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Biased dice probability question

4

2

$begingroup$

A biased six sided dice is rolled twice. Show that the probability that the two results are the same is at least $frac{1}{6}$.
(Hint: $(p_1 − a)^2 + . . . + (p_6 − a)^2 ≥ 0$ and choose suitable
$p_1, . . . , p_6$, a.)

probability

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edited 38 mins ago

mathpadawan

2,019422

asked 42 mins ago

mandymandy

211

New contributor

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: First try to show that if a coin is flipped twice, the probability that the two results are the same is at least $1/2$. This will help you figure out what to choose as $a$.
  $endgroup$
  – Lorenzo
  29 mins ago
  

  
  
  
  

add a comment | 

4

2

$begingroup$

A biased six sided dice is rolled twice. Show that the probability that the two results are the same is at least $frac{1}{6}$.
(Hint: $(p_1 − a)^2 + . . . + (p_6 − a)^2 ≥ 0$ and choose suitable
$p_1, . . . , p_6$, a.)

probability

share|cite|improve this question

edited 38 mins ago

mathpadawan

2,019422

asked 42 mins ago

mandymandy

211

New contributor

mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: First try to show that if a coin is flipped twice, the probability that the two results are the same is at least $1/2$. This will help you figure out what to choose as $a$.
  $endgroup$
  – Lorenzo
  29 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

A biased six sided dice is rolled twice. Show that the probability that the two results are the same is at least $frac{1}{6}$.
(Hint: $(p_1 − a)^2 + . . . + (p_6 − a)^2 ≥ 0$ and choose suitable
$p_1, . . . , p_6$, a.)

probability

share|cite|improve this question

edited 38 mins ago

mathpadawan

2,019422

asked 42 mins ago

mandymandy

211

New contributor

mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 38 mins ago

mathpadawan

2,019422

asked 42 mins ago

mandymandy

211

New contributor

mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 38 mins ago

mathpadawan

2,019422

asked 42 mins ago

mandymandy

211

New contributor

mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 38 mins ago

mathpadawan

2,019422

edited 38 mins ago

mathpadawan

2,019422

asked 42 mins ago

mandymandy

211

New contributor

mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 42 mins ago

mandymandy

211

1 Answer
1

active

oldest

votes

4

$begingroup$

Let $p_i$ be the probability of rolling $i$. Then $sum_{i=1}^6 p_i = 1$.

By Cauchy-Schwarz inequality,

$$begin{align*}
left(sum_{i=1}^6 1^2right) left(sum_{i=1}^6 p_i^2right) &ge
left(sum_{i=1}^6 1p_iright)^2\
6left(sum_{i=1}^6 p_i^2right) &ge 1\
sum_{i=1}^6 p_i^2 &ge frac16end{align*}$$

Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.

share|cite|improve this answer

answered 27 mins ago

peterwhypeterwhy

12.3k21229

$endgroup$

add a comment | 

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4

$begingroup$

Let $p_i$ be the probability of rolling $i$. Then $sum_{i=1}^6 p_i = 1$.

By Cauchy-Schwarz inequality,

$$begin{align*}
left(sum_{i=1}^6 1^2right) left(sum_{i=1}^6 p_i^2right) &ge
left(sum_{i=1}^6 1p_iright)^2\
6left(sum_{i=1}^6 p_i^2right) &ge 1\
sum_{i=1}^6 p_i^2 &ge frac16end{align*}$$

Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.

share|cite|improve this answer

answered 27 mins ago

peterwhypeterwhy

12.3k21229

$endgroup$

add a comment | 

4

$begingroup$

Let $p_i$ be the probability of rolling $i$. Then $sum_{i=1}^6 p_i = 1$.

By Cauchy-Schwarz inequality,

$$begin{align*}
left(sum_{i=1}^6 1^2right) left(sum_{i=1}^6 p_i^2right) &ge
left(sum_{i=1}^6 1p_iright)^2\
6left(sum_{i=1}^6 p_i^2right) &ge 1\
sum_{i=1}^6 p_i^2 &ge frac16end{align*}$$

Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.

share|cite|improve this answer

answered 27 mins ago

peterwhypeterwhy

12.3k21229

$endgroup$

add a comment | 

$begingroup$

Let $p_i$ be the probability of rolling $i$. Then $sum_{i=1}^6 p_i = 1$.

By Cauchy-Schwarz inequality,

$$begin{align*}
left(sum_{i=1}^6 1^2right) left(sum_{i=1}^6 p_i^2right) &ge
left(sum_{i=1}^6 1p_iright)^2\
6left(sum_{i=1}^6 p_i^2right) &ge 1\
sum_{i=1}^6 p_i^2 &ge frac16end{align*}$$

Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.

share|cite|improve this answer

answered 27 mins ago

peterwhypeterwhy

12.3k21229

$endgroup$

share|cite|improve this answer

answered 27 mins ago

peterwhypeterwhy

12.3k21229

answered 27 mins ago

peterwhypeterwhy

12.3k21229

answered 27 mins ago

peterwhypeterwhy

12.3k21229