# Probabilities for Fun and Profit

So it gets around after a time that I’m pretty good with probabilities. Strangely, the management at the casino I work at has asked me to start giving them “the answers” to various problems that their management just hasn’t been able to figure out. I am sometimes astounded as to the questions they’ve started bringing me.

For example, check this one out. If the casino gives away stamps for players playing blue chip games, and the stamps can be traded in for electronics goods at the rate of 3 stamps for every \$1 cost of the item, what is the dollar value of the stamps given? It’s obviously \$.33, and I feel that my 8-year old daughter could have solved this very simple math problem, however, various members of management came up with the answer of \$6 per stamp. I have no idea how they reached this conclusion. It’s funny that people who are in the business of advantage gambling don’t really seem to understand some of the basic math behind it. As I told my manager, these people could never make a living playing poker.

Two days ago I was approached to see if I could solve some more advanced probability questions that they were thinking of doing for a game very close to my heart, Blackjack. I said sure, since they offered to pay me. Here are “the answers”, which I had to figure for both a 5-deck shoe as well as for double deck black jack.

One idea was to pay more if you ever got three sevens on your first three cards. That is, your first hand was 77 and, when you hit, you receive another 7. The bonus would be larger if the sevens were all the same color, and largest if they were suited. I informed management that it was simply not possible to get three suited sevens in double deck Blackjack because there would be, at most, two available. This is, apparently, why they pay me.  They nixed that idea.

The next idea was if you received A-2-3-4-5-6 to make 21 in any order, plus a bonus for being the same color, and the largest bonus for being suited. Let’s take a look at the math. I used combinatorics to solve these problems, so first we need to know how many possible combinations there are of six cards drawn at random from a double deck shoe. In math speak, we’d say that’s 104 (two decks of 52 shuffled together) choose six, which is 1,517,381,580.  The total number of possible combinations of a six-card straight are 8 choose 1 for each of the card ranks, A through 6, which simply becomes 8 to the sixth power, or  262144. 262144/1517381580 which is .00017276, or 1 in 5788.

Now for an all black or red A-2-3-4-5-6 to make 21, it’s 4 choose 1 for each of the card ranks A through 6- four black card ranks for each rank in question. If we divide that by the total number of combinations, we will have the chance of getting either a black or a red six card straight. Since either are rewarded, we should double the result. So the math is 4 to the sixth power x 2 divided by 1517381580. The reciprocal of that number gives us the figure management wants, which is 1 in 185227.

Now for a six card straight of the same suit, that’s simply 2 choose 1 for each card rank A through 6 multiplied by four since there are four suits, and divided by the 1517381580. I’ll go ahead and give the reciprocal result which is 1 in 5,927,271. For a 5 card shoe, the total number of combinations are 260 choose 6 (which gives 404 830 840 960)- quite a few. Now let’s take a look at basically the same math for the 5 card shoe:

• A six card straight is 20 choose 1 for each rank, divided by the total number of combinations, and then take the reciprocal to give you 1 in 6326.
• A six card red of black straight is 10 choose 1 for each rank, multiplied by 2 (red OR black), divided by the total number of combinations, and then take the reciprocal, which is 1 in 202415.
• A six card straight flush is 5 choose 1 for each rank, multiplied by 4 (four suits), divided by the total number of combinations, and then take the reciprocal, which is 1 in 6477293.

Management decided that these events were too rare to make a suitable promotion, so they asked me to refigure it for only a 5 card straight A-2-3-4-5. Well, since we’re only selecting five cards instead of six, we need a new figure for total combinations, or 104 choose 5 for a two card deck (which gives 91,962,520) and 260 choose 5 (which gives 9,525,431,552). Since the number we’re really interested in is the 1 in whatever number figure instead of the base probability, I’m going to actually reverse the numerator and the denominator to save a step since that will give me that figure directly without having to then take the reciprocal.

For a two deck shoe, the math for:

• A five card straight A-2-3-4-5 is 8 choose 1 for each rank A through 5. The take the total number of combinations (91962520) and divide it by the total number of straights (32768) to give 1 in 2807.
• A five card red or black straight is 4 choose 1 for each rank with the result multiplied by 2. Then divide the total number of combinations with that result (2048) to give 1 in 44903.
• A five card straight flush is 2 choose 1 for each rank with the result multiplied by 4. Then divide the total number of combinations by the this result (128) to give 1 in 718457.

For a five deck shoe, the math for:

• A five card straight is 20 choose 1 for each rank. Take that result (3200000) and divide it into the total number of combinations (91,962,520) to give 1 in 2977.
• A five card red or black straight is 10 choose 1 for each rank with the total result multiplied by 2. The total number of combinations is divided by the result (200000) to give 1 in 47627.
• A five card straight flush is 5 choose 1 for each rank with the result multiplied by 4. The total number of combinations is divided by the result (12500) to give 1 in 762034.

The next Blackjack promotion idea was to pay out a bonus if you split Aces and received two Kings. For this we’re going to need to know the total number of 2 card combinations out of 104 cards (for your starting hand) and out of 102 cards (for when you’ve already removed two Aces and want to know the probability of getting two Kings). So we need 104 choose 2 (which is 5356) and 102 choose 2 (which is 5151). For double deck Blackjack the math for:

• Receiving two Aces for your starting hand is 8 choose 2 (28) divided by the total number of combinations (5356) to give us .005227. The chances of then receiving two Kings is 8 choose 2 (28 again) divided by the total number of combinations (5151) to give us .005436. Since these are independent events, the total probability of receiving two Aces and then receiving two Kings are simply the probabilities multiplied by each other (.005227 x .005436). The reciprocal of that number gives us 1 in 35189.
• Receiving two Aces of the same color  followed by two Kings of the same color 4 choose 2 (6) divided by 5356 multiplied by 4 choose 2 (6) divided by  5151 with the entire result doubled because you want to see the total chance of this happening for either red or black. Multiply it all together and that the reciprocal gives us chances of 1 in 383177.
• Receiving two suited Aces followed by two suited Kings is 2 choose 2 (1) times 2 choose 2 (1) divided by 5356 and then divided by 5151 with the entire result multiplied by 4 since there are four suits. The reciprocal of that number gives us 1 in 6,897,189.

Now let’s look at a five deck shoe. The total number of two card combos for starting hands are 260 choose 2 for 33670. Now for the Kings, we’ve again removed to cards already so the total number of combinations are 258 choose 2 for 33153. The math for a five card shoe becomes:

• the chances of starting with two Aces is 20 choose 2 (190) divided by 33670 to give .005643. To then receive two Kings is 20 choose 2 (190) divided by 33153 to give .005731. If we multiply these probabilities together and that the reciprocal we get the chances as being 1 in 30921.
• the chances of two same color Aces followed by two same color Kings is 10 choose 2 multiplied by 10 choose 2 (45 each time) divided by 33670 and then divided by 33153 with the entire result multiplied by 2. If we take the reciprocal we get chances of 1 in 275620.
• the chances of getting suited Aces followed by suited Kings is 5 choose 2 (10) multiplied by itself divided by 33670 and 33153 with the entire result multiplied by 4. If we take the reciprocal of that number we get 1 in 2,790,653.

For most of these problems, the chances are pretty similar, but for the suited Aces problems, we see that the results vary dramatically depending on whether it’s double deck or a five card shoe. That’s because if we’re trying to select two suited Aces, they are the only two Aces of that suit in the entire 104 cards, which makes it a much rarer event.

The next promotion idea was receiving a 6-7-8 on three cards to make 21. To solve these problems we’re going to need to know the total number of three card combinations drawn out of two decks (182,104) and the number of three card combinations drawn out of a five deck shoe- which is 260 choose 3 for 2,895,620. For double deck Blackjack, the number becomes:

• 8 choose 1 for each rank of 6-7-8 which is 8x8x8 for 512. If we divide the total number of 3 card combos 182,104 by the total number of 6-7-8s we get the results are 1 in 356.
• 4 choose 1 for each rank of 6-7-8 (because we’re going for black or red straights) which gives us 4x4x4=64. If we multiply it by 2 (because it’s red and black) to get 128 and then divide the total number of combinations by 128 to give us the results of 1 in 1423.
• 2 choose 1 for each rank of 6-7-8 (because we’re going for suited straights) which gives us 2x2x2=8. If we multiply the result by the 4 suits in the deck and divide the total number of possible combinations, we get the chances being 1 in 5690.

For a five deck shoe, the math becomes:

• 20 choose 1 for each rank of 6-7-8 (8000) divided into the total number of 3 card combinations (2,895,620) to give us the total chances of 1 in 361.
• 10 choose 1 for each rank of 6-7-8 (1000) multiplied by 2 (red or black) with the result divided into the total number of combinations to give us chances of 1 in 1448.
• 5 choose 1 for each rank of 6-7-8 then multiplied by 4 (for the 4 suits) to give us 500. If we divide the total number of combinations by 500, we get chances of 1 in 5791.

The last problem was the simplest, and that was simply a jackpot for number of consecutive Blackjacks. The odds of hitting a Blackjack in a five card shoe which is being continuously reshuffled is very straight forward. It’s simply the 20 choose 1 (for the total number of Aces) times 80 choose 1 (the total number of 10s) with the result divided by the total number of two cards combinations (33670 from the split Aces problem above). So  the chances of hitting Blackjack are simply (80×20)/33670 to give 4.75%. The chances of sitting down at a table and receiving two consecutive Blackjacks are .0475 squared. Three consecutive Blackjacks are .0475 cubed, and so on. If we take those results and express them in reciprocal form we get:

Number of Consecutive Blackjacks                             Chances of It Happening

2                                                                                      1 in 442

3                                                                                      1 in 9319

4                                                                                      1 in 196106

5                                                                                      1 in 4,126,821

For double deck, the math really strange because you need to know if the deck gets reshuffled at any point along the way. If it doesn’t then you’ve just removed an Ace and a Ten from the deck, so the chances of getting consecutive Blackjacks actually become harder as you go along.