Calc the odds of all 4 Qs

How do I calculate the odds that all 4 of a kind were dealt out to one of the players including the f/t/r?
Assume 10 people at the table.

I think it would be:
Chances of 1 Q dealt to a player (+f/t/r) = 4/(20 + 3 + 1 + 1) = 16%
Chances of 2nd Q dealt to a player (+f/t/r) = 3/(20 + 3 + 1 + 1) = 12%
Chances of 3rd Q dealt to a player (+f/t/r) = 2/(20 + 3 + 1 + 1) = 8%
Chances of 4th Q dealt to a player (+f/t/r) = 1/(20 + 3 + 1 + 1) = 4%
So multiply those all by eachother to get .006%

Is that the proper way to calculate it?

No, that's wrong. Here's a simple way:

there are 52 positions for the 4 queens to be in in the deck, giving:
52!/4!(52-4)! possible arrangements of 4 queens.

There are 25 positions for the 4 queens to be in for them all to be dealt in a hand, giving: 25!/4!(25-4)! possible arrangements.

the first combination is 52*51*50*49/4*3*2*1 =270725
the second combination is 25*24*23*22/4*3*2*1=12650

12650/270725 is about .047, or nearly 5% of the time.

There are many other ways to do it. When doing these types of probability problems, it is usually much easier to count up the ways for all possibilities than to try to find the union of the probabilities by calculating each one seperately.

Aye-yye-yye. Thank you. Ill try and digest this.

Edit: 5% is not nearly as low as I thought it would be.

Here's how to think of it ... there are 52 positions in the deck ... so there are 52 places the first q can be in ... after that is chosen, there are 51 other places the next can be in, 50 for the 3rd q and 49 for the 4th. So there are 52*51*50*49 different arrangements of places for the 4 queens to be in. Since we don't care about particular queens' positions, we divide by 4*3*2*1 or the 24 different ways you can arrange 4 queens.

It's the same with the 25 cards that are dealt.

This is an interesting question, but you should realize it gives you no information on how to play a particular hand. If you are in a hand, you have information, and therefore must calculate conditional probabilities. For example, you have Q2 and the flop comes with 2 queens. What is the probability someone else got dealt a Q? This is useful because if someone else has the q you are almost certainly outkicked. Turns out that probability is very simple, its 18/47. There are 18 places the queen could be among your opponents' hands and 47 places it could be among the cards you haven't seen.

Thanks again for your help.

The first paragraph makes sense.

The second one I am struggling with. You are thinking of the exact scenario I am. If we calculate out 18/47 we get 38% that an opponent has the other Q. But that seems completely at odds with the chance being 5% that all 4 queens were in the top 25 cards of the deck. It seems to me that the 5% sounds more correct, yet your logic is sound. I don’t know how to reconcile the two.

This is conditional probability. In the first case you are asking the question before the cards are dealt. In the second you are asking the question after you have seen 3 of the queens.

It's like the odds of Albert Pujols hitting 4 home runs in a game. If asked 'What are the chances Albert will get 4 home runs tonight?' You would answer 'extremely remote'. The second in this example would be ... you are watching the game and Pujols has hit 3 home runs already and is about to bat and someone asks if he will hit 4 home runs tonight. The odds are far greater right? You already have the information.

A poker example ... the odds of getting a royal flush are greater than 500,000 to 1. If you are dealt four cards and get AKQJ of hearts, aren't your odds now much greater of getting the royal flush? Of course they are ... 1 out of 48 times that card will bet the 10h.

Oh that makes total sense. Thanks man.