Crazy Hand

[snoopy1239]

 I just see it as only having 2 outs against a full deck that can screw you that doesn't sound like 4-1 to me.

Well it's nearer 4.5 to 1 to be fair (18.453%). Don't look at it as 2 outs, its 4.5 to 1, 4.5 to 1.

Keep saying that to yourself over and over again. fourpointfivetooone

you can safely assume one of your other queens would be gone aswell and that just means you're in realy trouble.

How on earth is this a safe asumption? Which of the other players played their hand like AQ?

ah and here in lies the problem my friends...how many players are at this table? I am assuming ten, therefore I would always assume one of my queens to have been folded. There are not just 3 players in the hand...there are 5 or 7 more and that means 10 or 14 other cards are out...

it's a fair bet that one of your queens will therefore be gone.

That logic doesn't work here, Boldie. The chances of the Queen being folded are the same regardless of the number of players.

Personally, I think it's a close call, although I think I'd opt for a fold because one guy's gonna win a monster pot, which can't be bad if it's a 'top 3 go through' type set-up.

[MrsLime]

ah and here in lies the problem my friends...how many players are at this table? I am assuming ten, therefore I would always assume one of my queens to have been folded. There are not just 3 players in the hand...there are 5 or 7 more and that means 10 or 14 other cards are out...

it's a fair bet that one of your queens will therefore be gone.

That logic doesn't work here, Boldie. The chances of the Queen being folded are the same regardless of the number of players.

Not wanting to be overly pedantic, but Boldie IS correct when he says that the more players who are dealt in, the more likely it is that a queen will have been folded.  The important point is that this statement is irrelevant.

Let me draw an analogy.  We start with a full deck and I deal you a card.  The chances that I deal you a red card is, of course, 50-50.

Now let's start again, but first I will throw half the deck into the bin.  Of course it is very likely that some red cards will get thrown into the bin.  But when I deal you a card, it is still 50-50 that you get a red one.

Now, suppose I throw 51 cards into the bin.  We can say for certain that there are some red cards in the bin.  But it is still 50-50 that the one remaining card is red.

So, in exactly the same way, the odds of a queen popping out are the same however many cards have been thrown in the bin/folded by other players.  It is correct to say that there are fewer card left in the deck which are queens, but don't forget that there are also fewer cards left in the deck which AREN'T queens, and as such the proportion stays exactly the same.

[snoopy1239]

ah and here in lies the problem my friends...how many players are at this table? I am assuming ten, therefore I would always assume one of my queens to have been folded. There are not just 3 players in the hand...there are 5 or 7 more and that means 10 or 14 other cards are out...

it's a fair bet that one of your queens will therefore be gone.

That logic doesn't work here, Boldie. The chances of the Queen being folded are the same regardless of the number of players.

Not wanting to be overly pedantic, but Boldie IS correct when he says that the more players who are dealt in, the more likely it is that a queen will have been folded.  The important point is that this statement is irrelevant.

Let me draw an analogy.  We start with a full deck and I deal you a card.  The chances that I deal you a red card is, of course, 50-50.

Now let's start again, but first I will throw half the deck into the bin.  Of course it is very likely that some red cards will get thrown into the bin.  But when I deal you a card, it is still 50-50 that you get a red one.

Now, suppose I throw 51 cards into the bin.  We can say for certain that there are some red cards in the bin.  But it is still 50-50 that the one remaining card is red.

So, in exactly the same way, the odds of a queen popping out are the same however many cards have been thrown in the bin/folded by other players.  It is correct to say that there are fewer card left in the deck which are queens, but don't forget that there are also fewer cards left in the deck which AREN'T queens, and as such the proportion stays exactly the same.

true

I meant in comparison to other cards tho

[boldie]

ah and here in lies the problem my friends...how many players are at this table? I am assuming ten, therefore I would always assume one of my queens to have been folded. There are not just 3 players in the hand...there are 5 or 7 more and that means 10 or 14 other cards are out...

it's a fair bet that one of your queens will therefore be gone.

That logic doesn't work here, Boldie. The chances of the Queen being folded are the same regardless of the number of players.

Not wanting to be overly pedantic, but Boldie IS correct when he says that the more players who are dealt in, the more likely it is that a queen will have been folded.  The important point is that this statement is irrelevant.

Let me draw an analogy.  We start with a full deck and I deal you a card.  The chances that I deal you a red card is, of course, 50-50.

Now let's start again, but first I will throw half the deck into the bin.  Of course it is very likely that some red cards will get thrown into the bin.  But when I deal you a card, it is still 50-50 that you get a red one.

Now, suppose I throw 51 cards into the bin.  We can say for certain that there are some red cards in the bin.  But it is still 50-50 that the one remaining card is red.

So, in exactly the same way, the odds of a queen popping out are the same however many cards have been thrown in the bin/folded by other players.  It is correct to say that there are fewer card left in the deck which are queens, but don't forget that there are also fewer cards left in the deck which AREN'T queens, and as such the proportion stays exactly the same.

true

I meant in comparison to other cards tho

yep....my bad ofcourse...also explains why i have a tough time getting the nhang of pot odds.....*hangs head in shame*

[JungleCat03]

ah and here in lies the problem my friends...how many players are at this table? I am assuming ten, therefore I would always assume one of my queens to have been folded. There are not just 3 players in the hand...there are 5 or 7 more and that means 10 or 14 other cards are out...

it's a fair bet that one of your queens will therefore be gone.

That logic doesn't work here, Boldie. The chances of the Queen being folded are the same regardless of the number of players.

Not wanting to be overly pedantic, but Boldie IS correct when he says that the more players who are dealt in, the more likely it is that a queen will have been folded.  The important point is that this statement is irrelevant.

Let me draw an analogy.  We start with a full deck and I deal you a card.  The chances that I deal you a red card is, of course, 50-50.

Now let's start again, but first I will throw half the deck into the bin.  Of course it is very likely that some red cards will get thrown into the bin.  But when I deal you a card, it is still 50-50 that you get a red one.

Now, suppose I throw 51 cards into the bin.  We can say for certain that there are some red cards in the bin.  But it is still 50-50 that the one remaining card is red.

So, in exactly the same way, the odds of a queen popping out are the same however many cards have been thrown in the bin/folded by other players.  It is correct to say that there are fewer card left in the deck which are queens, but don't forget that there are also fewer cards left in the deck which AREN'T queens, and as such the proportion stays exactly the same.

Impeccable logic and a cracking looking picture of a cat, you should post more often Mrs Lime!