The last to draw is the most unlucky?
Boredom is a great element for creativity, both in the service of mathematics and games. In this case, think of a very simple game with a deck of 5 red cards and 1 black card. After shuffling, the first player takes the top card, if it is black he loses, if it is red nothing happens and it is the next player’s turn. The process continues until someone draws a black card.
In the case of two players, a simple question we can ask is: which of the two players has a greater chance of winning? One way to solve this problem is to imagine that for each position in the pile there is the name of a player, therefore, for 6 cards and 2 players called player A and player B, we would have the following pile:
Player A;
Player B;
Player A;
Player B;
Player A;
Player B.
So, of the 6 cards in the pile, 3 are occupied by player A and 3 are occupied by player B. Since the card of defeat is placed in a random position, we have a 50% chance of player A winning and a 50% chance of player player B to win.
Now suppose, for example that the deck has 4 red cards and a black card, in this case, we would have:
Player A;
Player B;
Player A;
Player B;
Player A.
So, of the 5 positions, 3 are assigned to player A and 2 are assigned to player B. Since the black card is placed in a random position, we have a 60% chance of player A losing and a 40% chance of player B losing .
Interestingly, we can think of a situation with the same number of players as that of cards. Thus, with 5 cards, players A, B, C, D, E compete. We each have a designated position.
Player A;
Player B;
Player C;
Player D;
Player E.
Winning depends only on the position in which the black card is initially placed. Thus, each player has exactly 80% chance of winning. Regardless of whether you are the first or the last to play. The false belief that the last player will be doomed because when his turn comes, the remaining card is the black card, in fact it is erroneous, because we need to consider that his chance of drawing the card is:
P (A) .P (B) .P (C) .P (D),
in which P (X) is the chance of player X to draw the card and not lose
Turning this into numbers, we have:
(4/5). (3/4). (2/3). (1/2) = 1/5.
In other words, regardless of where we start the “game”, the conditions for victory are the same, what changes is, at most, our “impression of victory” when considering that the card we will draw is closer or more distant from us .
Another way to analyze the problem is to imagine a deck with 3 red cards and 2 black cards in a 5-player game. Now it’s not so trivial. We have the same initial list:
Player A;
Player B;
Player C;
Player D;
Player E.
The simplest case is to analyze player A, his chance of losing initially is 2/5, so his chance of winning is 3/5.
For player B, we have 4 situations.
Draw with the deck containing two black cards and win:
(3/5). (2/4) = 30%.
Draw with the deck containing a black card and win:
(2/5). (3/4) = 30%.
Draw with the deck containing two black cards and lose:
(3/5). (2/4) = 30%.
Draw with the deck containing a black card and lose:
(2/5). (1/4) = 10%.
Totaling 60% of the victory against 40% of the defeat.
For player C, we have 7 situations:
Draw with the deck containing two black cards and win:
(3/5). (2/4). (1/3) = 10%.
Draw with the deck containing a black card and win, case 1:
(2/5). (3/4). (2/3) = 20%.
Draw with the deck containing a black card and win, case 2:
(3/5). (2/4). (2/3) = 20%.
Draw with the deck containing two black cards and lose:
(3/5). (2/4). (2/3) = 20%
Draw with the deck containing a black card and lose, case 1:
(2/5). (3/4). (1/3) = 10%.
Draw with the deck containing a black card and lose, case 2:
(3/5). (2/4). (1/3) = 10%.
Win a card:
(2/5). (1/4) = 10%.
Totaling 60% of the victory against 40% of the defeat.
For player D, we have 10 situations:
Draw with the deck containing a black card and win, case 1:
(2/5). (3/4). (2/3). (1/2) = 10%.
Draw with the deck containing a black card and win, case 2:
(3/5). (2/4). (2/3). (1/2) = 10%.
Draw with the deck containing a black card and win, case 3:
(3/5). (2/4). (2/3). (1/2) = 10%.
Draw with the deck containing a black card and lose, case 1:
(2/5). (3/4). (2/3). (1/2) = 10%.
Draw with the deck containing a black card and lose, case 2:
(3/5). (2/4). (2/3). (1/2) = 10%.
Draw with the deck containing a black card and lose, case 3:
(3/5). (2/4). (2/3). (1/2) = 10%.
Draw with the deck containing two black cards and lose:
(3/5). (2/4). (1/3) = 10%
Win draw card, case 1:
(2/5). (1/4) = 10%.
Win draw card, case 2:
(2/5). (3/4). (1/3) = 10%.
Win draw card, case 3:
(3/5). (2/4). (1/3) = 10%.
Totaling 60% of the victory against 40% of the defeat.
For player E, we have 10 situations:
Win draw card, case 1:
(2/5). (1/4) = 10%.
Win draw card, case 2:
(2/5). (3/4). (1/3) = 10%.
Win draw card, case 3:
(3/5). (2/4). (1/3) = 10%.
Win draw card, case 4:
(2/5). (3/4). (2/3). (1/2) = 10%.
Win draw card, case 5:
(3/5). (2/4). (2/3). (1/2) = 10%.
Win draw card, case 6:
(3/5). (2/4). (2/3). (1/2) = 10%.
Draw with the deck containing a black card and lose, case 1:
(2/5). (3/4). (2/3). (1/2) = 10%.
Draw with the deck containing a black card and lose, case 2:
(3/5). (2/4). (2/3). (1/2) = 10%.
Draw with the deck containing a black card and lose, case 3:
(3/5). (2/4). (2/3). (1/2) = 10%.
Draw with the deck containing a black card and lose, case 4:
(3/5). (2/4). (1/3) = 10%.
Totaling 60% of the victory against 40% of the defeat.
Did you think the chances of winning and losing would be different between the order of the players now that we had two black and three red cards?