I was thinking... According to the Monty Hall problem, knowing that the sheep wasn't behind a certain door did not mean a change of the initial probability.

Wouldn't it be applicable in poker when everybody folds to the button? IE in a full ring game, if there was an X% chance that at least one opponent would have a Sklansky group 1 hand, and everybody folded to the hero on the button, doesn't it mean that the SB and BB now have the probability of the entire table to have the said hand?

Your thoughts on this would be highly appreciated.

It doesn't work that way.
If there is X% chance that player Z has a "Sklansky group 1 hand" and it's Y% chance that anyone at the table have it then Y% goes down everytime someone fold. X% stays the same for every new opponent.

A key component of Monty Hall is that the host can never reveal the door with the prize, and therefore must always open a door with a donkey.

Something similar might be if the dealer looked at the deck and removed a non-ace card from it.

There is a very small increase in probability. The average hand has 2/13 = 0.154 Aces in it (or any other rank). The average Sklansky Group 1 hand has 0.57 Aces. Therefore, the average non-SG1 hand has 0.145 Aces. Kings are the same as Aces, for Queens and Jacks the average non-SG1 hand has 0.148.

If we know that none of seven hands are SG1, then on average they contained 1.01 Ace, 1.01 King, 1.04 Queens and 1.04 Jacks between them. These are less than the 1.08 average for each card if we know nothing about the seven hands. So on average there are 0.07 extra Aces and Kings in the deck and 0.04 extra Queens and Jacks for the button and blinds. This slightly increases the chance of an SG1 hand.

The Monte Hall effect relies on a host who knows all the cards. Suppose the cards are dealt at random, but then he rearranges the cards so that any SG1 hands go first to the big blind, then counterclockwise around the table. In that case it's true that the big blind has almost all the probability of a table's worth of SG1 hands.

So, to get it straight, say there are two dark jars. In one there are 5 marbles, and in the other there are 4 marbles and one candy. Now, a kid has to choose one of the jars. He does so, opens it, and takes out one item which is found to be a marble. Is his initial chance of finding a candy in that particular jar changed, or is it still 50%??

It is less. Consider 10 kids, each of whom make different choices. Five choose the candy jar, four pick marbles and one picks a candy. Five choose the other jar, all five get marbles. Now, given that a kid picked a marble, what's the chance that he picked the candy jar? Four out of nine, not 50%.

In the Monte Hall problem, there is a host who knows all. After the kid picks a jar, but before he picks a candy or marble, the host pulls a marble out of the jar the kid picked. He always does this, whether the kid picked the candy jar or not, and he always pulls out a marble. In that case, there's still a 50% chance that the kid picked the candy jar.

Mathematically, this is a nice problem for using Bayes.

P(Candy Jar|Marble) = P(M|CJ) P(CJ) / P(M)

= (4/5)*(1/2) / (9/10)

= (4/10) / (9/10) = 4/9

Probability of picking marble from a 5 marble jar: 1
Probability of picking marble from a candy jar: 4/5

Odds that marble came from candy jar: 4/5:1 = probability of 4/9.

Thanks for the input, gents. It's much clearer to me now.

good point. Otherwise "Deal or No Deal" would be way too beatable. If Howise always knew where the million was and chose for you ones that weren't it would make the switch at the end obligatory. as it plays though with no knowledge and the player picking randomly, I doubt it changes the odds which of the final two the player opens.

Maybe I'm wrong though. Any slight odds difference there?

On another note: I'm no probability expert, but this question still screamed to me "Bayes"... Good call, all!