My Math Forum Two different way to play the lottery.

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 February 6th, 2013, 04:14 PM #1 Member   Joined: Jun 2012 Posts: 91 Thanks: 0 Two different way to play the lottery. Consider a lottery with a daily draw, where the odds of winning is exactly 1/200. Smith and Jones are about to buy 100 tickets each but will distribute those tickets differently. Smith buys 100 tickets for one draw and Jones buys 1 ticket every day on 100 consecutive days. Who has the biggest chance to win? Since the probability to win on each ticket is 1/200 and Smith buys 100 tickets, the chance that he will win is 100/200, or 1/2. So there is an equal chance that he will win or lose. Jones plays just one ticket for each draw, so he has always the chance of 1/200 to win, which means that his chances to loose is 199/200. The probability that he will lose all the 100 drawings is therefore 199^100 = 0,6057 So Jones chances to win is therefore 1 - 0,6057 = 0,3943 = 39,43 % while Smith had a 50% chance to win. Is this correct?'
 February 6th, 2013, 05:13 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond Re: Two different way to play the lottery. The chance of Jones winning is 1/200 + 1/200 + 1/200 + . . . + 1/200 = 100/200 = 1/2.
 February 6th, 2013, 05:17 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,274 Thanks: 1959 The draw isn't properly defined.
 February 6th, 2013, 05:28 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond Re: Two different way to play the lottery. What do you mean by "draw"?
 February 6th, 2013, 08:16 PM #5 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 634 Thanks: 96 Math Focus: Electrical Engineering Applications Re: Two different way to play the lottery. Hi [color=#0000BF]zengjinlian[/color], I think that your analysis is correct for Jones. I also think that your analysis is correct for Smith provided that there is exactly one winning ticket and exactly 199 losing tickets. Consider the following analysis for Smith along the same line of reasoning that you gave for Jones, first in the case of 1 winning ticket and 199 losing tickets. The probability of Smith losing for all tickets is: $\frac{199}{200} \cdot \frac{198}{199} \cdot \frac{197}{198} \ . \ . \ . \ \frac{102}{103} \cdot \frac{101}{102} \cdot \frac{100}{101}=\frac{100}{200}=\frac{1}{2}$ So the chance of Smith winning, is as you state, $\ 1-\frac{1}{2}=\frac{1}{2}$ Now consider the case where there are 2 winning tickets and 398 losing tickets. The probability of a ticket winning is still $\ \frac{1}{200} \$ but the probability of Smith losing on all tickets is now: $\frac{398}{400} \cdot \frac{397}{399} \cdot \frac{396}{398} \ . \ . \ . \ \frac{301}{303} \cdot \frac{300}{302} \cdot \frac{299}{301}=\frac{300 \cdot 299}{400 \cdot 399}=\frac{299}{532}$ So the probability of Smith winning is $\ 1-\frac{299}{532}=\frac{233}{532} \approx 0.438$ The probability for Jones does not change, it is $\ 1-\left(\frac{398}{400}\right)^{100}=1-\left(\frac{199}{200}\right)^{100}$ I wrote a Ruby script to verify the results. The averages of two runs of 10000 trials each are as follows: With 1 winning ticket out of 200, Smith won an average of 50.085% of the time, while Jones won 39.405% of the time. With 2 winning tickets out of 400, Smith won an average of 43.815% of the time, while Jones won 39.395% of the time. If anyone is interested, I will post the script.
 February 6th, 2013, 08:27 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,955 Thanks: 988 Re: Two different way to play the lottery. 1/200 * 100 = .5 199/200 * 100 = 99.5
February 6th, 2013, 08:42 PM   #7
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Quote:
 Originally Posted by skipjack The draw isn't properly defined.
O.k. I see now. Thanks, all.

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