Quote:
Originally Posted by pkdk
answered in red
pkdk if you honestly believe this then you need to go back to statistics 101.
Your first 4 answers are correct - 2/4 - which can be expressed as 1/2. There is a 50% chance that a box contains a prize.
Your last 2 answers however fail to grasp something very fundamental to probability - which is that the fact that some unknown choices have been removed has zero impact on the probability.
In the case where 4 boxes exist, 2 of them randomly selected, have a prize, and 2 of them, randomly selected independently of the first selection, are obliterated by launching them into the sun, the destruction of 2 of the boxes has no effect on the odds of the other 2 having a prize.
(This isn't a quantum mechanics problem - this is a "standard reality" problem)
4 boxes exist, each box has a 50% chance of having a prize. 2 boxes are destroyed. The other 2 boxes still have a 50% chance of having a prize.
Just like in poker - you get dealt 2 cards - you "know" 2 values. You do not know the other 50. Thus, calculations of the odds of what other players hold are based on the 50 remaining values. Lets say you have K
K
When the dealer burns a card, but hasn't dealt the flop yet, the probability of someone else having an ace is 4/50 or 12.5%. The probability of someone else having K
is 1/50 or 2%. The probability of someone else having K
is 0% (0/50) because you KNOW where that value is.
When the flop comes out, 3 new values are known. the unknowns drop to 47. Those unknowns are still all the players cards, the burn card, and the stub. The known cards are on the flop and in your hand.
That's all that matters when we talk about poker probability. None of the above has any bearing on the next hand we play, or any other hand we play, nor is it influenced by any hands that have previously been dealt or by any hands yet to come.