tag:blogger.com,1999:blog-4115025577315673827.post1279467212717937205..comments2019-10-23T18:01:26.365+05:30Comments on CSE Blog - quant, math, computer science puzzles: Coin Puzzle: Predict the Other's CoinUnknownnoreply@blogger.comBlogger18125tag:blogger.com,1999:blog-4115025577315673827.post-46516507587504504642018-05-22T23:42:29.995+05:302018-05-22T23:42:29.995+05:30Expected pay off for C is 1/2 -1/2*2 which is nega...Expected pay off for C is 1/2 -1/2*2 which is negative. So C should never playSaurabhhttps://www.blogger.com/profile/00869676391995913137noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-74221637074704674902013-01-29T19:11:32.759+05:302013-01-29T19:11:32.759+05:30Great. We all agree that C should not play.
Some ...Great. We all agree that C should not play.<br /><br />Some of you are missing the point that "A and B can initially talk to each other"<br />Some of you are calculating the payoff wrong. If C wins, it has 1 dollar. If C loses, it has to give 1-3=-2 dollars.<br /><br />The strategy that A and B follow:<br />A and B will just say the coin that they have. Hence, on HH and TT - they win. Otherwise, they lose. So, half probability C would win and half that C would lose. Since, C loses more when he loses but gains less when he wins, C should not play.<br /><br />Correct solution by: Takaki, Andre, Felix, JDGM<br />Correct strategy (but not so correct calculation) by Joshua, Shubham Mittal<br /><br />If you are just looking for the solution, Perfect solution by Andre. Thanks<br />Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-3690737309433665342013-01-02T15:20:29.471+05:302013-01-02T15:20:29.471+05:30You din't take in account the fact that "...You din't take in account the fact that "A & B may initially talk to each other".<br />This will allow them to make some strategy to guess correct answer with higher probability.<br />e.g.<br />Their strategy could be "They guess the same as their own coin toss."<br />A (Guess by A) B (Guess by B) Win<br />H H H H Yes<br />H H T T No<br />T T H H No<br />T T T T Yes<br />So now this strategy increases their winning probability to 0.5<br />so expected gain of C per round = 1*0.5-3*0.5 = -$1Shubham Mittalhttps://www.blogger.com/profile/08300602657432728321noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-49464142523781289462012-12-31T13:37:04.319+05:302012-12-31T13:37:04.319+05:30Typo: "C loses $3" should be "C los...Typo: "C loses $3" should be "C loses $2" in both places I wrote it. This is because C takes $1 from the players at the start.<br /><br />I did realise this when calculating my expectation of negative $0.5, just typed it wrong in the table.JDGMhttps://www.blogger.com/profile/11829357060109505064noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-60981861819820748282012-12-31T11:15:27.194+05:302012-12-31T11:15:27.194+05:30Nice one.
If A and B agree always to say what the...Nice one.<br /><br />If A and B agree always to say what they see on their own coin:<br /><br />Heads-Heads: C loses $3.<br />Tails-Tails: C loses $3.<br />Tails-Heads: C wins $1.<br />Heads-Tails: C wins $1.<br /><br />These outcomes are equally likely so C's expected return per round is negative $0.5. Ouch! Don't play!<br /><br />Note: If A and B agree to say the opposite of what they see on their own coin then it works the same.JDGMhttps://www.blogger.com/profile/11829357060109505064noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-61321409088762918232012-12-31T10:45:00.986+05:302012-12-31T10:45:00.986+05:30C should play,
as making a correct predict for a...C should play, <br /><br />as making a correct predict for a team is 1/2, it makes that C will loose(if both predict right)with probability with 1/4.<br /><br />so out of 4 matches C looses 1 and wins 3.<br />for wining 3 he get Rs3<br />and for loosing 1 he gets Rs 1 but then give back Rs3, so total loos of Rs2 in this match.<br />Finally after 4 matches C have Rs3-Rs2=Rs1.<br /><br />Niraj Kumarhttps://www.blogger.com/profile/02292877595504129823noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-88978341824640933952012-12-31T05:44:04.567+05:302012-12-31T05:44:04.567+05:30C should not play this game.
A and B can win with ...C should not play this game.<br />A and B can win with probability one half, simply by guessing that the other has the same toss (e.g., if A obtains tail, he guesses that B has tail as well, and vice-versa). Since both tosses are equal with probability one half, the expecte gain of C is<br />1 - 3 * 1/2 = -1/2<br />which is negative, so he should not play.<br />FĂ©lix de Chaumont Quitryhttps://www.blogger.com/profile/02724420621332283916noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-85965441962436200152012-12-31T05:03:46.584+05:302012-12-31T05:03:46.584+05:30How does the game end? Is it A and B who choose we...How does the game end? Is it A and B who choose went to stop playing?randombluehttps://www.blogger.com/profile/13515312228949377641noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-58596620671331170242012-12-31T02:44:50.525+05:302012-12-31T02:44:50.525+05:30Probability for C to win is 3/4.
Hence the expec...Probability for C to win is 3/4.<br /> <br />Hence the expectation for C is 0.<br /><br />So he should not playAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-40995525814179732752012-12-31T01:05:35.798+05:302012-12-31T01:05:35.798+05:30Consider the following strategy. Both A and B alwa...Consider the following strategy. Both A and B always guess that the result of the coin toss of the other player is the same as theirs (so e.g. if A gets heads, they guess that B also got heads). With two independent coin flips there is 0.5 probability that the two coins are either both heads or both tails. In that case A and B would both guess correctly, so therefore they have a 0.5 chance of winning and 0.5 probability of losing.<br /><br />So with probability 0.5 player C will earn 1 dollar and with probability 0.5 player C will lose 2 dollars (1 gained initially -3 given away in the end). Then the expected value of the game for C will be <br />E = 0.5*1 + 0.5*(-2) = 0.5 - 1 = -0.5<br /><br />Therefore if A and B are smart players (and decide on a good strategy like always guessing the same or always guessing the opposite of what they get) then C should NOT play this game.<br /><br />(Note that if A and B would guess randomly each time, then the probability that they would get each others coin toss results correct would be 0.5 * 0.5 = 0.25 and the expected value for C would be<br />E = 0.25*(-2) + 0.75*1 = -0.5 + 0.75 = 0.25,<br />so then C SHOULD play the game)Andrehttps://www.blogger.com/profile/16809250021969872384noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-75464391039819769162012-12-30T22:19:55.427+05:302012-12-30T22:19:55.427+05:30yes, even in the event that C is an honest player....yes, even in the event that C is an honest player. A has 1/2 chances of being correct and B has 1/2 chances of being correct. this is a total of 1/4 chances of both being correct. Each round C gains 1 dollar minus 3 dollars multiplied by the chance of A and B being correct (1/4). so C leaves each round with 1 - (3/4) = 1/4. on average C should gain 25 cents each round (1/4 of a dollar). i've been out of school for a while am i missing something obvious?Thomashttps://www.blogger.com/profile/01255348996885829648noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-51548525045231451432012-12-30T21:44:04.611+05:302012-12-30T21:44:04.611+05:30in point 4, did u mean C has to give back 3 dollar...in point 4, did u mean C has to give back 3 dollars to each team (A & B). so total is $6?Mark Thienhttps://www.blogger.com/profile/00509179305586931730noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-61537271429326373162012-12-30T21:43:06.024+05:302012-12-30T21:43:06.024+05:30in point 1, did you mean each team give C a dollar...in point 1, did you mean each team give C a dollar? mean total is 2 dollars?Mark Thienhttps://www.blogger.com/profile/00509179305586931730noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-20800511600847192112012-12-30T21:29:18.450+05:302012-12-30T21:29:18.450+05:30Its 50-50 ... for loss-gain oddsIts 50-50 ... for loss-gain oddsPyklerhttps://www.blogger.com/profile/04954477854058057203noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-16018504037284570592012-12-30T21:21:17.070+05:302012-12-30T21:21:17.070+05:30C should play, it's a fun game. Also the expec...C should play, it's a fun game. Also the expected value of their winnings is 25c, but C should be happy with the risk that they might lose. If the game can be played repeatedly C can expect to make a profit in the long run.<br /><br />Let A be the event that A guesses correctly; B that B guesses correctly. P(A) = P(B) = 1/2<br /><br />Events A and B are independent (no communication) so<br />P(A and B) = P(A) x P(B) = 1/4 <br />P(neither A nor B) = 1 - P(A and B) = 3/4<br /><br />Expected value of C's winnings:<br />P(A and B) x ($1 - $3) + P(neither A nor B) x $1<br />= 1/4 x -$2 + 3/4 x $1<br />= 25c<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-74392917263904654912012-12-30T20:26:22.720+05:302012-12-30T20:26:22.720+05:30Why not? He has a 50/50 chance of profiting 2 doll...Why not? He has a 50/50 chance of profiting 2 dollars and only stands to lose 1 dollar if the team gueses wrong.James Carringtonhttps://www.blogger.com/profile/13216994745698141309noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-33219812772475502282012-12-30T19:06:34.374+05:302012-12-30T19:06:34.374+05:30No,
A and B guess the same as their own coin toss....No,<br />A and B guess the same as their own coin toss.<br />A B GuessA GuessB Win<br />H H H H Yes<br />H T H T No<br />T H T H No<br />T T T T Yes<br />Expected Value for C = (2*1 - 2*3)/4 = -$1 per round<br />Joshua McKinneyhttps://www.blogger.com/profile/04446770219941589530noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-31895582915073447942012-12-30T17:23:20.838+05:302012-12-30T17:23:20.838+05:30No. A and B can win with probability 1/2 by saying...No. A and B can win with probability 1/2 by saying heads if their own coin is heads and tails if their coin is tails. Then both will guess correctly if the outcome is HH or TT.takakihttps://www.blogger.com/profile/01295985839611242377noreply@blogger.com