# Using normal approximation to estimate the probability of winning more than you lose in 100 plays

You play a game where if the two cards you pick from a deck of cards (without replacement) have consecutive rank (i.e. 2 and 3, A and K, A and 2, etc.), you win. The game pays 81 USD to 5 USD, meaning if you win, you have a net gain of 81 USD and if you lose, you lose 5 USD. You have two scenarios: play 10 times or play 100 times.

I calculated the chance of winning to be $\binom {2}{1} \frac {52}{52} *\frac {4}{51} =\frac {8}{51}$

I then calculated the winning E(X)=$100* \frac{8}{52} =15.6$ SE(X)= $\sqrt {(\frac{8}{51}*\frac{43}{51})} *\sqrt{100}=3.64$

The solution given for the P(win more than lose) is 0.997

How can the probability of winning more than losing be the case when we only win 15.6 games out of 100 plays?

What am I missing here?

Thank you!

• In the $15.6$ games you win you win $\$81\cdot15.6=\$1263.6$ and in the $84.4$ games you lose, you only lose $\$5\cdot84.4=\$422.$ Where can I go to play this game? – saulspatz Jul 9 '18 at 4:56
• lol only in the land of arbitrary math problems. How did they arrive the 0.997 approximation though? If we use the formula $\frac{Observed \, Value - 15.6}{3.64}$ what would we plug in for the Observed value though? – pino231 Jul 9 '18 at 6:36