The Paradox of Chevalier De Mere and Prop Bets

Update: A big thanks to everyone has a made this a top search for Chevalier De Mere. We had fun working out this problem, and I hope that my explanation helps you to learn the key concepts as well. If you enjoy this entry, please pay it forward by visiting my other blog where we are truly trying to start a movement...thanks again

The Thursday night game brought up a prop bet that didn't sound right to me. When I first heard it, I knew something smelled fishy and that there was an edge that nobody had really mathematically worked out. So, because I can't let things go, here's the bet and the mathematical edge:

A guy tells you that, while watching the flop in hold-em he'll take three cards as his. Let's say he takes the jacks, sevens and fours. You take the rest of the deck. You run the flops and pay even money to whomever can hit.

A quick look at the bet determines that he's taking 12 out of 52 cards (23% of the deck) and you are getting 40 out of 52 cards (77% of the deck). Okay, so the edge is to you so far. I mean, if this was a prop bet on every river card, you'd clean this guy after a couple or rounds...

But the bet isn't on the river card. It's on the flop which includes three cards. And I think the reason this hustle works is two fold:

1. the guy gives off the appearance of being a maniac at the table;
2. crunching the numbers on the basic probabilities give the taker such an edge that people will assume they will keep a good edge even peeling of three cards on a flop

At any rate, dialogue ensued, and at first we were having trouble working out the math...

I personally started to work this problem by adding the probabilities of each event (independently) together. blowpar68 stopped me in my tracks because, using this logic, I'd have over a 100% edge if we were running out six or seven cards. This incorrect form of logic is better known as The Paradox of Chevalier De Mere.

Quick note here: The chance that at least one of the three cards here COULD be determined by adding the sum of the individual chances IF the outcomes were mutually exclusive. The problem is that they are not, so you are double-counting. So, I was on the right track, but just couldn't remember that my equation was double-counting. Luckily, an OM Rounder popped into the conversation in just enough time to save me...

Using AltronIV's approach was, by far, the quickest way to arrive at the correct probabilities for both outcomes. Rather than trying to determine the probability of your outcome OCCURRING, work out the math on your outcome NOT occurring and subtract from 100%.

Alas, here is the math:

To calculate the probability of three events occurring sequentially (all together) you need to multiply the probability of each event together. (P1 X P2 X P3) So, the probability that we will NOT hit a J, 7 or 4 is:

40/52 (76.9%) missing on first card
39/51 (76.4%) missing on second card
38/50 (76%) missing on third card

Multiplying them together we get:

76.9% x 76.4% x 76% = 44.6%

Therefore, 44.6% of the time we will miss a jack, seven or four on all three cards dealt on a flop. Subtracting this from 100%, we realize that 55.4% of the time we WILL hit one of those cards on one of the cards from the flop.

The edge is 5.4% to the guy who takes the jacks,sevens and fours...

There's the math, and I guess the moral is (as always), if a bet seems too good to be true, it usually is...

Damn Donkeys,
Bmore