I've recently been teaching a few of my friends to play poker, and I've explained various facets of poker play to them, now I want to explain Expected Value.
It's not a simple concept and I don't expect them to understand it straight away, but I'm looking for a 'laymans terms' explanation, maybe with a very simple, but easy to understand analogy.
I have already found a number of different explanations, but none seem simple enough for beginners.
Say we have a $1 million raffle that has only two tickets: a winner and a loser.
I give you a random ticket and tell you that you MUST sell it. How much money should you sell it for? (I.e. how much is it worth?)
A. $1 million?
No, because it could just as easily be worth $0.
B. $0?
No, because it could just as easily be worth $1 million.
C. $0.5 million?
Yes. This is the expected value of the ticket.
Notice that the holder of the winning ticket will never actually get $0.5 million. They will either get $0 or $1 million. We just use $0.5M as the expected value. If you sell it for more than $0.5M, you made money on the deal. If you sell it for less than $0.5M, you lost money on the deal.
Note: We got $0.5 million by multiplying the odds of winning (0.5) by the amount of money you would win ($1 million).
You would use the same strategy in poker, where the "value of the winning raffle ticket" is the money in the pot, and the "odds of winning the raffle" are the odds of your poker hand winning.
Another way to think about this is from the point of view of the raffle creator.
Say I am giving away $1 million in my raffle again with two tickets. How much should I sell each ticket in order to break even?
Well, if I am giving away $1 million, I want to make sure that I am getting $1 million back, otherwise I will be losing money (if I sell it for less), or making money (if I sell it for more).
A. $1 million per ticket?
No, because then I will take in $2 million total (2 tickets * $1 million each).
I would make money in this case.
B. $0 per ticket?
No, because then I will take in $0 total (2 tickets * $0 each).
I would lose money in this case.
C. $0.5 million per ticket?
Yes, because now I will break even by getting back $1 million (2 tickets * $0.5 each).
When I tell my non-poker playing friends that I play poker they think I'm gambling - until I give them the following explaination.
Poker, played correctly, is not gambling in my opinion. And I think this is a good way to explain EV to non-poker players or new players. The idea in poker is to do two things - make correct decisions based on available information and to maximize your EV (at least make sure it's positive). If you do both, then you will eventually come out ahead. It's just a mathematical certainty. High end poker or WSOP tournament play poker is a skill-based competition that eventually focuses on the subtlties of decision making, reading players, throwing off your opponents ability to make the correct decision and EV maximization.
However, the above explaination leaves a puzzled look on the face of my non-poker playing friends. So I explain it this way:
When you go to the casino, do you think the casino is gambling? Certainly the casino customers/players are considered gambling. But no one will think the house is gambling. "Of course not, they have the odds in thier favor!" Bingo, step one complete. They know that the house has an edge and even though they loose a big hunk of cash every now and then (which is good for business, btw) they will eventually, over time, ALWAYS WIN. Once I get there, I explain that this same formula that poker player uses. And therefore, IMHO, is not gambling, but a game of skill. Gambling to me is constantly being on the short side of the odds and hoping to get lucky.
Then I explain that the poker player does some very basic math to determine if they have an odds advantage over another player based on thier cards and what they think the other player has. However, just because a player might be "behind" in a hand doesn't mean they are gambling. The other part of the decision is to see how much money/chips they stand to win as compared to their current odds of winning. If they have a 1 in 3 chance of winning (e.g. open end straight draw or flush draw), then as long as they are getting better than 2 to 1 from the pot, then it's a good decision to call a bet - it's not gambling. It's a positive EV. If you are ahead in the hand, make sure your "value" bet is big enough to not give your opponent, who is on a draw, the correct EV. Simple as that. Keep making good decisions like that and you will come out ahead in the long run. You might crash and burn occasionally and there is some "luck" involved, but poker is a game to be played over the long haul - like the casinos operate. Players who have a negative EV, like those that "chase" hands or are a "calling station", even though they will catch every now and then according to the odds, will eventually loose all thier money. Simple as that.
I have been successful with this explaination of EV, and poker strategy in general, to non-poker players.
I hope this helps.
Probability: the chance of a particular outcome.
More precisely, the probability of any given outcome is the ratio of all the favorable outcomes and every outcome that is possible. (so favorable / everything)
The probability of throwing 6 with a dice is exactly 1/6 because all the sides are perfectly equal, there are 6 of them but only 1 is favorable in this case.
The probability of randomly chosen card from a deck being an Ace is exactly 4/52 because every card looks exactly the same from the back, you have 52 of them and only an Ace (of which it has 4) is favorable.
Expected value is simply the product of the probability of something happenning AND the 'value' of that something. ( probability * value )
If i say I'll give you 100€ every time you manage to pick an Ace from a deck of cards, then your expected value is 4/52 * 100 = 7,69€. Why? Because you are virtually winning the 4/52nd part of the 100€s every time you pick a card.
You could say that probability is just theoretical, but the law of large numbers says (in plain english) that the more you try (approaching infinity, that is), the closer your results will get to your theoretical probability, or expected value if you will. That is quite mindblowing if you think about it, but also makes a lot of sense.
Just thought of something even more simple: Let's say there is only one card, face down: an Ace. Now I tell you that if you flip it over and it's an ace, then you get 100€. How much is your expected value? Of course it's 100€. Why? Because the probability of it being an ace is 1 (or 100%) your potential winning is 100€ so 1 * 100€. Now what if you had two cards face down, and only one is an Ace? Three cards face down? Four? Seventeen? What is your expected value in each case?
Say that we decide to bet on coin flips.
In the first case, I will give you $1 every time that it is heads, and you will give me $1 every time that it is tails. Simple logic will tell you that since 1/2 of the time I owe you $1 (heads) and the other 1/2 of the time you owe me $1 (tails) that if we flip the coin enough times, it will even out. In this case, the amount that you expect to win (on average) is $0.00. Now this will be wrong 100% of the time for one particular flip (someone always wins more than $0), but over time it will average out.
Now in the second case, I will give you $1 every time that it is heads, and you will give me nothing when it is tails. Again, simple logic will tell you that you should take this bet as many times as I am willing to offer it. Since 1/2 of the time you will win $1, and the other 1/2 of the time nothing, you end up winning an average of $0.50 each time that we flip the coin. That $0.50 which you win on average is your EV (expected value). Even though you never win exactly $0.50 (it is always $1 or $0), that would be the average win that you could expect to win in this situation.
Lastly, let's reverse the bets. I will give you nothing every time that it is heads, and you will give me $1 every time that it is tails. This means that on average, each time that we flip the coin, you will owe me $0.50. In this case, your EV is -$0.50 (or negative EV) and you probably won't want to do it very often. :-)
If you have a simple 6 side dice you can arrange a simple demonstration.
Tell your friends you'll give them 1$ in every throw if the result is 3-6, and they have to pay 1$ if it is 1-2.
Tell them you're going to play 5 rounds and what they expect in terms of money won / lost. Then play the game and compare the results to their estimation.
Intuitively they should perceive its EV+, even if the don't understand why. Then you can tell them even though most of the times they do win, it can happen that they lose some money, even if normally they'll be winning.
You can then repeat the exercise with 10, 20, 30 and 50 throws to show how every time it's more difficult for them to lose money. Then you can expand on how, the more you play, the better the chance of making money and compare that to poker... Even if a specific dice roll lost you money in the long run you expect to win money.
And additional advantage is you can compare what they expect to happen with what actually happens and explain concepts like deviation and probability. For example, in a 10 throw game the expected value is 3.33$ (mathematically)... But the actual result may be 2$... Then you can explain deviation and show how that "uncertainty" goes down as the number of throws go up.
I find a dice is complex enough, a coin flip is to simple, but not too much (as a 52 card deck) and most people have an intuitive understanding of the math and the probability involved.
First of all it's important to explain that EV is a concept rooted in the law of large numbers, and which poker players use to calculate risk and reward. I'll get back to this concept.
When calculating EV a poker player takes a few things into consideration: 1) The size of the pot 2) The probability of winning that pot 2) The size of the bet they're faced with
Let's say the pot is $100 and I'm on a flush draw after the flop. My probability of making the flush is 40%, which makes my current EV +$40 ($100 x 40%, facing a bet of $0).
However, my opponent leads out for $50. Pot size = $150. Probability of winning: Still 40% Potential loss: $50
In this scenario, my positive EV is calculated the same way ($150x40% = $60). However, we also must calculate the risk of losing the $50 by multiplying it by the Loss probability (-$50 x 60% = -$30). <-- Note the negative values
In this hand my overall EV is +$30 (the difference between Loss outcome and Win outcome).
What if my opponent had bet $100? The potential winnings increase, but so too does the risk. Here's how it plays out: Pot size = $200 Probability of winning =40% Bet size = $100 Probability of losing = 60%. ($200x40%)+(-$100x60%) = ($80)+(-$60) Yielding an overall, still positive EV of $20.
However, back to the theory of EV, this is not a calculation that says "Make the call-You'll win twenty bucks!!" Instead it means that if you were to find yourself in this situation a large number of times that, just like a series of coin flips will go closer and closer to a 50/50 heads/tails distribution, so too will your average takeaway per hand equate to $20. But for this one hand your 40% win probability will always mean that you will probably lose. EV helps you draw the line of when it's okay to proceed in a hand in which you know you're behind. Another way to put it is as a mathematical justification for drawing.
Obviously this doesn't take into account anticipated future bets, reverse-implied odds, re-draws, cash vs tourney scenarios, using EV to force your opponent to bad decisions, etc but seems like you/your friends aren't concerned about that depth at the moment.
this page does a very good job of explaining variance and expected value
Expected value is the amount that you can expect to win, on average, by taking a bet.
Something like this? A simple discussion of EV in gambling. http://youtu.be/rC0kiLqvG-Y
The notion of "expected value" in poker is a bit different from its concept in most casino games, because players place bets during play. In a heads-up pot-limit hold-em, if both players knew before the flop that the first player to act had pocket aces and the other had some other pair involving the same suits (but only the person with the second pair knew which pair), the player with the unknown pair could with the right strategy eke out an expected value over 50% even if their opponent knew their strategy and played optimally to counter it, and even though the player with aces could win 80% of the time merely by refusing to fold.
To understand how that could be, recognize that there are a number of outcomes for a hand if the player with aces doesn't fold. Among them:
Even though The player with a random pair may only be able to win about 17% of hands at showdown, the expected value from lose 17% of hands would be about 4.5x the initial pot. Losing hands will have a negative expected value, but most of them will be at lower stakes, and can be balanced out so as to give the player with the random pair an overall effective value of over 50%.
