Poker

In this discussion we will look at:

The Number of Possible Poker Hands

Probabilities For:  Royal Flush    Straight Flush    4 of a Kind    Full House    Flush

Expect Values for the Bonus Hand and the Contract Hand

Rules of Let It Ride

There are many variations of the game of poker, most too difficult to analyze in this informal discussion.  We will focus on one of the simplest games, Bonus Let It Ride.  Bonus Let It Ride is basically five card stud with three opportunities to make a wager per game.  Each player places three equal bets in place and may palace a \$1 side bet called the "bonus."  After looking at the first three cards, the player decides if he has a chance of making a good poker hand.  If not the player may take one of the bets back.  Then the dealer turns over one more card.  The player then can decide to take the second bet back.  The third bet and the bonus bet are wagered whether the hand looks promising or not.  The dealer then turns over the fifth card and the payoff for the three normal bets is a follows:

 Hand Regular Bet Bonus Bet Royal Flush 1000 to 1 \$20,000 Straight Flush 200 to 1 \$1,000 4 of a Kind 50 to 1 \$100 Full House 11 to 1 \$75 Flush 8 to 1 \$50 Straight 5 to 1 \$25 3 of a Kind 3 to 1 \$4 Two Pair 2 to 1 \$3 Pair of 10's or Better 1 to 1 \$1

It turns out that the mathematics behind the bonus bet is the easiest.  Since there is no decision making other than whether to play or not, we just need to compute the expected value.

The Number of Possible Poker Hands

We need the probabilities of each of the winning hands.  First we find out how many total hands there are.

There are 52 choices for the first card, 51 for the second card (it cannot be the same as the first card), 50 for the third card, 49 for the fourth card, and 48 for the fifth card.

The number of ways of selecting is

(52)(51)(50)(49)(48)  =  311,875,200

We have over counted since different orders of the same five cards represent the same hand.  We now calculate how many orderings of 5 cards there are.  Think about putting the five cards in five slots.  We have five slots to choose for the first card, 4 for the second, 3 for the third, 2 for the  fourth, and 1 for the fifth.  Hence the number of orderings is

(5)(4)(3)(2)(1)  =  120

Now to find the number of distinct possible hands we divide

311875200/120  =  2,598960

Now lets find the probability of each of the hands.

Royal Flush

There are 4 possible royal flushes, hence the probability is

P(Royal Flush)  =  2598960/4  =  0.000001539

Straight Flush

Notice that there are four suits for the straight flush.  For each suit, the low card can be led by the A,2,3,4,5,6,7,8, or 9.  Hence there are 9 straight flushes per suit.  The total number of straight flushes is

(4)(9)   =  36

Hence the probability of a straight flush is

P(Straight Flush)  =  36/2598960  =  0.00001385

4 of a Kind

There are 13 four of a kind's to choose from.  For each of these we have 48 cards to choose from for the singleton.  The total number of 4 of a kinds are

(13)(48)  =  624

Hence the probability of a 4 of a kind is

P(4 of a kind)  =  624/2598960  =  0.000240096

Full House

Their are 13 different choices for the three of a kind and 12 choices for the pair.  For the 3 of the kind, there are 4 ways for each number.  Consider selecting three queens.  This is the same as not selecting one of the queens,  There are 4 choices of the queen not selected.  Once we have selected the card value for the pair, we need to select the two suits of that value for this pair.  We can list the choices

SH, SD, SC, HD, HC, SC

where S, H, D, C represent spade, heart, diamond, and club respectively.  Hence there are choices for each pair value.  Putting this all together, we can find the total number of possible full houses there are.

(13)(12)(4)(6)  =  3744

Hence the probability of a full house is

P(Full House)  =  3744/2598960  =  0.00144058

Flush

There are 4 choices for suits.  Once a suit is selected we choose 5 cards from that suit.  There are

(13)(12)(11)(10)(9)  =  154440

ways of doing this.  However, just with the calculation of the number of total hands, we need to divide by

(5)(4)(3)(2)(1)  =  120

We have

154440/120  =  1287

Now multiply by the 4 suits to get

(1287)(4)  =  5148

However, we need to subtract the ones that would be counted as a straight or royal flush.  We know how many to subtract, since we have already made these counts when we looked at straight and royal flushes.  The total number of flushes is

5148 - 36 - 4  =  5108

The probability of a flush is

P(Flush)  =  5108/2598960  =  0.0019654

Straight

First, the choices for the low value of the straight are A,2,3,4,5,6,7,8,9,10.  There are 10 of these values.  Each of the five cards for the straight could be any of the four suits.  Hence there are

(4)(4)(4)(4)(4) =  45  =  1024

of these

Multiplying by 10 gives 10240.  However, we need to subtract the ones that are all of the same suit since these would be counted as a straight or royal flush.  We know how many to subtract, since we have already made these counts when we looked at straight and royal flushes.  The total number of straights is

10240 - 36 - 4  =  10200

Hence the probability of a straight is

P(Straight)  =  984/2598960  =  0.0039246

3 of a Kind

There are 13 choices for the value of the triple.  As we mentioned with the full house, once we have declared the value, there are 4 choices of for the given suits.  That leaves 2 junk cards.  There are 48 choices for the first junk card.  For the second junk card, we can choose between the 44 remaining cards that do not give us a full house or 4 of a kind.  We calculate

(13)(4)(48)(44)  =  109824

We now must divide by 2 since there are two ways of ordering the two junk cards.  Hence the number of 3 of a kinds is

109824/2  =  54912

Hence the probability of a 3 of a kind is

P(3 of a Kind)  =  54912/2598960  =  0.021128

Two Pair

First there are 13 choices for the first pair value and 12 choices for the second pair value.  Since the order of the pairs does not matter, we divide by 2 to get

(13)(12)/2  =  78 different pairs

Now as in the calculation for the full house, there are 6 choices of suits for each pair.  There are 44 choices for the junk card.  Putting all this together we get the total number of two pair hands is

(78)(6)(6)(44)  =  123552

Hence the probability of getting a two pair hand is

P(Two Pair)  =  123552/2598960  =  0.047539

Tens or Better

First we can choose 10,J,K,Q, or A for our pair.  There are 5 of these choices.  As with the full house calculation, there are 6 choices of suits for this pair.  There are 48 choices for the first junk card, 44 for the second junk card, and 40 for the third junk card.  We have to divide by the 6 ways of ordering the three junk cards.  Putting this together, we get that the total number of pairs with tens or better is given by

(5)(6)(48)(44)(40)/6  =  422400

Hence the probability of getting a pair of tens or better is

P(Tens or Better)  =  422400/2598960   =  0.162527

Losing Hand

To compute the probability of a losing hand we just subtract the total of the winning hands from 1:

P(Losing Hand)  =  1 - 0.000001539 - 0.00001385 - 0.000240096

- 0.00144058 -  0.0019654 - 0.0039246

- 0.021128 - 0.047539 - 0.162527

=  0.76122

The Expected Value

The table summarizes our findings

 Hand Royal Flush Straight Flush 4 of a Kind Full House Flush Straight 3 of a Kind Two Pair Tens or Better Losing Hand Probability 0.000001539 0.00001385 0.000240096 0.00144058 0.0019654 0.0039246 0.021128 0.047539 0.162527 0.761212 Winnings \$20000 \$1000 \$100 \$75 \$50 \$25 \$4 \$3 \$1 \$-1

To compute the expected value, we just add the products of the winnings and the probabilities:

EV  =  (20000)(0.000001539) + (1000)(0.00001385) + (100)(0.000240096)

+ (75)(0.00144058) + (50)(0.0019654) + (25)(0.0039246)

+ (4)(0.021128) + (3)(0.047539) + (1)(0.162527) + (-1)(0.761212)

=  0.001512

Since this number is very close to 0, you can expect to break even with this wager.  In fact since it is slightly positive, it is in your best interest to always play the Bonus whenever you are playing "Let It Ride."

The expected value for the bet that the player has no choice about, the result is not so great.  Suppose that there is \$1 wagered for this bet then the using the odds for the regular bet, we can compute the expected value in a similar way.  We have

EV  =  (1000)(0.000001539) + (200)(0.00001385) + (50)(0.000240096)

+ (11)(0.00144058) + (8)(0.0019654) + (5)(0.0039246)

+ (3)(0.021128) + (2)(0.047539) + (1)(0.162527) + (-1)(0.761212)

=  -0.3727

This bet loses on average about 37 cents per wager.

The mathematics of the strategy for when to let the other two bets ride is much too difficult for this discussion.  Instead, I will refer you to the site at

http://www.conjelco.com/faq/misc.html#M6

With the optimal strategy, you can increase you expected value to a point where you can expect to lose on average only about 3.5 cents per hand not including the bonus.  Since the bonus pays you an average of about 0.2 cents per hand, with the bonus you can decrease your losses to about 3.3 cents per hand

Always decide how much you are willing to spend before you begin.  Never go beyond this number.  If you find yourself pulling out more cash to get back what you lost, then you have a gambling problem and should call the National Council on Problem Gambling at 1-800-522-4700.