Suppose we play the following gambling game.we roll a die, if the roll is 1 or 6

Question

Suppose we play the following gambling game.we roll a die, if the roll is 1 or 6 we win, if the roll is 2 we lose.if the roll is 3,4 or 5 then we roll repeatedly until we either get that same number again (win) or roll a 1 or 6(lose).

a) Suppose we roll a 3.What is the conditional Probability of winning the game? (must roll a 3 again before rolling a 1 or 6).Even if you are able to intuitively guess the answer,you must show a correct mathematical calculation to jutify your answer. In fact, you must solve the problem in two different ways. First, by conditioning on wether the next roll is in the set{1,3,6}

b) Now compute your answer to the previous part in another way; condition on the number of rolls it takes to resolve the game. the result will be an infinit sum which you can evaluate with geometric series

c) Calculate the initial probability of wining the game (no longer assuming the first roll was 3)

Explanation / Answer

(a) We roll a 3 . so, probabilioty of winning the game will be when in next rolls, i should get a 3 before rolling a 1 and a 6.

Method 1

Probability of getting any value from set { 1,3,6} in any roll is 3/6 =1/2

so P ( 3) = (1/3 * 1/2 )/ (1/3 * 1/2 + 2/3 * 1/2) = 1/3

(b) P(win if first roll is 3) = P(win in first roll after first roll of 3) + P(Nothing happened in first roll after first roll of 3) * P(win in second chance) + ....

P(win) = 1/6 + 3/6 * 1/6 + (3/6)2 *(1/6) + .......= 1/3

(c) The initial probability of wining the game

P (win) = P(1,6 in first roll) + P (3 in first roll) * P( win in this case) = 2/6 + 1/6 * 1/3 = 7/18

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