Probability - Cards

First, I'm going to assume that you are not counting ties. In other words, you mean 75 bets resolved. It would be very unlikely to go 75 hands without a tie. The expected number of banker wins out of 75 bets resolved is 38.00913745. The standard deviation is the square root of the product of 75, the probability of a banker win, and the probability of a player win. The probability of a banker win, given that there wasn't a tie, is 0.506788499 and the probability of a player win is 0.493211501 . The standard deviation is thus 4.329727904 . Then you'll have to make a half point correction for a binomial distribution and look up the Z statistic in a standard normal table (this step is left to the reader). The final answer is that the probability of the banker getting 52 or more wins is .0009. Your question also allowed for the possibility of the banker winning 23 or fewer times (also a difference of 29 more more) which has a probability of .0004 . So the final answer is that the probability of a difference of 29 or more is .0013, or 1 in 769.

Thanks for the compliment. You are off by a factor of five because the dealer can have any one of 5 cards face up. In other words order does matter with the dealer's hand, since the first card is dealt face up. The correct derivation of the total combinations is combin(52,5)*47*combin(46,4) = 19,933,230,517,200.

There are two ways to get a full house in this situation: (1) draw a three of a kind or (2) draw one more to the pair and another pair. I'm going to assume you discard three singletons.

First, lets work out the number of combinations under (1). There are 3 ranks with only 3 suits left (remember you discarded 3 singletons) and 9 ranks with 4 suits left. The number of combinations is thus 3*combin(3,3)+9*combin(4,3) = 3*1 + 9*4 = 39.

Next, let's work out the number of combinations under (2). There are 2 suits left to add to the existing pair. There are combin(3,2) ways to form a pair from the 3 ranks with 3 cards left and combin(4,2) ways to form a pair to from the ranks with 4 cards left. So the total combinations under 2 is 2*(3*combin(3,2)+9*combin(4,2)) = 2*(3*3 + 9*6) = 126. The total number of ways to arrange a full house is the sum under (1) and (2), or 39+126=165. There are combin(47,3)=16,215 ways to arrange the 3 cards on the second draw. The probability of drawing a full house is the number of ways to draw a full house divided by the total combinations, or 165/16,215 = 0.0101758, or about 1 in 98.

For more information on the combin() function, please see my section on probabilities in poker page.

No, the probability of getting any given hand is the same regardless of how many other players are at the table. An unseen card is an unseen card, it doesn't matter if another player has it or it is still in the deck.

I'm going to give an approximate answer by assuming that each player was dealt a hand from a separate deck. This should not change the odds much. The probability of any one player drawing a straight flush as found in my section on probabilities in poker is 36/2,598,960. Lets call this probability p. The probability of two players drawing a straight flush is combin(6,2)*p²*(1-p)⁴ = .0000000028779. In other words, the odds against this happening are 347,477,740 to 1.

I'll just work out the king high first and briefly show the formula for the other two. The probability will be the number of king high hands divided by the total number of hands. The number of ranks less than king is 11. The king high hand must have two different of these ranks. The number of ways to arrange 2 out of 11 is combin(11,2) = 55. However, one of these combinations is king-queen-jack, resulting in a straight, so subtracting that combination there are 54 left that do not form a straight. Next, there are four suits for each rank, or 4³=64 possible combinations of suits. However, four of these 64 result in a flush so there are 64-4=60 combinations of suits left. So, the total number of king high combinations is 54*60=3240. There are a total of combin(52,3)=22,100 to arrange 3 cards out of 52. So, the probability of forming a king high is 3,240/22,100 = 0.1466063 . The probability for ace high is: (combin(12,2)-2)*(4³-4)/combin(52,3)=0.1737557. Note the -2 instead of -1 because of both the a-2-3 and q-k-a straights.

The probability for queen high is: (combin(10,2)-1)*(4³-4)/combin(52,3)=0.119457.

The probability of any royal flush is the number of possible royals, which is four (one for each suit), divided by the number of ways to choose 5 cards out of 52, which is combin(52,5)=2,598,960. So, the answer is 4/2,598,960 = 0.00000153908, or 1 in 649,740.

The probability of a sequential royal flush equals (number of suits) * (number of directions) / (total permutations of 5 cards out of 52) = 4 * 2 / permut(52,5) = 8 / 311,875,200 = 8 / number of possible royals, which is four (one for each suit), times the number of directions it can be in divided by the number of ways to pick 5 cards out of 52, which is permut(52,5)=311,875,200. So, the answer is 4/311,875,200 = 0.00000002565, or 1 in 38,984,400 .

First let me answer the unasked question on the probability that a specific number will show up n times on a random bill. There are 8 digits on a bill so the probability of n of a specific number is combin(8,n)*0.1ⁿ*0.9^8-n/10⁸. Here is a table showing the probability for 0 to 8 of a specific number.

The next table shows the probability of every possible type of bill, categorized by the number of each n-of-a-kind. For example, the serial number 66847680 would have one three of a kind, one pair, and three singletons, for a probability of 0.1693440.


o.a.k. = "of a kind"

For more information, see my page on liars poker.

Assuming you draw five cards, and count all hands with exactly two jacks, then the probability would be combin(4,2)*combin(48,3)/combin(52,5) = 6*17296/2598960 = 3.99%.

Not counting a three of a kind and two pairs, the following are the ways to get a three pair and number of combinations.

No wild card: combin(13,3)*10*6³*4 =2471040
Wild card used to compete pair of aces: combin(12,2)*10*6²*4² = 380,160
Wild card used as singleton ace: combin(12,3)*6³ = 47,520

The total number of combinations is 2,898,720. This is less than half of the 747,0676 combinations for a three of a kind.

You, are right that on average a royal flush will occur once in every 649,740 hands, and that in 1,299,480 hands the expected number of royal flushes is 2. However, this is only the average. With every hand that goes by you are no closer to getting a royal. Every game of independent trials has this memory-less property so a royal flush is never overdue.

The probability of zero royals in 1,299,480 hands is 13.53%.

There are two ways to arrange the ranks for form a 9 high hand, the one you mentioned and 2-3-4-6-7-8-9. The number of suit combinations without forming a flush is 4⁷-4*(combin(7,5)*3^2+6*3+1) = 15,552. So the probability of a 9 high hand is 2*15,552/combin(53,7) = 31,104/154,143,080, or 1 in 9,911. If you were to play just five times the probability of getting 2 9-high hands would be 1 in 9,826,685. That this happened is a coincidence I believe as opposed to a fault in the random number generator or the coding of the program.

Let's let d be the number of decks. The probability of a tie on the first round is (4*d-1)/(52*d-1)= 0.073955. The probability of a tie in the second round is 12*4*d/(52*d-2)*(4*d-1)/(52*d-3)+(4*d-2)/(52*d-2)*(4*d-3)/(52*d-3) = 0.073974. Lets call p₁ the probability of a tie in the first round and p₂ the probability of a tie in the second round. Then the player return is p₁*(2*p₂ +(1-p₂)/2*(1-2))= -0.023301. Multiply by -1 and you have the house edge of 2.33%. I hope I didn't go over this too quickly.

You correctly calculated the number of player combinations as combin(52,5)=2,598,960. From there, the dealer can have combin(47,5)=1,533,939 possible hands. Then any one of five dealer cards can be face up. So 2,598,960*1,533,959*5=19,933,230,517,200.

Thanks for your kind words. I agree that calculating the numbers for seven-card stud is hard. That is why I do it my computer. My program goes through all possible combinations and scores each one. The number of wild straights in pai gow poker is 11*(4⁴-4)+10*3*(4⁴-4)=10332. Combined with the 10200 natural straights the total is 20532.

Your method is a good approximation. However by that logic, when flipping a coin, the probability of at least one person in 3 flipping a heads would be 3*50%=150%. Assuming independent events the probability of at least one success out of n trials, where the probability of each success is p, is 1-(1-p)ⁿ. In the case of the coin flipping example this would be 1-.5³=0.875. In the case of four players of Caribbean Stud Poker the probability of at least one royal flush would be 1-(1-4/2598960)⁴ = 0.00000615629. However since all the cards are dealt out of the same deck the events are not independent. The math gets very complicated to determine the exact right answer and the approximation should be very close to the right answer.

There are 32 possible natural straight flushes (4 ranks times 8 possible spans of 7 cards). There are combin(53,7) = 154143080 possible ways 7 cards can be drawn out of 53. So the answer is 32/154143080, or 1 in 4816971.

The probability of getting a straight flush on the first hand is 4*12/combin(52,3) = 48/22100 =~ 0.0022. The probability that the next hand will be exactly the same is 1/22100. So the answer is (48/22100)*(1/22100) = 48/488410000, on 1 in 10,175,208. This is a 1.37 more likely than hitting a 6/49 lottery, which has a probability of 1 in 13983816.

(8²/combin(416,2))* (7²/combin(414,2)) = 0.00000043, or 1 in 2308093

Let’s define the probability of a flush of either getting one on the deal, or drawing to a 4-card flush. For the sake of simplicity we’ll assume a player will draw to a pat pair or straight with 4 to a flush. The probability of getting a flush on the deal (not including a straight/royal flush) is 4*(combin(13,5)-10)/combin(52,5) = 5108/2598960 = 0.0019654. The probability of being dealt a 4-card flush is 4*3*combin(13,4)*13/combin(52,5) = 111540/2598960 = 0.0429172. The probability of completing the flush on the draw is 9/47. So the overall probability of getting a 4-card flush and then completing it is 0.0429172*(9/47) = 0.0082182. So the total probability for a flush is 0.0019654 + 0.0082182 = 0.0101836. The probability that exactly 2 out 5 players receiving a flush is combin(5,2)* 0.0101836²*(1-00.0101836)³ = 0.001006, or about 1 in 994.

There are combin(52,7)=133,784,560 ways to arrange 7 cards out of 52. The number of 7-card sets including a four of a kind is 13*combin(48,3) = 224,848. The 13 is the number of ranks for the 4 of a kind and the combin(48,3) is the number of ways you can choose 3 cards out of the 48 left. So the probability is 224,848/133,784,560 = 0.0017, or 1 in 595.

1 in 52 factorial, or 1 in 80,658,175,170,943,900,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000.

First, there are 10*9/2=45 ways you can choose 2 players out of 10. The probability of two specific players getting four aces is 1/combin(52,4)=1/270725. So the probability of any two players getting a pair of aces is 45/270725=0.0001662.

The probability of a single player getting a 7-card flush is 4*combin(13,7)/combin(52,7) = 1 in 19491. The probability of at least one player out of 7 getting a 7-card flush is about 1 in 2785.

For the sake of simplicity let’s assume each hand is dealt from a fresh deck. The probability of a four of a kind is 13*48/combin(52,5) = 624/2598960. The probability of exactly 3 out of 40 four of a kinds is combin(40,3)*p³*(1-p)³⁷ = 1 in 7378135, where p = 624/2598960. So that is more like a 1 in 7 million shot.

You’re welcome, thanks for you compliments. The probability of being dealt 3 jokers in a row from a six deck shoe (plus the 3 jokers) is 1/combin(315,3) = 1 in 5,159,805. Another solution is (3/315)*(2/314)*(1/313).

The answer is (4¹³/COMBIN(52,13))* (3¹³/COMBIN(39,13))* (2¹³/COMBIN(26,13)) = 1 in 61,204,166,001.

There are four possible suits for the royal. There are five possible missing cards. The fifth card can be one of 47 other cards. So there are 4*5*47=940 ways to get four to a royal. There are combin(52,5) = 2,598,960 total combinations. So the probability is 940/2,598,960 = 1 in 2,765.

The easy way to solve the problem is that the probability the first pile has three aces is (4/52)*(3/51)*(2/50) = 1/5525. However each pile has an equal probability of having three aces so we multiply by 6, yielding 6/5525 = 0.001086

(12/52)*(11/51)*(10/50)*(9/49)*(8/48) = 0.00030474, or about 1 in 3282.

The probability of a four of a kind in any given hand is 13*48/combin(52,5) = 0.0002401. Let’s assume in two hours you can play 120 hands. The probability of exactly two four of a kinds would be combin(120,2) × 0.0002401² × (1-0.0002401)¹¹⁸ = 0.000400095 = 1 in 2499.41.

Not counting straight flushes and royal flushes the probability of a straight is 1.02% and of a flush is 1.04%. So a flush is slightly more likely.

It isn't often I say this but I tried for hours but the math on this one was simply over my head. So I turned to my friend and math professor Gabor Megyesi. Here is his formula for any "drought" problem.

1. Let p be the probability of winning any given hand.
2. Let d be the length of the drought.
3. Let n be the number of hands played.
4. Set k=dp and x=np.
5. If k=1 then let a=-1, otherwise find a such that k=-ln(-a)/(1+a). (a is a negative number, if k>1 then -1 < a < 0, if k < 1 then a < -1, and a needs to be calculated to high accuracy.) [Wizard’s note: This kind of solution can be easily found in Excel using the Goal Seek feature under the tools menu.]
6. if k=1 then let A=2, otherwise let A=(1+a)/(1+ak).
7. The probability of no drought of length d in n hands is approximately Ae^ax.

In this particular problem p=1/40391, d=200000, n=1000000, k=4.9516, x=24.758, a=-0.0073337, A=1.03007. So the probability of no drought is 1.03007*e^{-0.0073337*24.758} = 0.859042. Thus the probability of at least one drought is 1-0.859042 = 0.140958.

Here is Gabor Megyesi's full 5-page solution (PDF). Thanks Gábor for your help.

I did a random simulation of 32,095 sets of one million hands. The number with at least one drought was 4558, for a probability of 14.20%.

The following table shows the probability of 0 to 4 aces in a totally random hand.

Take the sum for 1 to 4 aces we see the probability of at least one ace is 0.341158. The probability of two or more aces is 0.041684.

The probability of there being at least one more ace, given there is at least one, can be restated per Bayes' theorem as probability(two more aces given at least one ace) = probability (two or more aces)/probability(at least one ace) = 0.041684/ 0.341158 = 0.122185.

For those rusty on Bayes' Theorem, it states that the probability of A given B equals the probability of A and B divided by the probability of B, or Pr(A given B) = Pr(A and B)/Pr(B).

The next table shows the combinations and probability for each number of other aces given that the ace of spades was removed from the deck.

This shows the probability of at least one more ace is 0.221369.

For fun, let's solve it the same question using Bayes' Theorem. Assume that random hands are dealt until one is found containing the ace of spades. The probability of at least one additional ace, given the hand contains the ace of spades, could be rewritten as probability(at least two aces given ace of spades is in hand). According to Bayes' theorem, this equals Probability(hand contains ace of spades and at least one more ace) / Pr(hand contains ace of spades). We can break up the numerator as Probability(2 aces including ace of spades) + Probability(3 aces including ace of spades) + Probability(4 aces). Using the first table this equals 0.039930×(2/4) + 0.001736×(3/4) + 0.000018 = 0.021285. The probability of the ace of spades is 5/52 = 0.096154. So the probability of at least two aces given the aces of spades is 0.021285/0.096154 = 0.221369.

So the probability of two or more aces given at least one ace is 12.22% and given the ace of spades is 22.14%.

Let’s look at another simpler situation. Suppose woman A says "I have two kids and at least one is a boy." Woman B says "I have two kids and the older one is named John." We can assume nobody named John is a girl and no woman gives the same name to more than one kid. Using conditional probability the probability of both kids being a boy of woman A is pr(both boys)/pr(at least one boy) = pr(both boys)/(1-pr(both girls)) = (1/4)/(1-(1/4)) = (1/4)/(3/4) = 1/3. However the probability woman B’s younger kid is a boy, or both kids are boys, is ?, because saying the older child is named John tells us nothing about the younger child.

To make another example suppose you go to Jiffy Lube and they offer two deals for the same price. Deal A is they will check four parts and replace only the first defective one found. Deal B is they will check only one problem and fix it if found. Wouldn’t you rather take deal A? Your car came in with the same number of expected bad parts but the probability of finding a problem is greater under deal A, and thus you will leave with a small number of expected defective parts under that plan. Likewise a test for any ace will probably turn up the only ace, while a test for the ace of spades does not check for the other three suits, leaving them more likely to be aces.

First I separated the straight flushes into two types, those with four consecutive suited cards and those with five. The number of five cards straight flushes is the number of suits * number of spans (ace to 10 as lowest card) = 4*10 = 40. Four the four card straight flushes there are 11 different spans (ace to jack as the low card). In the case of the A234 and JQKA straight flushes the fifth card can be one of 47 (52 less the 4 cards already removed and the fifth card which would form a 5 card straight flush, which were already accounted for). So there are 4*2*47=376 straight flushes of span A234 or JQKA. Of the other nine there are 46 possible cards for the fifth card (52 less the 4 cards already removed and two for cards that would form a five-card straight flush). So the number of straight flushes of span 2345 to TJQK are 4*9*46=1656. So the total number of 4-card straight flushes are 40+376+1656=2072.

A straight is guaranteed only with 45 cards. For example your could deal every A, 2, 3, 4, 6, 7, 8, 9, J Q, and K for a total of 44 cards and still not have a straight.

How can I refuse after you buttered me up so nicely? First there are combin(52,7) = 133,784,560 ways to choose 7 cards out of 52, without regard to order. There are 8 possible spans for a 7-card straight (the lowest card could be A to 8). If we had 7 different ranks there are 4⁷ = 16384 ways to arrange the suits. Note that this includes all the same suit, which would form a straight flush. So the probability is 8*16,384/133,784,560 = 1 in 1020.6952.

The probability of zero spades is (39/52)*(38/51)*(37/50)*(36/49) = 0.303818. So the probability of at least one spade is 1-0.303818 = 0.696182.

From probability 101 we know Pr(A or B) = Pr(A) + Pr(B) - Pr(A and B). So Pr(A and B) = Pr(A) + Pr(B) - Pr(A or B). Let's let A be getting an ace and B be getting a deuce. Pr(A) = Pr(at least one ace) = 1-Pr(no aces) = 1-combin(48,4)/combin(52,4) = 1-0.7187 = 0.2813. The probability of no deuces would obviously be the same. By the same logic pr(A or B) = Pr(at least one ace or deuce) = 1-Pr(no aces nor deuces) = 1-combin(44,4)/combin(52,4) = 1 - 0.501435 = 0.498565. So the probability of getting at least one ace and deuce is 0.2813 + 0.2813 - 0.498565 = 0.063962.

The probability of getting it any one hand is 6/52. The probability of getting it 27 hands in a row is (6/52)²⁷ = 1 in 20,989,713,842,161,800,000,000,000.

There are six ways to arrange two suits out of four for each pair. Then there are 44 cards for the singleton. So the number of successful combinations is 6*6*44 = 1584. There are 2,598,960 combinations in total, so the probability is 0.0609%.

There is only one way to get that exact hand. So the probability would be 1 in combin(52,5) or 1 in 2,598,960.

I’m very familiar with this concept. There are 134,459 unique five-card hands. The reason I know this is my first video poker program analyzed all 2,598,960 possible hands and took days to run through a pay table. However by running just one of each of the 134,459 classes of hands, weighting by the number of all hands in its class, you can cut down the run time by 95%. Sometimes when a game is suit specific like my recent analysis of Black Jack Bonus Poker, I have to dust off my old program and do it the slow way.

Combinations in Five Suit Poker

Hand	Combinations	Probability	Formula
Five of a kind	13	0.000002	13
Straight flush	50	0.000006	5*10
Four of a kind	3900	0.000472	13*12*COMBIN(5,4)*5
Flush	6385	0.000773	5*(COMBIN(13,5)-10)
Full house	15600	0.001889	13*12*COMBIN(5,3)*COMBIN(5,2)
Straight	31200	0.003777	10*(5^5-5)
Three of a kind	214500	0.025969	13*COMBIN(12,2)*COMBIN(5,3)*5^2
Two pair	429000	0.051938	COMBIN(13,2)*11*COMBIN(5,2)^2*5
Pair	3575000	0.432815	13*COMBIN(12,3)*COMBIN(5,3)*5^3
Nothing	3984240	0.48236	(COMBIN(13,5)-10)*(5^5-5)
Total	8259888	1

Note that I reversed the order of the full house and flush.

The two suits can be divided either 4 and 1 or 3 and 2. Let’s look at the 4/1 split first. There are 4 suits to choose from for the one with 4 cards, and 3 left for the one with 1 card. There are combin(13,4)=715 ways to choose 4 ranks out of 13. There are 13 ways to choose a single rank. So there are 4×3×715×13=111,540 ways to have a 4/1 split between the two suits. By similar logic there are 4×3×combin(13,3)×combin(13,2)=267,696 ways to have a 3/2 split. So the overall probability is (111540+267696)/combin(52,5) = 14.59%.

Thanks. (4⁵-4)/combin(52,5) = 1020/2598960 = 1 in 2,548.

A simple way to estimate the probabiity of winning is to assume that every card has a 12/13 probability of not matching the stated rank. To win this bet, the victim would have to do this successfully 52 times. The probability of 52 wins is (12/13)⁵² = 1.56%. A fair price to pay would be 63.2 to 1. At 20 to 1 Vince had a 67.3% advantage (ouch!).

Accoring to G.M., who is a better mathematician than I am, the actual probabiliy is 1.6232727%. The reason for the difference is the outcome of each pick is positively correlated to previous picks.

Before any cards are seen, the probability of any three ranks not appearing on the flop are combin(40,3)/combin(52,3) = 9880/22100 = 44.71%. So this guy had a 10.59% advantage.

The way I prefer to answer probability questions is to use the combinatorial function. Doing it that way, there are combin(48,4) = 194,580 ways to choose four cards that are not kings out of the 48 non-kings in the deck. There are combin(51,4)=249,900 ways to choose any four cards out of the remaining 51 cards in the deck. So, the probability of getting no kings in the next four cards are 194,580/249,900 = 77.86%. Thus, the probability of getting at least one kind is 100% - 77.86% = 22.14%.

Several people have said that the combinatorial function is probably over the heads of the type of people asking these kinds of simple probability questions. I don’t disagree with that, but a major reason for this site is to try to teach my readers something about math. The combinatorial function is extremely useful in probability, and saves a lot of time. However, the question at hand can be easily answered without it.

The probability the second card is not a king is 48/51. That is because there are 48 non-kings left in the deck, and 51 total cards left. If the second card is not a king, then the probability the third card is also not a king is 47/50 (47 non-kings divided by 50 cards left). Following this to the end, the probability that none of the other four cards are kings is (48/51)×(47/50)×(46/49)×(45/48) = 77.86%. The probability that this is not the case, in other words at least one king, is 100% - 77.86% = 22.14%.

Surprise, an odd total is more likely at 50.03%, despite 30 of the 52 cards being even. The following table shows the probability of each even/odd split.

Odd/Even Question

Evens	Odds	Combinations	Probability	Sum
0	5	15504	0.005965	Odd
1	4	155040	0.059655	Even
2	3	565440	0.217564	Odd
3	2	942400	0.362607	Even
4	1	719200	0.276726	Odd
5	0	201376	0.077483	Even
Total		2598960	1

The probability the first card is a spade, and the second two are not, is (13/52)×(39/51)×(38/50) = 14.53%. You should then multiply that by 3, because the spade could be any one of the three cards. So the answer is 3×14.53% = 43.59%. For those who prefer the combinatorial function, the answer is 13×combin(39,2)/combin(52,3) = 9,633/22,100 = 43.59%.

For two numbers, the answer is 1/3, and for n numbers it is 1/(n+1). I posted the solutions to my page of math problems, questions 194 and 195.

I hope you’re happy, I spent all day on the second part, and my answer was still wrong. Lest I deprive my readers of the same joy, I won’t just blurt out the answers here. I broke this into two problems, and posted answers and solutions at mathproblems.info, problems 196 and 197.

There would have to be one suit with two cards, and three with one card each. There are four possible suits for the one that is represented twice. For the suit represented twice, there are combin(13,2)=78 ways to choose 2 ranks out of the 13. For each of the other three suits, there are 13 possible ranks each. So, the total number of combinations is 4 × 78 × 13 × 13 × 13 = 685,464. There are combin(52,5)=2,598,960 ways to choose 5 cards out of 52. So the probability is 685,464/2,598,960 = 26.37%.

According to a random simulation, the probability is 48.64%. So, I would have taken that bet.

Yes. First assign each card a value from 0 to 51. Call the cards c1 to c5, ordering them with c1 as the lowest to c5 as the highest. Then call the following function:

int GetIndex(int c1, int c2, int c3, int c4, int c5)
{
return combin(c5,5) + combin(c4,4)+ combin(c3,3) + combin(c2,2) + combin(c1,1);
}

Where combin returns the traditional value, except if the first value is less than the second value, return 0, as follows:

int combin(int x, int y)
{
if (y>x)
return 0;
else
{
int i,n;
n=1;
for (i=x-y+1; i<=x; i++)
n*=i;
for (i=2; i<=y; i++)
n/=i;
return n;
}
}

If you are doing this to access an array element, load the array as follows.

count=0;
for (c5 = 4; c5 < 52; c5++)
{
for (c4 = 3; c4 < c5; c4++)
{
for (c3 = 2; c3 < c4; c3++)
{
for (c2 = 1; c2 < c3; c2++)
{
for (c1 = 0; c1 < c2; c1++)
{
index_array[count]=WhateverYouWish;
count++;
}
}
}
}
}

This is a binomial distribution kind of problem. The general formula is that if the probability of an event is p, and each outcome is independent, then the probability of it happening exactly w out of t trials is combin(t,w)×p^w×(1-p)^t-w.

In this case, there are 2 ways to make the straight flush. You need the 8 of diamonds and another card of either the 6 or J of diamonds. There are combin(47,2)=1,081 ways to draw 2 cards out of the 47 left in the deck. So, the probability of getting a straight flush in any one hand is 2/1,081 = 0.0018501. The probability of making 3 out of 10 is combin(10,3)×0.0018501³×(1-0.0018501)⁷ = 0.000000750178, or 1 in 1,333,017.

There are 4 red threes and 104 other cards. There is just one way to get all four red threes. There are combin(104,50)= 1.46691 × 10²⁸ ways the player could get 50 of the other 104 cards. The total number of combinations is combin(108,54)= 2.48578 × 10³⁰. combin(104,50)/combin(108,54) = 0.059012.

If you don't like dealing with such large numbers, here is an alternative solution. Number the four red threes 1 to 4. The probability that the first red three is in the player's stack is 54/108. Now remove the first three. The probability that the player has the second red three is 53/107, because the player has 53 cards left, and there are 107 remaining cards. Likewise, the probability that the player has the third red three is 52/106, and the fourth red three is 51/105. (54/108) × (53/107) × (52/106) × (51/105) = 0.059012.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Interesting. As I wrote, I estimate that hand will occur about once every 23.7 years anywhere in Nevada. I’d say that was one of those times.

I get 48.6279%. If you're looking to make a bet on it, fair odds on the "yes" would be 1.0564 to 1.

There are also rumors of that happening in a game of whist in Warwickshire in 2011. For the benefit of other readers, this is asking what is the probability of dividing a 52-card deck into four groups of 13 cards each, where each set is composed entirely of the 13 cards of one of the suits.

There are combin(52,13)*combin(39,13)*combin(26,13) = 53,644,737,765,488,800,000,000,000,000 possible ways to arrange 52 cards into four 13 card sets. 4! = 24 would be the number of winning combinations, since you could distribute the four suits to the four players any way you wish. So the odds are equal to 1 in 2,235,197,406,895,370,000,000,000,000. To give that number some perspective, if all 7.5 billion people on earth dealt bridge hands at a rate of one per second, the probability of someone dealing a so-called perfect hand before the sun explodes in 5 billion years is 1 in 16,558.

However, some define a "perfect" hand as just one player getting all 13 cards of any suit. I show the probability of this is 1 in 39,688,347,497 games. This probably does happen once in a while somewhere on earth.

First, let's determine the number of combinations of player and board cards where this could happen. There are obviously four suits. Then there are combin(13,4)=715 ways you can choose four out of 13 cards of the given suit.

Second, one way this could happen is with three cards of that same suit the players have on the board and the other two among the other 39 cards. There are combin(9,3)=84 ways the board can have three out of the 9 cards left of the chosen suit. Then there are combin(39,2)=741 ways to pick two more cards out of the other 39 in the other three suits. So, there are 84*741=62,244 ways with three of the suit in question on the board.

Third, another way this could happen is with four cards of that same suit the players have on the board and the other one among the other 39 cards. There are combin(9,4)=126 ways the board can have four out of the 9 cards left of the chosen suit. Then there are 39 ways to pick one more card out of the other 39 in the other three suits. However, not all of these will result in both player's using both hole cards. For that condition to be met, the lowest card in the suit in question must be on the board. The probability of that, out of the 8 cards of that suit in play, is 4/8 = 1/2. So, there are 126*39*(1/2)=2,457 ways with four of the suit in question on the board.

Fourth, the last way this could happen is with five cards of that same suit the players have on the board. There are combin(9,5)=126 ways the board can have five out of the 9 cards left of the chosen suit. However, not all of these will result in both player's using both hole cards. For that condition to be met, the lowest two cards in the suit in question must be on the board. The probability of that, out of the 9 cards of that suit in play, is combin(5,2)/combin(9,2) = 10/36 = 5/18. So, there are 126*(5/18)=35 ways with four of the suit in question on the board.

So, the number of combinations where this will happen is 715*(62,244 + 2,457 + 35) = 46,286,240.

The total number of combinations of ways to pick four cards for the player hole cards out of 52 and then 5 more out of the 48 left on the board is combin(52,4)*combin(48,5) = 463,563,500,400.

Thus, the probability is 46,286,240 / 463,563,500,400 = 0.000399395 = 1 in 2,504.

This question was asked and discussed in my forum at Wizard of Vegas.

This question is asked and discussed in my forum at Wizard of Vegas, beginning with this post.

Let's start with 100% and subtract out the probabilities that result in less than four suits.

What is the probability the 48 cards have no hearts, for example? There are 36 cards that are not a heart. The number of ways to choose 15 cards out of 36 is combin(36,15) = 5,567,902,560. The number of ways to choose 15 cards out of all 48 is 1,093,260,079,344. So the probability of the 15 cards without a heart is 5,567,902,560 / 1,093,260,079,344 = 0.005093.

Next, let's multiply that by four, to get the probability of missing any suit, not just hearts: 4 × combin(36,15)/combin(48,15) = 0.02037174.

However, this double-counts some situations. Consider getting 15 black cards. That would omit both hearts and diamonds. We would have double counted that situation. So, we need to correct for that. There are combin(4,2) = 6 ways to choose two suits out of four. The probability of all 15 cards being any two specific suits is combin(24,15)/combin(48,15) = 1307504/1,093,260,079,344 = 0.00000120. As mentioned, there are six ways to choose two suits out of four, so the number of ways all the cards will be of two suits is 6 × combin(24,15)/combin(48,15) = 0.00000718.

Subracting out what we double counted, we get a probability of two or three suits being represented of 0.02037174 - 0.00000718 = 0.02036456.

Note that we don't need to worry about one suit being represented, because it's impossible to choose 15 cards out of 12.

As a final step, subtract the probability of 2 or 3 suits from 100% to get the probability of all four suits being represented: 1.00000000 - 0.02037174 = 0.97963544.

This question is asked and discussed in my forum at Wizard of Vegas.

 

This question is asked and discussed in my forum at Wizard of Vegas.

To win, you opponent has to be incorrect with all three predictions. The first prediction has a 10/13 chance of being incorrect. The second guess has a 9/12 chance of being incorrect, because we can remove the first rank guessed as a possibility. The third guess has an 8/11 chance of being incorrect, because we can remove the first two ranks guessed as a possibility.

All three of these things would have to happen for you to win. Thus, the chances of you winning are (10/13) * (9/12) * (8/11) = 720/1716 = 41.96%.

At even money, the house edge on this bet, from your side, is 16.08% (ouch!).

This question was culled from The Book of Proposition Bets by Owen E'Shea (number 7).

The average waiting time between 12's is indeed 36 rolls, including the 12 roll itself. However, that doesn't mean they come along exactly every 36 rolls. The probability of NOT rolling a 12 is (35/36). The probability of not doing so over 27 rolls is (35/36)^27. So, the probability of at least on 12 is 1-(35/36)^27 = 53.26%.

The following table shows the probability of rolling at least on 12 for 20 to 36 rolls. Note that 25 are needed to have an advantage at even money.

Let's take this one type of hand at a time.

• Five of the same card: There are 52 different cards in a deck, for 52 combinations.
• Four of the same card: There are 52 possible cards for the four that are the same and 51 for the singleton. So, there are 52*51 = 2,652 possible combinations for a four of a kind.
• Three of the same card and two of another card: There are 52 possible cards for the three that are the same and 51 for the pair. So, there are 52*51 = 2,652 possible combinations for a full house.
• Three of the same card and two different singletons : There are 52 possible cards for the three that are the same and combin(51,2)=1,275 for the two singletons. So, there are 52*1,275 = 66,300 possible combinations for a three of a kind.
• Two pairs of the same card and one singleton : There are combin(52,2)=1,326 combinations for the two different cards that form pairs. Then there are 50 left for the singleton. So, there are 1,326*50=66,300 possible combinations for a two pair.
• One pair of the same card and three singletons: There are 52 possible cards for the pair. Then there are combin(51,3)=20,825 ways to choose 3 cards out of the 51 left for the singletons. So, the number of combinations for a pair is 52*20,825 = 1,082,900.
• Five singletons: There are combin(52,5)=2,598,960 ways to choose five cards out of 52.

The sum of all these combinations is 3,819,816.

This question is asked and discussed in my forum at Wizard of Vegas.