Is there a math concept that never made sense to you? Which one or ones do you wish someone would explain in new ways?
If none of these come close you can mention something as a comment.
Yes I think fractions cause people as many problems as these other "more advanced" ideas. I've seen students in calc 2 who still had messy ideas about fractions. It's not trivial and just because we teach some of it to 5th graders doesn't mean everyone knows how they work.
@futurebird I was in a "math/physics/informatics" profiled class, and our math teacher was an absolute legend. Friends who went on to study math easily coasted on what they learned in high school for a year or two.
That said, conditional probability remains Black Magic to me.
1/ Conditional probability explained to my kids:
Deck of cards. Shuffle. Now draw cards until you get an ace:
P(Ace) = (Number of favorable outcomes) / (Total number of possible outcomes)
Let's break it down:
Favorable outcomes: These are the outcomes we're interested in, which is drawing an Ace. There are 4 Aces in a standard deck.
2/ Total possible outcomes: This is the total number of cards in the deck, which is 52.
So, plugging in the numbers:
P(Ace) = 4 / 52
This can be simplified to:
P(Ace) = 1 / 13
Therefore, the probability of pulling an ace from a standard deck is 1/13, or approximately 7.69%.
This equation represents the basic foundation of probability.
Also card counting, which will get your picture in The Book and you'll never get in a casino again
@tuban_muzuru
One can follow up with the questions of :
-you want the ace of diamonds specifically
-you already tried twelve times, how likely is it you'll succeed the next?
-you don't put the (wrong) cards back in the deck
1/ You want the Ace of Diamonds specifically:
Favorable outcomes: Now there's only 1 favorable outcome (the Ace of Diamonds).
Total possible outcomes: Still 52 cards in the deck.
So, the equation becomes:
P(Ace of Diamonds) = 1 / 52
The probability of pulling the Ace of Diamonds specifically is 1/52, or approximately 1.92%.
2/ You already tried twelve times, how likely is it you'll succeed the next?
This is where conditional probability comes into play! Here's how it works:
Assuming you didn't put the cards back: Each failed attempt changes the deck's composition. After 12 failed attempts, there are only 40 cards left.
Assuming none of those were aces: All 4 aces are still in the deck.
So, the equation becomes:
P(Ace | 12 failed attempts) = 4 / 40 = 1/10
3/ The probability of pulling an Ace on the 13th attempt, given 12 failed attempts without replacement, is 1/10 or 10%.
3. You don't put the (wrong) cards back in the deck:
This is the key to understanding how conditional probability works. Each time you draw a card and don't replace it, you're changing the possible outcomes for the next draw. This is called "sampling without replacement."
4/4
It makes each draw dependent on the previous ones. The outcome of one draw affects the probabilities of the following draws.
It generally increases the probability of the desired outcome (in this case, drawing an Ace) with each failed attempt.
This is just the beginning. Probability is the real world.
@tuban_muzuru @rysiek @futurebird Invaluable in a game of bridge, though.
@benfulton @rysiek @futurebird
A friend of mine in high school ( with an Erdős Number of 1 ) went on to become a very serious quant. Also a phenomenal poker player.
But he and I both hated bridge.
@futurebird haha I still don't get it.
Draw until you get an ace. Probability is 1. You will get an ace if you keep drawing. Probability of drawing an ace before only 4 cards remain?
Is it
4/52
plus 48/52x4/51
plus 47/51x4/50
plus... Down to
Plus 2/6x4/5?
No.
What is conditional probability?
It's the probability of an event happening given that another event has already occurred. Think of it as narrowing down the possibilities. We're not looking at the whole universe of events anymore, just a specific subset.
@tuban_muzuru @japonica I'd like to not be mentioned in this thread, thanks