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Suppose that two players are playing a game, players select between two choices:

Scenario 1:

  • player $1$ chooses option $1$ with probability $60\%$, option $2$ with $40\%$

  • player $2$ chooses option $1$ with probability $40\%$, option $2$ with $60\%$

Scenario 2:

  • player $1$ chooses option $1$ with probability $50\%$, option $2$ with $50\%$

  • player $2$ chooses option $1$ with probability $50\%$, option $2$ with $50\%$

question: which scenario are both players most likely to choose the same option if the game is played repeatedly over several stages?

Intuitively, scenario $2$ would be the scenario for which both players are most likely to choose the same option, but how do I formally prove my intuition?

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  • $\begingroup$ In scenario 1 the probability that both players choose the same option at one stage is $0.6\cdot 0.4+0.4\cdot 0.6$. In scenario 2 the probability that both players choose the same option at one stage is $0.5\cdot 0.5+0.5\cdot 0.5$. Now consider more than one stage. $\endgroup$ – callculus Oct 31 '15 at 16:39
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Scenario 1

Probability that player $1$ chooses option $1$ and player $2$ chooses option $1$:
$0.6*0.4 = 0.24$

Probability that player $1$ chooses option $2$ and player $2$ chooses option $2$:
$0.4*0.6 = 0.24$

Probability that at least one of these (mutually exclusive) events happens is their union:
$0.24+0.24 = 0.48$

Scenario 2

Probability that player $1$ chooses option $1$ and player $2$ chooses option $1$:
$0.5*0.5 = 0.25$

Probability that player $1$ chooses option $2$ and player $2$ chooses option $2$:
$0.5*0.5 = 0.25$

Probability that at least one of these (mutually exclusive) events happens is their union:
$0.25+0.25 = 0.50$

$0.50 > 0.48$, so both players have a higher probability of choosing the same option in scenario 2.

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