# Probability: chances of winning

• acrimon86
In summary, the game discussed involves a player rolling two dice and winning if the sum is 7 or 11 on the first roll, or if the initial outcome is rolled again before a 7 appears. The probability of winning by throwing a 7 or 11 initially is 6/36. The probabilities of winning by throwing a 4, 5, 6, 8, 9, or 10 initially and then getting the same outcome again before a 7 appears can be expressed as a formula, but further clarification is needed.
acrimon86

## Homework Statement

Consider a game where a player rolls 2 dice. The player looses if the sum of the dice of the first roll is 2, 3 or 12, and wins if the sum of the dice is 7 or 11. If the outcome of the first roll is 4, 5, 6, 8, 9 or 10, the player continues to roll the dice until the initial outcome is rolled again (player wins) or the 7 appears (the player looses).
Find the probability of winning the game.

## Homework Equations

The probability of event E occurring on a sample space $$\Omega$$ is: P(E) = $$\frac{|E|}{|\Omega|}$$

## The Attempt at a Solution

The size of the sample space is |$$\Omega$$| = $$6 * 6 = 36$$, since we have two six-sided dice, so a total of 36 possible outcomes on a given throw.

I gather that if the player rolls a sum of 2, 3, or 12 on their first try, they automatically lose the game. Therefore, I let event $$E_{L}$$ denote the event that either of those sums is thrown. Therefore, $$E_{L}$$ is the set of all outcomes that result in one of those sums.

$$E_{L}$$ = { (1,1), (1,1), (2,1), (1,2), (6,6), (6,6) }
|$$E_{L}$$| = 6

Likewise, I let event $$E_{W}$$ denote that a player throws a 7 or 11 on their first try, meaning they automatically win.

$$E_{W}$$ = { (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) }
|$$E_{W}$$| = 6

I also worked out the size of events $$E_{4}$$, $$E_{5}$$, $$E_{6}$$, $$E_{7}$$, $$E_{8}$$, $$E_{9}$$, $$E_{10}$$, where each subscript represents the number of outcomes that could lead to that sum being thrown. I got:

|$$E_{4}$$| = 4
|$$E_{5}$$| = 4
|$$E_{6}$$| = 6
|$$E_{7}$$| = 6
|$$E_{8}$$| = 6
|$$E_{9}$$| = 4
|$$E_{10}$$| = 4

Ok, having that done, the chances of automatically winning the game by throwing a 7 or 11 initially are:
P($$E_{W}$$) = $$\frac{6}{36}$$

This is where I am stuck, however. How do I handle the probabilities of winning by throwing a 4, 5, 6, 8, 9, 10 and then getting that same outcome again without getting a 7?

I know that each of those sums have a certain probability of being thrown initially, and then the same probability of being thrown again before getting a 7. But I'm confused as to how I could express this in terms of an equation/formula. Any help would be greatly appreciated!

## What is probability?

Probability is a measure of the likelihood of an event occurring. It is expressed as a number between 0 and 1, where 0 represents impossibility and 1 represents certainty.

## How is probability calculated?

The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. For example, if you roll a six-sided die, the probability of rolling a 3 would be 1/6, since there is only one favorable outcome out of six possible outcomes.

## What is the difference between theoretical and experimental probability?

Theoretical probability is based on mathematical calculations and assumes that all outcomes are equally likely. Experimental probability is based on actual observations and can vary from the theoretical probability. For example, the theoretical probability of flipping a coin and getting heads is 1/2, but in a real-life experiment, you may get more or less than half heads depending on chance.

## What is the law of large numbers?

The law of large numbers states that as the number of trials or experiments increases, the experimental probability will approach the theoretical probability. This means that over a large number of trials, the actual results will become more consistent with the expected results.

## Can probability be used to predict the outcome of a single event?

No, probability cannot be used to predict the outcome of a single event. It only gives us information about the likelihood of an event occurring based on past data and assumptions. Each event is independent and has its own unique outcome.

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