and other

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.
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.
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?


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

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



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
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.



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,

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.


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.


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.


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.  :)




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.




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.




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




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.



Hello Edward,

I came across your site while looking for some help to a statistics assignment that has me all stressed out.  It is not all that difficult but I am having a hard time. I am taking virtual classes for college, this is my last semester to graduate, but my first doing online classes, since I just moved and I needed the flexibility. I miss having a teacher to ask questions to.  The system here is pretty weird - they just don't seem to care and when you have doubts it is kind of your own problem.  So here I am, reaching to you to see if maybe you can help.

This is the assignment:

Imagine yourself at a fair playing one of the midway games. Pick a game and calculate the expected value and post your results along with how you calculated them. For example, you may decide to throw a basketball to try to win a $10 bear. You paid $2.00 for three shots.  What is the expected value? (Please do not use this example in your answer)

And so I came up with this example (which if you honestly think it sucks, if you can help me change it, if needed, it would be great).
There is one game at the fair that you really want to try, the prize is cash! $500! So you approach the stand and ask how do you play this game in order to get the $500.
In order to win the prize you have to be able to insert 7 golf balls inside 7 cups displayed on a table, you have 3 tries (5 golf balls each try) and it costs $5 to get the 5 golf balls.
So, I need like a whole explanation, charts and calculations as I have to post this in a discussion forum and explain the whole thing. 

I would like to thank you in advance.

I hope to hear from you soon!


Hi Andrea,

I understand the question / problem.

The example you came up with, regarding the $500.00 and the golf balls, is not the type of example you should use.  Furthermore, I don't even like the example that was given in the assignment, regarding the $10.00 bear and having to pay $2.00 for three shots.

The reason why:  There is no way to determine the Expected Value, because it depends upon too many unknown factors.  I'll explain more about this more in a moment.

I THINK what your instructor / the assignment is looking for is something similar to the following example:

Imagine a coin flipping game.  It costs $1.00 to play the game.  The rules are simple:  Flip a coin three times and if you flip three heads in a row, you win! 

That's it!  If you win, you get back $10.00 (Meaning you will "gain" $9.00.)

What is the Expected Value (EV) of this game?

For this type of a game, the Expected Value can be determined EXACTLY.  And it's very easy to do so.

First we have to determine the odds of winning.  The odds of flipping a coin three times, and having it land on heads each time, is 1 chance out of 8, which can be expressed as 1/8, or 7 to 1 against, or as a probability, which is .125.

(There are eight different results that can occur if a coin is flipped three times:  H-H-H, H-H-T, H-T-H, H-T-T, T-H-H, T-H-T, T-T-H, T-T-T, and only one of these is H-H-H.  That's how we arrived at 1 chance out of 8.)

Thus, I can expect, that every 8 times I play this coin flipping game, on the average I will win one time. 

That's what I can EXPECT.  We both know, of course, that if I play maybe I will get lucky and win more often, or maybe I will be unlucky and lose more often than that.  But that's the expected number of wins.  That's the probability of me winning... .125

The 8 tries will cost me a total of $8.00 in fees, but I will get back $10.00 one of those times.  I will make a profit of $2.00 every eight times I play.  Thus, the expected value for this game is + 25 cents!.  The formula to determine this is easy.  (There are different methods to compute it, but I will give you just one method, my favorite.):


Expected Value = (Cost to play the game x the chance of losing) + (the "gain" if you win x the chance of winning)

That's it.

In our example it would be EV = ($-1.00 x 7/8) + ($9.00 x 1/8) which is +$0.25.

Thus, this coin flipping game would be areat game for the player.  It would be a terrible game for the operator.  This game would be known as +EV and every gambler searches for these types of games.  Every time the game is played, the person playing can expect to earn a quarter!  That's what "expected value" means.

You can see the math works.  Winning a quarter each time means my dollar turned into $1.25 each time.  And $1.25 x 8 games = $10.00, the same exact amount as before  As I said above, the 8 tries will cost me a total of $8.00 in fees, but I will get back $10.00 one time.

So 25 cents, or +.025, is the expected value of this game.

Here's further proof the formula works.  Maybe you're familiar with the casino game of roulette.  A quick Google search will tell you the house percentage for roulette is .0526.  That's the house edge, so expressed differently, the percentage against the player is - .0526.

Question:  How was that house edge determined?

Answer:  With the same formula we used above.

When you select a roulette number, the chances of winning are 1/38.  (There are 38 slots on the roulette wheel.)  So the chances of losing are 37/38.  When you win, the payoff is 35 to 1.  That's all we need to determine the Expected Value:

EV = (-$1.00 x 37/38) + ($35.00 x 1/38) which is -$0.0526.

Again, notice the answer is a negative number.  This means on the average, at the cost of a dollar to play, the player will LOSE 5.26 cents each time he she plays.  Because it's a negative number is how the casino makes its profit.

I said earlier I would explain more about why I didn't like the two games in your e-mail.  (The game you proposed and the game in the assignment example.  There is no way to determine the Expected Value.  There is no way to determine the exact odds of winning.

What are the chances/probability to sink a basketball in that basket?  Well, that depends upon many factors...  How far away the basket is, the size of the basket rim related to the size of the ball, the skill of the player, the weight of the ball, the amount of air in the ball, etc., etc., etc.  A pro-basketball player will most certainly have a better chance of sinking it than you or I, but there's just no way to determine the exact odds of that happening.

When you can't determine the odds of winning, you can't find the Expected Value.

The carnival/mid-way game operators probably can ESTIMATE the value, based upon all the people who have ever played at his booth.  If one person out of 20 wins, on the average, they can estimate and say it's 1/20.  And if that person is only going to win a $10.00 bear, then they know the game will be profitable for them. 

They certainly aren't going to offer a $10.00 bear if, on the average, one person out of TWO or one person out of THREE has been winning.  They wouldn't make any money at all that way, and wouldn't last very long as an operator.

Your golf balls example is the same.  What are the odds of sinking a golf ball in the cup?  Answer:  No one knows.  Too many unknown factors.  (Heck, maybe it's 0% if the size of the balls are LARGER than the size of the cups!)

Unless you state what the odds are of winning in the problem, then you can't determine the Expected Value.  (Unless the odds/probability is something that can be computed, like coin flips or dice rolls or roulette balls, etc.)

Now let's come up with a possible game for you.  You probably won't want to use this example, but it might give you some ideas. 

How about something with a deck of cards?  (Everyone is familiar with playing cards, so everyone will immediately understand the concept and the rules.)

Let's say it cost $3.00 to play our game.  You could really make it a lot more complicated, but for this example, lets say all the player has to do to win is to draw a king, a queen or a jack (any face card) from a shuffled deck of playing cards.  If they do that, they get back $13.00.  (Meaning, the win or "gain" $10.00.)

What is the Expected Value of this game?

Again, let's just use our formula.

The chances of drawing a face card is 12 out of 52, expressed as 12/52.  (12 face cards in the deck, and 52 total cards.)  The chances of NOT drawing a face card, therefore, is 1 - 12/52 or 40/52.

Expected Value = (Cost to play the game x the chance of losing) + (the gain if you win x the chance of winning)

EV = (-$3.00 x 40/52) + ($10.00 x 12/52) = $0.00

$0.00.   In this example, this is a fair game.  Neither side has an advantage.  The expected value is exactly zero.  On the average, I will win just often enough to get all of my money back that I paid to play.

Obviously, if the cost to play the game is
raised by any amount, or if the amount won when you do win is lowered, by any amount, the operator will then have an advantage.  The new Expected Value can be determined by just plugging in the new numbers.

If the
cost to play is lowered, or if the amount that is won is increased, then the player will have an advantage, and that Expected Value for that can be determined too.

You should now get 'A' on your assignment.




Ed, I think you're the only one who can help me.  I need an Excel spreadsheet.  I need an array that is 52 columns wide and 52 rows long. (A 52 x 52 array).  I need each row to contain each of the digits from 1 through 52 AND I need EACH COLUMN to ALSO contain the digits from 1 through 52.  That is, no number can be duplicated in a row OR any column.

I tried putting this together in Excel but I'm having problems.  Can you help me?

I'm an idiot.

I originally wrote a script to generate a random number.  After generating the number, I looped through all of the prior numbers generated in that row, and in that same column, looking for this same number.  If found, I tossed it out and tried again, generating another random number.

Actually, this method worked fine... until the final few rows when it took a long time to find the last few valid numbers.  And this was just with a test array of 20 or so rows.  It would have ran all night, at the rate I was going, to generate the size grid you wanted.

And then I realized how stupid I was.  I figured out an alternate way of doing it, that takes less than a second for my PC to generate:

Here's a link to your grid.  It's an Excel file.

P.S.   52 by 52... something to do with a deck of cards??  A different card needed in each row and column?



Ed, this isn't exactly a math question, but can you generate for me a text file, of every possible 5-card poker hand? 

Each line in the text file should be a different hand.  If you didn't know it, there are 2,598,960 possible 5-card poker hands, so this text file will have that many lines in it.  I need each card separated by a comma.  For example, the first few lines of the file might look like this:



No problem.  I wrote a short program to generate the text file.  You can find it here.  It is 40,609 KB in size.  For that reason, I recommend downloading this compressed file.  It is compressed with the wonderful 7z utility.  It is only 152 KB in size.