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Beckham and the reserve player's chance


Los Angeles Galaxy and San Jose Earthquakes meet in a league match of the Western Conference.


It's 5 minutes to finish the game, and reserve player of Los Angeles Galaxy team Jack McBean, asks coach Bruce Arena, to play the closing minutes of the match.

Bruce is not convinced to make any change, but at the insistence of the boy, decides to give him the chance to play the final minutes of the match, provided that Jack finds out which of the three players Bruce has decided to replace: David Beckham, Juninho or Landon Donovan. All three players have played just as well and are equally tired.

Jack replies that he believes Bruce will replace Beckham. Then Bruce tells him that Juninho will not be the player to be substitued, and offers Jack the possibility of switching the choice of Beckham by Donovan, but young Jack decides to keep his initial choice, because he thinks Bruce doesn't want him to play the match.

Will Jack succeed by keeping his first choice? Will he finally enter the football field?

Go to the second half to discover the solution


Although after having removed Juninho, it seems that, apparently, there's a 50% chance that the player selected by Bruce Arena to leave the playing field is David Beckham, in fact Jack McBean actually has only a 1 in 3 chance of playing the match.

Suppose that the coach had proposed Jack to choose which player would be replaced among 10 field players (we remove the goalkeeper from the choice). The possibility of the player chosen by Jack would be of 1/10. And the chances of the other 9 players, would be 9/10. So at first time, each player has a chance of 10% to be substituted: a 1/10 to the chosen one, and 9/10 for all the other 9 players:

1/10 for the chosen player A + ( 1/10 for player B + 1/10 for player C + 1/10 for the player D 1/10 for player E + ...)

As we remove players from this group, what we do is changing the percentage corresponding to each of these 9/10 of probability. So, with just 3 players, plus the one we have picked, the chances would be:

1/10 for the chosen player A + (3/10 for player B + 3/10 for player C + 3/10 for player D)

And when there are only 2 options, the probabilities are:

1/10 for the chosen player A + (9/10 for player B)

Thus, when the coach removes the other players, the player who stays 'inherits' the probabilities of all the eliminated players, so he has a 9/10 probability of being substituted.

Applying this reasoning to the present puzzle, Beckham has 1/3 of the chances of being substitued, and the other 2 players {Juninho and Donovan} have 2/3 of the chances. If you'd to choose among these two options, what would do? Will you choose either Beckham, (1/3), or {Juninho and Donovan} (2/3)? Obviously, if we would take the {Juninho and Donovan} set.

1/3 ---2/3--- ^ ^

When the coach tells that Juninho is not to be replaced, this fact does not alter the chances among Beckham (1/3) and the {Juninho and Donovan} set (2/3), but what it does is changing the odds in this last set. {Juninho and Donovan} still have together 2/3 of probabilities, a 0% of them for Juninho, and a 100% for Donovan , who 'inherits' the initial probability of the group: 2/3.

1/3 ---2/3--- ^ ^

We all knew that one of the two 'unchosen' players would not be changed. It doesn't matter either. This additional information doesn't bring us any advantage.

Thay way Jack would have been better switched his chance after Juninho's removal, and should have chosen Donovan as the player to be replaced.

This problem is known as the Monty Hall problem. Its name refers to the conductor of the famous American game show 'Let's make a deal'. In this type of competitions, which offer a good reward from 3 possible choices, it's always better to swich the first choice, once the conductor of the contest rules out one of the options, which contained a 'secondary' prize.

From a mathematical point of view, we have:
A = the event in which the contestant chooses the awarded option at first instance
A'= the contestant initially chooses a failed option
B = the contestant is right to maintain their initial choice
B'= the contestant succeeds changing their initial choice
P(A) = probability of succeeding the event A
P(A') = probability of succeding the event A'
P(B/A) = probability of hitting maintaining the initial choice, if you hit at first
P(B'/A') = probability of hitting by changing the initial choice, if you failed it

And so we apply the theorem of total probability, we have:

P(B) = P(B/A) x P(A) = 1 x 1/3 = 1/3
P(B') = P(B'/A') x P(A') = 1 x 2/3 = 2/3
Since, in this case:

P(A) = 1/3
P(A') = 1-1/3 = 2/3
P(B/A) = 1, P(B/A') = 0
P(B/A') = 0, P (B'/A') = 1

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