# Profit estimation with a dice: 10 dollars for 6, -1 dollar for anything else

I recently found the following question: What is your profit estimate throwing a dice in the long run if you get 10 dollars for each time you hit 6 and lose 1 dollar for any other number?

I tried to use probabilistic "common sense" to arrive to the following wrong answer:

Probability tells me that every 6 throws I get one 6 and 5 different numbers. So every 6 throws my profit will be:

10x1 - 1x5 = $5 So if i keep throwing the dice forever i will have a steady flow of 5 bucks and I will eventually become a millionaire. Where am I wrong in my thinking? - Should be move to Stats.SE, because it's off topic here. – SRKX Dec 13 '12 at 0:17 ## 3 Answers If you want to model this using probability theory, you can define a stochastic variable$X$as follows: $$X = \left\{ \begin{array}{l l} 10 & \quad \text{with probability p=\frac{1}{6}}\\ -1 & \quad \text{with probability 1-p=\frac{5}{6}} \end{array} \right.$$$X$models the payoff of a trow of dice. As Freddy said, this is an expectation problem, so let's compute the expectation of$X$: $$\mathbb{E}[X] = 10 \cdot p + (-1) \cdot (1-p) = \frac{10}{6} + \frac{-5}{6} = \frac{5}{6}$$ So, on average, you make on a single throw a profit of$\frac{5}{6}=0.83$USD. Now, the term "on the long run" is a bit ambiguous in the question. To stay abstract and general, assume we make$n$throws and we denote the$i$-th throw as$X_i$. We can model the result of$n$throws as: $$S_n= \sum_{i=1}^n X_i$$ To know the expected value of your wealth after$n$throws, you need to compute the expectation of the variable$S_n$: $$\mathbb{E} [S_n] = \mathbb{E} \left[ \sum_{i=1}^n X_i \right] = \sum_{i=1}^n \mathbb{E} \left[ X_i \right] = \sum_{i=1}^n \frac{5}{6} = n \cdot \frac{5}{6}$$ So, over the long run (assumed to be$n$throws), you can expect to make$ n \cdot \frac{5}{6}$USD on average. This means that you could very well throw$n$dices and get no 6, ending up with a wealth of$-n$. But this is basic statistics, so I'll ask the question to be moved to Stats.SE. - The key here lies in risk management. The dice player must survive the long strings of losing throws to make that money. Due to the high volatility of the expected return, even with Kelly's bet sizing, you wouldn't be able to put on huge bets with respect to bankroll. - agree with your comment, which kinda re-emphasizes the difference between deterministic and probabilistic outcomes. – Matt Wolf Dec 6 '12 at 7:43 Its a simple expected value question: Probability of throwing a 6 is 1/6 Probability of not throwing a 6 is 5/6 thus expected pay off per roll: 10 dollars * 1/6 + (-1 dollar) * 5/6 = 5/6 dollars Edit: And several of your above assumptions are plain wrong: • "Probability tells me that every 6 throws I get one 6 and 5 different numbers." -> That is NOT what "probability" tells you. The result is an expected value, meaning, something you can expect on average given a large enough sample set. (please google "law of large numbers") • Your profit will not necessarily be 5 dollars after 6 throws. -> please see above • "So if i keep throwing the dice forever i will have a steady flow of 5 bucks and I will eventually become a millionaire." -> No, you will end up with infinitely large wealth if you get to roll infinitely many times. - Oh. So my answer is right. Every 6 rows I get 5. So every row I get 5/6 dollars. :) – HFT Trader Dec 4 '12 at 8:51 Nope, read again Freddy's answer. Also perhaps it is worth mentioning that 5/6 dollars is$0.83333... –  Hugues Fontenelle Dec 4 '12 at 13:03
Naa, he's kinda right. Every 6 throws he'll get \$5, on average. He's just stating it in a convoluted way. –  eckesicle Dec 6 '12 at 5:28
@eckesicle, makes quite a difference, whether you state something is guaranteed or whether something occurs on average over a large sample set? If you disagree then you basically say that probability theory is useless and the law of large numbers makes no sense. –  Matt Wolf Dec 6 '12 at 5:32
I don't disagree, but I also think that OP is new to the subject and probably haven't been taught the mathematical vocabulary to express himself in the proper way. From OP's statements I think it's inferred that we're talking about expectation, or 'probabilistic common sense' as he put it. His lack of formalism does not make him wrong. –  eckesicle Dec 6 '12 at 6:43