Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**Fruityloop****Member**- Registered: 2009-05-18
- Posts: 135

Hello all!

I haven't posted in a while, but I've been working on a math problem and I'm kind of stuck. I wasn't sure whether to post this under Help! or not. I would like this to be a help/discussion. Anyways, I was at a casino and there is this game called four card poker. The player receives 5 cards from a standard 52 card deck and makes the best 4-card poker hand he can out of those 5 cards. The player also makes an initial 'ante' bet. The dealer receives 6 cards and makes the best 4-card poker hand out of those 6 cards. The extra card that the dealer receives is what gives the casino the advantage. Once the player receives his cards he can either fold and lose his 'ante' bet, or he can bet up to 3x his ante bet and continue against the dealer. If the player continues and wins, he gets paid 1-to-1 on both his ante bet and additional bet. It seems to me that the best strategy for the player would be to know which hand he receives that would give him a 50% or greater chance of winning. If his chances of winning are 50% or greater, then he should bet 3x his ante bet and continue, otherwise he should fold. Apparently there is a strategy that has been proposed by Stanley Ko in which he recommends...Raise 3x with pair of tens or higher...Raise 1x with a pair of twos to nines...fold all others. I really don't understand why somebody would raise 1x. So in order to figure out the best strategy for the player, we need to figure out the probability of the dealer winding up with various hands. Once I started doing the math, I began to realize that this is very, very complicated. Here's the ranking of the hands...

Four of a kind

Straight flush

Three of a kind

Flush

Straight (Ace counts both high and low)

Two pair

One pair

Nothing/High card

Here's what I have so far for the dealer.

These are the total number of ways for the dealer to receive these hands..

Four of a kind

Straight Flush

Three of a kind

Flush

Straight

Two pair

One pair

Nothing/High card

These total to 21,224,216, but they should total to _{52}C_6 = 20,358,520.

So I have nearly 1 million extra card combinations.

My mistake(s) is (are) probably in the high card/one pair/two pair totals since these are most complicated and difficult.

*Last edited by Fruityloop (2010-03-03 17:08:40)*

**The eclipses from Algol come further apart in time when the Earth is moving away from Algol and closer together in time when the Earth is moving towards Algol, thereby proving that the speed of light is variable and that Einstein's Special Theory of Relativity is wrong.**

Offline

**soroban****Member**- Registered: 2007-03-09
- Posts: 452

.

. .

. .

. .

. .

. .

. .

. .

. .

.

Offline

**Fruityloop****Member**- Registered: 2009-05-18
- Posts: 135

I've finally completed my analysis of Four Card Poker. The difficult part was figuring out the probabilities for the dealer to wind up with various hands.

The player initially makes a bet on the Ante spot. After the player receives his cards, he can either fold his hand and lose his ante, or he can bet from one to three times the ante bet.

If the player's hand ties or beats the dealer's hand the player wins and gets paid 1-to-1 on both his ante and raise bet.

If the dealer's hand beats the player's hand, the player loses both his ante and raise bet.

If the player gets Three of a kind, Straight Flush, or Four of a kind, the player receives a bonus based on the Ante bet as follows:

Three of a kind pays 2-to-1

Straight Flush pays 20-to-1

Four of a kind pays 25-to-1

The bonus pays regardless of the dealer's hand.

The dealer receives six cards and the player receives five, giving the casino an edge.

Hands are ranked as follows:

Fouf of a kind

Straight flush

Three of a kind

Flush

Straight

Two pair

One pair

High card

The Ace can count both high and low in making the player's or dealer's hand.

Here are the various ways that the dealer can wind up with various hands:

(Just divide by

Four of a kind

Straight Flush

Three of a kind

Flush

Straight

Now subtract out the straights which are also flushes

Phew!

Two pair

One pair

Subtract out the straights

Subtract out the flushes

Add in the straights which are also flushes

High card

The number of ways for the player to wind up with the hands are as follows:

Four of a kind

Straight Flush

Three of a kind

Flush

Straight

Two pair

One pair

High card

The best strategy for the player is to fold if he only has high card, bet 1x the ante for a pair of 2's through 9's, bet 3x the ante for a pair of 10's or higher.

We are now in a position to estimate the player's mathematical expectation.

In order to simplify the calculation, I ignored the situations where both the player and the dealer have the same hand.

High card

pair of 2's through pair of 9's

pair of 10's through Aces

Two pair

Straight

Flush

Three of a kind

Straight Flush

Four of a kind

Now we need to add the bonus which is as follows:

So,

So the casino has an advantage of on Four Card Poker.

According to James Grojean, the casino has an advantage of 3.396%.

So my analysis is very close.

*Last edited by Fruityloop (2010-04-19 17:58:42)*

**The eclipses from Algol come further apart in time when the Earth is moving away from Algol and closer together in time when the Earth is moving towards Algol, thereby proving that the speed of light is variable and that Einstein's Special Theory of Relativity is wrong.**

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Fruityloop;

Mama Mia! No wonder I haven't seen ya in a while. I am getting cross eyed just looking at it. Good work!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**Fruityloop****Member**- Registered: 2009-05-18
- Posts: 135

Thank You BobbyM!

Yeah, I've been working on this problem for about two months!

It is amazing that a simple game can have such complicated math associated with it.

**The eclipses from Algol come further apart in time when the Earth is moving away from Algol and closer together in time when the Earth is moving towards Algol, thereby proving that the speed of light is variable and that Einstein's Special Theory of Relativity is wrong.**

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Yea, tedious to do all the casework. And you know what, your problem isn't any easier with a computer.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline