1 in 4, two chances, only one needs to hit Currently in disagreement with a friend over the probability of a specific situation.
My friend believes that if you have a 1 in 4 chance of something happening, getting two opportunities at it makes it 2 in 4, or 50%.
Perfect example - a deck of cards. If you flip the top card the chances of a heart are 1 in 4. If you flip the second card, this is another 1 in 4 (waiving the small adjustment made for whichever suit the first card is. For the sake of argument, why not make it a second deck of cards and flip the top card there).
My friend is an accountant, and therefore probably better with numbers than me. I haven't done the maths, but my gut tells me that flipping another card definitely doesn't make it a 2 in 4.
Without getting hung up on cards because of the slight tweak to probability (removing 1 of the 13 cards of 1 of the 4 suits) :
If I have a 1 in 4 chance of something happening, and I get a second chance at it, does this double the probability to 2 in 4?
 A: Your friend is wrong. If we followed his reasoning, flipping a coin twice would give you 100% chance to get heads (each flip has a 50% probability). This is obviously not true. Then how should you calculate the probability in your example?
Suppose we have two independent trials, each with $1/4$ probability of success and therefore $3/4$ probability of failure. Then the probability that neither are successfull is $(3/4)\cdot (3/4)=9/16$. This means that the probability of at least one success is $1-9/16=7/16$, which is slightly less than 50%.
A: This depends entirely on the context.
Let $A$ be the event that you succeed on the first attempt.  Let $B$ be the event that you succeed on the second attempt.  You tell us in the problem statement that $Pr(A)=Pr(B)=\frac{1}{4}$.  You are asking how to calculate $Pr(A\cup B)$
In general, $Pr(A\cup B)=Pr(A)+Pr(B)-Pr(A\cap B)$.  The answer will depend on $Pr(A\cap B)$.  If $A$ and $B$ are independent events then $Pr(A\cap B)=Pr(A)\times Pr(B)$ in which case 
$$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)=\frac{1}{4}+\frac{1}{4}-\frac{1}{16}=\frac{7}{16}=0.4375$$
This happens for example in the situation where you have four cards, one of which is the ace of spades, and you draw a card at random trying to get the ace of spades.  After your first attempt, you put the card you drew back, mix the cards up well, and draw again.

It could have been the case that $A$ and $B$ are mutually exclusive events in which case $Pr(A\cap B)=0$ and you would have
$$Pr(A\cup B)=Pr(A)+Pr(B)-Pr(A\cap B) = \frac{1}{4}+\frac{1}{4}-0=0.5$$
This would have happened for example in the situation where you have four cards, one of which is the ace of spades, and you draw a card at random trying to get the ace of spades.  After your first attempt, you do not put the card you originally drew back and you instead draw a random card from the remaining three cards.

Further, it could have been in the other extreme where $A$ and $B$ are actually the same event in which case $Pr(A\cap B)=Pr(A)=Pr(A\cup B)$ and you would have
$$Pr(A\cup B)=Pr(A)=\frac{1}{4}$$
This would happen for example in the situation where you have four cards, one of which is the ace of spades, and you draw a card at random trying to get the ace of spades.  After your first attempt, you put the card you just drew onto the very top of the pile and draw the top card from the pile which is of course the very same card you just drew so you guarantee that you draw the same card twice in a row.

You can show as a result that being told that $Pr(A)=Pr(B)=\frac{1}{4}$ that the probability $Pr(A\cup B)$ is some number between $0.25$ and $0.5$ and without additional information about the overlap between $A$ and $B$ every number in that range is possible.

For the specific problem of one full 52-card deck, flipping the top two cards, and looking to see if at least one of them is a heart this would be $\frac{1}{4}+\frac{1}{4}-\frac{1}{4}\times\frac{12}{51} = \frac{15}{34}\approx 0.44118$.
Meanwhile, if you have two full 52-card decks, and you flip the top card from each this falls under the first scenario I talked about where the events are independent yielding the answer of $\frac{7}{16}=0.4375$ that we had earlier.
A: Your gut feeling is right and your friend is wrong. With four (independent) events, each with probability $\frac14$ of happening, your friend's reasoning would conclude that it is guaranteed that at least one event happens – false, since there's a $\frac{3^4}{4^4}\approx\frac13$ chance that none of the events materialises.
With smaller probabilities, like that of winning the lottery, such simple addition is a useful approximation to the real probability, but it's just that – an approximation, which gets worse as the summands increase.
A: You have two decks of cards. You draw card $1$ from the first deck and card $2$ from the second deck.
The probability that you draw a least one heart is $$P(\text{card 1 is a heart & card 2 is not a heart})$$ $$+ P(\text{card 1 is not a heart & card 2 is a heart})$$ $$+ P(\text{card 1 is a heart & card 2 a is heart}),$$ which is $$\frac{1}{4} \cdot \frac{3}{4} + \frac{3}{4} \cdot \frac{1}{4} + \frac{1}{4} \cdot \frac{1}{4} = \frac{3}{16} + \frac{3}{16} + \frac{1}{16} = \frac{7}{16}.$$
