Settle question re deck odds

[Tajmari]

Hi! My husband and daughter -- both of whom are not very good poker players -- are insisting that it is harder to make a royal flush (or any kind of flush) when you are playing in a full game, say, with 9 players rather than in a heads up game.

They claim that since a lot of cards are "missing" it makes it "harder" to make your hand.

I say that is nonsense reasoning since the deck is always 52 cards, and whether the cards are mucked or still in the deck, they are still unknown and should be used to calculate odds.

If anything, it seems to me that it would be more likely to make a royal flush in a full game with a bunch of muckers since it's reasonable to assume that people who fold are most likely folding at least some low cards, making the remaining cards in the deck probably higher in value than those coming from a fresh deck.

Thanks!

[uDevil]

You are right. It's not more likely in a full game, however, unless you're referring to a bunching effect and you are looking at your cards after some people have folded. That effect is small.

[SunyD]

[ QUOTE ]

I say that is nonsense reasoning since the deck is always 52 cards, and whether the cards are mucked or still in the deck, they are still unknown and should be used to calculate odds.


[/ QUOTE ]

I think this is interesting, and something I was thinking about the other day. When we calculate odds, we are pretty much calculating our "best-case scenario" to win the hand. Of course we do not know if our outs are sitting in the deck or in the muck pile, and so we still consider them.

So if I have 12 outs, with one card to come, I am AT-BEST going to win 25% of the time... not necessarily 25% of the time. Is my reasoning flawed here? Comments welcome.

[uDevil]

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So if I have 12 outs, with one card to come, I am AT-BEST going to win 25% of the time... not necessarily 25% of the time.

[/ QUOTE ]

If the cards are turned up so that you know you have 12 outs on the turn, you will lose 36 times for every 12 times you win. You'll win 25% of the time on average. Simple as that.

Suppose you take 2 cards, say the A[img]/images/graemlins/spade.gif[/img] and A[img]/images/graemlins/club.gif[/img], and shuffle them. Now pick up the top card and throw it out the window. What is the probability that the remaining card is the A[img]/images/graemlins/spade.gif[/img]?

It either is or is not the A[img]/images/graemlins/spade.gif[/img], but until you look, the probability is...?

Suppose you had many decks and did this many times. How frequently did the A[img]/images/graemlins/spade.gif[/img] get thrown out? So how frequently is the remaining card the A[img]/images/graemlins/spade.gif[/img]?

[AngusThermopyle]

[ QUOTE ]
[ QUOTE ]

I say that is nonsense reasoning since the deck is always 52 cards, and whether the cards are mucked or still in the deck, they are still unknown and should be used to calculate odds.


[/ QUOTE ]

I think this is interesting, and something I was thinking about the other day. When we calculate odds, we are pretty much calculating our "best-case scenario" to win the hand. Of course we do not know if our outs are sitting in the deck or in the muck pile, and so we still consider them.

So if I have 12 outs, with one card to come, I am AT-BEST going to win 25% of the time... not necessarily 25% of the time. Is my reasoning flawed here? Comments welcome.

[/ QUOTE ]

Your cards + His cards + board = 8 cards so 44 "unseen"

So, instead of 12, let's say 11 outs (to get your 25%).

10 handed play. Before the dealer burns and turns the river, there are 26 cards left in the 'stub'. (20 hole cards, 4 board, 2 burns, 26 stub).

"Best case", all 11 outs are in the stub, and you have 43% chance. (actually, "best case" is one of your 11 outs is sitting as the second card in the stub, right after the burn, so you win 100% of the time).

"Worst case", all 11 outs are in the muck/burns. You win 0% of the time (unless dealer drops deck, and mixes all the muck, burns and stub together before burning and turning. Then you have, amazingly, 25% chance of winning.)

[SunyD]

[ QUOTE ]

If the cards are turned up so that you know you have 12 outs on the turn, you will lose 36 times for every 12 times you win. You'll win 25% of the time on average. Simple as that.

Suppose you take 2 cards, say the A[img]/images/graemlins/spade.gif[/img] and A[img]/images/graemlins/club.gif[/img], and shuffle them. Now pick up the top card and throw it out the window. What is the probability that the remaining card is the A[img]/images/graemlins/spade.gif[/img]?

It either is or is not the A[img]/images/graemlins/spade.gif[/img], but until you look, the probability is...?

Suppose you had many decks and did this many times. How frequently did the A[img]/images/graemlins/spade.gif[/img] get thrown out? So how frequently is the remaining card the A[img]/images/graemlins/spade.gif[/img]?

[/ QUOTE ]

The probability the first time is 1:2 (50%), and the probability of it over a large sample size is still 1:2 (50%). Hence, whether one of our outs is sitting in the muck doesnt change the probability "on average".

The little light bulb just went off in my head [img]/images/graemlins/smile.gif[/img] Thanks.

[AKQJ10]

[ QUOTE ]
I think this is interesting, and something I was thinking about the other day. When we calculate odds, we are pretty much calculating our "best-case scenario" to win the hand. Of course we do not know if our outs are sitting in the deck or in the muck pile, and so we still consider them.

[/ QUOTE ]

You also don't know if the blanks that you don't want to hit are in the stub or the muck. But we can safely assume that the expected ratio of outs to blanks in the stub is the same as the ratio outs:blanks in the muck. That's why we use 47 or 46 as our denominator (i.e., the total number of unseen cards in muck, opponents' hands, stub, and burn cards).

I suppose instead of saying the ratio of spades in the stub on the river is 9/46, we could say it's 9*(35/46)/35, where 35 is the number of cards left in the stub (assuming 9 other hands, three burn) and the 9 * 35/46 means that the stub has its fair share of the nine remaining spades.

But that's ridiculous because it's exactly the same result. The 35s in the expression cancel out, so you may as well think of stub + other hands + burn + muck as one entity.

Same with the OP -- the royal flush cards could well be in people's hands, but you don't have any reason to expect that (except for the bunching effect uuDevil mentioned).

[WCGRider]

On a random note, you are more likely to hit a royal flush heads up beacuse the hands in that range are always played heads up, whereas at a full ring table less often.

For example, you are in MP with AJs, UTG who has been playing tight opens up for 5x the BB. You can pretty safely fold here, whereas in heads up folding AJs to anyhting but a monster raise is pretty much incorrect.

So you see more flops, more chances, etc.

[Small Fry]

And on a more random note...... Unless you're playing small stakes in a casino with bonus payouts for specific hands. Then players will play any combination of hands possible that might get them that 1:10,000 shot of hitting and turning a $2 call into a $400 jackpot...

[uDevil]

[ QUOTE ]
[ QUOTE ]

If the cards are turned up so that you know you have 12 outs on the turn, you will lose 36 times for every 12 times you win. You'll win 25% of the time on average. Simple as that.

Suppose you take 2 cards, say the A[img]/images/graemlins/spade.gif[/img] and A[img]/images/graemlins/club.gif[/img], and shuffle them. Now pick up the top card and throw it out the window. What is the probability that the remaining card is the A[img]/images/graemlins/spade.gif[/img]?

It either is or is not the A[img]/images/graemlins/spade.gif[/img], but until you look, the probability is...?

Suppose you had many decks and did this many times. How frequently did the A[img]/images/graemlins/spade.gif[/img] get thrown out? So how frequently is the remaining card the A[img]/images/graemlins/spade.gif[/img]?

[/ QUOTE ]

The probability the first time is 1:2 (50%), and the probability of it over a large sample size is still 1:2 (50%). Hence, whether one of our outs is sitting in the muck doesnt change the probability "on average".

The little light bulb just went off in my head [img]/images/graemlins/smile.gif[/img] Thanks.

[/ QUOTE ]

Miraculous, since I mangled the argument. The idea of course was to compare the result to what happens with a "complete" 2-card deck. Insomnia is no excuse for posting gibberish, but hey, whatever works.... [img]/images/graemlins/smile.gif[/img]