

Looking for advice on a random pub argument. I'll try and keep it short:
You are about to play high Vs low 1 on 1. The deck is shuffled, You cut the deck revealing the bottom card as your number. You keep hold of the cards you cut off the top. Your opponent cuts the remaining cards.
The question is; When you go first, does taking 1 card off the top leaving your opponent with 51 chances against taking 51 cards leaving your opponent with a single card to pick up make a difference? Are you more likely to win leaving your opponent a single card, or does it all boils down to a 1 in 52 chance?

wayne mathias
United States Niceville Florida

The chance of the card chosen being any particular thing is not altered by randomly eliminating some choices ahead of time  it is still pure unskewed random all the way with the chosen card being 1 of 51 unknowns.
Equivalent to chits well shaken in a bag and grabbing a chit from the top of the bag vs grabbing from the bottom vs grabbing from wherever in the bag.



Even with the one revealed card?

wayne mathias
United States Niceville Florida

Think chits in the bag  you take one out and know that chit only thus drawing from the remaining unknown pool.



Ok. How can I convince my nonboard game playing friend of this? That it's always a 1 in 52 chance because it's always random? Is there factual maths to show this?



Logan8 wrote: Ok. How can I convince my nonboard game playing friend of this? That it's always a 1 in 52 chance because it's always random? Is there factual maths to show this?
https://www.mathsisfun.com/data/probability.html



goldspire wrote:
This means we are both wrong, actually, based on the marbles in a bag. Revealing the first card after cutting the deck means a single number has been removed from the 52 card deck, so you'll have an informed opinion if you are likely to win or lose based on that one card.
But, my question is, if you take 4 out of 5 marbles out of the bag as opposed to 1, is there any more chance that the red one is left, or is it all the same?



Logan8 wrote: Even with the one revealed card?
The revealed card changes the odds of winning drastically, but doesn't change the odds of any specific card being drawn.

Larry L
United States Stockton California
He who games with the most dice wins.
I + I = 0

Before beginning the game as you described, you are equally likely to win or lose, regardless of how many cards you leave behind. You can discover, during the middle of the game if you have an advantage or not, but revealing or not revealing that card does not in anyway change the overall odds of the game.
Before beginning the game, the bag of marbles (4 blue, 1 red?) the chances are exactly equal that the first marble you pull will be red, the second, and so on. As you pull more marbles from the bag, you can gain some information about the odds of the next marble, but that doesn't change the original odds.
For example, you have pulled 4 marbles out of the bag. 80% of the time you have already pulled the red marble and the probability the marble left is red is 0%. 20% of the time you have pulled 4 blue marbles and the probability that the red marble is left is 100%. Those two facts combine to demonstrate that 20% of the time you will pull the red marble last.



Logan8 wrote: goldspire wrote: This means we are both wrong, actually, based on the marbles in a bag. Revealing the first card after cutting the deck means a single number has been removed from the 52 card deck, so you'll have an informed opinion if you are likely to win or lose based on that one card. But, my question is, if you take 4 out of 5 marbles out of the bag as opposed to 1, is there any more chance that the red one is left, or is it all the same?
If you look at the marbles each time you draw and you haven't pulled the red yet the chance goes up with each draw. If you pull 4 marbles without looking the odds haven't changed.

Niko
Canada Calgary Alberta
[This space is intentionally left empty]

Logan8 wrote: Ok. How can I convince my nonboard game playing friend of this? That it's always a 1 in 52 chance because it's always random? Is there factual maths to show this? Maybe try the following: Does it make a difference whether they picked their card randomly out of all available ones, or if you make the choice for them by randomly removing all but one card? The answer is of course no. In either case one card was randomly chosen, whether they pick it or you doesn't make a difference. Might be they are getting tripped up on their agency in the whole thing, but that is irrelevant.



Right. So let's simplify this. 4 marbles, 2 red, 2 blue. You only ever reveal 1 marble regardless or how many you draw and you aren't allowed to draw them all. Does the initial action of taking 1 version taking 3 l, before the reveal, given you any more chance if winning?
My arguement has always been no, in terms of cards. It's always 1 in 52 of what you leave behind, and then you reveal your one card changing the probability.

Larry L
United States Stockton California
He who games with the most dice wins.
I + I = 0

Logan8 wrote: Right. So let's simplify this. 4 marbles, 2 red, 2 blue. You only ever reveal 1 marble regardless or how many you draw and you aren't allowed to draw them all. Does the initial action of taking 1 version taking 3 l, before the reveal, given you any more chance if winning?
My arguement has always been no, in terms of cards. It's always 1 in 52 of what you leave behind, and then you reveal your one card changing the probability.
You are correct. If you can pull any number of marbles (13) and reveal one randomly that you have pulled while leaving the rest for your opponent to pick from, the odds stay exactly the same, no matter how many you pull.

Mark C
United States Ypsilanti Michigan

I believe what you are talking about is often referred to in statistics as "dependent" and "independent" events. If you're interested, you can google that or look on youtube and get some nice explanations that may help.
For marble/card draws, the underlying event of an additional draw is dependent on the first draw, but if you set up the question such that you don't know what happened, the dependent part doesn't come into play.

wayne mathias
United States Niceville Florida

OK  let us say you draw a 6  we are looking for a 7,8,9,10,j,q,k,a (32 cards) and not a 2,3,4,5,6 (19 cards since you took a 6 already)
Now, of the 51 cards remaining I decide to take the bottom card by setting all the others to one side before I pick it up (same as you keeping all but one card) instead of just taking the card off the bottom the stack.
The odds for the bottom card being a card you want remain unchanged. Same odds for top card after a random cut. Same odds to just take the next card in the deck.
Let us say we have 3 chips  2 white and 1 black
Your goal is to get me to draw the black chip.
First you draw 0 or 1 or 2 chips blindly without looking at or displaying them, then I draw 1 chip from the remaining chip(s).
What are my odds of drawing a black chip and can you change them? (I will lose 1/3 of the time and win 2/3)
You take 0 one of three are black coming to me or 1/3
You take 1 = 2/3 chance of me having one of each to pick from so 2/3 of 1/2 = 1/3 for me to pick the black one this adds to 1/3 chance of me getting all white with no chance to draw black and 1/3 + 0 = 1/3
You take 2 = 1/3 chance of black being the one remaining for me
No matter how you divide a random set, the odds remain unchanged as long as the entire set remains unknown.
Just like dealing 8 cards including 1 joker to 8 people sitting around a table and then looking to see who got the joker  odds are 1 in 8 no matter where you sit in the sequence.

Russ Williams
Poland Wrocław Dolny Śląsk

It sounds like you are possibly confused about (or unaware of the distinction between) the absolute probability that you will win this game (which is obviously 50%) and the conditional probability that you will win this game given the knowledge of what the first player's number is.
So a bit of web research about that will hopefully be helpful.
As to how you can convince your friends: if you don't want to learn or explain the math, just do it experimentally. If they think it should make a difference, then repeat the game a few dozen times each way (first player takes the top 1 card only; first player takes the top 51 cards) and they'll probably start to see that there's no obvious difference.

wayne mathias
United States Niceville Florida

The question is not about what the initial odds are, which is determined by the original card drawn and is known.
The question is about any intentional changes to those original odds caused by blindly subdividing the remaining set and then picking from one of the subsets with no additional knowledge.
Think of a table that will seat 51. You draw a card  does not matter if you look at it now or not because when you DO look at it you will know the odds of any seat having a winning card  and place a card face down in front of each seat so that you do not know what they are.
I come in and will choose a seat with a card. If you walk with me and I cannot pick any seat we pass until you shout yahtzee does that change the "odds" of the seat I choose having a winning card?
Please note: this is an exact functional equivalent of the cut the deck question  shouting yahtzee equals the cut, which leaves some seats/cards remaining to choose from.

CARL SKUTSCH
United States New York New York
Agricola, Sekigahara, Concordia, Innovation, COOKIE!!! (and Guinness)
SANJURO: You're all tough, then? GAMBLER: What? Kill me if you can! SANJURO: It'll hurt.

Going into the game I hope we can agree it's totally random who wins or loses.
You go first. You pick the top card, leaving 51 for your opponent. Your card is a random card but will average an "8" (assuming you play Ace as high). With 51 cards left, 3 will be 8s for a tie, 24 will be below an 8, 24 will be above an 8. Even odds.
You go first, you pick all the cards but one. Now your poor opponent has no choice, he can only pick that one card. Doesn't matter. Your card on average will be an 8. Of the remaining 51 cards 3 will still be 8s, 24 will be below an 8, 24 will be above an 8. Sure, 50 of those 51 cards are already out of play (because of your weird cut) but which cards are in the deck of 50 and which card is remaining is totally random. The odds of that one remaining card being a winner remain the same.
The point is not how many cards remain but how many cards remain unknown, that determines the odds of whether your opponent wins or loses. In both examples, 51 card remain unknown. The odds don't shift. All that changed is that in the first example HE got to choose the random card that would win or lose, the second example YOU got to choose the random card that would win or lose.

Bryan Thunkd
United States Florence MA

So think about it this way... if you randomly shuffle a deck of cards then the card on the bottom of the deck is equally as likely to be an Ace of Spades as it is to be the Two of Clubs, or any other card in between, right?
When you draw all the cards except the last card, you haven't really constrained the other player much at all. It's true that he can't have the card that you've revealed certainly, but that's true no matter how many cards you left him. And he can't have any of the 50 cards that you picked up behind your revealed card, only the one card that you left him... but as it is completely random which cards are in the fifty you denied him and in the one you left him, you haven't taken away his chance to have any card in the deck (other than your card).
In fact you haven't done anything to make it less likely he'd have any other card in the deck. If that doesn't click for you, tell me which card he's less likely to have than any other card. Which card do you think you've kept him from having? I mean, sure it's unlikely that he has the Ace of Spades, but it's just as unlikely for him to draw the Five of Hearts right? Both cards are more likely in the 50 you denied him. But that's true of every card in the deck (except the one you revealed).
Another way of saying this is asking which card you think it's most likely he has. If you can't predict a particular card that's more likely than any other card to appear, then you can't possibly predict that any card is less likely to show up either. So whatever cards you're trying to deny him are just as likely to appear as any other card.


