Understanding and Misjudging Probabilities
Many times the likelihood of a particular event turning out a certain way doesn’t always seem like possible at first glance. Do you know how to calculate the probability of a certain result? How do you think you can use probability in maximising your sports betting returns?
You cannot ignore the importance of value identification when it comes to successful betting in the long term. Anyone who doesn’t give ample weightage to identifying value while placing bets is more likely to lose his/her betting money in the long term. And you can’t correctly identify value if you don’t have a good understanding of the probability of a certain event happening. Furthermore, you also need to pay attention to the exact probability used by bookmakers for the same outcome, and if those bookmakers are right or wrong in their assessments.
Nevertheless, you may face a problem with this. Human mind is pretty complex in nature and it may often succumb to certain tricky situations, and the same happens with probability. Actually, it happens to a great extent.
About the Monty Hall problem
This phenomenon can be explained very well using the example of the Monty Hall problem. Let’s go over it in good detail now.
A game show host who’s completely unbiased, has strategically places a car behind one of the 3 doors at a demonstration place. While there’s a car behind one of these 3 doors, the other 2 doors have goats behind them. There’s no prior knowledge that can enable you to differentiate between these doors.
The host asks you to first point towards a particular door, and when you do, he says he’ll open one of the other 2 doors and reveal a goat. Once he has revealed that goat, he gives you the option to make the final choice between the door you had originally selected and the last remaining door. You stand to win whatever emerges from behind the door (a car or a goat). Let’s say you chose door number 1 at the start, and the host then picks up door number 3 to reveal a goat.
Now the important question here is, whether the probability of you winning a car would increase by selecting door number 2 or not? Or do you think that the probability would remain the same no matter which of the last 2 doors you select. Intuitively a large majority of us tend to believe that the probability doesn’t change whether we make the switch or not. Most of us assume that the probability remains 50% between the last 2 options.
However, mathematically speaking, the probability of revealing a car is only 33.3% if you don’t opt for a switch. In the similar manner, if you always opt for a switch, the probability of getting the right decision increases to a whopping 2/3 or 66.7%!
We must be kidding, right?!
To tell you the truth, no. There are several different ways in which you can show the possible outcomes in this scenario. In fact, you can visit the Wikipedia website, and look up ‘Monty Hall Problem’ to go through the article which talks about this problem and provides you with all relevant explanations at good length (https://en.wikipedia.org/wiki/Monty_Hall_problem). However, the simplest way you can put it is in the form of the following table which takes into account all possible arrangements involved in this experiment. As per this table, it is assumed that you always select door number 1. However, needless to say, it’s applicable to other door selections too. So, here is the table:
1 / 2 / 3
Car / Goat / Goat
Goat / Car / Goat
Goat / Goat / Car
Switch / No Switch
Goat / Car
Car / Goat
Car / Goat
Please note: In this table it is assumed that you pick door number 1 initially
Alright, but why does this happen?
An important point to remember while understanding this problem is that it’s important for the TV show host to not reveal the car. If the first door you select has a goat behind it (which may happen two out of three times), there will only be one goat left for the TV show host to reveal. Hence, you’ll end up with a correct decision 66.7% of the times if you decide to make a switch. What the TV show host does essentially is that he offers you more information. In the event that you pay no heed to that information and decide against the switch, it’d mean that you choose to stick with the same probabilities that were applicable before the host revealed any goat at all.
Another good method of understanding the Monty Hall Problem at a more intuitive level is by significantly increasing the number of doors in this scenario. Imagine a situation where there are 1000 doors and you need to pick just one. Once you do that, the TV show host then opens up 998 of the remaining doors that have goats behind them. All you’re left with in the end are 2 doors from the 1000 that you started with. What do you think? Wouldn’t it be better for you to make a switch in this scenario?! Or would you prefer going with your original choice?
The answer is anyone’s guess! You’d obviously make a switch!
About the Birthday Paradox
This is another excellent example of people often misjudging probabilities, and quite significantly on some occasions. Also commonly referred to as the Birthday Problem, this is about the probability of any 2 or more people in a certain group having their birthdays on the same day. What do you think would be the size of the group in order to make such probability touch 99% and 50% respectively?
The answer to this problem is as phenomenal as in Monty Hall Problem! In order to make the likelihood of any people sharing their birthday on the same date in a certain group (with none of them being twins) would reach 50% if you have 23 people in the group. You’d require 57 people in total to take it up to 99%.
Why you may ask?
What mostly happens in this problem is that people start guessing numbers which are way higher compared to the actual answer. This is because we’re prone to making wrong assumptions. We fail to realise that we are actually looking for the likelihood of any 2 people sharing their birthday on the same day. Taking one specific person from a group of 23 whose birthday needs to match with another member belonging to that group, the chances of this happening are only 22. In the event that you’re looking for the likelihood of any 2 members belonging to this group sharing their birthdays on the same day, you’d need to take into account 253 pairs in total (11 pairs x 23), which makes it easy to understand the real probabilities. If you wish to dig deeper into this, you can visit the Wikipedia website and look up the Birthday Problem and read a detailed article on this. However, you’d need to have some statistical background to understand it better.
About Hole-in-One gang
You should know that misjudging probabilities isn’t a problem which affects only the sports bettors. Many bookmakers get impacted by it as well, resulting in occasional exceptional value bets. Hole-in-One gang is a famous and excellent historical example of this. It comprised of two extremely sharp and crafty sports bettors known as John Carter and Paul Simmons.
In the year 1991, these two punters worked out the chances of any given golfer taking part in a tournament, scoring a hole-in-one. As you may be aware, this isn’t as unlikely to happen as many people think. In fact the actual probability of any given player scoring a hole-in-one is an impressive 50% for any particular tournament. These two punters travelled all over the United Kingdom and placed as many wagers as they possibly could (please note, there was no Internet in those times). Quite obviously, sports books were also pretty forthcoming in giving them major odds on such a bet. Their odds ranged anywhere from 4.00 to 101.00 (based on decimal odds). Simply put, they got exceptional value!
It’s pretty obvious that majority of those sports books were too lazy to study any statistics. However, you should also notice the stark similarity of these bets with the Birthday Paradox. Majority of the bookmakers who offered such odds were obviously depending on their intuition for quoting the odds. And as in the case of the Birthday Problem, committed the error of confusing odds of one specific golfer scoring a hole-in-one in a tournament with that of any golfer participating in the tournament to score a hole-in-one. This is exactly the result of not doing the math properly.
Eventually what happened is that 3 of the 4 golf Majors in 1991 witnessed hole-in-ones. Needless to say that Carter and Simmons won a good deal of money that year! In fact, they reportedly earned around half a million pounds in profits from that endeavour alone! It was a jackpot of sorts back in those times!
There are many valuable lessons you can learn from this. First and foremost is that human intuition is capable of playing cruel tricks at times. However, you should know that this isn’t necessarily a problem that’ll always hurt your betting endeavours. Provided that you place your bets strategically, the same problem can help you beat the bookmakers and score major betting profits!