In one casino I play at, for seat changes instead of a seat change list, the dealer will deal a card face up to all players who are vying for the seat, in order from seat 1 to seat 9. The highest card wins the seat change.

Example: Two players, seated in seats 4 and 8, wish to move to an open seat. The dealer deals one card to seat 4 first, then seat 8. The highest card wins.

What are the odds of winning for each player in a 2 person race?

It may seem like its 50/50 but its not because the first player can draw the Ace of Spades and the game ends immediately.

Then what are the odds for 3 player, 4 player, 5 player etc?

I don't even know where to start on this one!

Thanks!

Why do you think it's not 50/50 just because the first person drawing the ace of spades would end it? The first person drawing the deuce of clubs would also end it....with the opposite outcome. And guess what, he's equally likely the draw the deuce of clubs as the ace of spades.

Yes, in situations like this it is best to ignore "interim" results and simply focus on the "overall" probability.

Suppose all 52 cards are numbered from 1-52 (say deuce of clubs = 1 up to ace of spades = 52). Then it should be clear that if two people each choose a card at random from a 52-card deck and high card wins, then it is 50/50 who will draw the higher card.

Another way of looking at it is the following. Consider your approach of one person showing his card first. Of course, he has a 1/52 chance of drawing each of the 52 cards. If he chooses deuce of clubs (1), then he will lose for sure (win 0/51) no matter which of the 51 remaining cards his opponent draws. If he chooses deuce of diamonds (2), then he will win with prob 1/51, etc., up to the case where he chooses the ace of spades (52) in which case he will win for sure (51/51).

If you work through the math, you will find that the odds of the first player winning is given by:

= (1/52)*(52*51/2)/51
= 1/2

You want to ask the question where suits don't count and aces end the game.

Make our posters work a tiny bit.

Ok; When I though about this I though of the extreme case where the game is played with 52 players.

The last player has the same chance of winning as every other player, but as far as winning the game (his odds vs the field) would he do better being up front or does it not matter?




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It doesn't matter. If he has the same chance as every other player, then every other player has the same chance as he does. Wherever he is, he can offer to switch seats with anyone else, and whether they accept the switch or not, the probabilities don't change. If he doesn't get the ace of spades, he loses - and his chance is the same no matter where he sits.

What do you think the chances are that someone else will win before it gets to the 52nd player?

51/52 unless im missing something?!

I understand for 52 players the first person has the same chance as the last but im having trouble wrapping my head around a 9-player game where the player could win with the As OR the highest card






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I think the problem is that these are usually dealt face up. Imagine if the cards are dealt face down and everyone turns over at exactly the same time. Who has the greater chance of winning?

My friend and I just made a bet. The terms being “In a draw of 3 or more players, does being first to draw confer an advantage, no matter how slight”

He took the ‘yes’ side I took No, there is no advantage. He is really emphatic that there is an advantage but I dont know how to convince him.

Is there any math or a formal proof we could do to show theres no advantage?


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Yes if they are face down its obvious but....




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I showed the math (or rather the results of the math) for the two-person case.

Maybe it helps if you think of the 2-person game with a 2-card deck. It should be obvious in that case, right? (Get out a deck of cards and deal out a few deals if you need, say a deck with just one Ace and one King.)

Now add another card to the deck (say a Queen).

Now, if you want, add another player to the game.

Eventually the light bulb will go on.


P.S. As Didace and other have said above, the probabilities are governed by the deal (how the cards are dealt). So it does not matter how or in which order the cards are revealed. Maybe play around with revealing the cards in these toy games in different ways to convince yourselves of that fact of how probabilities work.


P.P.S. With 3 players you can actually deal out all the possibilities yourselves. There are clearly 6 different equally-probable "orderings" of three cards (AKQ, AQK, KAQ, KQA, QAK, QKA). So I suggest getting out two decks of cards and pull out 18 cards (3 aces, 3 kings, and 3 queens from each deck). Then arrange all 6 possible orderings for the three players on a table. You will quickly see that "going first" confers no advantage.

Here’s the math.

Assume 3 cards, 1 ace. Whoever gets dealt the ace wins.Player 1 is dealt first.

1st Player. P(win) = P(dealt the ace) = 1/3

2nd Player. P(win)= P(player 1 no ace) * P(layer 2 get ace given no ace for P1) = 2/3 * 1/2 = 1/3

3rd. Player. P(win) = 1- Player 1 wins – Player 2 wins = 1- 1/3 – 1/3 = 1/3

Same question different context
Is seat 9s chances of receiving the button the same as seats 1-8?
At the casino I play in the house race off for the button.Seat 1 is dealt the 1st card face up to seat 2nd all the way to seat 9.The high card gets the button but once a player is dealt the ace of spades it ends he gets the button and there will be no more cards dealt out.My friend believes seat 9 might be at a microscopic disadvantage of receiving the button compare each individual players who received cards before him.
He's reasoning is the times seat 1 receives the ace of spade 1/52 times or when seat 2 receives it 1/51 seat 3 1/50 and so on.seat 9 will never get a chance to receive a card.
All and any posters ?
Att. David Skalansky ?

Question is not the probability of drawing the ace of spade but of receiving the button. Seat 1 vs seat 9

Note high card by suit counts for the button

Why do you believe drawing for the button is any different than drawing for a seat change?




ETA: Also, you couldn't have waited 5 months for your first post to make it a nice round 10 years?

Your right.No difference between drawing for the button and seat change.
So you agree with the original poster that all 9 players have a equal chance of winning the botton

There is nothing special about the top card in the deck, nor whichever card is picked first. It has the same chance of being an ace as any other card.

Also, there's a sticky for this. This is no different than thinking AK is less likely on the button because 8 other players were dealt before you, or that your flush draw has fewer than 9 outs because of random folded cards.