bonanova Posted February 7, 2009 Report Share Posted February 7, 2009 May said to Moe, take all the coins out of your pocket, and I'll get all the coins I have in my purse. We'll throw them all, at the same time, and whoever gets more heads buys the other a chocolate malt, OK? Moe agreed, but was disappointed to find that May had 6 coins to his 5. What's the probability that May won the bet? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 7, 2009 Report Share Posted February 7, 2009 (edited) Assuming that there are no two-headed coins (or two-tailed, I suppose) May would, on average, flip 3 heads, but Moe would, on average, only flip 2.5 heads. So the probability of May winning would be 3/5.5, or about 54.5%. Of course, this is only a guess, most of bonanova's puzzles contain some sneaky trick so I'll have to think about it some more. EDIT: I just looked at it another way, and somehow... is a mere .5; May and Moe have an equal chance of winning. I have to go right now, but I'll be back later- I can't believe that that's right! Maybe I messed up somewhere. Edited February 7, 2009 by rossbeemer Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 7, 2009 Report Share Posted February 7, 2009 (edited) Okay, being lazy, I did a Brute Force solution. If my code is correct, there are 2^11 (2048) possible outcomes. In 1024 of those, May has more heads, in 562 Moe has more heads, in 462 it's a tie. So, May wins 50% of the time. It is unclear what happens for the 20.5% of the time it's a tie. Oops, error in the code. I've corrected the answers in the above. Edited February 7, 2009 by MrAscii Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 7, 2009 Report Share Posted February 7, 2009 Double oops, I missed a number in my correction and the post is now locked. 2048 possible outcomes. 50.0% (1024) May > Moe 27.4% (562) Moe > May 22.6% (462) Tie Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted February 7, 2009 Report Share Posted February 7, 2009 Mary wins about 55% of the time Moe wins about 21% of the time They tie about 24% of the time Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 7, 2009 Report Share Posted February 7, 2009 50/50? Her first 5 coins would have an equal chance of being heads, so that leaves a 50/50 chance on the last (sixth) coin. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 7, 2009 Report Share Posted February 7, 2009 Double oops, I missed a number in my correction and the post is now locked. 2048 possible outcomes. 50.0% (1024) May > Moe 27.4% (562) Moe > May 22.6% (462) Tie Yeah, I'm pretty sure that's right, or something close to that. When I did my analysis, I calculated that May would win 50% of the time. However, I then made the incorrect statement that it was 50-50 that each of them win, because it is unquestionably not. While May wins 50% of the time, Moe only wins 27.4% of the time (assuming that MrAscii's calculations are actually correct, I haven't checked them ) So, the odds that May WINS are 50-50, however the odds that Moe wins are only 27.4-72.6, since they tie 22.6% of the time. So, if we assume that they flip again if they tie, May will win about twice as much as Moe does. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 7, 2009 Author Report Share Posted February 7, 2009 The puzzle is cleaner if it simply asks the probability that May wins. My bad, that's the way I intended to word it. The entire problem has been calculated by MrAscii. fleminator has offered way of looking at the puzzle that has a natural extension to N and N+1 coins. Can the proof be reduced to two statements of certainty, i.e. that don't reference probability at all? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 7, 2009 Report Share Posted February 7, 2009 I'm not really sure what you mean by two statements of certainty. But it's true that when dealing with N and N+1 coins, the odds of the person with N+1 coins winning is 50% because as fleminator said, on average, each person will flip the same number of heads for N coins, and so it is only the N+1 coin that makes the difference. Since it's either heads or tails, the odds of the person with N+1 coins winning is 50%. That said, if dealing with N and N+2 coins, I would assume that the person with N+2 has a 75% chance of winning, for the same reason. So perhaps, when dealing with an excess of X coins, the probability that the person with X additional coins would be (1 - probability that all X coins are tails), which is the equivalent of (probability that at least 1 of the additional X coins is heads). Quote Link to comment Share on other sites More sharing options...
0 Prime Posted February 8, 2009 Report Share Posted February 8, 2009 (edited) The puzzle is cleaner if it simply asks the probability that May wins. My bad, that's the way I intended to word it. The entire problem has been calculated by MrAscii. fleminator has offered way of looking at the puzzle that has a natural extension to N and N+1 coins. Can the proof be reduced to two statements of certainty, i.e. that don't reference probability at all? So in case of a tie they both go hungry? If they tossed 5 coins each, they would have equal chances. The 6th coin could break a tie if landed heads up at the probability of 1/2. There are 32*64 = 2048 cases for 5 against 6 coin toss. There are 1+52+102+102+52+1 = 252 cases where 5 against 5 coin toss results in a tie. (2048 -252)/2 = 898 wins for each side in 5 against 5 toss. Half the time, or 126 cases, May would add those to her win with the 6th coin toss. So the probability is (898+126)/2048=1024/2048=1/2. This tells us there must be a simpler way to state the solution. More to the point, are 5 or 6 coins enough to buy a chocolate malt? Edited February 8, 2009 by Prime Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 8, 2009 Author Report Share Posted February 8, 2009 More to the point, are 5 or 6 coins enough to buy a chocolate malt? Excellent question. Where do you get all these questions ... ??? All I can say is, I recall a day when a regular malt was $0.25; with extra scoop $0.35. Maybe my puzzles are getting outdated like me. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 8, 2009 Author Report Share Posted February 8, 2009 I'm not really sure what you mean by two statements of certainty. A is true. [probability is not mentioned or estimated or inferred in the statement.] B is true. [probability is not mentioned or estimated or inferred in the statement.] Therefore May's chance of winning is C. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 9, 2009 Report Share Posted February 9, 2009 The problem is indifferent to whether more heads or more tails is considered the winner. If the problem is reversed, the answer is the same. Therefore, the probability must be 50% That what you're looking for bonanova? Quote Link to comment Share on other sites More sharing options...
0 Prime Posted February 9, 2009 Report Share Posted February 9, 2009 That what you're looking for bonanova?The problem is indifferent to whether more heads or more tails is considered the winner. If the problem is reversed, the answer is the same. Therefore, the probability must be 50% Whether they bet on heads, or tails has no bearing. However, the winning probability for May (6 coins) is 1/2; for Moe (5 coins) its is less than that. There are 3 possible outcomes: win, loss, draw. And draw has its probability too. Here is another attempt at a simple explanation without using any actual numeric calculations: Let's consider 5 vs. 5 coin toss. The number of cases where May wins is M. The number of cases where Moe wins is also M (their names start with the same letter). And the number of cases where it is a draw is D. Thus probability for May to win is M/(2M+D). (Obviously less than 1/2). Now add the 6th coin toss by May. The number of all cases doubles (heads or tails with each case). So total cases are 4M+2D The M cases where May won with 5 coins will reamain a win with either head, or tail. So they become 2M cases. D cases with additional "head" convert to win. D cases with additional "tail" stay draws, and all the losses either remain so, or become draws. So May's total winning cases with 6 tosses number 2M+D. Thus her probability of winning is (2M+D)/(4M+2D) = 1/2 I suspect, Bonanova may have even simpler formulation. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 9, 2009 Report Share Posted February 9, 2009 A is true. [probability is not mentioned or estimated or inferred in the statement.] B is true. [probability is not mentioned or estimated or inferred in the statement.] Therefore May's chance of winning is C. 1. With N coins each, Mae wins half of the time there isn't a draw (by symmetry). 2. If Mae has N+1 coins, she also wins half of the time there is a draw after N coins. Because these statements are exclusive and cover all outcomes, she wins half of the time overall. Notwithstanding Prime's point about inflation, where do you get a real malt anymore? I haven't seen a malt or soda shop in what must be at least twenty years. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 11, 2009 Author Report Share Posted February 11, 2009 Agree with voltage, prime and d3k3. The statements that do it most simply for me areHaving only one excess coin, May cannot throw more heads and more tails.Therefore she will throw more heads or more tails.I was going to add "with equal probability" to statement 2. But the terms were to not say that word. So unsaid, but still true. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 11, 2009 Report Share Posted February 11, 2009 Notwithstanding Prime's point about inflation, where do you get a real malt anymore? I haven't seen a malt or soda shop in what must be at least twenty years. Good point. These days, malt=shake+"malted milk powder", where shake=soft serve ice cream+"shake mix". That's a disappointing equation. Lucky for me, I have a quaint drive-in restaurant straight out of the fifties within walking distance. The shakes there are absurdly delicious and made by hand with a recipe as old as the establishment. Unless those are mostly dollar coins, you're out of luck though. <_< Quote Link to comment Share on other sites More sharing options...
Question
bonanova
May said to Moe, take all the coins out of your pocket,
and I'll get all the coins I have in my purse. We'll throw
them all, at the same time, and whoever gets more
heads buys the other a chocolate malt, OK?
Moe agreed, but was disappointed to find that May had
6 coins to his 5.
What's the probability that May won the bet?
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