6-card poker play: dealt 4oak

[Minn. Fatz]

In 6-card poker with a bonus for 4 of a kind with a pair, you are dealt 4oak and an odd card. Which play has the higher EV, holding all five dealt cards and trying to pair the kicker or holding only the 4oak and trying to draw a pair? The solution is left as an exercise for the reader.

[Galeygoo]

I would have no idea; but I'm very happy to see you back in the forum, Fatz.  Hope it's not a short visit....

[billryan]

I'll play.
If you keep the card,there are three cards out of forty six possibilities that will give you a pair.
Discard it and your fifth card will be any one of thirteen possibilities. Your sixth card will be one of forty six, three of which makes your pair.
I'd say your EV is equal, either way.
Final answer, but far from certain it's correct.

[Vman96]

Hold all five cards:

Make pair: 3/47 = 0.06383 (you only are dealt five cards in this game bill. The sixth card is on the draw only.)

Toss 5th card:
3 combinations to make pair with discarded rank
6 combinations to make pair with other 11 ranks
1081 combinations to choose 2 cards from 47 remaining

(3 + 6*11)/(47*46/2) = 69/1081 = 3/47.

So in games where drawing to a kicker doesn't matter, you can do either one. But if you get a bonus pay with a kicker, drop or hold the kicker as appropriate.

[Minn. Fatz]

(3 + 6*11)/(47*46/2) = 69/1081 = 3/47.

Correct, as is:

(3 + 6*11)/(47*46/2) = 69/(47*23) = 3/47.

Another point is that throwing a non-kicker card to increase the probability of drawing a kicker bonus to your quads has no effect on the overall probability of getting a quad + pair.

[billryan]

I'm confused. How was my answer incorrect?

[Vman96]

I'm confused. How was my answer incorrect?

It lead you to the same answer of "either decision has the same probability", but you assumed the game dealt out 6 cards instead of 5, so you used denominators of 46 vs. 47.

But also you make the comment that when throwing away the card that the probability of the next card is 1 of 13 of any rank, and that's definitely not true because when your holding a quad, one rank is totally blocked out. I'm thinking the fact you came up with the same final answer is coincidence. :(

[Minn. Fatz]

Maybe not quite coincidence...

Holding only the 4oak and having drawn the first card, there are indeed only 3 possibilities for a second card to make the pair, regardless of how many ranks are "live" or what the denominator of the fraction is...

Similarly there are only 3 possibilities (that is, combin(3,2) ) to make a pair out of the 3 cards left in the rank of the discard...

So, good instincts, and the mistaken assumptions didn't affect the "final answer," or as The Man famously said, "sometimes you make the right move for the wrong reason."

[billryan]

I do the conceptualizing and hire folks to do the actual goezintas if you follow my drift.

[Galeygoo]

The Man THAT...was too cool!!