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How do you calculate this against 2 opponents or more than 3 opponents, etc?

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  • also, we do not have the 4 – Brian Mar 31 '18 at 9:49
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Let's first calculate the odds of 1 opponent having a four. I will then give the general formula for 'x' opponents.

There are 3 cards on the board and you have 4 cards. This means there are 45 cards left in the deck. There are therefore a total of C(45,4) different combinations of hands your opponent can hold. Chances of your opponent having a four can be calculated by substracting the chances of your opponent not having a four from 1. Chances of your opponent not having a four simply means there are now only 43 cards your opponent is allowed to draw (since there are 2 fours left). Total combinations of hands without a four is therefore C(43,4). So the odds of your opponent having a four is 1 - ( C(43,4) / C(45,4) ) = 0.171717...

As general formula we get: 1 - ( C(43,4x) / C(45,4x) ), where 'x' is the number of opponents.

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  • Thanks very much Raymond. So against 2 opponents the probability that one opponent has a 4 is 32.7% and against 3 opponents it is 46.7%. – Brian Mar 31 '18 at 21:05
  • What if we want the probability of an opponent having a 4 or a set of 3's? – Brian Mar 31 '18 at 21:06
  • @Brian those percentages are correct yes. You then ask for the probability of an opponent having a 4, however it is identical to your question. Your last question, proability of a set of 3's occuring is harder to answer than your original question. I suggest you open a new question for this and I'd be happy to answer it. – Raymond Timmermans Mar 31 '18 at 22:05

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