The Birthday Bet

4335 /www/nrich/html/content/id/4335/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"><mdo:image alt="Birth dates" height="254" src="Birth%20Dates70.JPG" width="346"></mdo:image> Here's a puzzle : The next ten people coming into a store will be asked what day in the year their birthday is. For example 17th October. If the prize is £20, would you bet £1 that two of these ten people will have the same birthday ? If not, what's the lowest prize value for which you would take this bet and why? No matter how big the prize or how easy it looks to win, it isn't smart to bet if I can't stand the loss. However, lots of things are not certain and we often need to make decisions in the face of that uncertainty. Probability is how mathematicians quantify uncertainty. Puzzles and games can be an excellent way to explore this. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This solution comes from Tom Ridley: Assuming that the problem stated means that you win if AT LEAST two people have the same birthday, then the following method is appropriate. For the purpose of simplifying the problem, people with their birthday on the 29th of February have been excluded. Instead of looking at the problem in the stated way (i.e. the probability that two or people will have the same birthday) it is simpler to look at it in terms of the probability that no one will have the same birthday. Once this has been calculated, the probability that two or more people will have the same birthday can be calculated by subtracting this value from 1. There are 365 days in the year. Assume that primarily we only have two people. The probability that they have different birthdays is 365/365 $\times$364/365. This is because if the second person's birthaday is not the same as the first person's birthday then his/her birthday can be on any of the 364 other days of the year. Therefore for ten people, the probability that nobody will have the same birthday is 365/365 $\times$364/365 $\times$363/365 $\times$362/365 $\times$ 361/365 $\times$360/365 $\times$359/365 $\times$358/365 $\times$357/365 $\times$356/365. This approximately equates to: 0.883.The probability therefore that at least 2 people will have the same birthday is 1 - 0.883 = 0.117. So, the question still remains: Is it worth it? If you're wrong, you lose £1, if you're right, then you gain £20. As the potential loss is 1/20 of the potential gain, I would argue that for the bet to be worth making, the chance of winning should be greater than 1/20. 1/20 = 0.05 and 0.116948177> 0.05 therefore I believe that the bet is worth making. Even if the prize were only £10, probability would still be in favour of you making youre money back as 0.116948177 > 0.1 Several other correct solutions to this tricky puzzle were received -- well done everyone! </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> <a href="https://nrich.maths.org/admin/edit.php?obj_id=4335">This problem</a> deliberately involves the consequences of an event (the prize size) not just the chance of the event occurring. <h3>Possible approach</h3> Progress with the problem requires stage-by-stage thinking - and also may require starting with smaller numbers (eg. three people) and then building up from there. <h3>Key questions</h3> <ul> <li>What is the chance that one person's birthday is different from that of another person chosen at random ?</li> </ul> <ul> <li>So what is the chance that their birthdays do match ?</li> </ul> <ul> <li>Given that the first two did not match, what is the chance that a third person will not match with either of those first two ?</li> </ul> <ul> <li>Can you now calculate the probability that all three birthdays are different ?</li> </ul> <h3>Possible extension :</h3> <div>This problem allows a wider exploration and research of 'systems' to win, at Roulette for example.</div> The concept of gambler's ruin is useful to include, where the winning system cannot be continued because a series of losses has caused the situation where there is nothing left to bet with. <div>There is also lots to discuss in the end note'</div> <div>No matter how big the prize or how easy it looks to win, it isn't smart to bet if I can't stand the loss.</div> However, lots of things are not certain and we often need to make decisions in the face of that uncertainty. Probability is how mathematicians quantify uncertainty.' <h3>Possible support :</h3> Explore the situation where two people each roll a dice and ask how likely is it that they roll the same number. Consider the complementary situation where the numbers rolled must be different. Extend that to three people and onwards, asking at each stage what prize for a £1 stake would be a good-odds bet. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">What is the chance that one person's birthday is different from that of another person chosen at random ? So what is the chance that their birthdays do match ? Given that the first two did not match, what is the chance that a third person will not match with either of those first two ? Can you now calculate the probability that all three birthdays are different ? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> The prob that a second person has a different birthday from the first is 364 / 365 The prob that a third person is different from these two is 363 / 365 Below are calculated results from Excel - four values per row. The values are : Number of people P(that the last person is different from those already present) P(everyone's different) : product of all probs above P(a match somewhere) : 1 - P(everyone's different) With ten people there's a little above 11% chance of a match So if the bet is taken 100 times I "expect" around 11 successes each worth Â£20. Â£220 for Â£100 outlay is worth taking provided I can stand any losses that may occur in the chance of things. The lowest prize value that is worth taking is Â£8.55 (1 / 0.116948 ) <mdo:image width="380" height="572" src="Excel results for Birthday Probability.JPG" alt=""></mdo:image> </mdoxml> 2 4 0 0 0 1 0 The Birthday Bet The next ten people coming into a store will be asked their birthday. If the prize is £20, would you bet £1 that two of these ten people will have the same birthday ? Probability Theoretical probability Probability Conditional probability