Theory of Expectation :: Problems on Throwing/Rolling Dice : Probability Distribution

Problem Back to Problems Page
 
A and B play for a prize of Rs.99. The prize is to be won by a player who first throws a ‘2’ with one die. "A" first throws, and if he fails "B" throws. and if he fails "A" again throws, and so on. Find their respective expectations.

Net Answers :
[Expectation(A): 54 ; Expectation(B):45; Variance(A): 2,430 ; Variance(B): 2,430 ;
Standard Deviation(A): +49.3; Standard Deviation(B): +49.3]

Solution  
 

Let "x", "y" indicate the amount of prize won by A and B respectively

[Since you are required to find the expectated prizes won by A and B, the variables would represent the amount of prize won.]

The amount of prize won

  • By "A" would be
    • Rs. 99 if he throws a "2" before "B" throws it
    • Rs. 0 if "B" throws a "2" before "A" throws it
  • By "B" would be
    • Rs. 99 if he throws a "2" before "A" throws it
    • Rs. 0 if "A" throws a "2" before "B" throws it

⇒ The values carried by the

  • variable ("x") would be either 0, 99
    ⇒ "X" is a discrete random variable with range = {0, 99}
  • variable ("y") would be either 0, 99
    ⇒ "Y" is a discrete random variable with range = {0, 99}

"X" and "Y" represent random variables and P(X = x) and P(Y=y) represent the probability that the value within the range of the random variable is a specified value of "x" or "y" respectively.

In a single throw with a dice, Probability of:

Getting a "2" ⇒ P(2) =
1
6
Not Getting a "2" ⇒ P(2c) = 1 − P(2)
=
1 −
1
6
=
6 − 1
6
=
5
6

In the experiment of tossing a dice, there are six possible elementary events, the events of the die showing up either ONE, TWO, THREE, FOUR, FIVE or SIX all of which are mutually exclusive, equally likely and exhaustive.

Therefore the probability of occurance of each elementary event is 1/6 Probabilty that the dice would show up

  • 2 ⇒ P(2) =
    1
    6

    Probabilty of

    • "A" Getting a "2"
      • On his first throw
        ⇒ P(2A1) = P(2)
        =
        1
        6
      • On his second throw
        [For this event to happen "A", "B" should not get a "2" on the first throw and "A" should get a "2" on the second throw]
        ⇒ P(2A2) = P((2c)A ∩ (2c)B ∩ 2A)
        = P((2c)A) × P((2c)B) × P(2A)
        = P((2c)) × P((2c)) × P(2)
        =
        5
        6
        ×
        5
        6
        ×
        1
        6
        =
        1
        6
        × (
        5
        6
        )2
      • On his third throw
        [For this event to happen "A", "B" should not get a "2" on their 2 throws and "A" should get a "2" on his third throw]
        ⇒ P(2A3) = P((2c)A ∩ (2c)B ∩ (2c)A ∩ (2c)B ∩ 2A)
        = P((2c)A) × P((2c)B) × P((2c)A) × P((2c)B) × P(2A)
        = P((2c)) × P((2c)) × P((2c)) × P((2c)) × P(2)
        =
        5
        6
        ×
        5
        6
        ×
        5
        6
        ×
        5
        6
        ×
        1
        6
        =
        1
        6
        × (
        5
        6
        )4
      • ...
      • ...
    • "B" Getting a "2"
      • On his first throw
        [For this event to happen "A" should not get a "2" on his first throw and "B" should get a "2" on his first throw]
        ⇒ P(2B1) = P((2c)A ∩ P(2)B
        = P((2c)A × P(2)B
        =
        5
        6
        ×
        1
        6
      • On his second throw
        [For this event to happen "A", "B" should not get a "2" on their first throw, "A" should not get a "2" on the second throw and "B" should get a "2" on his second throw]
        ⇒ P(2B2) = P((2c)A ∩ (2c)B ∩ (2c)A ∩ 2B)
        = P((2c)A) × P((2c)B) × P((2c)A) × P(2B)
        = P((2c)) × P((2c)) × P((2c)) × P(2)
        =
        5
        6
        ×
        5
        6
        ×
        5
        6
        ×
        1
        6
        =
        1
        6
        × (
        5
        6
        )3
      • On his third throw
        [For this event to happen "A", "B" should not get a "2" on their 2 throws, "A" should not get a "2" on his third throw and "B" should get a "2" on his third throw]
        ⇒ P(2B3) = P((2c)A ∩ (2c)B ∩ (2c)A ∩ (2c)B ∩ (2c)A ∩ 2B)
        = P((2c)A) × P((2c)B) × P((2c)A) × P((2c)B) × (2c)A) × P(2B)
        = P((2c)) × P((2c)) × P((2c)) × P((2c)) × (2c)) × P(2)
        =
        5
        6
        ×
        5
        6
        ×
        5
        6
        ×
        5
        6
        ×
        5
        6
        ×
        1
        6
        =
        1
        6
        × (
        5
        6
        )5
      • ...
      • ...

    The probabilty of

    • "A" winning the game
      P(A) = P(2A1) + P(2A2) + P(2A3) + ... + ...
      =
      1
      6
      + {
      1
      6
      × (
      5
      6
      )2} + {
      1
      6
      × (
      5
      6
      )4} + ... + ...
      Hide/Show Calculations for simplification

      ≡ a + ar + ar2 + ar3 + ... to ∞ terms.

      Comparing the above series with the general form we get
      a =
      1
      6
      and r =
      1
      6
      × (
      5
      6
      )4
      1
      6
      × (
      5
      6
      )2
      ⇒ r =
      (
      5
      6
      )2
      ⇒ r =
      25
      36
      Sum of infinite terms of a geometric series, where "r ≤ 1"
      is given by S =
      a
      (1 − r)
      Therefore, Sum of infinite terms of the above geometric series,
      ⇒ S =
      1
      6
      1 − (
      25
      36
      )

      =
      1
      6
      1 − (
      25
      36
      )
      a
      (1 − r)
      =
      1
      6
      36 − 25
      36
      =
      1
      6
      ×
      36
      11
      =
      6
      11
    • "B" winning the game
      P(B) = P(2B1) + P(2B2) + P(2B3) + ... + ...
      =
      {
      1
      6
      ×
      5
      6
      } + {
      1
      6
      × (
      5
      6
      )3} + {
      1
      6
      × (
      5
      6
      )5} + ... + ...
      Hide/Show Calculations for simplification

      ≡ a + ar + ar2 + ar3 + ... to ∞ terms.

      Comparing the above series with the general form we get
      a =
      1
      6
      and r =
      1
      6
      × (
      5
      6
      )3
      1
      6
      × (
      5
      6
      )
      ⇒ r =
      (
      5
      6
      )2
      ⇒ r =
      25
      36
      Sum of infinite terms of a geometric series, where "r ≤ 1"
      is given by S =
      a
      (1 − r)
      Therefore, Sum of infinite terms of the above geometric series,
      ⇒ S =
      1
      6
      ×
      5
      6
      1 − (
      25
      36
      )

      =
      1
      6
      ×
      5
      6
      1 − (
      25
      36
      )
      a
      (1 − r)
      =
      5
      36
      36 − 25
      36
      =
      5
      36
      ×
      36
      11
      =
      5
      11

    Alternative for finding probability Hide/Show

    The probabilty of

    • "A" winning the game
      P(A) = P(2A1) + P(2A2) + P(2A3) + ... + ...
      =
      1
      6
      + {
      1
      6
      × (
      5
      6
      )2} + {
      1
      6
      × (
      5
      6
      )4} + ... + ...
      =
      1
      6
      { 1 + (
      5
      6
      )2 + (
      5
      6
      )4 + ... + ... }
    • "B" winning the game
      P(B) = P(2B1) + P(2B2) + P(2B3) + ... + ...
      =
      {
      1
      6
      ×
      5
      6
      } + {
      1
      6
      × (
      5
      6
      )3} + {
      1
      6
      × (
      5
      6
      )5} + ... + ...
      =
      (
      1
      6
      ×
      5
      6
      )2
      { 1 + (
      5
      6
      )2 + (
      5
      6
      )4 + ... + ... }

    Therefore
    P(A) : P(B) =
    1
    6
    { 1 + (
    5
    6
    )2 + (
    5
    6
    )4 + ... + ... }
    :
    (
    1
    6
    ×
    5
    6
    )2
    { 1 + (
    5
    6
    )2 + (
    5
    6
    )4 + ... + ... }
    =
    1
    6
    :
    (
    1
    6
    ×
    5
    6
    )
    =
    1 :
    5
    6
    =
    6
    6
    :
    5
    6
    = 6 : 5
    (Or) =
    6
    11
    :
    5
    11

    These terms which represent ratios add up to 1 and can be taken to represent probabilities only if "A", "B", are mutually exclusive and exhaustive events.

    Since one of the two would/should win the events "A" and "B" are exhaustive events.
    ⇒ P(A ∪ B) = 1     → (1)

    Since if one wins the other would not win the two events "A" and "B" are mutually exclusive
    ⇒ P(A ∪ B) = P(A) + P(B)     → (2)

    Probabilty that the amount won by

    • By "A" would be
      • Rs. 99 = P(A)
        ⇒ P(X = 99) = P(A)
        =
        6
        11
      • Rs. 0 = P(Ac)
        ⇒ P(X = 0) = P(Ac)
        = 1 − P(A)
        =
        1 −
        6
        11
        =
        11 − 6
        11
        =
        5
        11
    • By "B" would be
      • Rs. 99 = P(B)
        ⇒ P(X = 35) = P(B)
        =
        5
        11
      • Rs. 0 = P(Bc)
        ⇒ P(Y = 0) = P(Bc)
        = 1 − P(B)
        =
        1 −
        5
        11
        =
        11 − 5
        11
        =
        6
        11

    The probabilty distribution of "x" would be
    x 0 99
    P(X = x)
    5
    11
    6
    11

    The probabilty distribution of "y" would be
    y 0 99
    P(Y = y)
    6
    11
    5
    11

    Expected amount of winnings

    • of "A"
      ⇒ Mathematical Expectation of "x"
      ⇒ E (x) = Σ px
      = p1x1 + p2x2
      =
      5
      11
      × 0 +
      6
      11
      × 99
      = 0 + 54
      = 54
    • of "B"
      ⇒ Mathematical Expectation of "y"
      ⇒ E (y) = Σ py
      = p1y1 + p2y2
      =
      6
      11
      × 0 +
      5
      11
      × 99
      = 0 + 45
      = 45

    Variance of

    • of "A"'s Winning
      ⇒ Variance of "x"
      ⇒ var(x) = E(x2) − {E(x)}2
      = Σ px2 − {Σ px}2
      = [p1x12 + p2x22] − [{p1x1 + p2x2}2]
      =
      [
      5
      11
      × (0)2 +
      6
      11
      × (99)2 ] − [ {
      5
      11
      × 0 +
      6
      11
      × 99 }2 ]
      =
      [0 +
      6
      11
      × (9,801)] − [ { 0 + 54 }2 ]
      = [0 + 5,346] − [(54)2]
      = 5,346 − 2,916
      = 2,430
    • of "B"'s Winning
      ⇒ Variance of "y"
      ⇒ var(y) = E(y2) − {E(y)}2
      = Σ py2 − {Σ py}2
      = [p1y12 + p2y22] − [{p1y1 + p2y2}2]
      =
      [
      6
      11
      × (0)2 +
      5
      11
      × (99)2 ] − [ {
      6
      11
      × 0 +
      5
      11
      × 99 }2 ]
      =
      [0 +
      5
      11
      × (9,801)] − [ { 0 + 45 }2 ]
      = [0 + 4,455] − [(45)2]
      = 4,455 − 2,025
      = 2,430
    Standard Deviation of
    • of "A"'s Winning
      ⇒ SD (x) = + Var (x)
      = + 2,430
      = + 49.3
    • of "B"'s Winning
      ⇒ SD (y) = + Var (y)
      = + 2,430
      = + 49.3

    Credit : Vijayalakshmi Desu

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