Probability question: Chances of winning

Question

A contest will award 100 winners with a voucher each. There are 300 participants. Elly joined the contest 3 times, but entering using her details, her sister's details, and her mother's details. What is the chances that Elly wins 1, 2, ,or 3 vouchers?

 

 

Answer 1

Hi Tee,   * Update to previous answer *

  Sorry for only providing an approximate answer a few days ago, here is the full/correct answer.

 

We can think of the 300 participants entries arranged in a line: the first 100 will get a voucher, the remaining 200 will not.

The total ways the entries can be arranged is 300!

 

How many arrangements will have exactly one of Elly's entries winning?

For her entry to win it must be in one of the 100 winning locations,

and her sister and mothers entries must be in the 200 non-winning locations.

The remaining 297 entries can be in any location, so all together the chance her entry wins is:

 

 chance her entry wins =    100 * 200 * 199 * 297!  / 300!

                                  =   100 * 200 * 199 / (300 *299 * 298 )

 

The chance her sister's only wins or her mother's only wins is the same, so:

 

  chance 1 win = 3 * 100 * 200 * 199 / (300 *299 * 298)

 

In a similar way we get:

 

  chance 2 wins = 3 * 100 * 99 * 200 /  (300 *299 * 298)

 

and

 

  chance 3 wins = 100 * 99 * 98 / (300 *299 * 298)

 

For completeness,

 

  chance of 0 wins = 200 * 199 * 198 / (300 *299 * 298)

 

These do add up to 1 as expected:

 

  3*100*200*199  +  3*100*99*200  +  100*99*98  +  200*199*198

        =  26730600

which is the same as:  300 *299 * 298  =  26730600 

 

- - - Here is the simple approximation I presented at first: - - -

A simple (slightly approximate) answer comes from assuming each ticket has

1/3 probability of winning, and

2/3 chance of not being chosen.

 

With these probabilites the chance she gets one voucher is:

(1/3)*(2/3)*(2/3) + (2/3)*(1/3)*(2/3) + (2/3)*(2/3)*(1/3)

  = 4/9     (= 12/27)

where the three pieces of the expression are the chance that her ticket only wins, her sisters only, or her mothers only.

 

The chance she gets 2 vouchers is, using similar logic:

(1/3)*(1/3)*(2/3) + (1/3)*(2/3)*(1/3) + (2/3)*(1/3)*(1/3)

  =  2/9    (=6/27)

 

The chance she wins with all 3 is then:

(1/3)*(1/3)*(1/3)

  = 1/27

 

For completeness the chance she wins nothing is:

(2/3)*(2*3)*(2/3)

  = 8/27

 

Note that these all add to 27/27 = 1 as they should.