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# probability and favourable conditions in dice games?

if the probability of winning a game of dice (after accounting for the favourable conditions), is 0.4 with 1 die, then what will be the probability of winning with 3 dice if it is fine as long as at least one of the dice has the favourable condition?

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so the probability of 1 dice is 0.4

the probability of 3 dices are :

dice 1 succeed + dice 2 succeed +dice 3 succeed but we have to substract some cases that we double count : dice 1 and dice 2 succeed+ dice 1 and dice 3 succeed + dice 2 and 3 succeed , but this will make a case hasn't been counted : dice 1 and 2 and 3 succeed

so we have 0.4 + 0.4 + 0.4 - (0.4*0.4) - (0.4*0.4) - (0.4*0.4) + (0.4*0.4*0.4)=1.2 - 0.48 + 0.064 = 0.784

- 0.4*0.4 because we want both die succeed, so we need to multiply both probability, same with the 0.4*0.4*0.4

To make it clearer, you can try to make 3 circle each representing dice 1, 2, or dice 3 succeed, with some area interception between tow or three circles.

• If you win with just one die showing the favourable result, then the only way you lose is if all three dice show a losing result.

This probability (all three showing a loss) is 0.6 x 0.6 x 0.6 = 0.216

So the probability of winning = 1 - 0.216 = 0.784.

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