Fantasy Football - Footballguys Forums
Welcome to Our Forums. Once you've registered and logged in, you're primed to talk football, among other topics, with the sharpest and most experienced fantasy players on the internet.
Odds Question (1 Viewer)
- Thread starter otello
- Start date
AAABatteries
Footballguy
Are we counting an Ace as 1?
If so
If not
If so
I don't know
If not
I don't know
Ilov80s
Footballguy
This is a permutation without repetition. There are 10x9x8x7x6x5x4x3x2x1 possible outcomes when flipping over the cards. Am I correct in saying that you want to know what are the odds that none of the cards get shuffled into the position their number represents (the card labeled 6 ending up in the 6th spot for example)?
IvanKaramazov
Footballguy
(9/10)^10?
Ketamine Dreams
Footballguy
(.9 X 8/9 X 7/8 X 6/7 X 5/6 X 4/5 X 3/4 X 2/3 x 1/2) I believe. You can't do the last card because it will be the only option left to you so you will never avoid it.
ETA: Assuming you can only pick from the numbers remaining.
ETA: Assuming you can only pick from the numbers remaining.
Last edited by a moderator:
IvanKaramazov
Footballguy
This was my very first thought too, but I'm pretty sure it's wrong. Consider Card 10. There is a 9-in-10 chance that it has already been drawn by the time you get down to the final card.
Ketamine Dreams
Footballguy
It depends if you see and know every card pulled along the way. I took it to mean you guess and then pull a card. Then once you see that card you guess again for the next draw.
AAABatteries
Footballguy
Pretty sure he means that the number called out corresponds to the times he's flipped the card. He's not guessing a random card.
IvanKaramazov
Footballguy
Ah, I see. I'm visualizing the problem differently. I'm thinking it's like the cards have been dealt out into ten positions, and what are the odds that Card 1 is not in Position 1, and Card 2 is not in Position 2, and Card 3 is not in Position 3, and so on.
Ilov80s
Footballguy
Right, I took it as the 5th card flipped is the one that has "5" written on it but I'm not entirely sure
AAABatteries
Footballguy
culdeus
Footballguy
Thorpe
Footballguy
abbottjamesr
Footballguy
I get (n-1) / n * 1/2 where n is the number of cards. So for 10 cards 9/10 * 1/2 = 9/20 = 45%
It works for sure on n = 2, 3, 4, 5 cause I tried wrote in out in xcel. I assume it carries on.
There are n! ways to arrange the cards. If any one of the numbers on the cards alligns to the order you lose.
so for n = 3 you have 6 possible orders of which none of the ones that start with 1 will be right so that is where we get the (n-1)/n of the remaining half will be winning and half wont regardless of n.
1 2 3 - x
1 3 2 - x
2 1 3 - x
2 3 1
3 1 2
3 2 1 - x
It works for sure on n = 2, 3, 4, 5 cause I tried wrote in out in xcel. I assume it carries on.
There are n! ways to arrange the cards. If any one of the numbers on the cards alligns to the order you lose.
so for n = 3 you have 6 possible orders of which none of the ones that start with 1 will be right so that is where we get the (n-1)/n of the remaining half will be winning and half wont regardless of n.
1 2 3 - x
1 3 2 - x
2 1 3 - x
2 3 1
3 1 2
3 2 1 - x
Last edited by a moderator:
otello
Footballguy
10 cards shuffled at random
I turn over 1st card, and say 1, the card is a 3. I continue on.
I turn over the second card, and say 2, and the card is an 8. I continue on.
I turn over the third card, say three, and the card is 5. I continue on.
4th card, say 4, and I turn over a 4. Game over.
What are the odds I turn over all ten cards, counting to ten as I do it, and don't turn over the card with a number that matches the xth card I am turning over?
is that clearer?
I turn over 1st card, and say 1, the card is a 3. I continue on.
I turn over the second card, and say 2, and the card is an 8. I continue on.
I turn over the third card, say three, and the card is 5. I continue on.
4th card, say 4, and I turn over a 4. Game over.
What are the odds I turn over all ten cards, counting to ten as I do it, and don't turn over the card with a number that matches the xth card I am turning over?
is that clearer?
abbottjamesr
Footballguy
Bronco Billy
Footballguy
Pretty complicated little problem. You have 10 events without replacement complicated by dependent elimination. Might be a little above my capability.
Chemical X
Footballguy
am I eating a steak sandwich while shuffling?
Ketamine Dreams
Footballguy
Haha. I definitely interpreted it wrong this morning in my sleep-deprived haze.
-OZ-
Footballguy
Ilov80s
Footballguy
No, there are 36628800 possible permutations in this problem, but many more than just 1 that has would cause him to "lose". Just look at the most simple level: card 1 could be in the 1st spot or card 2 could be in the 2nd spot. That is already more than 1 possible losing scenario. What you calculated are the odds that the cards are in the exact order of 1,2,3,4,5,6,7,8,9,10.
parasaurolophus
Footballguy
I enjoyed this. Since I purposely waited until reading all responses and thought about it a bit.
Ignoratio Elenchi
Footballguy
1/e. (Or at least, pretty close to that.)
Longer answer, provided without proof:
Given a(1) = 0 and the recurrence relation a
= n*a(n-1) + (-1)^n, then if you have n cards numbered 1 to n in random order, the probability that no card's rank will match its place in the deck is a / n! That sequence converges to 1/e pretty quickly.
Longer answer, provided without proof:
Given a(1) = 0 and the recurrence relation a
= n*a(n-1) + (-1)^n, then if you have n cards numbered 1 to n in random order, the probability that no card's rank will match its place in the deck is a / n! That sequence converges to 1/e pretty quickly.
Last edited by a moderator:
OrtonToOlsen
Footballguy
Would it work with pizza slices?
Buckychudd
Footballguy
I got 0.0000276%.
Buckychudd
Footballguy
Buckychudd
Footballguy
Ilov80s
Footballguy
How does your post make my post wrong? I was commenting to the number of possible permutations and showing that there are better odds than 1/ 10!
I think 10% is right myself
Last edited by a moderator:
Ignoratio Elenchi
Footballguy
What would the answer be if there were only 4 cards, instead of 10?
Bucks
Footballguy
Short Corner
Footballguy
Ignoratio Elenchi
Footballguy
Yeah, I assume that's what he meant - and with 4 cards there are few enough permutations that it's easy to check if this is, in fact, the right answer.
It isn't.
abbottjamesr
Footballguy
Lets try it for n=4. The problem states you want to know the odds of not turning the card which matches the turn it is.
1, 2, 3, 4 - x
1, 2, 4, 3 - x
1, 3, 2, 4 - x
1, 3, 4, 2 - x
1, 4, 2, 3 - x
1, 4, 3, 2 - x
2, 1, 3, 4 - x
2, 1, 4, 3 - o
2, 3, 1, 4 - x
2, 3, 4, 1 - o
2, 4, 1, 3 - o
2, 4, 3, 1 - x
3, 1, 2, 4 - x
3, 1, 4, 2 - o
3, 2, 1, 4 - x
3, 2, 4, 1 - x
3, 4, 1, 2 - o
3, 4, 2, 1 - o
4, 1, 2, 3 - o
4, 1, 3, 2 - x
4, 2, 1, 3 - x
4, 2, 3, 1 - x
4, 3, 1, 2 - o
4, 3, 2, 1 - o
24 possible orders the cards can be laid out. 9 times you get through with out matching the turn number out of 24 or 3/8.
The formula I end up with is (n-1) / n X 1/2 or 3/4 x 1/2 for this example. The reason this works is that all the permutations that start with 1 are no good so that leaves (n-1)/n of those half are the good ones.
1, 2, 3, 4 - x
1, 2, 4, 3 - x
1, 3, 2, 4 - x
1, 3, 4, 2 - x
1, 4, 2, 3 - x
1, 4, 3, 2 - x
2, 1, 3, 4 - x
2, 1, 4, 3 - o
2, 3, 1, 4 - x
2, 3, 4, 1 - o
2, 4, 1, 3 - o
2, 4, 3, 1 - x
3, 1, 2, 4 - x
3, 1, 4, 2 - o
3, 2, 1, 4 - x
3, 2, 4, 1 - x
3, 4, 1, 2 - o
3, 4, 2, 1 - o
4, 1, 2, 3 - o
4, 1, 3, 2 - x
4, 2, 1, 3 - x
4, 2, 3, 1 - x
4, 3, 1, 2 - o
4, 3, 2, 1 - o
24 possible orders the cards can be laid out. 9 times you get through with out matching the turn number out of 24 or 3/8.
The formula I end up with is (n-1) / n X 1/2 or 3/4 x 1/2 for this example. The reason this works is that all the permutations that start with 1 are no good so that leaves (n-1)/n of those half are the good ones.
Ilov80s
Footballguy
