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A throws 4 dice and wins if he gets 6 at least once. B throws 8 dice and wins if gets 6 at least twice. Who has the greater probability to win?

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by chokeh

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by chokeh

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Probability of A getting a 6 on throwing one die : 1/6 as there are 6 faces and six numbers.

Probability of A getting at least one 6 on throwing 4 dies =

= 1 - probability of getting no six at all

= 1 - 5/6 * 5/6 * 5/6 * 5/6 = 0.5177

Probability of A winning = 0.5177

probability of B getting a 6 on throwing one die : 1/6

Probability of B getting no six at all : (5/6)^8 = 0.23256

Probability of B getting 1 six only of 8 dice:

= P(first die = 6, remaining all 1 to 5) + P(2nd die=6, remaining all =1 to 5) + .....

= 8 P(first die = 6, remaining all 1 to 5)

= 8 * (1/6) * (5/6)^7 = 0.3721

**Total probability of B getting at least two 6: 1 - 0.23256 - 0.3721 = 0.3853**

**Probability of B winning = 0.3853 < P(A wins) = 0.5177**

**A has a higher probability to win the game.**

Probability of A getting at least one 6 on throwing 4 dies =

= 1 - probability of getting no six at all

= 1 - 5/6 * 5/6 * 5/6 * 5/6 = 0.5177

Probability of A winning = 0.5177

probability of B getting a 6 on throwing one die : 1/6

Probability of B getting no six at all : (5/6)^8 = 0.23256

Probability of B getting 1 six only of 8 dice:

= P(first die = 6, remaining all 1 to 5) + P(2nd die=6, remaining all =1 to 5) + .....

= 8 P(first die = 6, remaining all 1 to 5)

= 8 * (1/6) * (5/6)^7 = 0.3721