/24 x 19/23 x 18/22 x 17/21 = .45595... about 45.59%.

 Three marbles are drawn at random from an urn containing 4 black, 7 yellow, and 5 green marbles? What is the probability that none of them are yellow? The probability that none of them are yellow is the product of 9/16 x 8/15 x 7/14, which is .15. (15%)

 2 playing cards are drawn from a standard deck of 52 cards.  What is the probability that the two cards drawn are either both aces or both red cards? The chances both cards are aces is small... .004525   (4/52 * 3/51) The chances both cards are red is nearly 1 in 4... .245098.  (26/52 * 25/51) The chances of EITHER event happening is the sum of the two events.  .004525 + .245098 = .249623... or about 24.96%.   Thus, slight less than 1 time in 4 will you draw two cards from the deck, and they will either both be aces or both be red.

 If a roulette wheel is spun 190 times, about how many times the ball can be expected to land on 0 or 00?   Oh oh, it sounds as if you're working on a roulette system.  Be careful... there is no way to beat the house at roulette. Including 0 and 00, there are a total 38 different numbers on an American roulette wheel.  Thus, the odds of a 0 or a 00 coming up are exactly 2 out of 38... which is reduced to 1 out 19.  Your 190 times fits into this nicely... multiply each by 10.  If the ball is expected to come up 0 or 00 1 out of 19 spins, then it can be expected to come up 10 times out of 190 spins.

 Ed! How are you at logic problems?  I need to know the answer to this RIGHT AWAY! Jordan, Jesse, Jake, Joey, Justin, Jane, Julie, and Jessica are going to the movies on Saturday night.  How many different seating arrangements are possible given the following restrictions: - Jordan & Jessica just broke up so they can't sit next to each other. - Jake always has to use the bathroom so he must sit on one of the ends. - Justin & Jane don't get along and need to have at least two seats between them. - Jane & Julie are best friends and insist on sitting next to each other. - Jesse has a crush on Jessica and wants to sit next to her, but he can't sit next to Jordan (after all, Jordan & Jessica did just break up.) - Joey needs to sit in one of the two middle seats. I decided this would be easier to solve if I wrote a computer program to figure out all the different permutations.   There are exactly 108 different seating arrangements: 1 4 2 7 8 5 6 3     1 4 2 7 8 6 5 3     1 4 7 2 8 5 6 3     1 4 7 2 8 6 5 3     1 5 6 8 2 7 4 3     1 5 6 8 4 2 7 3     1 5 6 8 4 7 2 3     1 5 6 8 7 2 4 3     1 6 5 8 2 7 4 3     1 6 5 8 7 2 4 3     2 7 4 1 8 5 6 3     2 7 4 1 8 6 5 3 2 7 4 8 1 5 6 3     2 7 4 8 1 6 5 3     2 7 4 8 6 5 1 3     2 7 5 6 8 1 4 3     2 7 5 6 8 4 1 3     2 7 6 5 8 1 4 3     3 1 4 8 2 7 5 6     3 1 4 8 2 7 6 5     3 1 4 8 6 5 2 7     3 1 4 8 6 5 7 2     3 1 4 8 7 2 5 6     3 1 4 8 7 2 6 5 3 1 5 6 8 2 7 4     3 1 5 6 8 4 2 7     3 1 5 6 8 4 7 2     3 1 5 6 8 7 2 4     3 1 6 5 8 2 7 4     3 1 6 5 8 7 2 4     3 2 7 4 8 1 5 6     3 2 7 4 8 1 6 5     3 2 7 4 8 6 5 1     3 2 7 8 4 1 6 5     3 2 7 8 5 6 1 4     3 4 1 8 2 7 5 6 3 4 1 8 2 7 6 5     3 4 1 8 5 6 2 7     3 4 1 8 5 6 7 2     3 4 1 8 6 5 2 7      3 4 1 8 6 5 7 2     3 4 1 8 7 2 5 6     3 4 1 8 7 2 6 5     3 4 2 7 8 1 5 6     3 4 2 7 8 1 6 5     3 4 2 7 8 5 6 1     3 4 2 7 8 6 5 1     3 4 7 2 8 1 5 6 3 4 7 2 8 1 6 5     3 4 7 2 8 5 6 1     3 4 7 2 8 6 5 1     3 5 6 1 8 2 7 4     3 5 6 1 8 4 2 7     3 5 6 1 8 4 7 2     3 5 6 1 8 7 2 4     3 5 6 8 1 4 2 7     3 5 6 8 1 4 7 2     3 5 6 8 2 7 4 1     3 5 6 8 7 2 4 1     3 6 5 1 8 2 7 4 3 6 5 1 8 4 2 7     3 6 5 1 8 4 7 2     3 6 5 1 8 7 2 4     3 6 5 8 1 4 2 7     3 6 5 8 1 4 7 2     3 6 5 8 2 7 4 1     3 6 5 8 7 2 4 1     3 7 2 4 8 1 5 6 3 7 2 4 8 1 6 5     3 7 2 4 8 6 5 1     3 7 2 8 4 1 6 5     3 7 2 8 5 6 1 4 4 1 6 5 8 2 7 3     4 1 6 5 8 7 2 3     4 2 7 8 1 5 6 3     4 2 7 8 1 6 5 3 4 2 7 8 5 6 1 3     4 2 7 8 6 5 1 3     4 7 2 8 1 5 6 3     4 7 2 8 1 6 5 3 4 7 2 8 5 6 1 3     4 7 2 8 6 5 1 3     5 6 1 4 8 2 7 3     5 6 1 4 8 7 2 3 5 6 1 8 2 7 4 3     5 6 1 8 4 2 7 3     5 6 1 8 4 7 2 3     5 6 1 8 7 2 4 3 5 6 2 7 8 1 4 3     5 6 2 7 8 4 1 3     5 6 7 2 8 1 4 3     5 6 7 2 8 4 1 3    6 5 1 8 2 7 4 3     6 5 1 8 4 2 7 3     6 5 1 8 4 7 2 3     6 5 1 8 7 2 4 3 6 5 2 7 8 1 4 3     6 5 2 7 8 4 1 3     6 5 7 2 8 1 4 3     6 5 7 2 8 4 1 3     7 2 4 1 8 5 6 3     7 2 4 1 8 6 5 3     7 2 4 8 1 5 6 3     7 2 4 8 1 6 5 3     7 2 4 8 6 5 1 3     7 2 5 6 8 1 4 3     7 2 5 6 8 4 1 3     7 2 6 5 8 1 4 3 (1 = Jordan, 2 = Jessica, 3 = Jake, 4 = Justin, 5 = Jane, 6 = Julie, 7 = Jesse, 8 = Joey) Notice that every criteria is met. Jake (3) is always on one of the ends, Joey (8) is always in seat #4 or #5, Jesse (7) is never next to Jordan (1), etc.

 Given that a throw of 3 dice show 3 different faces what the probability if the sum of this number = 8   There are 216 different ways three dice can be rolled. The following 12 rolls are the only rolls that fit the conditions of the problem. (Sum = 8 and three different faces.) 1 2 5     1 3 4     1 4 3     1 5 2 2 1 5     2 5 1     3 1 4     3 4 1 4 1 3     4 3 1     5 1 2     5 2 1 Thus, the probability is just 12/216... or 5.55%.

 Hi.  I was reading your Dice and the Laws of Probability essay and I was wondering if you could apply a similar statistic to poker, for instance.  If you are playing Hold'em and you have four cards of the same suit from the flop, let's say that the probability that you will hit that 5th card to make your flush is 3 to 1.  (It's not, but let's say it is.)  Is there any way to prove that the probability would be expanded if you played that similar instance more than once?  For instance, if you came across that example once and didn't hit the card, another time and didn't hit the card, would there be any more of a probability that you might hit it on the third try than would have been present on the first or the second given that you've had a three in one chance the entire time?  Would that example be the same as rolling three dice to make a six, acheiving a greater probality than with one dice?  I've been racking my brain about this and everyone disagrees with me, but I seem to think your Dice and the Laws of Probability article shows that their is an expanded probability in my similar instance as there is an expanded probability with multiple dice.  I'd love to be proven right or wrong, so any of your input is welcome.  Thanks for your time.   There's a reason why everyone disagrees with you.  It's because you're dead wrong.   From one hand to the next, cards don't have a memory either, just like dice rolls or spins of the roulette wheel, or lottery numbers, etc.  If you missed your flush ten times in a row, that doesn't mean it is any more likely that you will hit it the next time.  The odds will still be exactly the same.

 Hi Ed,   While playing blackjack on Yahoo, I use this betting strategy:  At first I may bet \$5.  If I lose, then the next hand I'll bet \$10.  If I win, then my total amount of money is as if I won the first \$5 hand.  If I lose again, then I'll bet \$20, then \$40, then \$80, and so on until I win my original \$5.   As long as I had enough money to keep going, then even if I got on a bad streak I can ride it out to at least break even in one good hand.   Have you heard of anyone else using this strategy?  I thought of it myself, but I'm sure others have and it's probably in hundreds of blackjack books.   Thank you,   Jonathan Congratulations Jonathan!  You are the one billionth person to re-invent the Martingale System!  And no, it doesn't work.   Eventually you WILL lose enough hands in a row that you will either exceed the table maximum or run out of money.

 Ed,   Found your website while doing a search on probabilities.  I have a question for you.  Recently, after a night of playing poker with the guys, one of my friends offered this proposition:  From a shuffled deck, he said a Queen always appears next to either a Jack or a King.  He offered even money on this bet.  He was taken up on it and won ALL three times we tried it!  After seeing it occur all three times, I suspect he had an edge.  Can you confirm if he did and if so, by how much?  Thanks. Rick This is another one of those problems I just couldn't do without a computer simulation. I wrote a program to simulate this and ran it 10,000,000 times.  It appears as if a Queen is next to either a Jack or a King (or any chosen card next to any other two chosen cards) 75% of the time!  Your friend had a whopping 50% edge.  To make it fair your sneaky friend should have offered 3 to 1 odds!

 What is the probability of rolling 3-of-a-kind in Farkel? First, I'm going to assume you mean Farkle, an application you can play on Facebook.  (I play Farkle all the time.) Second, I'm going to assume you meant what is the probability of rolling 3-of-a-kind WHEN ROLLING ALL SIX DICE. Obviously, if you have just three dice left to roll, and what to know what the probability is of rolling a 3-of-a-kind is, the answer is going to be different than if you had 4, 5, or 6 dice. Probably the easiest way to determine this (for me) is to write a quick and dirty program to loop through all possible ways to roll six dice, and then simply count how many of them contain a triple. So I did. There are exactly 46,656 ways (6 to 6th power) to roll six dice. (There are 36 different ways to roll two dice, 216 ways to roll three dice, etc.) Of those 46,656 different ways, exactly 14,700 of them contain at least one 3-of-a-kind. (In fact, some rolls contain two 3-of-a-kinds. For example, 1-1-1-2-2-2.) So the probability of rolling at least one triple, when rolling all six dice, is 14,700/46,656 or .315072 or 31.5072 %.

 What is the probability of drawing at least one ace, when drawing 13 cards?  Note: my deck is missing one non-ace.  So I am drawing from a deck of 51 cards. Um... okay. The odds of NOT drawing an ace on the first card (47/51) is multiplied by the chances of NOT drawing an ace on the second card (46/50) multiplied by then chances of NOT drawing an ace on the third card (45/49), multiplied by the chances of NOT drawing an ace on the fourth card (44/48) etc., etc., etc., all the way to the 13th card. (35/39) The product of all multiplying all 13 of these fractions is: .2953782 That's the chances of NOT drawing an ace. Subtract this from 1 to find the probability OF drawing an ace. 1 - .2953782 = .7046218. That's your answer. 70.46218%.   A computer simulation (50,000,000 runs) confirms the number.  (I get 70.45979% as an answer.)

 Bengals ML Falcons ML Chargers ML \$2,000.00 to win \$1,000.00.   Isn't this a 95% winner? Yes, all three teams are heavily favored over their opponents, but I don't know how you come to the conclusion this bet is a 95% winner. Assume each of your teams has an 87% chance of winning.  Collectively, that's only about a 66% chance of all three winning.  (.87 x .87 x .87 = .6585) Assume you make this bet 100 times:    65.85% of the time you will win \$1,000    34.15% of the time you will lose \$2,000 \$65,850 - \$68,299 = - \$2,449 That's a negative EV.  You've lost money overall. Of course, if you believe each team has a greater than 87% chance of winning, you're okay.  And yet if the chances of one or more of your teams winning are less than 87%, you're even worse off. Trust me, it's the not great bet you think it is.  There's NO WAY all three of those teams win 95% of the time, either individually or collectively.

 Steelers over Browns Packers over Lions Jags over Rams Jets over Bills Pats over Titans   100 gets ya 120...What you think? Let's estimate each of your teams has approximately an 85% chance to win.  (90% is just too high... I don't think any of those teams win 9 times out of 10 in this spot, yet I'll concede that 8 times out of 10 seems too low.) If you had just one team, you would win 85% of the time.  (8.5 / 10) With each additional game you add, the chances of you winning all games are reduced. For example, if you had two teams, each with an 85% chance to win, you would win both of them 72.25% of the time.  (8.5 / 10  x  8.5 / 10) By the time you get to five teams, there's just a 44.37% chance that all five will win.  (8.5 / 10  x  8.5 / 10 x 8.5 / 10  x  8.5 / 10  x  8.5 / 10) At that percentage, the bet wins \$120 44 times (+\$5,280), and loses \$100 56 times, (-\$5,600) for a net loss of \$320 over 100 such bets. Of course, if you DO believe that each game by itself has, say, a 90% chance of winning, your chances of winning all five are 9/10 to the 5th power which is about 59%.  Then the bet becomes a positive EV bet.  (60 x \$120) + (40 x -\$100) Edit:  (For the record, the Bills upset the Jets that weekend, but the other four favorites all won.)

 Hi Ed, Will you please help me with a question about playing craps?  I am going to Las Vegas and will be playing craps with this system: When a 4 comes out I will be playing \$2.00 to win \$1.00 on the don’t 4, betting it will not come out again for a 2nd time. If a 4 does come out before the 7 then I will be playing \$10.00 to win \$5.00 on the don’t 4, betting that it will not come out for a 3rd time before a 7.  If a 4 comes out before a 7 then I will be playing \$50.00 to win \$25.00 on the don’t 4, betting that it will not come out for a 4th time before a 7. That means a 4 has to come out 4 times before a 7 for me to lose all the bets. What are my odds of winning with the \$50.00 bet? Is it your opinion that I am suffering from gamblers fallacy? Or would you say I am being an advantage player. Thank you for your opinion. John Hi John, Sorry for not getting back to you earlier. I've been busy in a chess tournament all weekend, out of town. I just got home. I know the basics of Craps, of course, but it's been a long time since I've played. It sounds like you're playing a Martingale type of system, where after you lose you increase your bet, hoping to recoup the losses from that bet, and more. EVERYONE, at one time or another comes up with this "idea" / "system," yours truly included. Often it's for Roulette, or Blackjack, but it can be used for any game, including Craps. As I understand it, after a point is established, in your example a 4, you are betting it won't happen again. Meaning a 7 will be rolled before a 4. If so, you have 6 "winning" rolls and just "three" losing rolls: Winning rolls: 1-6, 6-1, 2-5, 5-2, 3-4, 4-3 Losing rolls: 1-3, 3-1, 2-2 There are 36 ways two dice can be rolled, but for our purposes, the other 27 ways are meaningless. Only the 9 rolls above are important. So even though you have twice as many winners as losers, you have to lay 2 to 1 for this privilege. But that's the exact odds of the event occurring... 2-1... so there is no house edge or house percentage against you. 6 winning rolls, 3 losing rolls reduces to 2-1, of course. To answer your exact question, the odds of you winning that final bet is 2 to 1 in your favor... but again, you're laying 2 to 1 for the privilege. Before any of the runs start, the odds of you losing four consecutive times is small... approximately 1.2%. (3/9 x 3/9 x 3/9 x 3/9.)  But the dice have no memory... after three consecutive losses, the chances of losing that fourth time are right back at approximately 3/9, or .3333 repeating. If you're going to play any kind of a Martingale System, it's best to do it with a game that has a low house edge, or in your case, a zero house edge. So yes, in this respect you are an Advantage Player. Roulette Martingale Systems, for example, suffer from a 5.26% house edge... because that's the house edge in Roulette, and nothing can overcome this. 5.26% is terrible. But eventually yes, you will lose that 4th time in a row... and a 5th time in a row... and a 6th time in a row, etc. All Martingale Systems suffer from the same "problem"... many small session wins followed by the eventual one large loss, wiping out all of the wins. Many times you will walk away from the table a winner, having met your goal. But if you play long enough you will lose. As far as systems go, it's no better or no worse than any other type of system. But there is no way to bet, or manner of betting that will put the odds in your favor. In your case, you will break even in the long run... again, because there is no house edge, when laying 2 to 1 on a true 2 to 1 proposition. The one problem with the above, and correct me if I'm wrong, is that these bets are made AFTER a point is established. But you're not allowed to just walk up to a table and wait for a point to be established and do this.  You have to first bet on the pass or don't pass bet. Only then can you bet on the 4 point making it or not, and taking the odds. (True odds, in this case.) And that amount that you can bet on these rolls after a point is established is limited to that particular casino. (That's the triple odds and quadruple odds and ten times odds and stuff I can recall seeing on marquees.) Hope this helps. at least a little bit. Good luck in your Crap playing. May the dice be with you.  :) Ed

 Hi Ed. What is the probability of a "kill" taking place: Assume the following: A total of ten, twelve-sided dice are rolled... seven attacking dice and three defending dice. In order for a kill to occur, three (or more) 12s must be rolled from the seven attacking dice.  However, for every 1 that is rolled on the attacking dice, this cancels out a 12. For every 12 that is rolled on the defending dice, this also cancels out a 12 on the attacking dice. To clarify, there must be at least three 12s remaining on the seven attacking dice, after subtracting one 12 for every 1 that is rolled on the seven attacking dice, and for every 12 that is rolled on the three defending dice. The probability is exactly 0.00877970848425242.  This is verified by first a computer simulation, over 1 billion trials, and then simply by examining all possible ways to roll ten, 12-sided dice. The computer simulation, which takes just a few seconds to run, indicates the probability is .00878024. There are exactly 61,917,364,224 ways to roll ten 12-sided dice.  (12^10)  According to a program I wrote that examines each roll and determines if a kill occurred, (that takes several minutes to run), exactly 543,616,408 of these rolls have "three or more" 12s remaining.  543,616,408 divided by 61,917,364,224 is exactly .00877970848425242, which verifies the simulation.

 Ed. I'm devising a strategy to bankrupt casinos. What is the probability of the sum of a pair of six-sided dice equaling either 2, 3, 4, 9, 10, 11, or 12 at least once in ten rolls? There are 36 different ways two, 6-sided dice can be rolled.  There a total of 16 suitable outcomes and 20 non-suitable outcomes.  The chances of a non-suitable outcome happening with one roll is 20/36. The chances of a non-suitable outcome happening with two rolls is 20/36 x 20/36.(20/36)^10 subtracted from 1 (to find the chance of a suitable result)  is .997199246.  This verifies a computer simulation of of .997199058.

 Ed! Quick!  I need your help!  I need to know a two-digit number and a three-digit number that when multiplied together, gives a four-digit number, with the stipulation that each the digits 1 through 9 are represented just once in the nine total digits of the three numbers!  Can you help me? The only way I know of to solve this problem is to write a program to loop through all possible two and three digit numbers, multiplying them together, looking for the given criteria.  I did so and the program returns the following: 12 x 483 = 5,796 18 x 297 = 5,346 27 x 198 = 5,346 28 x 157 = 4,396 39 x 186 = 7,254 42 x 138 = 5,796 48 x 159 = 7,632

 Ed! Thanks. What about a two-digit number and a three-digit number that when multiplied together, gives a FIVE-digit number, with the stipulation that each of the digits 0 through 9 are represented just once in the TEN total digits of the three numbers? 27 x 594 = 16,038 36 x 495 = 17,820 39 x 402 = 15,678 45 x 396 = 17,820 46 x 715 = 32,890 52 x 367 = 19,084 54 x 297 = 16,038 63 x 927 = 58,401 78 x 345 = 26,910 Notice only one of the products is odd.  That might make for a good math puzzle.