House edge calculation

I'm trying to get my head around the elusive (to me, anyway) math of gambling, specifically blackjack. I'm interested, for starters, in the calculation of "house edge", stated as a percentage.

As a start along this long and painful road, consider Match The Dealer side bet. It pays 4 to 1 for a single non-suited match of one of your cards to the dealer's up card. I know that the odds of this occurring are 6.5 to 1 (13 possible cards with two chances for your first two cards. 13 / 2 = 6.5.

So what's the house edge? There are a lot of threads out there, on a variety of games, most more complicated than this one. But one that looks pretty straightforward is:

he = (true odds - payout odds ) / (true odds + 1)

Given the example above: he = (6.5 - 4.0) / (4.0 + 1) = .3333 = 33.3%

Is this true? I see so many people playing this side bet so often, hand after hand after hand, that I can't imagine the house edge on it is this huge. So this formula must be wrong. But I hope it is, since it's simple enough for me to begin this process, and it also gets me away from EVER, EVER, EVER playing Match The Dealer. I know that all the sidebets are horrible, but 33%? That's immoral.

Not sure if I understand this completely.

First of all, are you sure it's un-suited match? Or can it be suited?

It's not 2 in 13 or 1 in 6.5.

If you were dealt 13 cards, you wouldn't have a 100% chance of getting a queen, for example.

I'd say it's more like: (1/13) + (12/13)*(1/13) = 0.14792899408 or 14.79%

That's 1 in 6.76.


Every 6.76 rounds you WIN $4 and lose $5.76. Overall you lose $1.76 per 6.76 rounds (or $6.76 in action).

$1.76 / $6.76 = 0.26035502958 or 26.03%


I looked it up.

Quote: http://www.wsgc.wa.gov/activities/game-rules/match-the-dealer-blackjack.pdf

Blackjack
Win Frequency for each
Match The Dealer
bonus wager
Unsuited
Match
Payout
Suited
Match
Payout
House
Advantage
2 deck 13.2%, 1 in 7.6 hands 4:1 19:1 3.31%
4 deck 14.0%, 1 in 7.1 hands 4:1 12:1 4.84%
5 deck 14.2%, 1 in 7.1 hands 3:1 15:1 4.76%
6 deck 14.3%, 1 in 7.0 hands 4:1 11:1 4.06%
8 deck 14.4%, 1 in 6.9 hands 3:1 14:1 3.67% 

You're going to have to figure out how frequently you win 4x, 19x, and lose 1x.

I meant this as an example to demonstrate the formula. I don't think the situation is quite so complex.

There are 13 possible cards - A-2-3-4-5-6-7-8-9-10-J-Q-K.
The dealer's up card is some one of those.
You have two cards. They also must be, each, one of those.
So, the possibility of one of your two cards matching the dealer's up card is 2 in 13 or 1 in 6.5.
If one of your cards matches the dealer's, they pay you 4 times your wager.
4 (what they pay you) is less than 6.5,(the odds of getting the winning situation), so there is some advantage to the house.
What is the formula to convert that into a house percentage?

There are more complex situations where the odds would be different based on what cards have come out - maybe there are no more threes left in the deck - so the odds would be different.

But in a situation where the odds of an event are 1 in 6.5 and the payout is 4 times your wager, what's the house advantage as a percent?

Quote: racquet

I meant this as an example to demonstrate the formula. I don't think the situation is quite so complex.

There are 13 possible cards - A-2-3-4-5-6-7-8-9-10-J-Q-K.
The dealer's up card is some one of those.
You have two cards. They also must be, each, one of those.
So, the possibility of one of your two cards matching the dealer's up card is 2 in 13 or 1 in 6.5.
If one of your cards matches the dealer's, they pay you 4 times your wager.
4 (what they pay you) is less than 6.5,(the odds of getting the winning situation), so there is some advantage to the house.
What is the formula to convert that into a house percentage?

There are more complex situations where the odds would be different based on what cards have come out - maybe there are no more threes left in the deck - so the odds would be different.

But in a situation where the odds of an event are 1 in 6.5 and the payout is 4 times your wager, what's the house advantage as a percent?

AFAIK, there isn't some "one size fits all" formula.

If something happens 2 in 13 times and you're paid out 4:1 (4 to 1), then you determine how often you win, multiplied by how much you win, then subtract how frequently you lose and the amount you lose, divided by the total amount of money in action.

ie: Wagering $1 at a time........You win 2 wagers out of 13. Each wager wins you $4.

2*$4 = +$8.

11 wagers out of 13 are losers. Each wager loses you $1.

11*(-$1) = -$11


Now add together:

+$8 - $11 = -$3


Divide by amount of money wagered:

-$3 / $13 = 3/13 = -0.230769, or about 23% HE.


Of course, this isn't applicable to the side bet you're talking about, since not all winners pay 4:1, and the odds of winning are NOT 1 in 6.5 (or 2 in 13).

An easy way to check your work is to google "blackjack side bet match the dealer" and then click the wizards links..

https://wizardofodds.com/games/blackjack/appendix/8/

6 decks = 4.06%
8 decks = 3.67%

Also, really study the tables that result in this analysis and you'll see a repeat of variables every time the Wizard calculates the house edge on something... Event, Pays, Combinations, Probability, and then the Return (Pays * Probability). Sum up all of the different events that could happen and you'll get your house edge. I like this because you can see where the bulk of the return, and the bulk of the house edge, comes from.