A very simple problem
 

This one should be a breeze, but for some reason, I'm having trouble with it.

What are the odds of being dealt the same card (rank and suit) as one of your hole cards in two consecutive hands of holdem.

I figured that it was 1-[(1/52*1/51)*(1/52*1/51)] to get the percentage and then

odds = (1/%)-1

Is this correct or am I missing something?

Thanks,

Dov

Re: A very simple problem
 

On the first deal, you get dealt two cards, any two cards. That fixes your target for the second deal.

On the next deal, you have a 1-(50C2/52C2) = 101/1326 chance of getting at least one of them again.

(101/1326)^2 would be the chance for naming two cards ahead of time, then being dealt at least one of them on the next two deals you played, or, equivalently, of playing three deals and having a card from the first deal show up on both the second and third.

Re: A very simple problem
 

[ QUOTE ]
What are the odds of being dealt the same card (rank and suit) as one of your hole cards in two consecutive hands of holdem.


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I am not sure what you are asking for? Are you looking for the same card or either of your hold cards? I will answer specifically one of your hold cards. Lets say you have AK of spades. The probability of getting Ace of spades on the next two deals is.

=(1*51/(52c2))^2 = .00148 or 1 in 676 hands or 1 to 676

Cobra

Re: A very simple problem
 

This is another question that can be answered with super easy math (read: non-combinations)

It's 1 in 26 that you get a specific card in holdem. Multiply this by itself, and you've got the odds of getting it two hands in a row. 26x26 = 676.

Rob

Re: A very simple problem
 

[ QUOTE ]


This is another question that can be answered with super easy math (read: non-combinations)

It's 1 in 26 that you get a specific card in holdem. Multiply this by itself, and you've got the odds of getting it two hands in a row. 26x26 = 676.

Rob


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The one problem I have with that is that you are only accounting for one of the two cards. The probability of a repeat from either hole card is the probability that either the first card drawn the second time matches either of the previous hole cards or the second card matches either of the previous.

P(A OR B) = P(A) + P(B) - P(A AND B)

P(A) is the probability the first card matches (2 in 52)
P(B) is the probability the second card matches (2 in 51)
P(A AND B) is the probability both cards match (2 in 52 * 1 in 51)

So,
P(A OR B) = 2/52 + 2/51 - (2/52 * 1/51).

Is there a flaw in this reasoning?

Re: A very simple problem
 

It's late, and I wish I was drunk so I'd have more excuses for not thinking clearly, but I'm definitely neither drunk nor thinking clearly. Damn.

I think you're answering a question he didn't ask. I answered his question the most simple way I could. I could do different math, so you'd understand where I'm coming from, and what question I'm answering.

Probability of getting the ace of spades (for example) in two consecutive hands is done like this;

The probability of NOT getting the ace of spades in hand one is as follows;

when you draw the first card, 51 of 52 times it's not the ace of spades. when you draw the second card, 50 of 51 times it's not the ace of spades. So the probability of not getting the ace of spades is;

(51/52) * (50/51), which is = 25/26. Since the probability of NOT getting the ace of spades is 25/26, the probability of getting it is 1/26.

Rinse, repeat. Two times in a row = (1/26) * (1/26) = 1/676

Again, this kind of math gets kinda tedious, and you may as well learn the combination math anyway, but it's not for everyone.

Can I go to bed now?

Rob

Re: A very simple problem
 

I wanted to say thank you to everyone who took the time to explain this. I appreciate it.

You guys are the best.

Good Luck to everyone.

Dov

Re: A very simple problem
 

When we read the post we were assuming he wanted the probability of a specific one of his two cards being dealt again, you are answering the probability that either or both of his two cards are dealt again.

There is a small error in your answer.

P(a or b)=P(a)+P(b)-P(a and b) this is correct.

Your answer is P(a)+P(b/a)-P(a and b)

Your terms should be 2/52+2/52-(2/52*1/51) you figure the probability of hitting either of your two cards irrespective of each card. I don't have much time to explain in detail maybe someone else will.

Cobra