If you are playing a game of Five Card-Draw with a friend out of a standard deck 52 card deck and you have the following hand (neglecting suits for this example) 3,3,3,5,7. You decide to discard the "5" and draw from the deck. What is the probability that you upgrade to Four of a Kind ( 3,3,3,3,7)?

I'm having a bit of trouble wrapping my head around the following... Does the fact that your opponent has 5 of the 47 remaining out of the draw deck ( leaving 42 in the deck from which you can choose) effect the probability of you drawing the remaining "3" or not?

I may be perpetually inept in probability, but I do "enjoy" the frustration.

Thanks, Eric

Note by Eric Roberts
3 years, 9 months ago

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Since your opponent is equally likely to have any card, you can ignore their hand. Their hand is only important if you have some way to get knowledge of what it might be. You don't have a way, so their hand is just completely random to you. If you look through every permutation of cards they could have, you will see that the probabilities will not change, and this should always be true.

- 3 years, 8 months ago

P(opponent has 3)
= (1C1)(46C4) / 47C5
= 5 / 47

P(upgrade to four of a kind)
= (0/42)(5/47) + (1/42)(42/47)
= 1 / 47

So it's 1/47 if you have not peek at the other hand and 1/42 if the opponent wearing reflective accessories and your 20/20 vision caught the no-3 hand in their hand.

- 4 months, 3 weeks ago

I sort of agree with saying “you can ignore their hand”, but the probability of choosing a three is 1/47, not 1/42. There are 47 cards that are not in your hand, and only 1 of them is a three - it may be in the opponents hand. That is why the probably is somewhat smaller.

- 11 months, 3 weeks ago

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