Counting Outcomes - Probability Question

Click For Summary

Homework Help Overview

The problem involves calculating the probability of winning a game based on independent coin flips, specifically determining the likelihood of achieving two consecutive heads within a maximum of five flips. Additionally, there is a question regarding the probability of two games resulting in the same outcome when played twice with a fair coin.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the total number of possible outcomes from the coin flips and the conditions under which Z wins or loses. There is an attempt to calculate the probability of winning by considering the outcomes that lead to a loss. Questions arise about whether all possible outcomes have been accounted for, particularly regarding the inclusion of zero heads.

Discussion Status

The discussion is ongoing, with some participants confirming the correctness of the initial approach while others express uncertainty about the completeness of the outcome count. There is a recognition of the need to clarify the total number of outcomes considered.

Contextual Notes

Participants are working within the constraints of a maximum of five flips and are exploring the implications of different configurations of heads and tails in relation to winning and losing the game.

Legendre
Messages
59
Reaction score
0

Homework Statement



Z plays a game where independent flips of a coin are recorded until two heads in succession are encountered.

Z wins if 2 heads in succession occurs.
Z loses if after 5 flips, we have not encounter two heads in succession.

1) What is the probability that Z wins the game?

2) Suppose coin is fair. Z plays twice. What is the probability that both games have the same outcome?

Homework Equations



N.A.

The Attempt at a Solution



I get very confused by questions related to probability. :(

There are 5 flips at most so the sample space has 2^5 = 32 possible outcomes. When seen as 5 "slots", as long as two adjacent slots are filled with H (for "heads") then Z wins.

P(Z wins) = 1 - P(Z loses)

Let X = number of heads. Then, Z loses when X = 1. Z loses in some cases when X = 2,3.

X = 1 : There are 5 possible outcomes.

X = 2 : The two heads are either 1 slot apart (3 choose 1 = 3 outcomes) or 2 slots apart (2 choose 1 = 2 outcomes) or 3 slots apart (1 outcome)

X = 3 : The three heads need to be in slot 1,3,5 => 1 outcome.

So total outcome that results in a loss = 1 + 5 + 3 + 2 + 1 = 12.

Therefore, P(Z win) = 1 - 12/32 = 20/32 = 5/8.

Question : Is this method correct? Are there "cleaner" methods? I have no idea how to start part 2.
 
Physics news on Phys.org
Are you sure you have accounted for all 32 outcomes?
 
D H said:
Are you sure you have accounted for all 32 outcomes?

Doh!

X = 0.

Thats one more outcome. :(

But my approach is correct?
 
Yes.
 

Similar threads

Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
31
Views
7K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K