claim: Your odds of correctly guessing the outcome of a flipped coin are 1 in 2, but your odds of guessing correctly twice in a row are only 1 in 4 —
evidences.evidence_id: Gambler's fallacy:12
evidences.article: Gambler's fallacy
evidences.evidence: Since the first four tosses turn up heads, the probability that the next toss is a head is: Pr ( A 5 | A 1 ∩ A 2 ∩ A 3 ∩ A 4 ) = Pr ( A 5 ) = 1 2 {\displaystyle \Pr \left(A_{5}|A_{1}\cap A_{2}\cap A_{3}\cap A_{4}\right)=\Pr \left(A_{5}\right)={\frac {1}{2}}} .
evidences.votes: NOT_ENOUGH_INFO SUPPORTS
evidences.evidence_id: Gambler's fallacy:20
evidences.article: Gambler's fallacy
evidences.evidence: Assuming a fair coin: The probability of 20 heads, then 1 tail is 0.520 × 0.5 = 0.521 The probability of 20 heads, then 1 head is 0.520 × 0.5 = 0.521 The probability of getting 20 heads then 1 tail, and the probability of getting 20 heads then another head are both 1 in 2,097,152.
evidences.votes: NOT_ENOUGH_INFO SUPPORTS
evidences.evidence_id: Probability:102
evidences.article: Probability
evidences.evidence: If two events, A and B are independent then the joint probability is P ( A  and  B ) = P ( A ∩ B ) = P ( A ) P ( B ) , {\displaystyle P(A{\mbox{ and }}B)=P(A\cap B)=P(A)P(B),\,} for example, if two coins are flipped the chance of both being heads is 1 2 × 1 2 = 1 4 {\displaystyle {\tfrac {1}{2}}\times {\tfrac {1}{2}}={\tfrac {1}{4}}} .
evidences.votes: SUPPORTS SUPPORTS
evidences.evidence_id: Probability:106
evidences.article: Probability
evidences.evidence: P ( A  or  B ) = P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) = P ( A ) + P ( B ) − 0 = P ( A ) + P ( B ) {\displaystyle P(A{\mbox{ or }}B)=P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-0=P(A)+P(B)} For example, the chance of rolling a 1 or 2 on a six-sided die is P ( 1  or  2 ) = P ( 1 ) + P ( 2 ) = 1 6 + 1 6 = 1 3 .
evidences.votes: NOT_ENOUGH_INFO REFUTES
evidences.evidence_id: Random variable:70
evidences.article: Random variable
evidences.evidence: {\displaystyle Y(\omega )={\begin{cases}1,&{\text{if }}\omega ={\text{heads}},\\[6pt]0,&{\text{if }}\omega ={\text{tails}}.\end{cases}}} If the coin is a fair coin, Y has a probability mass function f Y {\displaystyle f_{Y}} given by: f Y ( y ) = { 1 2 , if  y = 1 , 1 2 , if  y = 0 , {\displaystyle f_{Y}(y)={\begin{cases}{\tfrac {1}{2}},&{\text{if }}y=1,\\[6pt]{\tfrac {1}{2}},&{\text{if }}y=0,\end{cases}}} A random variable can also be used to describe the process of rolling dice and the possible outcomes.
evidences.votes: NOT_ENOUGH_INFO SUPPORTS