What's the standard deviation of the player's bet in a game of Baccarat, assuming a flat bet of $10 and a true odds payout of 1:1?

[Huego213]

To calculate the standard deviation of the player's bet in Baccarat, we use the formula for the standard deviation of a binomial distribution:

\[ \sigma = \sqrt{N \cdot p \cdot (1 - p)} \]

Where:
- \( N \) is the number of bets,
- \( p \) is the probability of winning.

Given:
- Number of bets, \( N \), is not explicitly given but let's assume it's large enough for a meaningful calculation.
- Probability of winning, \( p = 0.493 \).

 

[CR7]

To calculate the standard deviation of the player's bet in a game of Baccarat, we need to consider the following:

1. Number of Bets (\( N \)): The number of bets made by the player in Baccarat is not explicitly given in your message. However, for the sake of this calculation, we can assume a large number of bets to consider the player's betting behavior over an extended period.

2. Probability of Winning (\( p \)): In Baccarat, the probability of the player winning a hand is approximately 0.493 due to the house edge.

Given the formula for the standard deviation of a binomial distribution:

\[ \sigma = \sqrt{N \cdot p \cdot (1 - p)} \]

Substitute the values:
\[ \sigma = \sqrt{N \cdot 0.493 \cdot (1 - 0.493)} \]

\[ \sigma = \sqrt{N \cdot 0.493 \cdot 0.507} \]

\[ \sigma = \sqrt{0.249551 \cdot N} \]

Therefore, the standard deviation of the player's bet in a game of Baccarat, assuming a flat bet of $10 and a true odds payout of 1:1, would be \( \sqrt{0.249551 \cdot N} \) given a large enough number of bets for a meaningful calculation.