{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2026-05-26T00:16:20.000Z","description":"Problems submitted by members of the MATLAB Central community.","is_default":true,"created_by":161519,"badge_id":null,"featured":false,"trending":false,"solution_count_in_trending_period":0,"trending_last_calculated":"2026-05-26T00:00:00.000Z","image_id":null,"published":true,"community_created":false,"status_id":2,"is_default_group_for_player":false,"deleted_by":null,"deleted_at":null,"restored_by":null,"restored_at":null,"description_opc":null,"description_html":null,"published_at":null},"problems":[{"id":44286,"title":"Compute average gain for some bets.","description":"Given a vector containg the odds of some events in decimal format (e.g., |odds=[1.3 2.5 1.5]| ) and a vector of the same dimensions containing the amount of dollars that you are betting for each of those (e.g., |bets=[100 10 80]| ), the function |avg_gain(odds,bets)| should return the average of the gain for each possible outcome. The average is a simple average and should not weight the events by their actual probability. Moreover it needs to be rounded to the second decimal digit.\r\n\r\n\r\nExample:\r\nodds= [1.1]\r\nbets= [100]\r\n\r\nthere are only two possible outcomes in this simple case:\r\n\r\n\r\n# Item one I win the bet: I gain 100*1.1 -100 dollars= +10 dollars\r\n# Item two  I lose the bet: I just lose the bet of 100 dollars = -100 dollars. \r\n\r\n\r\nSo I need to perform an average of -100 and 10 that is -45.\r\n\r\nDisclaimer: a positive average does not mean that betting is a good idea, since  we are only performing a simple average, ignoring the probabilities of the different events. I want in no way encourage betting since it is risky and sometimes addictive. It's just a simple Cody problem, take it for what it really is.","description_html":"\u003cp\u003eGiven a vector containg the odds of some events in decimal format (e.g., \u003ctt\u003eodds=[1.3 2.5 1.5]\u003c/tt\u003e ) and a vector of the same dimensions containing the amount of dollars that you are betting for each of those (e.g., \u003ctt\u003ebets=[100 10 80]\u003c/tt\u003e ), the function \u003ctt\u003eavg_gain(odds,bets)\u003c/tt\u003e should return the average of the gain for each possible outcome. The average is a simple average and should not weight the events by their actual probability. Moreover it needs to be rounded to the second decimal digit.\u003c/p\u003e\u003cp\u003eExample:\r\nodds= [1.1]\r\nbets= [100]\u003c/p\u003e\u003cp\u003ethere are only two possible outcomes in this simple case:\u003c/p\u003e\u003col\u003e\u003cli\u003eItem one I win the bet: I gain 100*1.1 -100 dollars= +10 dollars\u003c/li\u003e\u003cli\u003eItem two  I lose the bet: I just lose the bet of 100 dollars = -100 dollars.\u003c/li\u003e\u003c/ol\u003e\u003cp\u003eSo I need to perform an average of -100 and 10 that is -45.\u003c/p\u003e\u003cp\u003eDisclaimer: a positive average does not mean that betting is a good idea, since  we are only performing a simple average, ignoring the probabilities of the different events. I want in no way encourage betting since it is risky and sometimes addictive. It's just a simple Cody problem, take it for what it really is.\u003c/p\u003e","function_template":"function avg = avg_gain(odds,bets)\r\n  avg=0;\r\nend","test_suite":"%%\r\nodds = [1.1] ;\r\nbets= [100] ;\r\navg_correct = -45.00;\r\nassert(isequal(avg_gain(odds,bets),avg_correct))\r\n\r\n%%\r\nodds = [1.1 1.3 2.5] ;\r\nbets= [100 200 300] ;\r\navg_correct = -40.00;\r\nassert(isequal(avg_gain(odds,bets),avg_correct))\r\n\r\n%%\r\nodds = [1.33 1.3 2.5 5.13] ;\r\nbets= [100 200 300 10] ;\r\navg_correct = -12.85;\r\nassert(isequal(avg_gain(odds,bets),avg_correct))","published":true,"deleted":false,"likes_count":1,"comments_count":4,"created_by":88062,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":32,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2017-08-12T17:34:09.000Z","updated_at":"2026-05-29T03:45:43.000Z","published_at":"2017-08-12T17:43:06.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven a vector containg the odds of some events in decimal format (e.g.,\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:rFonts w:cs=\\\"monospace\\\"/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eodds=[1.3 2.5 1.5]\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e ) and a vector of the same dimensions containing the amount of dollars that you are betting for each of those (e.g.,\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:rFonts w:cs=\\\"monospace\\\"/\u003e\u003c/w:rPr\u003e\u003cw:t\u003ebets=[100 10 80]\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e ), the function\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:rFonts w:cs=\\\"monospace\\\"/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eavg_gain(odds,bets)\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e should return the average of the gain for each possible outcome. The average is a simple average and should not weight the events by their actual probability. Moreover it needs to be rounded to the second decimal digit.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eExample: odds= [1.1] bets= [100]\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ethere are only two possible outcomes in this simple case:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"2\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eItem one I win the bet: I gain 100*1.1 -100 dollars= +10 dollars\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"2\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eItem two I lose the bet: I just lose the bet of 100 dollars = -100 dollars.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eSo I need to perform an average of -100 and 10 that is -45.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eDisclaimer: a positive average does not mean that betting is a good idea, since we are only performing a simple average, ignoring the probabilities of the different events. I want in no way encourage betting since it is risky and sometimes addictive. It's just a simple Cody problem, take it for what it really is.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"},{"id":1147,"title":"Possible Outcomes of American Roulette","description":"The payout for American roulette can be calculated by:\r\n\r\npayout = (38/n)-1\r\n\r\nwhere n is the number of squares the bet covers.\r\n\r\nGiven n and the amount bet, determine the possible outcome of one play.\r\n\r\nExample:\r\n\r\nx = [n amount_bet] = [4 25]\r\n\r\ny = [amount_win amount_loss] = [212.5 -25]","description_html":"\u003cp\u003eThe payout for American roulette can be calculated by:\u003c/p\u003e\u003cp\u003epayout = (38/n)-1\u003c/p\u003e\u003cp\u003ewhere n is the number of squares the bet covers.\u003c/p\u003e\u003cp\u003eGiven n and the amount bet, determine the possible outcome of one play.\u003c/p\u003e\u003cp\u003eExample:\u003c/p\u003e\u003cp\u003ex = [n amount_bet] = [4 25]\u003c/p\u003e\u003cp\u003ey = [amount_win amount_loss] = [212.5 -25]\u003c/p\u003e","function_template":"function y = all_on_red(x)\r\n  y = x;\r\nend","test_suite":"%%\r\nx = [1 100];\r\ny_correct = [3700 -100];\r\nassert(isequal(all_on_red(x),y_correct))\r\n\r\n%%\r\nx = [2 15];\r\ny_correct = [270 -15];\r\nassert(isequal(all_on_red(x),y_correct))\r\n\r\n%%\r\nx = [4 5];\r\ny_correct = [42.5 -5];\r\nassert(isequal(all_on_red(x),y_correct))\r\n\r\n%%\r\nx = [4 -5];\r\ny_correct = 'N/A';\r\nassert(isequal(all_on_red(x),y_correct))\r\n\r\n%%\r\nx = [18 20];\r\ny_correct = [200/9 -20];\r\nassert(isequal(all_on_red(x),y_correct))","published":true,"deleted":false,"likes_count":0,"comments_count":2,"created_by":9554,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":121,"test_suite_updated_at":"2012-12-28T16:24:59.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2012-12-28T16:14:17.000Z","updated_at":"2026-04-17T13:28:38.000Z","published_at":"2012-12-28T16:14:17.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe payout for American roulette can be calculated by:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003epayout = (38/n)-1\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ewhere n is the number of squares the bet covers.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven n and the amount bet, determine the possible outcome of one play.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eExample:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ex = [n amount_bet] = [4 25]\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ey = [amount_win amount_loss] = [212.5 -25]\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"}],"problem_search":{"problems":[{"id":44286,"title":"Compute average gain for some bets.","description":"Given a vector containg the odds of some events in decimal format (e.g., |odds=[1.3 2.5 1.5]| ) and a vector of the same dimensions containing the amount of dollars that you are betting for each of those (e.g., |bets=[100 10 80]| ), the function |avg_gain(odds,bets)| should return the average of the gain for each possible outcome. The average is a simple average and should not weight the events by their actual probability. Moreover it needs to be rounded to the second decimal digit.\r\n\r\n\r\nExample:\r\nodds= [1.1]\r\nbets= [100]\r\n\r\nthere are only two possible outcomes in this simple case:\r\n\r\n\r\n# Item one I win the bet: I gain 100*1.1 -100 dollars= +10 dollars\r\n# Item two  I lose the bet: I just lose the bet of 100 dollars = -100 dollars. \r\n\r\n\r\nSo I need to perform an average of -100 and 10 that is -45.\r\n\r\nDisclaimer: a positive average does not mean that betting is a good idea, since  we are only performing a simple average, ignoring the probabilities of the different events. I want in no way encourage betting since it is risky and sometimes addictive. It's just a simple Cody problem, take it for what it really is.","description_html":"\u003cp\u003eGiven a vector containg the odds of some events in decimal format (e.g., \u003ctt\u003eodds=[1.3 2.5 1.5]\u003c/tt\u003e ) and a vector of the same dimensions containing the amount of dollars that you are betting for each of those (e.g., \u003ctt\u003ebets=[100 10 80]\u003c/tt\u003e ), the function \u003ctt\u003eavg_gain(odds,bets)\u003c/tt\u003e should return the average of the gain for each possible outcome. The average is a simple average and should not weight the events by their actual probability. Moreover it needs to be rounded to the second decimal digit.\u003c/p\u003e\u003cp\u003eExample:\r\nodds= [1.1]\r\nbets= [100]\u003c/p\u003e\u003cp\u003ethere are only two possible outcomes in this simple case:\u003c/p\u003e\u003col\u003e\u003cli\u003eItem one I win the bet: I gain 100*1.1 -100 dollars= +10 dollars\u003c/li\u003e\u003cli\u003eItem two  I lose the bet: I just lose the bet of 100 dollars = -100 dollars.\u003c/li\u003e\u003c/ol\u003e\u003cp\u003eSo I need to perform an average of -100 and 10 that is -45.\u003c/p\u003e\u003cp\u003eDisclaimer: a positive average does not mean that betting is a good idea, since  we are only performing a simple average, ignoring the probabilities of the different events. I want in no way encourage betting since it is risky and sometimes addictive. It's just a simple Cody problem, take it for what it really is.\u003c/p\u003e","function_template":"function avg = avg_gain(odds,bets)\r\n  avg=0;\r\nend","test_suite":"%%\r\nodds = [1.1] ;\r\nbets= [100] ;\r\navg_correct = -45.00;\r\nassert(isequal(avg_gain(odds,bets),avg_correct))\r\n\r\n%%\r\nodds = [1.1 1.3 2.5] ;\r\nbets= [100 200 300] ;\r\navg_correct = -40.00;\r\nassert(isequal(avg_gain(odds,bets),avg_correct))\r\n\r\n%%\r\nodds = [1.33 1.3 2.5 5.13] ;\r\nbets= [100 200 300 10] ;\r\navg_correct = -12.85;\r\nassert(isequal(avg_gain(odds,bets),avg_correct))","published":true,"deleted":false,"likes_count":1,"comments_count":4,"created_by":88062,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":32,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2017-08-12T17:34:09.000Z","updated_at":"2026-05-29T03:45:43.000Z","published_at":"2017-08-12T17:43:06.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven a vector containg the odds of some events in decimal format (e.g.,\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:rFonts w:cs=\\\"monospace\\\"/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eodds=[1.3 2.5 1.5]\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e ) and a vector of the same dimensions containing the amount of dollars that you are betting for each of those (e.g.,\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:rFonts w:cs=\\\"monospace\\\"/\u003e\u003c/w:rPr\u003e\u003cw:t\u003ebets=[100 10 80]\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e ), the function\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:rFonts w:cs=\\\"monospace\\\"/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eavg_gain(odds,bets)\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e should return the average of the gain for each possible outcome. The average is a simple average and should not weight the events by their actual probability. Moreover it needs to be rounded to the second decimal digit.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eExample: odds= [1.1] bets= [100]\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ethere are only two possible outcomes in this simple case:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"2\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eItem one I win the bet: I gain 100*1.1 -100 dollars= +10 dollars\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"2\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eItem two I lose the bet: I just lose the bet of 100 dollars = -100 dollars.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eSo I need to perform an average of -100 and 10 that is -45.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eDisclaimer: a positive average does not mean that betting is a good idea, since we are only performing a simple average, ignoring the probabilities of the different events. I want in no way encourage betting since it is risky and sometimes addictive. It's just a simple Cody problem, take it for what it really is.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"},{"id":1147,"title":"Possible Outcomes of American Roulette","description":"The payout for American roulette can be calculated by:\r\n\r\npayout = (38/n)-1\r\n\r\nwhere n is the number of squares the bet covers.\r\n\r\nGiven n and the amount bet, determine the possible outcome of one play.\r\n\r\nExample:\r\n\r\nx = [n amount_bet] = [4 25]\r\n\r\ny = [amount_win amount_loss] = [212.5 -25]","description_html":"\u003cp\u003eThe payout for American roulette can be calculated by:\u003c/p\u003e\u003cp\u003epayout = (38/n)-1\u003c/p\u003e\u003cp\u003ewhere n is the number of squares the bet covers.\u003c/p\u003e\u003cp\u003eGiven n and the amount bet, determine the possible outcome of one play.\u003c/p\u003e\u003cp\u003eExample:\u003c/p\u003e\u003cp\u003ex = [n amount_bet] = [4 25]\u003c/p\u003e\u003cp\u003ey = [amount_win amount_loss] = [212.5 -25]\u003c/p\u003e","function_template":"function y = all_on_red(x)\r\n  y = x;\r\nend","test_suite":"%%\r\nx = [1 100];\r\ny_correct = [3700 -100];\r\nassert(isequal(all_on_red(x),y_correct))\r\n\r\n%%\r\nx = [2 15];\r\ny_correct = [270 -15];\r\nassert(isequal(all_on_red(x),y_correct))\r\n\r\n%%\r\nx = [4 5];\r\ny_correct = [42.5 -5];\r\nassert(isequal(all_on_red(x),y_correct))\r\n\r\n%%\r\nx = [4 -5];\r\ny_correct = 'N/A';\r\nassert(isequal(all_on_red(x),y_correct))\r\n\r\n%%\r\nx = [18 20];\r\ny_correct = [200/9 -20];\r\nassert(isequal(all_on_red(x),y_correct))","published":true,"deleted":false,"likes_count":0,"comments_count":2,"created_by":9554,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":121,"test_suite_updated_at":"2012-12-28T16:24:59.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2012-12-28T16:14:17.000Z","updated_at":"2026-04-17T13:28:38.000Z","published_at":"2012-12-28T16:14:17.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe payout for American roulette can be calculated by:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003epayout = (38/n)-1\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ewhere n is the number of squares the bet covers.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven n and the amount bet, determine the possible outcome of one play.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eExample:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ex = [n amount_bet] = [4 25]\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ey = [amount_win amount_loss] = [212.5 -25]\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"}],"errors":[],"facets":[[{"value":"Combinatorics I","count":1,"selected":false}],[{"value":"easy","count":1,"selected":false},{"value":"medium","count":1,"selected":false}]],"term":"tag:\"odds\"","page":1,"per_page":50,"sort":"map(difficulty_value,0,0,999) asc"}}