Probability and gambling math

That is just an obvious extension of the rule that log(a^b)=b×log(a). It is not worthy of any special term. I suppose the formula might be helpful in answering some questions about the probability of a succession of losses. For example, suppose a video poker player wants to know how many hands he would have to play, such that the probability of a royal drought is exactly 5%. The probability of a royal per hand in 9/6 Jacks or Better, with optimal strategy, is 0.00002476. The degree of certainty that at least one royal will appear is 95%. So, the number of hands in a 5% royal drought would be log(1-.95)/log(1-0.00002476) = 120,989.

However, you don’t need to use that formula to solve that problem. It could be set up as:

.05 = (1-0.00002476) ⁿ
log(.05) = log(1-.00002476)ⁿ
log(.05) = n × log(1-.00002476)
-1.301 = n × -0.000010753
n = 120,989 January 12, 2009

No. Using this actuarial table from the CDC (Centers for Disease Control), the probability of a 72 year old white male making it to age 76 is 85.63%. That is about a 1 in 7 chance of death. The survival rate can be found by dividing the birth cohort at age 76 of 57,985, by birth cohort at age 72 of 67,719, from the white male table on page 14. The table used is a called a "period life table," which assumes 2003 rates of mortality will not change in the future, and is the most commonly used kind of actuarial table. A perfectionist might want to use a 1936 cohort life table, but I don't think it would make much difference.

p.s. After posting this answer I have received several comments that my response did not take into consideration John McCain’s individual health situation. Working against him is being a cancer survivor. Working in his favor is access to the best medical care money can buy, he is obviously still in good shape mentally and physically for a 72-year old, and longevity, as evidenced by the fact that his mother is still alive. However, I never intended to factor in this information. It was Matt Damon who quoted actuarial tables, which is what I was referring to. All I am saying is that for the average 72-year old white male, the probability of surviving four more years is 86%. If forced, I would predict John McCain’s odds are even better than that. October 7, 2008

For two numbers, the answer is 1/3, and for n numbers it is 1/(n+1). I posted the solutions to my page of math problems, questions 194 and 195. March 4, 2008

Assuming that no numbers are skipped, the probability depends very little on the number of participants, as long as that number is fairly large. The greater the number of participants, the more the probability of at least one match will approach 1-(1/e) = 63.21%. November 4, 2007

If the probability of winning is 1/n, and you play n times, as n approaches infinity, the probability of winning at least once approaches 1-(1/e), where e = 2.7182818..., or about 63.21%. The exact answer can be expressed as 1-(999,999/1,000,000)^1,000,000 = 0.63212074. My estimate is 1-(1/e) = 0.63212056, which agrees to six decimal places. September 1, 2007

What are the odds a deck could be shuffled back into starting order, with either a totally random shuffling method or with a perfect rifiling shuffle, and how many times would it take? - Andrew from Pewaukee,WI

The probability of a random shuffle resulting in starting order is 1 in 52!, or 1 in 8.06582*10⁶⁷. If you did a perfect shuffle, in which last card was the first to come down, thus remaining last, it would only take 8 shuffles to be back to the starting order. If the 26th card was the first two come down then it would take 72 shuffles to back to the starting order. Nov. 22, 2005

Two people are playing rock paper scissors. It is presumed the game doesn't involve strategy. If you are playing 'best of 3', and player A wins the first round, what are the odds that player B will win the game? - Andrew from Pewaukee,WI

Player B would need to win the next two (not counting ties) so the probability is (1/2)*(1/2) = 1/4. Nov. 22, 2005

The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a standard deviation of $45. What is the probability that a randomly selected teacher earns more than $525 a week? I can't remember how to calculate a probability from just the mean and SD without the population. - Sue from Queen Creek

That would be $35 above average, or 7/9 standard deviations. The probability of being more than 7/9 standard deviations above expectations would be 1-Z(7/9) = 1- 0.78165 = 0.21835. Nov. 2, 2005

I remember that if 22 people are in a room the odds are even that 2 will celebrate the same birthday [month and day, not year]. I have forgotton how to do the math to prove this. Could you please provide it. Thanks, Dean from Bainbridge Island, WA

I think I have answered this before but the 50/50 point is closer to 23. To make things simple let's ignore leap years. The long answer is to order the 23 people somehow. The probability that person #2 has a different birthday from person #1 is 364/365. The probability person #3 has a different birthday from persons #1 and #2, assuming they are different from each other, is 363/365. Keep repeating until person 23. The probability is thus (364/365)*(363/365)*...*(343/365) = 49.2703%. So the probability of no match is 49.27% and of at least one match is 50.73%. Another solution is the number of permutations of 23 different birthdays divided by the total number of ways to pick 23 random numbers from 1 to 365, which is permut(365,23)/365²³ = 42,200,819,302,092,400,000,000,000,000,000,000,000,000,000,000,000,000,000,000 / 85,651,679,353,150,300,000,000,000,000,000,000,000,000,000,000,000,000,000,000 = 49.27%. Nov. 2, 2005

Which do I have better odds of winning:
A. one shot at 1 in 4
B. five shots at 1 in 20
- Mike from Lansing

The probability of A is obviously 25%. The probability of getting zero shots out of five is 0.95⁵=77.378%. So the probability of getting at least one out of five is 100%-77.378%=22.622%. So A has the higher probability. Oct. 18, 2005

Would you tell me the probability of a 19% chance coming in exactly 18 out of 34 trials?

That would be combin(34,18)*.19^18*(1-.19)^(34-18) = 0.000007880052468. Sept. 4, 2005

Just for simplicity, let's say there are 322 cups on a table and one has a ball under it. What are the chances that I will pick the ball if i pick a cup 75 times (and the cups don't go away after I pick it up, it's always a random pick with 322 cups). At first I just thought to say 75/322, but I realized that was incorrect, as 322 picks does not result in a 100% chance of getting the ball because I could pick a million times and not get the ball. - John from Miami

Your answer would be correct if you removed cups after an incorrect pick. Since you leave the cups on the table each pick as a 1/322 chance of being right, or 311/322 of being wrong. The probability of 75 picks being wrong is (311/322)⁷⁵ = 78.6022%. So the probability of getting at least one correct in 75 picks is 100% - 78.6022% = 21.3978%. Aug. 21, 2005

Imagine a island that is inhabited by 10 people, and the politics is such that each day an islander is chosen at random to be chief for exactly one day; after the day has elapsed another islander is chosen at random (so the same islander who was just chief has a 1/10 chance of being chief again). The question to be solved: on average, how many days would have to elapse before each islander would have been chief at least once?

It will only take 1 day so that 1 person has served as chief. For the second day the probability of a new chief is 0.9. The expected number of days it will take to get a new chief, if the probability each day is 0.9 is 1/0.9 = 1.11. This is true for any probability: the expected number of trials until a success is 1/p. So after 2 people have served the probability of a new chief on the next day is 0.8. So the waiting period for a 3rd chief is 1/0.8 = 1.25 days. The answer is the sum of the waiting periods, which is 1/1 + 1/.9 + 1/.8 + ... + 1/.1 = 29.28968 days. March 10, 2005

How many eggs do you start with if each day you sell 1/2 the eggs plus 1/2 an egg; after 3 days you have zero eggs? At the end of each day, the number of eggs is a whole number.

Let's let d (for day) be the number of eggs at the beginning of the day and n (for night) be the number at the end. The problem tells us that d/2 - ½ = n. So, let's solve d in terms of n.

d/2 = n + ½
d= 2n + 1
So on the third day n=0, so d=1.
On the second day n=1, so d=3.
On the third day n=3, so d=7.
So there you have it, you started with 7 eggs. Jan. 16, 2005

If you have 30 people, all born in the same 365-day calendar year, what is the probability that any two of them will have the same birthday? Please explain the formula in your response. Thank you very much, Scott in Madison, Indiana

Think of the 30 people as lined up. The probability the second person doesn't match the first person is 364/365. Then, assuming they didn't match, the probability the next person does not match either of the first two is 363/365. Then keep going one person at a time. The overall probability no two people match is (364/365)*(363/365)*...*(346/365) = 29.3684%. It is often asked what is the fewest people you need for the probability of a match to be at least 50%. The answer is that with 23 people the probability of at least one match is 50.7297%. Jan. 16, 2005

I saw in the paper last week that the latest earthquake that ravaged Indonesia hit on December 26th. It also showed that of the eight deadliest earthquakes to hit over the last 100 years, three of them have been on December 26th. I was wondering what the odds are of having three massive quakes hit on the same day knowing these facts: Earthquakes of this magnitude (8.0 or larger) happen only once per year. The last big quake was exactly one year ago, 12/26/03 in Iran (back to back probabilities?) I look forward to hearing from you. Steve A., Fort Collins, Colorado

After discovering the claim that the Florida hurricanes only hit Bush voting counties was a hoax (see the October 17, 2004 column) I am going to be more skeptical about such alleged coincidences. According to the National Earthquake Information Center of the top 11 earthquakes since 1990 only the recent one of 2004 hit on a December 26. The Iranian earthquake you mention was only 6.7 in magnitude, which is far from making the top eight. Jan. 9, 2005

With five different toppings to choose from, how many different pizzas can you make, with any number of toppings?

There is 1 way with 0 toppings, 5 ways with 1 topping, 10 ways with 2 toppings, 10 ways with 3 toppings, 5 ways with 4 toppings, and 1 way with 5 toppings. So the answer is 1+5+10+10+5+1 = 32. Another way to solve is either topping can be used or not. So the total is 2⁵ = 32. Dec. 27, 2004

If a university's football team has a 10% chance of winning game 1 and a 30% chance of winning game 2, and a 65% chance of losing both games, what are their chances of winning exactly once?

If we assumed the games were independent then the probability of losing both would be 90%*70%=63%. But since you say the probability of losing both is actually 65% (which is more than the 63%), that means the two events are correlated. If the probability of losing both is 65% and just losing game 2 is 70%, then the probability of winning game 1 and losing game 2 must be 5%. Using the same logic the probability of losing game 1 and winning game 2 must be 25%. That only leaves 5% for winning both games. So the probability of winning exactly once is 25%+5% = 30%. Sept. 23, 2004

I am about to take a professional licensure examination. The regulations provide that:

1. The examination shall consist of 7 subjects.
2. For each subject, 60 multiple choice questions shall be asked.
3. Each multiple-choice question shall have four possible answers, but only one correct answer.
4. In order to pass, an examinee must obtain a general average of at least 75% and must not have a grade lower than 65% in any subject.

My question is, if an examinee merely guesses all his answers, what is his chance of passing the exam? Stated differently, what is the probability of passing the exam by sheer luck?

To satisfy the 75% requirement the student must get at least 315 out of the 420 questions right. The expected number of correct answers from guessing is 420*0.25=105. The standard deviation is (420*0.25*0.75)^0.5 = 8.87412. So the candidate must exceed expectations by 210 questions, or 210/8.87412=23.66432 standard deviations. The probability of doing this is way off the charts. If every living thing on earth took this test, answering randomly, I doubt anyone or anything would pass. I won't even get into the other requirement. Aug. 23, 2004

In a St. Louis Post-Dispatch article, the reporter says, "A 500-year flood is one that has a 1-in-500 chance of happening in any given year. Stated another way, that would be a 1-in-10 chance of happening over 50 years, or a 1-in-5 chance of happening over a century." After reading through all your gambling pages, I believe this is not a correct way to put it, right? Extrapolating their assertion, it would mean that there is a 1-in-1 chance that a flood will occur every 500 years, and that can't possibly be right.

You are right, that article is incorrect. The probability of a 500-year flood in a period of x years is 1-e^-x/500. So the probability of at least one 500-year flood in 50 years is 9.52% and in 100 years is 18.13%. Aug. 7, 2003

Could you please help me understand why it's a "fallacy" that a win becomes more probable after a series of losses? The way I see it, since the expected outcome is an approximately 99.5% return, then if after 1000 hands, you're at 78%, then, by definition, it would necessitate that the next hand being a win must be more likely to occur. People say that cards "don't have a memory", but isn't the natural curve, in essence, its memory? Please help me understand this point! Thanks a lot. - Steve From Canada

The bell curve is a forward looking estimate of the sum of many random variables. You can not mix together past and future events. Once an event has happened it is no longer a random variable but a cold hard outcome. If you played 1000 hands of blackjack with a return of 78% then you fell on the tail end of the bell curve during that play. Starting from hand 1001 your results could fall anywhere on a new bell curve. I hope this helps, but it really takes a course in statistics to truly understand. Mar. 6, 2002

My Grandmother was born on October 28, 1912, she recently passed away on October 28, 2001 (her 89th birthday). My cousin asked me what the statistical odds were of this occurring. I know there is about a 1 in 365 chance of dying on any day throughout the year. But what are the odds of that day being a person's birthday? - Loren Thompson from Petersburg, Alaska

You should have asked me this while I was still an actuary at the Social Security Administration. I could have easily have done a nationwide query on death records. I would say the answer is close to 1 in 365. It probably is a little less because infant mortality rates are disproportionately high after birth. For births in the year 2000 the probability of death within the first year is 0.71% of male infants and 0.59% for female infants. In other words those infant deaths are unlikely to occur on the birthday because once the first birthday arrives the child is outside of the danger period. Also, and I don't know this to be a fact, but on the show 'Six Feet Under' they said that the business for funeral directors picks up in January, evidently because people try to hold on for just one more Christmas holiday and then let go. This same logic might apply to reaching a birthday. Consider George Burns, he died 48 days after his 100th birthday. Nov. 11, 2001

If multiple choice questions have possible answers of a,b,c,d,and e: what is the probability that among 100 guesses, there will be at least 25 correct responses? - Daniel Gould from Portales, USA

The probability of getting exactly x right in your example is combin(100,x)*(1/5)^x*(4/5)^(100-x). To get the exact answer you have to calculate this for all values of x from 0 to 24, add them up, and take the difference from 1. The answer is 13.14%. April 15, 2001

I'd really like to know how to read odds like 12 to 1, or 3 to 2. Which one shows the best chances of winning? 12 to 1 or 3 to 2? - Louis-Philippe from Montreal, Canada

I don't like using probabilities in that form but they are generally used in this kind of syntax, "The odds against drawing a royal flush are 649739 to 1." That means there are 649739 ways you can't draw a royal flush and 1 way you can. In your examples 12 to 1 is a probability of 1/13, or 7.69%, and 3 to 2 is 2/5, or 40.00%, so the 3 to 2 is the better chance of winning. April 15, 2001

I have been mulling over some money management techniques, and wanted to thank you for some sound advice (more to the point, probability theory) and it seems that win probabilities are based roughly on your starting bankroll. For example, estimating the probability of winning $100 on a $200 buy-in without going broke first. This is very helpful, but without doing the math (shame on me, I know) I get the feeling win limits should be based more on your betting unit, i.e. $1, $5, $10, etc. Basically the idea you will have smaller fluctuations over time using a smaller bet than a larger one. I guess my actual question is this: if I have a given bankroll (say $100) and a stated win limit (say $50), what (if any) betting unit would give the greatest possibility of success? I am thinking too small of a bet would give lower my chances of getting much above the mean and too large would run the risk of bankruptcy. Any advice or comments? - Scott Miller from Saline, Michigan

In your example, assuming a negative expectation game, the best bet size to maximize the probability of reaching your win goal is $50. In a positive expectation game the best bet size is as small as possible. The reason is that the more you play the more the house edge will grind you down, or the more you will grind the casino down if you have the edge. March 11, 2001

This is a question regarding fixed odds betting. If you say that the odds are 4 to 1 on something happening is that equivalent to saying the probability is 1 in 4 i.e. 0.25? If you consistently bet on 4 to 1 shots would you simply break even over time? Therefore could you not beat fixed odds betting by doubling up after every loss since you would expect a winner every fourth bet? - Calvin Broaddus from Long Beach, USA

If the odds against something are 4 to 1 then there are 4 chances it wonít happen and one chance that it will. So in this example the probability would be 1/5. It doesnít matter what the probability is, if the events are independent then the past does not matter. Jan. 20, 2001

Can you please explain what the term "Law of mathematical averages" mean? Thank you, and keep up the good work. ñ Dennis from Canada

I think what you are referring to is actually called the "Law of Large Numbers." This states that for a random sample of n random variables with mean x, that the sample mean x_n converges to x as the size of the sample approaches infinity. We can think of the outcome of a bet as a random variable. This law tells us that as the number of bets becomes very large the average result will get closer to the house edge. Dec. 2, 2000

Great site Mike! Often times I hear the word, "binomial distribution" being used in gambling. Can you explain to me what it means? Thanks in advance. -- Dennis from Toronto, Ontario

Thanks for the compliment. Any introductory probability and statistics book should give good treatment to the binomial distribution. Briefly the binomial distribution is the probability that any given number of events will happen given a specific probability for each event and a specific number of trials. Specifically if the probability of each success is p, the number of success is s, and the number of trials is n then the probability of s successes is p^s * (1-p)^n-s * combin(n,s). The combin function is explained in my section on probabilities in poker. For example suppose you want to know the probability that in 100 spins of a roulette wheel the number of reds will be exactly 60. According to the binomial distribution the probability is (18/38)⁶⁰ * (20/38)⁴⁰ * combin(100,60) = 0.003291. Nov. 11, 2000

Can it actually be true that what I experience has a statistical base? It seems to me that it takes a lot longer to win X number of chips that to lose the same amount (I only play blackjack). For example, if I start with 300 chips, it might take hours to double my money (my goal), yet I can lost that number in what seems like almost no time at all. Can this really be true? Also, do you have a rule of thumb about when to leave the table when you are winning? ñ Chris in Gaithersburg, Maryland

What you have experienced is likely the result of some very bad losing streaks. It may also be the result of progressive betting or mistakes in strategy. The basic strategy flat bettor should have a roughly symetrical expectation in terms of steep ups and downs, slightly favoring steep downs due to the house edge and 48% chance of a losing hand compared to 43% chance of winning. Personally I don't set win limits on myself but do set loss limits. Nov. 4, 2000

I disagree with a statement you made regarding random number generation in computers. While it is true that a sequence will appear and repeat after time, it is not true that this is unavoidable. The trick is setting a correct seed. If you are using a UNIX-based architecture, one method is to set the seed to the seconds passed since 1/1/70, which is a constantly updated variable inside the system. Since you are using Visual C++ and J++, they should reset themselves to some random seed each time they are run, but it would be wise to set the seed yourself during the program. I think you would be wise to set the random seed each time a new deck is 'dealt' to the current time on the machine or something similar. In this way, yes, you will be using the same loop of numbers, but at least you will be picking moderately 'random' points along the way, so as to not make it a total loop. ñ Joe B. from Pittsburgh, Pennsylvania

When using Visual C++ the seed is evidently always the same. If I give the program the same input the output will always be the same after a random simulation. It is my understanding that this is what Microsoft intended, so that experiments could be replicated exactly. Visual J++ is evidently different based on my games, otherwise the same hands would occur in the same order every time. I would be interested to know how to assign the seconds since 1/1/70 as the seed. Is this a function and if so what is the name of it? Sept. 26, 2000

Postscript: Since this writing I have a slower but much better way of calling random numbers.

How can I convert your probabilities into the x to y format? - Ralph Harpster from Turlock, USA

Saying the odds of something happening are x to y means that the event in question will happen x times for every y times it doesn't happen. To make the conversion let p be the probability of some event. The odds could also be expressed as (1/p)-1 to 1. Lets look at an example. The probability of drawing a full house in five-card stud is 0.00144058. This could also be represented as 693.165 to 1. Sept. 10, 2000

Q: I have read a couple of articles about 'Parrando's Paradox'. Is there a way that you could explain what is going on as it is extremely counter-intuitive that two losing games played in a certain sequence could add up to a winner - I thought I understood the mathematics of gambling / probability! I can see that there is a subtle link between games A & B as game B is dependent on capital that is affected by game A; I am unable to carry the logic any further. Does Parrando's Paradox have any implications for negative expectation casino gamblers? I doubt it but would love to hear it from someone with a greater understanding. - Gavin Parnell from Burt St Edmunds, England

A: Parrando's Paradox states that two suboptimal games of chance can show a long term gain if played alternately. However the games can not be independent of each other, which eliminates any comparison to casino games. To learn more there is a New York Times article here: http://oll.temple.edu/economics/Econ_92/Parrando/012500sci-statistics-paradox.html. July 30, 2000

Q: How can I determine the odds of flat betting ( no counting, no progressions , etc ) of being ahead in a negative game such as blackjack w/o counting with a .5% disadvantage after 45000 or so hands? Is it even possible? - Kevin

A: This is a typical question one might encounter in an introductory statistics class. Because the sum of a large number of random variables always will approach a bell curve we can use the central limit theorem to get at the answer. From my section on the house edge we find the standard deviation in blackjack to be 1.17. You won't understand this if you haven't studied statistics but the probability of being at a loss in your example will be the Z statistic of 45000*0.005/(45000^1/2*1.17) =~ 0.91. Any basic statistics book should have a standard normal table which will give the Z statistic of 0.8186. So the probability of being ahead in your example is about 18%. June 18, 2000

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