A Chance to Win?

5611 /www/nrich/html/content/id/5611/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> Imagine you were given the chance to win some money... and imagine you had nothing to lose... Imagine you arrive in a room where you are given £128 and six cards (3 red winners and 3 black losers). You are asked to choose and lay the cards down, one at a time. You can decide in which order to lay them down. At each stage you must bet exactly half the money that you have available. If you select and play a black card you lose the money you bet. If you select and play a red card you receive double the money you bet (ie. you get the money you bet back, plus that amount again, so if you bet £64 and win, your total will increase by £64). If you end up with more money than you started with you get to keep the profit. What's the best order for laying down the cards? What will your strategy be when you are offered 4 or 5 red winning cards? Use the interactivity below to test out some possibilities. Draw some conclusions on what strategy to adopt and try to justify your findings. <mdo:flash height="700" width="530"><param name="movie" value="/content/id/5611/Cards.swf" ></param><param name="flashplayerversion" value="8" ></param><param name="height" value="700" ></param><param name="width" value="530" ></param></mdo:flash> We are indebted to Rob Eastaway for introducing us to this problem. </mdoxml> <mdoxml version="1.0"> Tim explained clearly why you can never win: It doesn't matter what order you play the cards! Each black card makes you lose half your money - which is the same as multiplying what you have by 0.5. Each red card multiplies what you have by 1.5 since you win half of the cash you have. As there are 3 reds and 3 blacks, the total amount you will finish with will be: "starting amount" x 0.5 x 0.5 x 0.5 x 1.5 x 1.5 x 1.5 = starting amount x 0.42 (approx). Since you can multiply in any order, you will always end up with the same amount! (£54) Charlie, Gemma, Griselda, Tom, Jason and Greg from Colyton Grammar School also worked out what was happening: It does not matter in what order the cards are put down because the result of using a win card is that your money is multiplied by 1.5 and the result of you using a lose card is that your money is multiplied by 0.5. As multiplication is commutitive whatever order the cards are placed the results will be the same! A lose followed by a win results in a return of 3/4 (0.5 x 1.5) of your stake. This is with a ratio of 1:1 red:black. Given the 6 cards the return is 27/64 of your stake. In order to win, the ratio of red:black cards has got to be at least 1.71:1. This is because in order to cancel the effect of one black card which is multiplying by 0.5 you need to multiply it by at least 2 as 0.5 x 2 = 1. Since each red card multiplies by 1.5 you need to find the number (power) of 1.5's you need to multiply to get 2. That is, what is x if $1.5^x$ is going to be greater than or equal to 2? (Answer: x has to be greater than or equal to 1.71) </mdoxml> <mdoxml version="1.0"> Teachers may choose not to use the interactivity, and instead try out the activity in the classroom with actual cards, challenging students to find a winning combination. This may be a good problem to introduce when students are learning about percentage increases and decreases. </mdoxml> <mdoxml version="1.0"> If you start by choosing a black card and then choose two red cards this will be the state of play at the half-way stage: <mdo:image width="520" height="532" alt="half way through" src="Chance%20to%20win.gif"></mdo:image> Keep a record of the outcome from laying down the cards in different orders. What do you notice? How is your total amount of money affected by a win? How is it affected by a loss? </mdoxml> <mdoxml version="1.0">Ideas for July 2012 revamp of this problem Spreadsheet <a href="/content/id/5611/a%20chance%20to%20win.xls">a chance to win.xls</a> start with 4red 3black in prob and interactivity. surprised? why do we all end up with the same? with less than we started? what would happen with five red and 3 black? Given n black cards how many red cards do I need to turn a profit? questions to consider if I have equal numbers of red and black what happens? twice as many red as black? can I ever exactly break even? What if I bet amounts other than 0.5? what if I could have two more reds than blacks</mdoxml> 2 5 0 0 1 0 0 A Chance to Win? Imagine you were given the chance to win some money... and imagine you had nothing to lose... Calculations and Numerical Methods Commutative, associative and distributive laws Calculations and Numerical Methods Multiplication & division Fractions, Decimals, Percentages, Ratio and Proportion Calculating with fractions