Mathematics and Gambling Poker
- Page ID
- 218945
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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
Straight 3 of a Kind Two Pair 10's or Better Losing Hand
Expect Values for the Bonus Hand and the Contract Hand
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.
There are 4 possible royal flushes, hence the probability is
P(Royal Flush) = 4/2598960 = 0.000001539
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
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
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
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
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
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
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
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
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 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.
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