What is wrong with my two fair dice probability question reasoning? There is a game where you are asked to roll two fair six-sided dice. If the sum of the values equals 7, then win £21. However, must pay £5 to play each time both dice are rolled. Do you play this game?
One way to think about this is that getting a 7 comes with 1/6 chance, and to make money we need to get 7 at a rate of 1/4, so the answer is not to play.
Another way to think about it is: what is my chance of throwing a 7 at least once in every 4 throws? In which case I would calculate a probability of not throwing a 7 4 throws in a row (5/6)^4, and then subtract this from 1 to get a probability of throwing at least one 7. Which is 1 - (5/6)^4 = 0.52. By this logic I would play the game.
Both of these answers cannot be correct. Could someone explain to me which one is incorrect and why? Thanks!
EDIT: wow, this is the first time I asked a question on StackOverflow, did not expect to get so many responses. Thank you all, I am very grateful!
 A: Your first approach is correct, at least approximately. (If you won £20 on a 7, then you'd need to roll 7's at a rate of $\frac14$ to break even. Since you win £21, you can afford a slightly lower rate, though not as low as $\frac16$.)
Your second approach is wrong. It's true that more than half the time, you win at least once over a sequence of four throws. However, all that tells you is: more than half the time, you don't lose money. This doesn't mean that you break even on average! In fact, it tells you that:

*

*Approximately $48\%$ of the time, you roll no 7's, and so you lose £20 over the course of four throws.

*Approximately $52\%$ of the time, you roll at least one 7, and you make a net profit of at least £1. Most of the time, this will be exactly £1.

This doesn't look as good anymore. In general, just because a random variable is positive more than half the time, doesn't mean its expected value is positive.
A: You are correct about the probabilities involved here. The probability of rolling a total of $7$ on a pair of fair dice is indeed $1/6$. Likewise, the probability of rolling a $7$ at least once in four rolls is $1-\left(\frac{5}6\right)^4$ which is about $52\%$ - and this might equally well be interpreted as "the probability of earning money after four repetitions of the game". We have to be careful of how we interpret such quantities, however.
Imagine the following variation of the game, to which the logic of your second answer applies:

You pay $£20$ to play a game where you get to roll a pair of dice up to four times. If you get a $7$ on any roll, you earn $£21$ back.

Or, equivalently, by looking at the net gains for either outcome:

Roll a pair of dice up to four times. If you get a $7$ on any roll, you gain $£1$. Otherwise, you lose $£20$.

While it's true that you will win this game $52\%$ of the time, it's probably not a game you want to play - it's basically flipping a coin between "earn $£1$" and "lose $£20$". Your second argument posits a world where we'd play this game because we'll probably win - but it ignores that the consequence of losing far outweighs the benefit of winning. This game is more pessimistic than the original one (since it doesn't allow you to win multiple prizes), but more cleanly illustrates why your second reasoning is misleading*.

The typical way to evaluate this sort of game rigorously is by looking at expected values rather than probabilities. Expected value is the answer to the following question:

If I played this game repeatedly, how much would I earn or lose in an average round?

This can be computed by multiplying the possible outcomes by the respective probabilities - in the case of this problem, you have a $1/6$ chance of earning a net of $£16$ and a $5/6$ chance of earning $-£5$ (i.e. losing $£5$). You can calculate the expected value as:
$$\frac{1}6 \cdot £16 - \frac{5}6 \cdot £5 = -£1.50$$
Showing that you expect to lose $£1.50$ in an average round of this game - meaning it's not worth playing. We could also rewrite this as
$$\frac{1}6\cdot £21 - 1\cdot £5 = -£1.50$$
to capture the equivalent idea that we earn $£21$ with some probability, but always must pay $£5$ to enter - and this expression aligns well with your intuition: the amount we expect to be rewarded with (after paying to play) is $£21$ times the probability that it occurs - and for this to outweigh the entrance fee, it needs to happen about a quarter of the time, as you correctly say.
You could also imagine this as playing the game six times - and considering that you expect to win once (earning $£21$) but pay $£30$ to enter, leaving a loss of $£9$ - which is just six times the previously calculated loss for a single game.

As a matter of curiosity not directly implicated in your question: these numbers are somewhat related. Your second calculation can be described as:

The probability that I earn more money than I spend while playing $4$ rounds of the game.

This turns out to be greater than $50\%$, but you could imagine asking for a different probability such as:

The probability that I earn more money than I spend while playing $100$ rounds of the game.

I won't go into details, but this probability comes out to only $3\%$ - and if you play $200$ rounds, the probability of coming out ahead drops to $0.4\%$ and all the way down to $0.07\%$ after $300$ rounds. As a general rule, if you play a game with negative expected value, the probability of coming out ahead decreases at an exponential rate with the number of games played. The fact that you're more likely than not to come out ahead after $4$ games is true but misleading - the situation only gets worse with the more games played, which is one of the things that the expected value tells you (and one reason why expected value is a very good tool for evaluating games of chance).

*Note: I say "misleading" rather than "wrong" because it is a correct calculation of something - but it'd be a stretch to conclude from that probability that this would be a good game to play if you were just walking down the street. However, there are natural contexts where the second kind of calculation is useful - for instance, in the last round of many board games or card games, you might be able to directly calculate a probability of a certain move causing you to win the game via similar logic - where the expected value calculation of "what would happen if I did this a bunch and averaged it" isn't actually relevant (and might lead to different - and therefore incorrect - results). That said, expected value is usually the right thing to think about except when you have a compelling reason to consider something else.
A: Let $\$X$ be the nett winning in one game. Then $X$ has two possible values: $16$ or $-5.$
In one game, the probability of winning is
$$P(X=16)=\frac6{36}=\frac16.$$
Over the long run, the average value of $X$ is $$\sum_x xP(X=x)\\=16\left(\frac16\right)+(-5)\left(1-\frac16\right)\\=-1.5.$$
(This quantity is called the expected value of $X$. It is essentially a weighted average, where the the probability $P(X=x)$ is the "weight" of its associated outcome $x.$
In other words, the average winning per game is the signed sum of the average \$profit and the average \$loss; the latter, for example, equals the negative profit per unsuccessful game multiplied by the proportion of the games that are unsuccessful.)
Thus, I expect to lose $\$1.50$ per game. So, I would not play this game.
A: There are 36 possible combinations when you roll two dice. How many of these add up to 7? (6+1, 5+2, 4+3, 3+4, 2+5, 1+6) --> 6. So there is a 1/6 chance that you will win.
Let $X$ be the random variable that denotes the amount you win. With probability 1/6, you win 21 and with probability 5/6 you win 0. So your expected winnings are $21 \times 1/6 + 0 \times 5/6 = 21/6$. However, you also have to pay 5 to play (regardless of outcome), so your net expected winnings are $21/6 - 5 < 0$. Since the net winnings are expected to be negative, I would not play the game!
Let's say the probability of a winning combination was $p$. The net expected winnings would be $21p - 5$ based on the above calculations. For this to positive, you must have $p > 5/21$. So only play the game if the winning combinations have a probability > 5/21 (close to 1/4, though I am not sure how you concluded that it should be exactly 1/4).
