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In calculating probabilities for Roulette, my text book asks me to calculate the probability of winning after 1,000 bets.

Where spending \$1 has a $\frac{1}{38}$ probability of winning \$35 and a $\frac{37}{38}$ probability of loosing \$1.

The answer uses the formula

$P(S>0)=P(S>0.5)$ $\approx$ 1-$\phi$ $(\frac{.5+0.526 n}{\sqrt(33.21 n)} ) $

I understand from this question that the mean works out to be -0.0596 and the variance is 33.21

Why is S>0.5 the same as S>0?

Where does the 0.5 come from?

Why is 0.5 added to the mean?

Kirsten
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    minor typo : ...has a $\frac{1}{38}$ probability of winning $35... – user2661923 Oct 01 '20 at 20:53
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    Since you can only win or lose an integer number of dollars, $P(S>0)=P(S≥1)=P(S>.5)$. The reason you include the $.5$ is because of the so-called continuity correction that you need when you approximate a discrete distribution by a continuous one. – lulu Oct 01 '20 at 20:54
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    Should say, the mean for a single game is not $.596$ It's $\frac 1{38}\times 35-\frac {37}{38}\times 1=-.0526$. – lulu Oct 01 '20 at 20:57
  • @lulu Thank you. What does S stand for here? Why $P(S$ $\ge$ 1) ? – Kirsten Oct 01 '20 at 21:20
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    The $S$ is from your post. I assume you meant for $S$ to denote the amount of winnings. The problem asks, then, for the probability that $S>0$. – lulu Oct 01 '20 at 21:27
  • Thankyou @lulu, I have copied from the textbook. I wondered if it was S for Sample. Winnings makes sense. In your comment above why P(S≥1)=P(S>0.5) ? For example why would P(2) match P(1) ? – Kirsten Oct 01 '20 at 21:36
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    Nobody said $P(2)=P(1)$. Since you can only win an integer number of dollars, If I tell you you got more than fifty cents, you know you must have won at least $$1$. – lulu Oct 01 '20 at 22:16

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