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2 votes (50%)
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2 votes (50%)
2 votes (50%)
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3 votes (75%)
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4 members have voted

Romes
Romes
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August 30th, 2017 at 9:55:08 AM permalink
This is fun, we're having fun... Anyways "Bruce" and I were flying back from a recent playing trip. On the plane we of course brought a deck of cards to mess around and burn some time. At some point somehow the following question got brought up:

Given a completely random shuffled deck of cards. What are the odds you will draw ANY QUAD within the first half of the deck (re: 26 cards)? Not quads or better, no, just QUADS of any rank.

When we were younger we made up all kinds of card games. One of them we called "4 best." It's a 2 person poker variant. You randomly shuffle, get half the deck each, and make 4 five card poker hands. You set your hands and then compare one at a time, with each being worth different points. Hand 1 worth 1 point, 2 worth 2, etc, etc, up to Hand 4 worth 4 points. So clearly there was also strategy as to which points to go after (usually 4 and 2, but if someone knew that they could counter). During our fun of making this game up and playing it occurred to me that I almost always had a straight flush or quads. For some reason I feel like this might be influencing my rational, but none the less... I really wish I could remember more of the conversation and math we tossed back and forth, but it was unfortunately a couple weeks prior and I forgot to post this when we got home... as we made a bet on it =).

I figured since I've seen a bunch of these pop up as fun little math exercises I'd push it out to the masses to have some fun with. Please put any answers in spoilers, to not spoil the fun for others that want to take a guess and check back later!
Playing it correctly means you've already won.
ThatDonGuy
ThatDonGuy
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August 30th, 2017 at 10:30:14 AM permalink

There are (48)C(22) sets of 26 cards that have four Aces, the same number with four 2s, and so on.

However, this counts, for example, the (44)C(18) sets of 26 cards with both four Aces and four 2s twice, so these have to be subtracted.

However, this now undercounts the (40)C(14) sets of 26 cards with four Aces, four 2s, and four 3s. The adding and subtracting continues until getting down to the (28)C(2) sets of 26 cards with a particular set of six quads.

The total number of sets of 26 cards with at least one quad is:
13 x (48)C(22) - (13)C(2) x (44)C(18) + (13)C(3) x (40)C(16) - (13)C(4) x (36)C(10) + (13)C(5) x (32)C(6) - (13)C(6) x (28)C(2)
Dividing this by the number of sets of 26 cards in the 52-card deck, (52)C(26), returns about 56.8978%, which has been confirmed in simulation.

Romes
Romes
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August 30th, 2017 at 11:33:59 AM permalink
ThatDonGuy always ruining these questions for everyone ;-). I kid... A very detailed and simulation backed answer =P. Always love seeing you work these out DonGuy, thanks for your input!

I'd like to see some others with "off the top of their head" solutions. I started down the road of every single quad, etc, etc, on the plane... but of course looking for a quick answer, I took the over on 50%, whereas Bruce took the under.
Last edited by: Romes on Aug 30, 2017
Playing it correctly means you've already won.
billryan
billryan
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August 30th, 2017 at 11:59:45 AM permalink
I guesstimated 55%.
The difference between fiction and reality is that fiction is supposed to make sense.
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