Explaining EV to a derranged gambler

For someone who's a total degenerate gambling addict nothing's going to make a difference except multiple times going down to the felt and not just in terms of chips, I mean all money. Roulette is the best example because it's a simple game that doesn't rely on game play and magically thinking that the game/odds can be beaten through perfect gameplay. But you can simplify the game using lower odds to make it more understandable. How about rolling a 6 sided die with him about 50 times. He gets to bet on one of the numbers coming up. The probability is 1/6. But you can pay him 4 units everytime he wins and he keeps his laid bet. There's a 5/6 chance that he loses each roll. EV= -1x5/6+4x1/6= -16.67. It's clearly a 1/6 loss each time EV. A number small enough that someone who didn't go to school may understand. After 50 rolls at $1 each bet, he should have lost a little over $8. Have him sit down and roll the die 100 times, 200 times, etc. Show him the -EV. You can do it with flipping a coin too. But he'd have to bet 100 to win 80 to make sense, and that doesn't show EV as much as it shows vig on a sports bet. When it's all said and done you can explain to him that poker is the only game where you can bet with a +EV (some sports bets too, but that's a bit more complicated.) Then don't teach him pot odds or implied pot odds and I'll let you know when and where I'm going to play. He can think he's making +EV bets when he's not and I'll take all his money.

This might work. ^^^

Its hard to explain probability to college students, let alone someone with no formal education.

Going by your user name, Im guessing the problem is poker - which would be a whole different story though. Is this right?

A table has three poker cards on it, flipped face down. One ace, one king, and one queen. If he flips the ace, he wins $1,000. If he flips the king or queen he wins nothing. 2 in 3 times he will see zero dollars and 1 in 3 he will see $1,000. He can play as many times as he wants, but each time the cards are mixed up on the table. How much should he be willing to pay to play the game each time? Explain that every three times he plays he will see the ace once on average, or $1,000 in his pocket every three pulls. The most he would be expected to pay would be $333.33 each pull or he will be losing money over time. Might work to explain it that way - fairly simple, I think?

Actually, he could pay up to $499.50 each hand and still win money over time. Because for every two he loses he will still win $1 on every third hand. On average.

I assumed the fee to play was a sunk cost. In your scenario, it cost him 1498.5 to win 1000.00. A net loss of 498.5. You are talking about blackjack (or a similar table game) where he'd get his bet back plus the 1k, I think.

Fair enough.

A better question is, why would you want to? Someone has to take his money so it may as well be you!

posting a 2p2 question on styleforum = -EV

Look, the answer could have been read Skansky's Theory of Poker's chapter on deuce-seven triple draw, but I don't think that will explain it to anyone.

+ 1.

OP, stop playing like an idiot. Very -EV. If you're going to shove over someone's raise, at least make a real opening bet! Also, that's a bad idea in most live games. Fish don't reraise with anything less than a premium hand (except maybe at the hardrock in Vegas).