# Question on Odds, Pot Odds and Expected Values

I am reading about Odds, Pot Odds and Expected Values. As I understand Odds, they are a way to compare how many times you win vs how many times you lost. For example flipping a coin gives you odds of 1:1 and throwing a dice of 1:5 for one specific side. But in the Article Pot Odds and Expected Value they give two examples:

1. Example with the nut flush draw:

You have the nut flush draw (nine outs) on the flop and the pot is $4. Your opponent bets$1. There is now $5 in the pot ($4 + $1), and it is$1 to call. The pot odds are therefore 5:1.

According to the chart above, your odds are 4:1 to hit your flush draw. The pot odds are higher. You should therefore call.

You can see why this call is correct by looking at the long-term picture. If you make this call four times, mathematics says that you will hit your draw once. That means you will win $5 for every$4 (4 * $1) you invest. That is good business. Here they interpret Odds of 4:1 to mean "If you make this call four times, mathematics says that you will hit your draw once", which is quite opposite to my interpretation above, so what interpretation is correct, am I wrong or has the article an error? Also in the second example they wrote for Odds of 10:1 that "In this instance, you would need to play ten times in order to win$30", also an opposite interpreation to mine? Could someone please explain?

The article is correct in the way it uses 4:1 and 5:1. Under their assumptions (actual value given their example is more like 4.2:1), you are "4 to 1" to make it while you are getting "5 to 1" on your money. I'd say that this is precisely because both are written / pronounced / thought of this way that it's convenient.

If you check the Wikipedia article on odds, you'll see that they're explaining it like in the article you quoted:

en.wikipedia.org/wiki/Odds

For example, a probability 0.20 is represented as "4 to 1 against" (written as 4-1, 4:1, or 4/1), since there are five outcomes of which four are unsuccessful.

However the article you quoted is not correct or, at least, poorly worded. The article talks about what happens if you play 4:1 odds (doing an EV+ call by getting 5 to 1 on your call) and says that, mathematically, you'll hit your draw once if you run four deals like that.

Odds of 4:1 corresponds to a probability p = 0.20.

Well, if you play four deals like with p = 0.20, there's still a p = 0.4096 (40.96%) that you lose the deal all four times (0.8 exp 4 gives 0.4096).

So you "only" have (1 - p), which gives 0.5904 (59.04%) chance you'll win at least one of the four deals.

Now it is correct to say that you should always (at least) call when you have explicit odds in your favor: this is immensely EV+. (let alone the implied odds and discussions about fold equity if you are the one pushing all in semi-bluffing your draws but I digress)

But it is not correct to say that "If you make this call four times, mathematics says that you will hit your draw once". That's not what probabilities says.

Probabilities says that with p = 0.20 and four runs you have:

• 40.96% chance to lose all four deals
• 40.96% chance to win one the four deals
• 15.36% chance to win two of the four deals
• 2.56% chance to win three of the four deals
• 0.16% chance to win all four deals
• Thanks for your detailled answer! Why you deleted your further explanations? I liked them... and even had a question on them. There you said your average win per deal is 0.8 units from the stack, meaning if I play four times I earn 3,2 units, was that right? Cause seems quite high to me with 4:1 odds and 5:1 pot odds? – StefanH Mar 23 '14 at 19:52
• @Stefan: Indeed, I noticed I did a silly computation mistake, leading to values being to high (which you noticed too). So I deleted the buggy part of my answer and will repost it when I find the time to post the correct values : ) (probably tomorrow or so) – TacticalCoder Mar 23 '14 at 20:02

Since 4:1 are the correct odds for that scenario, it seems that the sentence "If you make this call four times, mathematics says that you will hit your draw once." was a mistake, rather than an error. Their intention was probably to say that for every four times you lose you'll win one, or for every five times you play you'll win one. In any case, your understanding of the concept "odds" is correct.

To be precise, the probability of hitting an out is 9 / (52 - 2 - 3) = 0.19 that expressed as odds is 4.22:1.

Also worth mentioning this:

Your opponent bets $1. There is now$5 in the pot ($4 +$1), and it is \$1 to call. The pot odds are therefore 5:1.

is a simplification and it's not an entirely valid number to compare to your cards odds. To calculate the actual pot odds you'd have to predict if the players after you will call or even reraise. That is, you need to predict the odds of the total amount you'll end up contributing to the pot in the current street to the size of the pot at the end of the street plus the possibility that you collect the pot if you reraise and the rest of the players fold (if it's you who folds then there's nothing to calculate).