Novice Probability Question

[Monty Cantsin]

I am trying, purely as an exercise, to figure the odds for dealing out 7 cards and having the 4 kings among them.

I believe this is equivalent to the odds for dealing out 45 cards and having no kings among them.

It occured to me that it might be easier to figure out this second number, because you can figure out the odds of getting no king on the first card, and then the second card, and then the third card, and so on, and multiply these all together to arrive at the number.

Is this approach correct?

The odds of card number 1 not being a king are 48/52 or .923

The odds of card number 2 not being a king are 47/51 or .921

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The odds of card number 45 not being a king are 4/8 or .5


Multiplying all these numbers together gives us .0001

So, if you are dealing out 7 cards, you expect to find 4 kings once every 10,000 trials.

Am I correct in my method or result?

Please feel free to point me towards existing threads if there is already a good introductory treatment of this general question.

Thanks,

/mc

[SumZero]

Given you don't care about order of the 7 cards the number of ways to get any 7 card hand is 52 choose 7. The number of ways to get 4 kings is 4 choose 4 * 48 choose 3. That is:

(1 * (48!/(45!*3!)))/(52!/(45!*7!))
=(48*47*46/(3*2))/(52*51*50*49*48*47*46/(7*6*5*4*3*2))
=(7*6*5*4)/(52*51*50*49)
=1/(13*17*5*7)
=1/7735

So if you round poorly once every 7735 trials might seem like once every 10,000 trials.

You are right that this is the same as dealing 45 cards with no king and if you look at your fractions you get:

48/52 * 47/51 * 46/50 * 45/49 * 44/48 * ... * 5/9 * 4/8

and if you note the terms cancel so you are left with:

(7*6*5*4)/(52*51*50*49)

which is the same as above.

[Monty Cantsin]

Thanks. Looks like I screwed up the calculations but got the main idea right. I blame society.

/mc