card cut odds?

Sucker: Is this a game of chance?
W. C. Fields: Not the way I play it, no.

Well, just for the fun of it, allowing the possibility of ties with no tie-breaker…

Whatever rank the first person cuts, the other person has 3 chances in 51 of cutting the same rank for a tie. That works out to approximately 6% chance of a tie, or 94% chance of no tie. Therefore, each player has a 47% chance of winning outright.

I am having trouble reconstructing my thoughts on this. I freely admit I may be wrong but something about the puzzle makes me think that it’s not a simple 2-from-52 problem. The first draw changes the odds of the second; I guess it boils down to whether you are comparing two blind draws or figuring the specific odds of the second draw beating the first.

If it’s two blind draws, the odds are 50-50. (Or 47-47-6 if ties are considered.)

But if I draw and show, say, an ace on the first pull, the odds of a winning card are 0 in 48 and a tie would be 3 in 51. The odds would change for each successively lower first draw.

As with so many statistical puzzles, it’s all in the semantics. Choose… wisely.

Amateur Barbarian - before the game begins, the odds are 50% (assuming a tiebreaker) - and that’s the purpose of a card cut, as well as the question posed by the OP.

Okay. No argument. But should I take Bob’s offer of door number two?

So that everyone has a chance for the excitement of suspense.

While OP’s question might be answered 50% mathematically, when actually presented with such a chance to gamble a large sum, you might do well to recall the advice Sky Masterson gave Nathan Detroit.

  1. I think you mean ‘the odds of a losing card are 48 in 51 and a tie would be 3 in 51; there is no chance of winning.’

  2. If you draw and show, say, an two on the first pull, the odds of a winning card are 48 in 51 and a tie would be 3 in 51; there is no chance of losing.

All these chances balance out exactly, giving the correct answer of an even chance.
Where do you get your idea of ‘microscopic odds favoring the first card drawn’ from?

All semantic and grammatical issues aside, from posting before I’d had enough coffee.

Astronomy club: sounds like they’re trying to build the suspense. I.e. everyone has a chance to feel they could be the winner up until the end.

Car Talk: I think you’re misunderstanding. They take all the correct answers they received and randomly select one as the winner. This ensures the contest focuses on the puzzle and not on a race to the telephone/mailbox. A couple of my favorite podcasts employ this method. One of them started doing it because even though they have 200,000 listeners every week, the same three people would win the puzzle every week. A few dedicated people would download the podcast the instant it came out, fast-forward to the puzzle section, and post the answer in the forums. Discourages participation from other people.

They could do that by just drawing for the Grand Prize last. I agree it sounds like they have a confused notion of odds, not that I’m presently anyone to talk…

Astronomy Club: It truly makes no difference. When you learn that you won a lessor prize then you know immediately that you didn’t win the grand prize, whether it was drawn first and set aside or is still waiting to be drawn.

Car Talk: Not suggesting that they go through the mail as it comes in. I believe that they wait until the cutoff date, mix all the submissions in a box and start opening them. There’s no need to sort out all the correct answers as the first correct answer will be just as random. This is different from the NPR Sunday Puzzler where they actually announce how many winning submissions they (Will Shortz) received. They probably track that to gauge the difficulty of the puzzle. It’s certainly possible that the Car Talk guys did this as well,and didn’t announce how many correct submissions they got. My point was just that it makes no difference, mathematically.

P.S. Car Talk Puzzlers > Sunday Edition Puzzler. IMO.

I can see their logic. It doesn’t change the overall odds at all, but it does change the individual odds for the Grand Prize once information starts being added. If there are 100 people, drawing in this order, Grand Prize -> 5th place -> 4th place -> 3rd place -> 2nd place, gives all 100 people a chance for the GP, 99:1 for 5th, 98:1 for 4th, 97:1 for 3rd and 96:1 for 2nd. But of course - once names start being read off, all the winners prior to the GP announcement will know their odds dropped to 0 - but at least they had non-zero odds at the time of the GP drawing.

Just like at the beginning of cutting cards the odds are 50:50, they change significantly once the first person goes. If the first slip out of the hat for the 5th place prize is yours, your odds for the Grand Prize are now 0.

And yet for someone who couldn’t make the event because they were sick, the odds remain the same.

I think you mean Amateur Barbarian, unless I’m misunderstanding your post.