Suppose any one round of a gambling game pays 3 to 1 and the odds are 4 to 1 against you.

if you play for 100 rounds and bet one dollar on each round, what is the chance you will come out ahead?

There is a 10.5% chance of having a positive payoff.

Step-by-step explanation:

The odds are 4 to 1 against, so we can estimate the probability of success (p) as

The expected pay for every success is 3 to 1, so we lose $1 for every lose and we gain $3 for every win.

The number of winnings in the 100 rounds to be even can be calculated as:

We have to win at least 25 rounds to have a positive payoff.

As the number of rounds is big, we will approximate the binomial distribution to a normal distribution with parameters:

The z-value for x=25 is

The probability of z>1.25 is

There is a 10.5% chance of having a positive payoff.

NOTE: if we do all the calculations for the binomial distribution, the chances of having a net payoff are 13.1%.

answer: i think it is (1,-2)

1. a^4+8a^3+24a^2+32a+16

2.y^8+8y^7+28y^6+56y^5+70y^4+56y^3+28y^2+8y+1

3.x^6-3x^4y^2+3x^2y^4-y^6