What are the exact odds of getting a royal flush with spades?

Factual Questions
1

I think that the question above should suffice. Well, what are the odds, big shot?

2

High Hands Approximate Deals
per Pat Hand Hands Possible
Total hands 1 2,598,960
No pair 2 1,302,540
One pair 2.5 1,098,240
Two pair 20 123,552
Three of a kind 50 54,912
Straight 250 10,200
Flush 500 5,108
Full house 700 3,744
Four of a kind 4,000 624
Straight flush 70,000 36
Royal straight flush 650,000 4
Five aces (with joker)* 3,000,000 1

  • A fifty-three card deck with the joker has 2,869,685 possible hands.
3

That came out messy


High Hands Approximate Deals
per Pat Hand Hands Possible
Total hands 1 2,598,960
No pair 2 1,302,540
One pair 2.5 1,098,240
Two pair 20 123,552
Three of a kind 50 54,912
Straight 250 10,200
Flush 500 5,108
Full house 700 3,744
Four of a kind 4,000 624
Straight flush 70,000 36
Royal straight flush 650,000 4
Five aces (with joker)* 3,000,000 1
* A fifty-three card deck with the joker has 2,869,685 possible hands.
4

Damn. Here’s the link http://www.neo-tech.com/poker/part1b.html

5

I’m guessing you really don’t play much poker, or you’d have a book with this data handy, but in case you are…

Lots of good poker links on this page http://www.conjelco.com/faq/poker.html

6

The odds of pulling those cards off the top of a shuffled deck are (5/52)(4/51)(3/50)(2/49)(1/48), or about 1/2,599,000. The odds change if you’re actually playing poker, as some cards will go to the other players, but I’m not sure exactly how to go about figuring that out.

7

If you decipher all that stuff you end up with about a one in 2.6 million chance of getting a spade royal flush in a five card deal.

8

Makes no difference. It doesn’t matter whether you get the 1st-5th card, or the 4th, 9th, 14th, 19th, and 24th cards, (In a 5 hand game where you are sitting to the right of the dealer.) the odds are exactly the same.

9

Had a question on a math test which was about the odds of getting a full house. I figured (52/52 * 3/51 * 2/50 * 48/49 * 3/48), which is 1 in 6941.666… Obviously not right.

I did (52/52 * 3/51 * 2/50 * 49/49 * 3/48), 1 in 6800, and that’s what I put.

But doesn’t my 4th card have to be 48/49, since it can be ANY card, except the one remaining in the deck with the value of the first 3 cards, which would give me four of a kind?

10

What you figured out was the odds of pulling three of a kind, and then a pair. The difference is between combinations and permutations.

The calculation should go…

(52/52)(51/51)…

The first two cards can be anything at all. Then you are left with two options: either the second card had the same value as the first (the chance of this is 3/51) or it was different (48/51). The probabilities make a kind of branching out tree, with each branch leading two a full house, but in a different way.

The easier way to figure this out is to figure out how many different poker hands there are (52 choose 5), and then how many ways there are to get a full house. I would do this for you, but I take Prob-Stat next semester. I’m sure someone will explain it more clearly if you need more explanation.