Jump to content
BrainDen.com - Brain Teasers
  • 0

Cut To The Chase


Question

Vernon is the most conservative player at our weekly friendly game of dealer's choice. So much so, five card draw is now simply referred to as "The Vern" at our table and most other games, well, "that's not (effin) poker." I mention this only to emphasize the pleasant surprise last week when Vern offered this little diversion. "Okay Huck, cut a card. Ace is one; king is thirteen. Good, the eight of hearts. So I take that card and the next seven unseen. I then give you the next eight cards plus two more. So I'll always have one face-up card and a total number of cards equal to the value of that card shown and you'll have two more cards than me. Five bucks says you don't have more cards than I do of the color of the cut card." Rob, Coop, and I begged in as well and Vern went around the table several times. We added a little drama by alternately flipping cards, with the challenger's last three cards flipped in turn by themselves. A pretty good game, actually. But of course I was wondering and have to ask, what are the odds of beating Vern at his own (other) game?

Link to post
Share on other sites

10 answers to this question

Recommended Posts

  • 0

Probabilities are not my strong point but quick guess:

There are 52 cards, I hope thats a given, the odds of vernon having that same suit as the known card is number of cards * 12/52 so in this example it would be 7*(12/52) plus the 1 card that he already has makes 2 and (8/13) or approx 2.615

any other player has 10*(12/52) chance - 2 and (4/13) or approx 2.308

So i reckon Vernon has the best chance of getting the most of the known suit.

Feel free to pick my knowledge of stats and odds etc to pieces as you wish...

Link to post
Share on other sites
  • 0

<p>

<br />

Probabilities are not my strong point but quick guess:<br />

<br />

There are 52 cards, I hope thats a given, the odds of vernon having that same suit as the known card is number of cards * 12/52 so in this example it would be 7*(12/52) plus the 1 card that he already has makes 2 and (8/13) or approx 2.615<br />

any other player has&nbsp;&nbsp;10*(12/52) chance - 2 and (4/13) or approx 2.308<br />

So i reckon Vernon has the best chance of getting the most of the known suit.<br />

Feel free to pick my knowledge of stats and odds etc&nbsp;&nbsp;to pieces as you wish...

<br />

</p>

Also not my strong suit. But I think your assumptions are wrong. First, we're working off "same color" not "same suit". And one card is already known. So instead of...

multiplying by (12/52), you should use (25/51). That would put Huck at 4.9 and Vern at 4.43.

That said...I'm not sure that's the right way to figure it.

Edited by bobbyplump
Link to post
Share on other sites
  • 0

Added considerations

Not only probability for the case of an 8 being turned up, but need to average the probability for each possible card from ace (1) to king (13). With Vern having the advantage of 1 guaranteed card in that color and winning in the case of even numbers of the color, it is quickly obvious that the edge goes to Vern. I'll give others a chance to develop the math before giving an answer.

Link to post
Share on other sites
  • 0

I think if you take the simplest case (drawing an A), then the probability would look something like this:

The odds of Vern getting at least one suit card are 100%. You get dealt 1 + 2 extra. Your three cards must beat his 1 card (ie, you must have at least two, as a draw Vern still wins). Essentially you get three draws to produce two cards with odds of something like 25/51. Thus your chance of doing so are ~24%. I believe the odds get slightly better in your favor the more cards that are drawn (a king let's say), but you are still poorly mismatched.

Link to post
Share on other sites
  • 0

So sorry for the delay in getting back to this one. That darn pesky work thing compounded by holiday stuff.

only briefly scanned your expression bushindo. impressive as always. personally, had to do it longhand on a spreadsheet with a bunch of cut and paste so dont doubt I may have flubbed somewhere. came up with pretty much the same - .4927945. just tried plugging in for a cut ace in your expression and get .38896? by my methodology Vern wins when all three of his cards are the same color (25/51)3 and when two of his cards are the same color 3*(25/51)2(1-25/51) or .48530. Please feel free to point out my wayward thinking.

did not mean for this problem to be such an exercise in combinatorics. was hoping perhaps initially the effect of the revealed card might be ignored and the (what I thought was unexpected) odds of 50/50 with Vern having two more cards guessed as if the game was played with infinite decks of cards. a better problem would probably have just been to ask who, if anyone, had the best chance of winning and why.

Link to post
Share on other sites
  • 0
So sorry for the delay in getting back to this one. That darn pesky work thing compounded by holiday stuff.
only briefly scanned your expression bushindo. impressive as always. personally, had to do it longhand on a spreadsheet with a bunch of cut and paste so dont doubt I may have flubbed somewhere. came up with pretty much the same - .4927945. just tried plugging in for a cut ace in your expression and get .38896? by my methodology Vern wins when all three of his cards are the same color (25/51)3 and when two of his cards are the same color 3*(25/51)2(1-25/51) or .48530. Please feel free to point out my wayward thinking. did not mean for this problem to be such an exercise in combinatorics. was hoping perhaps initially the effect of the revealed card might be ignored and the (what I thought was unexpected) odds of 50/50 with Vern having two more cards guessed as if the game was played with infinite decks of cards. a better problem would probably have just been to ask who, if anyone, had the best chance of winning and why.

First of all, let me say that this is an excellent problem. Thanks for sharing it.

For the case of the cut ace, then the face card f = 1, the additional same-color cards in Vern's hand c = 0, and the player's required same-color cards to win are p = 2 or p = 3, so according to the expression in the last post the probability of beating Vern when the cut card is an ace is

P = 25C2 * 26C1 /51C3 + 25C3 * 26C0 /51C3

= 0.3745498 + 0.1104442 = 0.484994

We can also compute the chance of Vern losing in this case using another way. As you pointed out, Vern wins when all three of his cards are the same color (25/51)*(24/50)*(23/49)= 0.1104442 and when two of his cards are the same color 3*(25/51)*(24/50)*(26/49) or .0.3745498. The total probability is then .484994, which is identical as the result from the other method.

As for the problem, the result is unintuitive to me as well. Obviously, the dependencies in card drawing would reduce Vern's chance of losing from 50% in the case of infinite decks. I thought that the dependency effect would be larger and that Vern's losing chance (P) should be in the mid 40 (and that this would be a lucrative game to play with my friends), so I was surprised to find P = 49.29124%. Again, thanks for the problem.

Link to post
Share on other sites
  • 0

doh, now I see the error of my ways which I should have noticed earlier with a little more studying of your answer. Thanks for the comments and fine solve, bushindo. I too was disappointed and surprised that the advantage was not greater. Tho the problems are made up the poker night is real (nicknames kept the same to embarrass the guilty) and this might otherwise have been a lucrative game.

Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
  • Recently Browsing   0 members

    No registered users viewing this page.

×
×
  • Create New...