Royal flush
 

This may (or not) help Mini Geek with his card probability query.

If we deal 7 cards form a single shuffled deck (without replacement),
what is the probability that there is a royal flush (A,K,Q,J,10)?

David

[QUOTE]If we deal 7 cards form a single shuffled deck (without replacement),
what is the probability that there is a royal flush (A,K,Q,J,10)?[/QUOTE]

I assume you mean 5 cards, since you wrote A,K,Q,J,10.
[spoiler]
There are 4 possible royal flush = 4 different suits. The total number of ways you can pull 5 cards out of 52 is:

52! / (47! * 5!) = 2.598.960

So the odds of royal flush is 4 in 2.598.960 or 1 in 649.740 or 0.0001539 %.
[/spoiler]

I think David meant 7 cards ;)
He could be playing Hold'em poker.
(2 cards in hands, 3 on the flop, 1 turn, 1 river... = 7)

[quote=ATH;115057]I assume you mean 5 cards, since you wrote A,K,Q,J,10.
[spoiler]
There are 4 possible royal flush = 4 different suits. The total number of ways you can pull 5 cards out of 52 is:

52! / (47! * 5!) = 2.598.960

So the odds of royal flush is 4 in 2.598.960 or 1 in 649.740 or 0.0001539 %.
[/spoiler][/quote]
I meant 7 but I dispute your probability for 5 cards.

There are 4^5 differsnt royal flush combinations.

[QUOTE]I meant 7 but I dispute your probability for 5 cards.

There are 4^5 differsnt royal flush combinations.[/QUOTE]

I missunderstood what you meant, but I'm pretty sure I'm right about the 5 card royal flush, I found this link now:
[url]http://mathworld.wolfram.com/Poker.html[/url]



About your original question: Royal Flush 5 cards out of 7 card Hold'em. For each of the 4 suits of royal flush the 2 remaining of the 7 cards can be whatever out of the 47 remaining cards, so total number of 7 card hands which has a 5 card royal flush in them:

4 * 47 * 46 / 2! = 4.324 (divided by 2! since we don't care about the order of the 2 last cards).

Total number of 7 card distrinct hands out of 52 cards is again the binomial coefficient:

52! / (45! * 7!) = 133.784.560 hands.

So the probablility is: 4324 in 133784560 or 1 in 30.940 or 0.00323 %.

Are the cards in a royal flush all the same suit?
If so I agree with you.

[quote=ATH;115063]I missunderstood what you meant, but I'm pretty sure I'm right about the 5 card royal flush, I found this link now:
[URL]http://mathworld.wolfram.com/Poker.html[/URL]



About your original question: Royal Flush 5 cards out of 7 card Hold'em. For each of the 4 suits of royal flush the 2 remaining of the 7 cards can be whatever out of the 47 remaining cards, so total number of 7 card hands which has a 5 card royal flush in them:

4 * 47 * 46 / 2! = 4.324 (divided by 2! since we don't care about the order of the 2 last cards).

Total number of 7 card distrinct hands out of 52 cards is again the binomial coefficient:

52! / (45! * 7!) = 133.784.560 hands.

So the probablility is: 4324 in 133784560 or 1 in 30.940 or 0.00323 %.[/quote]

Maybe I'm misunderstanding something in your analysis, but it seems to me that since the 2 cards we don't care about don't necessarily have to be the last two drawn, then we can compute the total number of possible royal flushes per suit in 7 cards as nPr = 7!/2! = 2520. So there are actually 2520*4 = 10080 ways of getting a royal flush out of 133,784,560 possible hands, for a probability of 0.00753%.

[quote=bsquared;115072]Maybe I'm misunderstanding something in your analysis, but it seems to me that since the 2 cards we don't care about don't necessarily have to be the last two drawn, then we can compute the total number of possible royal flushes per suit in 7 cards as nPr = 7!/2! = 2520. So there are actually 2520*4 = 10080 ways of getting a royal flush out of 133,784,560 possible hands, for a probability of 0.00753%.[/quote]
I think you are counting permutations where he was
computing combinations.

However, I think an accurate calculation requires separate
treatment of
A,A,K,Q,J,10,x
A,A,K,K,Q,J,10
A,A,A,K,Q,J,10

as well as A,K,Q,J,10,x,x, where x is <=9.

On second thoughts, not if the flush must be one suit.

Royal flush is 10,J,Q,K,A in same suit.

[QUOTE=bsquared;115072]Maybe I'm misunderstanding something in your analysis, but it seems to me that since the 2 cards we don't care about don't necessarily have to be the last two drawn, then we can compute the total number of possible royal flushes per suit in 7 cards as nPr = 7!/2! = 2520. So there are actually 2520*4 = 10080 ways of getting a royal flush out of 133,784,560 possible hands, for a probability of 0.00753%.[/QUOTE]

We don't care which order we draw which card, since we look at unordered sets of 7 cards chosen from 52. So for each suit of royal flush there is only 47 cards to choose the remaining 2 cards in. I'll try to show here:

[URL="http://www.hoegge.dk/mersenne/royalflush.txt"]royalflush.txt[/URL]
(H = Hearts, S = Spades, C = Clubs, D = Diamonds)

So Royal flush hearts uses up H1 (ace) and H10-13. So 1st extra card H2 and then the 2nd extra can be the 46 remaining cards. Then we choose 1st extra to be H3 and now there is only 45 choices for 2nd card, since H3 H2 is same as H2 H3 when we use unordered set of cards.

So the total for royal flush hearts: (46+45+44+43+...+1) = 47*46/2, and then the same for royal flush spades,clubs and diamonds, so in total: 4*47*46/2 = 4324.

[quote=davieddy;115085]I think you are counting permutations where he was
computing combinations.

However, I think an accurate calculation requires separate
treatment of
A,A,K,Q,J,10,x
A,A,K,K,Q,J,10
A,A,A,K,Q,J,10

as well as A,K,Q,J,10,x,x, where x is <=9.

On second thoughts, not if the flush must be one suit.[/quote]

The flush must be one suit. So the possible royal flushes are {As,Ks,Qs,Js,10s,x,x}, {Ad,Kd,Qd,Jd,10d,x,x}, {Ac,Kc,Qc,Jc,10c,x,x}, {Ah,Kh,Qh,Jh,10h,x,x}, where s,d,c,h are spades, diamonds, clubs, and hearts, and x is don't care.

We must get the 5 elements we care about in any of the 4 sets, in any order, among the possible 7 card hands. So for each suit it is a permutation of 5 elements in a 7 element set or 7!/2!. There are 4 different suits, so there are 4*7!/2! ways of getting the royal flush out of 52!/(45! * 7!) hands.

[quote=ATH;115087] So for each suit of royal flush there is only 47 cards to choose the remaining 2 cards in. I'll try to show here:

[/quote]

But this assumes you get your royal flush in the first 5 cards, right? (52 - 5 = 47). And then pick two more. What if you pick two non-royal-flush cards and then get your 5 A,K,Q,J,10 from the remaining 50 cards?

52!/(45! * 7!) gives all possible choices of 7 card hands. So all possible royal flushes are in there somewhere. We don't care what order, but we need 5 specific elements out of the 7, for each suit, which is a permutation...

I'm not an expert, but this makes sense to me. What am I doing wrong?