0
$\begingroup$

I'm trying to solve the following problem :

There is man in a casino that players makes a bet on the number $13$ at the roulette ( roulette has numbers between $1$ and $36$ ). He continues to make this bet until he finally wins; After he wins he then makes the same numbers of bets as he made on 13 but this time on 36.

What is the mean of the lost bets from the second round of betting ?

Well, i tried to solve it the in following way:

I've considered that this bet is a geometrical distribution.

Let $k$ be the number of times that he bet on number $13$ until he won;

$p = \frac{1}{36}$ , the probability of him winning a bet;

But i know that the mean of a geometrical distribution is the following :

$E(X) = \frac{1}{p};$

Can this be interpreted that, as an average he wins $ \frac{1}{36}$ games ; So that in k games the number of his wins is $\frac{k}{36}$;

Which means that the mean number of lost bets is $k- \frac{k}{36}$ ?

Is this approach correct ? If not can someone guide me to the correct one ?

Thank you!

$\endgroup$
1
$\begingroup$

Everything is correct until the last statement, which should be: the number of lost bets is $k-\frac{k}{36}$ (it seems that you mixed up the number of bets and probabilities). To get the final answer, you have to calculate the average of this quantity over the geometric distribution for $k$.

$\endgroup$
  • $\begingroup$ That is a typo. I'll correct it right now. I have to do something more? But isn't $ k- \frac{k}{30} $ is the mean number of lost bets when the total number of bets made was k? Shouldn't this be the final answer? $\endgroup$ – Eduard6421 Dec 30 '17 at 23:00
  • $\begingroup$ @Eduard6421 There's no "$k$" in the problem text, so I assume it's not the final answer. Since $k$ is the number of times that the man bet on number 13 until he won, $k$ is a random variable itself which has a (geometric) probability distribution. $\endgroup$ – colt_browning Dec 31 '17 at 1:10
  • $\begingroup$ K is indeed a random variable with geometric probability distribution. But isn't the average number of bets made equal to the mean of the probability distribution? ( $k=\frac{1}{p} = 36$) And we know that he loses $k-\frac{k}{36}$ on average? So the final answer would be $36 - \frac{36}{36} = 35$ ? $\endgroup$ – Eduard6421 Dec 31 '17 at 14:30
  • $\begingroup$ Yes, that's correct. $\endgroup$ – colt_browning Dec 31 '17 at 14:43

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.