Simple Probability Question

[msoccer29]

I'm sure this is a beginner question so I apologize in advance. I haven't taken Calculus in school yet nor Statistics so I'm doing my best here to go on what I know.

The argument is about the probability of flipping two coins in a row and them both coming up the same side (regardless of heads or tails).

If I were to set this up would it be ((2/2))*((1/2)), because it wouldn't matter what the first flip would be, but only that the second would have to match it. Using this thinking would it be correct to say that 50% of the time the coins will come up different and 50% they will be the same? Or am I missing something fundamental and the percentage is only 25% that they will come up the same.

On the same lines, I read that the odds of hole cards (or simply being dealt two cards) being a pair is 221:1.
((4/52))*((3/51)) = 221:1

What I don't understand is this - isn't the above only relevant for a specific pair - example the odds of getting 2 Kings. What about ANY pair regardless of what it is. Would the math change so that:
((52/52))*((3/51)) = 17:1

My thinking is the same here, shouldn't the first card not matter but rather its simply having the second card pair up with the first - or am I missing something fundamental here.

Any help is greatly apperciated. Thanks guys.

[Jordan Olsommer]

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I haven't taken Calculus in school yet nor Statistics so I'm doing my best here to go on what I know.

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You don't need that - combinatorics is just basically advanced arithmetic; it's not covered in calculus, and it might be covered in statistics, but that depends a lot on the particular course you are in. If you can find a "Discrete Math" textbook at a book sale, that'll point you in the right track.

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If I were to set this up would it be ((2/2))*((1/2)), because it wouldn't matter what the first flip would be, but only that the second would have to match it.

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Yup. Or another way to look at it, since the number of possibilities is so small, is to simply enumerate them all:

HH
HT
TH
TT

So there are four possibilities, two of which fit your criterion of "same result twice in a row" = 2/4 = 1/2. Sketching out all the possibilities for a very small case of something you want the probability for is a very good way of double-checking your methods when you're not sure whether or not your calculations are correct.

[ QUOTE ]
On the same lines, I read that the odds of hole cards (or simply being dealt two cards) being a pair is 221:1.
((4/52))*((3/51)) = 221:1

[/ QUOTE ]

Right, but with a small correction - this gives you 1/221, which is not exactly the same as 221:1 (221-to-1 would be equal to 1/222). 1/221 is equal to 220-to-1.

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What about ANY pair regardless of what it is. Would the math change so that:
((52/52))*((3/51)) = 17:1

[/ QUOTE ]

Same thing here - it's 1/17, or 16:1. Other than that, you're correct.

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My thinking is the same here, shouldn't the first card not matter but rather its simply having the second card pair up with the first - or am I missing something fundamental here.

[/ QUOTE ]

Nope, you're right on.

[TomCollins]

Welcome to the forums. You have a much better grasp of these topics than you thought. In fact, even much better than a great deal of posters.

[msoccer29]

Thanks for the indepth answer Jordan. I'm taking Discrete Math next year, but I think I'll get a leg up and take a look at it. Thanks for the confirmation of my assumptions and I see what you mean in terms of how I'm writing my odds. That makes more sense now too.

[msoccer29]

Thanks for the welcome Tom. I've always played Hold 'Em but never realized how much math could be used in conjunction with it. Instead of cheating I thought I'd build up on probability and odds and then move into more complicated things like EV. Nice place, I'm definetly gonna stick around. Thanks again guys.