4/52 * 1/51 * 1/50 * 1/49 * 1/48 * 8/47 * 3/46 * 2/45 * 1/44 * 9!/4!

4/52 is because there are four suits of 10. 1/51 is because there is only one J with the same suit. Same for 1/50 (Q), 1/49 (K) and 1/48 (A). 8/47 is because there are 8 choices of a card that can be a 4 of a kind (2-9) since 10,J,Q,K,A are already taken. 3/46 is one of the three remaining cars of the same rank, same for 2/45 and 1/44. Finally 9! (9 factorial = 9*8*7*6*5*4*3*2*1) is because the cards can appear in any order, and divide by 4! because the four of a kind involves 4 cards that are equivalent (i.e. the suits are not relevant unlike royal flush).

The numerator starts at 52 and goes down by 1 for each card because there are that many left in the card from which to choose.

Do the arithmetic and you get 2.17446E-09, or 1 in 459,884,425.

https://youtu.be/sPTHNr8s76w

I think the pattern would beQuote:jrblackieI have an equation question. If nine cards are dealt from a regular deck of cards what is the math to get a royal flush and four of a kind. What would the odds be. How is this equation written in a formula?

ZZZZ 10x,J,Q,K,A (I turned it around to be different)

4oak and a Royal

the ZZZZ is from 8 ranks only.

2 thru 9 as the 4oak can't B a 10 to A

8Choose1rank * 4suitsChoose4(4 suits and choose all 4)

C(8,1)*C(4,4) (in Wolfram Alpha)

8 * 1 = 8

now to the 10x

there are 4(four) 10s to select from.

so (4 choose 1) * (1choose1)^4 (the 4 other Royal cards)

4 * 1 = 4

4 * 8 = 32

32 / 52choose 9

32 / 3 679 075 400 =

4 / 459 884 425

about 1 in 114 971 106.3

different from another post by a factor of 4

maybe a 3rd will try

OF course, if this happened playing at home after they were set up that way

then that is a different story

Sally

Quote:mustangsallyI think the pattern would be

ZZZZ 10x,J,Q,K,A (I turned it around to be different)

4oak and a Royal

the ZZZZ is from 8 ranks only.

2 thru 9 as the 4oak can't B a 10 to A

Why not? Is, say, 10, J, Q, K of spades, four Aces, and any other card not both a royal flush and four of a kind?

Otherwise, I get what you get.

Would the order in which you pick, either 4oak or royal first matter?

the order does not matter.Quote:jrblackieSo, if you pick a royal in spades first, you still have the other suits available for the 4oak. Right?

Would the order in which you pick, either 4oak or royal first matter?

do you want the royal as 5 cards and the 4oak as 4 cards? that is 9 cards total.

Don suggested that a royal with the 4oak being one rank of the royal would be 8 cards. that would be more calculations

btw,

why the question?

Sally

I appreciate your help. I have a new slot game and thought that a 9 card combo progressive might be interesting. It looks like it would be on the order of MegaBucks. It might be too much for the NGC to allow.

Quote:jrblackieI appreciate your help. I have a new slot game and thought that a 9 card combo progressive might be interesting. It looks like it would be on the order of MegaBucks. It might be too much for the NGC to allow.

If it's a slot machine, that's different; either you have to allow for the same card to appear on two or more reels, or you need to explain the layout of the cards on what I assume are nine reels. You cannot make the value on a reel dependent upon the values on one or more other reels - at least, not in Nevada.

If it's a straightforward nine-card deal, and you do allow something like a royal flush and four Aces (which would be only 8 cards) to count, then the total number of winning deals is:

All nine cards used: 4 (suits for the royal) x 8 (ranks for the 4OAK) = 32

Eight cards used: 4 (suits for the royal) x 5 (ranks for the 4OAK) x 44 (possible cards for the "ninth card") = 880

The probability would be 912 / 3,679,075,400, or about 1 / 4,034,074. This is much better than the reported chances of winning the Megabucks jackpot.

Keep in mind that Nevada will allow odds greater than 100,000,000 - 1; however, if they are that high, the odds must be displayed on the machine.