YESHA SHAH
2013039
GROUP 6
Today, we played a card game in group. The game we played is Kruskal’s Count.
This trick may be perform to one individual or to a whole audience, and involves the spectators counting through a pack of cards until they reach a final chosen card. Yet, despite this seemingly random choice of cards, the magician is still able to predict the spectator’s chosen card. The trick is known as ‘Kruskal’s Count’ and was invented by the American mathematician and physicist, Martin Kruskal and described by Martin Gardner. Although this trick will not work every time, we will show that the probability of success is around 85%.
The Trick
A spectator is invited to shuffle a pack of cards as many times as they like. The spectator is then asked to secretly pick a number between 1 and 10 and to count along as cards from the deck are displayed. The magician may choose to display the cards one at a time, or he may choose to display all 52 cards together. The magician explains that the card in the position of the spectator’s secret number becomes the spectator’s first chosen card. The spectator is then told to use the value of that chosen card as his new number, and to repeat the process until the magician runs out of cards. Here, aces are worth 1; Jack, Queen, King are worth 5; and all other cards take their face value.
Yet, despite this seemingly random path through a shuffled pack of cards, the magician is able to predict
the spectator’s last chosen card. Watch and interact with a video of the trick being performed here.
The Secret
How is this done?
Well, unknown to the spectator, the magician also picks an initial number between 1 and 10, and proceeds to go through the same process. And although the magician may not have picked the same number as the spectator, there is a high probability they will land on the same final card. This is because, even though the magician and the spectator begin on different paths, there will come a point, simply by coincidence, when the two players land on the same card. And from that point on the two paths will become synchronized, meaning both players end on the same final card.
If we assume the initial numbers are equally likely to be chosen, then the probability
of success is 84%. And we can increase that chance slightly, to 85%, if the magician chooses 1 as his initial
number.
Furthermore, we will now show that, if N is the number of cards, and x is the mean average card value,
then the probability of success may be approximated with the simple formula
Source: http://www.singingbanana.com
2013039
GROUP 6
Today, we played a card game in group. The game we played is Kruskal’s Count.
Kruskal’s Count
The Trick
A spectator is invited to shuffle a pack of cards as many times as they like. The spectator is then asked to secretly pick a number between 1 and 10 and to count along as cards from the deck are displayed. The magician may choose to display the cards one at a time, or he may choose to display all 52 cards together. The magician explains that the card in the position of the spectator’s secret number becomes the spectator’s first chosen card. The spectator is then told to use the value of that chosen card as his new number, and to repeat the process until the magician runs out of cards. Here, aces are worth 1; Jack, Queen, King are worth 5; and all other cards take their face value.
Yet, despite this seemingly random path through a shuffled pack of cards, the magician is able to predict
the spectator’s last chosen card. Watch and interact with a video of the trick being performed here.
The Secret
How is this done?
Well, unknown to the spectator, the magician also picks an initial number between 1 and 10, and proceeds to go through the same process. And although the magician may not have picked the same number as the spectator, there is a high probability they will land on the same final card. This is because, even though the magician and the spectator begin on different paths, there will come a point, simply by coincidence, when the two players land on the same card. And from that point on the two paths will become synchronized, meaning both players end on the same final card.
If we assume the initial numbers are equally likely to be chosen, then the probability
of success is 84%. And we can increase that chance slightly, to 85%, if the magician chooses 1 as his initial
number.
Furthermore, we will now show that, if N is the number of cards, and x is the mean average card value,
then the probability of success may be approximated with the simple formula
This is an illustration of the game, where two different situations are showed. The first situation is when the player initially starts with number 1 in his mind and then plays the game, which is denoted by yellow dots. Similarly in the second situation the player starts the game with number 7 in his mind and then plays the game, which is denoted by blue dots. The last card which is left is 8 of hearts, in both the situations, which is 85% probability.
Source: http://www.singingbanana.com
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