a man considers 2 roulette strategies: betting $20 on red and hoping for best, or betting $1 at a time on red or black until he has won or lost $20. which strategy is better?

Question

anonymous · Answer

i think the second, since you have like a possibility of winning 17 dollars instead of losing  it all, u see

anonymous · Answer

?????/

anonymous · Answer

i dont get it

anonymous · Answer

how do you win $17

anonymous · Answer

anyone help please?

anonymous · Answer

please help

anonymous · Answer

where did al the maths genius go?

kropot72 · Answer

With a single bet of $20 the probability of winning is 0.5.
If $1 bets are placed, at least 20 games will be played before $20 can be lost.
By means of the binomial distribution it can be shown the the probability of winning 10 or more games out of 20 games is 0.5881.
Therefore it seems the second strategy is slightly better than the first.

anonymous · Answer

can you please elaborate on binomial part.

kropot72 · Answer

The probability of winning a game is 0.5.
According to the binomial distribution, if there are n trials with a probability p of success on each trial, then the probability of exactly x successes is
$$P(X=x)=\left(\begin{matrix}n \ x\end{matrix}\right)p ^{x}(1-p)^{n-x}$$
In the case in question, the probability of winning exactly 10 games out of 20 games is
$$P(X=10)=\left(\begin{matrix}20 \ 10\end{matrix}\right)0.5^{10}0.5^{10}=0.1762$$
The calculation is repeated for the probabilities of winning 11, 12, 13, 14, 15, 16, 17, 18, 19 and 20 games out of 20 and then the 11 values of probability are added to find the probability of winning 10 or more games out of 20.