# At what odds would it make sense to play a dice rolling game?

Assume that there is a game that drunk people are obsessed with winning. Once the drunk person wins, then s/he will quit. For simplicity sake, it's a simple dice game where the basic probability for getting any specific number is 1/6. Each roll will cost the person $1 to play. Before the game is played. The odds of winning on roll n is equal to p(n) = 1 - (5/6)n The first question then is on average how many rolls will the drunk make until they win? $$\mathrm E(n)=\sum\limits_{n=1}^{+\infty}n\cdot p(n)=$$ Assume that offering a bigger payout means that you will attract more people to play this game. What are the highest whole number odds that a game presenter can offer and still make money on average? Example: if the average number of times that someone plays the game (remember they play until they win) is 30 -then the game presenter can offer a payout of 29 to 1. - How do you define winning? Just guessing the correct number once, or getting into the black? – Arkamis Aug 22 '13 at 18:55 Winning is defined as rolling a 6. – Eric Raunig Aug 22 '13 at 19:03 The mean number of tosses is$6$. Standard fact about geometric distribution. – André Nicolas Aug 22 '13 at 19:06 The probability of winning on roll$n$is$(5/6)^{n-1}(1/6)$. – André Nicolas Aug 22 '13 at 19:29 Andre, if you get (5/6)<sup>n−1</sup>(1/6) then your odds of winning decrease as you play: – Eric Raunig Aug 22 '13 at 19:50 ## 1 Answer If$r$is the probability of FAILURE and$1-r$the probability of winning, the probability of winning on round$k$is equal to$(1-r)r^{k-1}$, as one has to lose all prior rounds before winning. Therefore, the expected value of the "winning round" should be: $$\sum_{k=1}^{\infty} k(1-r)r^{k-1}$$ Now$\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$for$|r| < 1\$. Using the trick of taking derivatives of both sides, we see that $$\sum_{k=1}^{\infty} kr^{k-1} = \frac{1}{\left(1-r\right)^2}$$ Thus the answer to your question should be: $$\frac{(1-r)}{(1-r)^2} = \frac{1}{1-r} = \frac{1}{1-\frac{5}{6}} = 6$$

As mentioned above, this is a geometric distribution which is the distribution of the number of failures before the first success (or vice versa).

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