{"id":1221,"date":"2023-08-30T17:37:18","date_gmt":"2023-08-30T16:37:18","guid":{"rendered":"http:\/\/www.cir-strategy.com\/blog\/?p=1221"},"modified":"2023-11-23T11:25:09","modified_gmt":"2023-11-23T11:25:09","slug":"generalised-game-show-problem-using-bayes-theorem","status":"publish","type":"post","link":"http:\/\/www.cir-strategy.com\/blog\/?p=1221","title":{"rendered":"Generalised Game Show Problem"},"content":{"rendered":"

We generalise a result in Professor David Mackay\u2019s book on inference. Bayes theorem also plays a crucial role in decision-making:<\/p>\n

P<\/mi>(<\/mo>A<\/mi>|<\/mo>D<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>D<\/mi>|<\/mo>A<\/mi>)<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo><\/mrow>P<\/mi>(<\/mo>D<\/mi>|<\/mo>A<\/mi>)<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo>+<\/mo>P<\/mi>(<\/mo>D<\/mi>|<\/mo>\u00ac<\/mi>A<\/mi>)<\/mo>P<\/mi>(<\/mo>\u00ac<\/mi>A<\/mi>)<\/mo><\/mrow><\/mfrac><\/math><\/span><\/p>\n

Let us consider a worked example which demonstrates how unintuitive results following from this 260-year-old theorem can be.<\/p>\n

In the Game Show example, we can simplify Bayes\u2019 Theorem by using a form that expands out all individual doors, and then cancelling off the unconditional probability of each door in numerator and denominator as they are each P(any door)<\/mtext>=<\/mo>1<\/mn>n<\/mi><\/mfrac><\/math><\/span>, where there are n doors. We consider the two distinct representative cases:<\/p>\n

P<\/mi>(<\/mo>prize behind my chosen door<\/mtext>\u2223<\/mo>m<\/mtext>)<\/mo>=<\/mo>P<\/mi>(<\/mo>m<\/mtext>\u2223<\/mo>prize behind my chosen door<\/mtext>)<\/mo><\/mrow>P<\/mi><\/mfrac><\/math><\/span><\/p>\n

P<\/mi>(<\/mo>prize behind another door<\/mtext>\u2223<\/mo>m<\/mtext>)<\/mo>=<\/mo>P<\/mi>(<\/mo>m<\/mtext>\u2223<\/mo>prize behind another door<\/mtext>)<\/mo><\/mrow>P<\/mi><\/mfrac><\/math><\/span><\/p>\n

where here<\/p>\n

P<\/mi>=<\/mo>1<\/mn>\/<\/mo><\/mrow><\/mrow>(<\/mo><\/mrow>n<\/mi>\u2212<\/mo>1<\/mn><\/mrow>m<\/mi><\/mfrac>)<\/mo><\/mrow>+<\/mo>(<\/mo>n<\/mi>\u2212<\/mo>1<\/mn>)<\/mo>\/<\/mo><\/mrow><\/mrow>(<\/mo><\/mrow>n<\/mi>\u2212<\/mo>2<\/mn><\/mrow>m<\/mi><\/mfrac>)<\/mo><\/mrow><\/math><\/span><\/p>\n

and where \u2223<\/mo>m<\/mi><\/math><\/span> means that any m doors were removed at the usual intermediate stage after I chose a door. When we start with n doors, and one has been chosen, and that door happens to have the prize behind it, then the Game Show Host is free to remove m doors from the set of n<\/mi>\u2212<\/mo>1<\/mn><\/math><\/span> doors and so there are <\/mrow>(<\/mo><\/mrow>n<\/mi>\u2212<\/mo>1<\/mn><\/mrow>m<\/mi><\/mfrac>)<\/mo><\/mrow><\/math><\/span> available ways for the host to do so.<\/p>\n

The probability P(any<\/mtext><\/mspace><\/mstyle>m<\/mtext>\u2223<\/mo>1<\/mn>)<\/mo><\/math><\/span> is therefore 1<\/mn>\/<\/mo><\/mrow><\/mrow>(<\/mo><\/mrow>n<\/mi>\u2212<\/mo>1<\/mn><\/mrow>m<\/mi><\/mfrac>)<\/mo><\/mrow><\/math><\/span>, since we equivocate between all the host\u2019s options. For the other case, there are only n<\/mi>\u2212<\/mo>2<\/mn><\/math><\/span> doors for the host to choose m from, so the number of ways is <\/mrow>(<\/mo><\/mrow>n<\/mi>\u2212<\/mo>2<\/mn><\/mrow>m<\/mi><\/mfrac>)<\/mo><\/mrow><\/math><\/span>, and the probability P(any<\/mtext><\/mspace><\/mstyle>m<\/mtext>\u2223<\/mo>not 1<\/mtext>)<\/mo><\/math><\/span> is therefore 1<\/mn>\/<\/mo><\/mrow><\/mrow>(<\/mo><\/mrow>n<\/mi>\u2212<\/mo>2<\/mn><\/mrow>m<\/mi><\/mfrac>)<\/mo><\/mrow><\/math><\/span>.<\/p>\n

after some algebra I find that<\/p>\n

P<\/mi>(<\/mo>prize behind my chosen door<\/mtext>\u2223<\/mo>m<\/mtext>)<\/mo>=<\/mo>n<\/mi>\u2212<\/mo>1<\/mn>\u2212<\/mo>m<\/mi><\/mrow>n<\/mi>\u2212<\/mo>1<\/mn>\u2212<\/mo>m<\/mi>+<\/mo>(<\/mo>n<\/mi>\u2212<\/mo>1<\/mn>)<\/mo>2<\/mn><\/msup><\/mrow><\/mfrac><\/math><\/span><\/p>\n

and<\/p>\n

P<\/mi>(<\/mo>prize behind another door<\/mtext>\u2223<\/mo>m<\/mtext>)<\/mo>=<\/mo>n<\/mi>\u2212<\/mo>1<\/mn><\/mrow>n<\/mi>\u2212<\/mo>1<\/mn>\u2212<\/mo>m<\/mi>+<\/mo>(<\/mo>n<\/mi>\u2212<\/mo>1<\/mn>)<\/mo>2<\/mn><\/msup><\/mrow><\/mfrac><\/math><\/span><\/p>\n

Thus, our probability factor is given by: P(prize behind another door<\/mtext>\u2223<\/mo>m<\/mtext>)<\/mo><\/mrow>P(prize behind my chosen door<\/mtext>\u2223<\/mo>m<\/mtext>)<\/mo><\/mrow><\/mfrac>=<\/mo>n<\/mi>\u2212<\/mo>1<\/mn><\/mrow>n<\/mi>\u2212<\/mo>1<\/mn>\u2212<\/mo>m<\/mi><\/mrow><\/mfrac><\/math><\/span><\/p>\n

This expression gives us back, from the original game, our game strategy factor of 2 times better if we shift door when n<\/mi>=<\/mo>3<\/mn><\/math><\/span> and m<\/mi>=<\/mo>1<\/mn><\/math><\/span>. The factor rises to n<\/mi>\u2212<\/mo>1<\/mn><\/math><\/span> better for shifting to another door for any n<\/mi>\u2265<\/mo>2<\/mn><\/math><\/span> and m<\/mi>=<\/mo>n<\/mi>\u2212<\/mo>2<\/mn><\/math><\/span> doors (all but one other than the one the player chose), and for the shift strategy a probability factor that tends to unity from above when n<\/mi>\u2192<\/mo>\u221e<\/mi><\/math><\/span> and m<\/mi>=<\/mo>1<\/mn><\/math><\/span>, i.e. only one door is removed.<\/p>\n

An informal survey by Student-b has shown that quite often, intuition among a random sample of people asked, is lacking. Some will believe it is better to stick, and some say in the standard three-door game that it is slightly better to shift. The mathematics show that for positive n and m: it is always better to shift<\/em>!<\/p>\n","protected":false},"excerpt":{"rendered":"

We generalise a result in Professor David Mackay\u2019s book on inference. Bayes theorem also plays a crucial role in decision-making: P(A|D)=P(D|A)P(A)P(D|A)P(A)+P(D|\u00acA)P(\u00acA) Let us consider a worked example which demonstrates how unintuitive results following from this 260-year-old theorem can be. In the Game Show example, we can simplify Bayes\u2019 Theorem by using a form that expands … Continue reading Generalised Game Show Problem<\/span> →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[882],"tags":[891,889,884,890,892],"class_list":["post-1221","post","type-post","status-publish","format-standard","hentry","category-probability-as-logic-for-decisionmaking","tag-bayesianism","tag-business-strategy","tag-decisionmaking","tag-probability-logic","tag-rational-degree-of-belief"],"_links":{"self":[{"href":"http:\/\/www.cir-strategy.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/1221","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/www.cir-strategy.com\/blog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/www.cir-strategy.com\/blog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/www.cir-strategy.com\/blog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/www.cir-strategy.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1221"}],"version-history":[{"count":10,"href":"http:\/\/www.cir-strategy.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/1221\/revisions"}],"predecessor-version":[{"id":1272,"href":"http:\/\/www.cir-strategy.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/1221\/revisions\/1272"}],"wp:attachment":[{"href":"http:\/\/www.cir-strategy.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1221"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.cir-strategy.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1221"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.cir-strategy.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1221"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}