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# Poker Game for math geeks

## Question

Here's one for you math geeks.

Writersblock is going to a poker game tonight. 5 other players will be there. Writersblock is a very, very good poker player and the other 5 players are really bad. 1000 hands of Texas Hold'em will be dealt. Assuming Writersblock doesn't make any mistakes and plays every hand dealt him perfectly as against the other hands which he also reads perfectly, what are his odds of winning every hand?

For those who don't know, Texas Hold'em Consists of 2 cards dealt from a 52 card deck to each player. One is then "burnt", and Three common cards are dealt. Another is burnt and one more common card is dealt. Then a third is burnt and a final common card gets dealt. We can assume all 1000 hands make it to the "river" or last card dealt, so that every hand will have 20 cards dealt. No wilds and standard poker hands apply.

## 7 answers to this question

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There's no answer as it depends on circumstances. Are these players so poor that they fold as soon as soon as someone raises regardless of the hands they're holding? Then a good player should win every hand. Or maybe they play so poor that they stay in every hand regardless of how bad their hands are and how much someone raises. Then you will win less hands but should make far more money.

You may play 'by the book' regarding when to hold and when to fold and win far more hands playing a thousand hands tonight then you will playing a thousand hands tomorrow night against the same opponents.

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HA! You got that pretty quickly. I was hoping to have weeks of statistics discussion about this, but what I was looking for is that if I play perfectly and they play poorly, it all depends on the betting, not the cards and I should win 100% of the hands.

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it all depends on the betting, not the cards and I should win 100% of the hands.

I said you would win 100% of the hands if your hypothetical opponents folded every time someone raised. You would win simply by knowing this fault in your opponents and raising every hand.

In your OP however, you stated that we assume all hands go to the river and there's no assumption made about how easily your opponents fold. In this case, playing perfect, by the book poker, does not mean you will win all 1000 hands. That just ain't gonna happen.

Someone goes all-in post flop. Flop is 2h 7h 10h. You have 8h Ah. Statistically speaking, you should call in a heart beat. Are you guaranteed to win?

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Since poker is designed with enough betting to counteract all the random chances, a very very very good player can usually almost always beat a very very bad player.

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I guess even a really bad player would not fold the nuts on the river to any bet, so I do stand corrected. The nuts being the best possible hand for the 5 common cards dealt.

Just out of curiousity, what are the odds of getting the "nuts" on any given hand if it always goes to the river? I'm not strong enough in statistics to figure that out.

For example, the odds of getting the nuts if the 5 cards showing are AH KH JH ANY ANY would be the odds of getting the QH and 10H dealt to you from the deck. But what would be the odds of getting the nuts if the shown cards are 10H 4C 2S 8S KD? The nuts there would be any 2 of the 3 kings left. Is there a way to figure out the odds of getting dealt a "nut" hand regardless of what 5 common cards are dealt?

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For example, the odds of getting the nuts if the 5 cards showing are AH KH JH ANY ANY would be the odds of getting the QH and 10H dealt to you from the deck.

Well, since you already have your hole cards, I'm guessing you just haven't looked at them yet.

There are 47 unknown cards and before looking at one of your hole cards, there are 2 outs, so:

2/47

After looking at one of your hole cards and you see you have either the QH or 10H, there is 1 out and 46 unknown cards:

2/47 x 1/46 = 0.000925 = 0.925%

Percentage to odds: (1/Percentage) -1

So the odds are about 1110 to 1. For probability, don't subtract 1. (1111 to 1)

But what would be the odds of getting the nuts if the shown cards are 10H 4C 2S 8S KD? The nuts there would be any 2 of the 3 kings left. Is there a way to figure out the odds of getting dealt a "nut" hand regardless of what 5 common cards are dealt?

3/47 x 2/46...

It's essential to calculate pot odds and implied odds if you want to be a winner in the long run. See this short article and the basic rule of thumb on page 2.

Here are two more helpful articles:

<!-- m --><a href="http://www.powerwurks.com/pages/texas-h" target="_blank">http://www.powerwurks.com/pages/texas-h ... t-odds.php</a><!-- m -->

<!-- m --><a href="http://www.powerwurks.com/pages/texas-h" target="_blank">http://www.powerwurks.com/pages/texas-h ... d-odds.php</a><!-- m -->

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Yeah, I know about pot odds, implied odds and position betting. I usually can bang them out dirty and quick, but close enough to keep me winning.

I think I was confusing on my second point. Here's what I wanted to know: I have any 2 cards; what are the odds that I will have the nuts after the flop, turn, and river come?

This is what I am trying to figure out and am pathetically unequiped mathmatically to do so. My illustration with the two hands was only to illustrate that the number of cards to make the nuts will change with whatever the 5 common cards are. Is there a way to figure this for any 2 cards with any 5 commons - not a specific hand?

Any statistics gurus out there?

EDIT: Let's figure this for a 6 player table. After the hole cards are dealt, there are 40 left in the deck. Burn 1 and turn 3, burn 1 and turn 1, burn 1 and turn 1. 32 cards in the deck after all are dealt.

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