Imagine a slot machine with $N$ reels.

I want to calculate the probability $P$ that a player hits a certain sequence $A$, if the player has given the possibility to spin again (and only once again), keeping the symbols belonging to $A$ (if any has come out).

No problem with only one spin:

$$P(A) = \prod_{n=1}^NP(a_n)$$

where $P(a_n)$ is the probability that a symbol belonging to $A$ comes out on the reel $n$ (with $a_n$ potentially different from $a_{n+1}$).

But how to introduce the second spin into the calculation?


$$\mathbb{P}(A) = \prod\limits_{n=1}^{N}P(a_n) + (1-P(a_n))P(a_n) = \prod\limits_{n=1}^{N} 2P(a_n) - P(a_n)^2$$

Because for each wheel, either we succeed in the first attempt (with probability $P(a_n)$) or we fail (with probability $1-P(a_n)$) and then succeed in the second attempt (with probability $P(a_n)$).

  • $\begingroup$ Nice. I did not think it was possible to elaborate the solution by calculating each reel probability individually for the two spins and then putting all together, but the way you explained it makes absolutely sense. It will help me also for future problems. Thanks! $\endgroup$ – Luca Fagioli May 9 '13 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.