I love logic puzzles, and a friend showed me some particularly good ones the other day. These are based on puzzles written by the Mathematician from AskAMathematician.com.

If you’re the first to get all three of them correct **with correct explanations for all of your answers**, you’ll win a prize! Skepticon will send you either 4 Skepticon pint glasses or 6 Skepticon shot glasses. To be eligible for the prize, include the work “banana” in your comment.

And as a bonus just for fun – these questions were from a larger set of 8, and I included the only question I got wrong. Which one did I mess up on?

**Question 1:** In front of you are four cards. You know that each card has a photo of a famous person on one side, and a photo of an animal on the other. The four sides that are visible to you are as follows: Ken Ham, Richard Dawkins, a narwhal, and a T-Rex. I let you know that all of these cards follow the same rule – that if a card has a religious person on it’s famous person side, it has a dinosaur on its animal side. What’s the lowest number of cards you’d need to flip to determine if this rule is true or false for these cards, and which cards would you flip?

**Question 2:** Because I’m super nice, I give you a giant one hundred pound watermelon as a gift. You determine that this giant watermelon is ninety-nine percent water by weight. Unfortunately you let the watermelon sit out in the sun, and some water evaporates. Now the watermelon is only ninety-eight percent water by weight. To the nearest pound, what does the watermelon now weigh?

**Question 3:** While you were at TAM9, you decided to suspend skepticism and gamble – specifically, by playing roulette (you can actually go and play roulette online if you like the game of odds, even if you were to do something like play cards online in Vietnam there can be a large number of roulette games offered). But since you want to have some sort of strategy, you decide to flip a coin before each bet to decide whether to place a bet on red or on black (which should have a 50/50 chance of happening). Sadly, you lose sixty seven times in a row – that is, the ball always lands on the opposite color that you pick. That’s bad luck, pure and simple – it would be unfortunate if this were to happen to you when playing roulette online but hopefully you’ll have more luck on your side when you check out the games over on this website where there are plenty of games to try your hand at, even if roulette’s not your thing. If you turned your skepticism back on, it would be most rational to think:

A. You just have bad luck

B. It’s terrible strategy to flip a coin to pick what color to bet on in roulette

C. You should keep up this strategy because you’ve really likely to win the next bet

D. The roulette table is obviously broken, but you can’t assume that’s intentional

E. The casino or the staff are dirty crooks who have rigged the game against you somehow

F. You can’t reasonably decide which of the listed options are more likely

Good luck!

**EDIT:** Katie from Skepticon adds:

“Keep posting! Even if someone has already given your response, I’m strongly considering a small consolation prize (for the first 10 correct folks), so it’s in your best interest to post. Unless you don’t like getting special prezzies in the mail, of course. :)”

This is post 10 of 49 of Blogathon. Pledge a donation to the Secular Student Alliance here.

## 85 comments

Robin says:

Jul 23, 2011

I’m surprised to see this answer to (3) coming up more than once.It’s wrong because you are not betting on something like “heads and red”. Your bet on the table DEPENDS on the outcome of the coin.Suppose you’ve decided to bet on red if the coin comes up heads, and on black if the coin comes up tails. The possible outcomes are:- The coin comes up heads, so you bet on red. The table comes up red. You win.- The coin comes up heads, so you bet on red. The table comes up black. You lose.- The coin comes up tails, so you bet on black. The table comes up red. You lose.- The coin comes up tails, so you bet on black. The table comes up black. You win.In two cases out of four, you win; in two out of four, you lose. 50% chance.(Ignoring the green slot on the table, and assuming a fair coin and a fair table.)

MidwayPete says:

Jul 23, 2011

1. 2 cards. The Ham and the narwhal cards. 2. 50 pounds, as explained above.3. A banana

jerry anning says:

Jul 23, 2011

banana.1) (assuming that you consider ham a religious person, not a complete charlatan) you need to turn over ham (to verify a dinosaur) and the narwhal (to verify not a religious person). dawkins and the t. rex are irrelevant to the stated rule.2) one percent of the original one hundred pounds is one pound. that one pound is two percent of the postevaporation weight. 1/.02=50. the watermelon now weighs fifty pounds.3) this puzzle is broken, in my opinion. you could argue that d is the best of the given possibilities, but it ignores the possibility that a fellow bettor is gaffing the wheel, perhaps with an electromagnet. indeed, since the ball never landed on green, which maximizes the house edge, you could argue that this is more likely than cheating by the house or the croupier. for the other possibilities,a) luck is bullshit, of course, and hitting .5 to the 67th odds is ridiculously unlucky. way way beyond “shitty”. b) with a slight tinker to allow for the green slots, random chouce is the best strategy, not the worst, for a fair wheel.c) that is the oldest gambler’s fallacy in the book. pascal would be ashamed of it. fair roulette spins are independent trials, and previous results have no effect on the odds of the current spin.e) false. as i showed above, you can reasonably decide that d is the best of the given options.i’m guessing that 3 gave you problems, since the other two are so trivial.

Bryn Dickinson says:

Jul 23, 2011

Editted for formatting. Did not touch content except in one place where I’ve marked.Second edit corrected a misread of the original question. I read 57 losses instead of 67. The argument isn’t changed so hopefully that edit is OK.Third edit: I stupidly wrote ‘does not’ instead of ‘does or does not’. With this many edits, I probably don’t qualify any more. Sorry :(1. Religious person implies dinosaur. Therefore, not-dinosaur implies not-religious-a. Ken Ham’s card is a useful test. Flip it.–i. If you do not see a dinosaur, the rule applies.–ii. If you see a dinosaur, this is consistent with the rule. Depending on the other test, the rule does or does not apply.-b. The narwhal is a useful test, as it is not a dinosaur. Flip it.–i. If you see a non-religious person, the rule applies.–ii. If you see a religious person, this is consistent with the rule. Depending on the other test, the rule does or does not apply.-c. Richard Dawkins is non-religious, so he is irrelevant to the rule, as the rule does not mention non-religious people.-d. T. rex is a dinosaur and so it’s irrelevant to the rule, as the rule does not say dinosaurs are only on the back of religious people.2. Initially, the melon has mass M of water and mass C of Stuff.-a. We know M+C=100 lb, and (M/(M+C))=0.99. This implies M=99 lb. Therefore C=1lb.-b. After some evaporation, there is now mass W of water.-c. We know (W/(W+C))=0.98. Rearranging, we get W=0.98W+0.98C, so 0.02W=0.98C=0.98lb and so W=(0.98/0.02)lb=49lb.-d. The watermelon now weighs W+C=50lb.3. Let’s consider…-a. If winning and losing are equally likely, the probability that somebody would experience any specific sequence of wins and losses of length 57, such as 57 losses in a row, on an unbiased game, is 0.5^67, or 6.78*10^-21. This is astonishingly unlikely. I think in this case, it is true that winning and losing are equally likely. Additionally, there is only one way to achieve total 0 wins, which is the sequence of 0 wins at each point, so you would expect that somebody going up and getting 0 wins in 67 [added: independent equal-probability] trials is the same astonishingly unlikely value. You can calculate this directly from a binomial distribution with trial probability 0.5. Will discuss this further.-b. Not a gambler, but flipping a coin to decide which colour to bet on could be a reasonable strategy if you decide to quit immediately when your random walk earns you some money and stop immediately if you’re spending too much.-c. If the game is unbiased, the probability of winning or losing is not determined by your history of play. Consequently, you would still have a 50% chance of winning on your next game.-d. It is hard to imagine a mechanism by which the roulette table could be biased to give the result that you didn’t randomly flip on your coin, without an intentional effort. There would not appear to be a causal link between your coin and the roulette table.-e. Dirty play by the casino staff would appear to be plausible.-f. How could we figure out whether or not they’re probably cheating? Well, it depends on our prior probabilities for cheating and for 0 wins if they are cheating. Not sure how to pick those.-g. Given some prior predictions of how likely they are to be cheating, and what the probable result is if they cheat, we could however reasonably calculate how likely they are to be cheating.If somehow I’m in the first 10, I will banana unless prize is also some glasses in which case I will banano :)

Robin says:

Jul 23, 2011

Wait… question 1: *if you got lucky* it would sometimes be possible to falsify the rule after one card flip – say if you flipped the narwhal and got the Pope. Are you asking for the minimum amount in any circumstances? Do you have to flip both the cards before looking at either? Is the question ambiguous, or did I just miss the point?

Ryan Schultz says:

Jul 23, 2011

1. 1 is the lowest number that I need to flip, provided that the card I flip does not adhere to the rule (flip Ken Ham or the narwal). This of course depends on the assumption that the statement “- that if a card has a religious person on it’s famous person side, it has a dinosaur on its animal side” is all true or all false. If the statement can be all true OR can be partially true, then 2 cards are necessary as described in many posts above (those two cards are Ham and narwhal).2. 50 lbs. For 1% non water weigh to change to 2% non water weight, with only the water weight in flux, then half the original weight must be present3. B. In American roulette betting even/odd or red/black gives a 47% chance of winning. However, by flipping a coin before the bet is placed, you are DRASTICALLY decreasing your odds of winning. By betting just red/black you are getting 47%, but by introducing the coin flip you are betting that two random events line up (.5*.47) resulting in a measly 24% (.236) chance of winning per roll. So while going o for 67 is certainly unlucky, the probability states that out of 100 turns, 77 should be expected to lose. Banana4. The roulette problem was likely your downfall. Probability problems suck.

ckitching says:

Jul 23, 2011

Well, for 3, the game of roulette is by-definition rigged. The zeros that appear on the board don’t belong to either betting color, so even if you weren’t getting 67 losses in a row, you’d most likely still be losing.

Jared says:

Jul 23, 2011

To begin with, the banana was obviously designed for the human hand.1. Two cards: Ken Ham and narwhal2. 50 pounds3. E: The only choices that can’t be eliminated right off the bat are A, E, and F; and the probability of rigging has to be vastly greater than 2^-67 so that’s sufficient evidence to assume E.Bonus: I’ll guess you missed 1 since a lot of people get that wrong.

Bryn Dickinson says:

Jul 23, 2011

Going with an equal-probability roulette table, there are four equally likely outcomes:Coin chooses red, roulette is red: WINCoin chooses black, roulette is red: LOSECoin chooses red, roulette is black: LOSECoin chooses black, roulette is red: WINI think winning is still 50% if you choose by coin.

Garrett Severson says:

Jul 23, 2011

1) You need to flip Ken Ham and the narwhal. This is based on a conditional statement: “If R[eligious person], Then D[inosaur]”. If Ken Ham’s card showed a bear or the narwhal flipped over to reveal the pope, you would show that R occurred and was not accompanied by a D, which is the only way to show that a conditional is not true. As an aside, the T-Rex card could have a banana on the other side and it wouldn’t mean anything for this purpose, since the statement “If R, Then D” remains true when the D occurs in the absence of an R.2) It now weighs 50 pounds. It starts at 100 lbs, 99% of which by weight are water. That means that 1% of it is solid stuff. This works out to be 1 lb solid stuff and 99 lbs water. After it shrinks, it’s 98% water, which means it’s 2% solid stuff. However, since the solid stuff stays the same, that means that the same 1 lb of solid is 2% of the new weight. Multiply that by 50 (2% of something is 1/50 of it) and you get your answer.3) The answer here is A. While it is phenomenally improbable that a streak like this would occur, it’s important to remember that each spin is independent and its outcome has absolutely no bearing on later spins. The assumption that they’re “linked” in some way is the root of the gambler’s fallacy (which, coincidentally, is choice C). Most importantly, this explanation requires nothing outside of the nature of the game and my own bad decision-making to explain what happened. There’s a distinct lack of evidence for external actors here, so there’s no reason to assume one exists.Choice B is wrong because flipping a coin is not a terrible strategy, it’s just not a perfect one. The green space means that I have just under a 50/50 shot if I were to bet on black (or red). If I flip a coin, I have a 50/50 chance of getting heads (or tails). Assuming heads from the coin will lead me to pick black on the board, I have 5 possible outcomes:1) Coin shows heads and the ball lands on black. I win.2) Coin shows tails and the ball lands on black. I lose.3) Coin shows heads and the ball lands on red. I lose.4) Coin shows tails and the ball lands on red. I win.5) The ball lands on green. I lose.Option 5 is unlikely, and the other two are 50/50. My chances of winning wind up being the same as if I had omitted the coin and just chosen red or black each round.Choice C is the gambler’s fallacy. The fact that I’m on a losing streak doesn’t affect my chances of winning later.I don’t have evidence to support choices D or E, so they’re not likely. They posit the interference of something external to the system, so I need some reason to believe they’re true to overcome the null hypothesis that my losing streak is a product of nothing more than my own bad decision to play roulette.A is the most likely, so F isn’t true.As for the bonus: My guess is you missed question 3.

zyxek says:

Jul 23, 2011

Yeah, I realized that about an hour after submitting my answer. I over-think logic puzzles all the time. This is the first one I’ve tried in a couple of years.

Ross says:

Jul 24, 2011

1) You would have to flip over two cards (Ken Ham and the narwhal). Ken Ham to make sure there’s a dinosaur on the back, and the narwhal to make sure there isn’t a religious person on the back. The others don’t matter.2) 50 lbs. 1% of 100 lbs is 1 lb of non-water, and if 1 lb of non-water is then equal to 2% of the total weight, multiply by 50 to find the total weight.3) E. The coin flip doesn’t make a difference, since there’s still a 50% chance of the roulette table matching your bet. The chances of legitimately losing a 50-50 bet 67 times in a row are so abysmally small that they can effectively be ignored. Assuming the coin didn’t produce the same result for all 67 flips (which has the same abysmally small chance of happening as legitimately losing 67 times in a row), the table would have produced at least some red and some black spins, which indicates either a normal table or one that is intentionally rigged. The chances of the table being legitimate have already been shown to be effectively 0, so the most likely option is that the casino has rigged the table against you.By the way, did you know bananas were designed to fit perfectly into the human hand?

Ross says:

Jul 24, 2011

I forgot to guess which one you got wrong. My guess would be A, since the others are just math.

Ross says:

Jul 24, 2011

Make that A a 1. I spent a lot of time typing the answer to number 3 and got the letter system imprinted in my mind.

Paul Wright says:

Jul 24, 2011

1. Ken Ham, to make sure there’s a dinosaur on the other side, and the narwhal to make sure there isn’t a religious person. Flipping over Dawkins and the T-Rex is unnecessary because the rule could still be true regardless of what’s on the other side. I could explain this better with a truth table for the “implies” sign, but I think this is enough.I’m guessing this is the one you got wrong.2. The watermelon now ways 50 pounds. The part of the watermelon that wasn’t water weighs 1% of 100 pounds, which is 2% of 50 pounds.3. B & C are right out, those are well known fallacies. If the table is broken such that it always turns up the same colour, that wouldn’t explain anything unless the coin is also fixed. But I think you’d notice if you were betting on black every single time and red came up every single time. I don’t think there’s any way it could be broken in a way that would cause 67 losses in a row.So either it’s shitty luck, or someone is controlling the roulette table to make you lose. If you make one bet a minute, it would take roughly the age of the universe to get 67 losses in a row. So my money’s on E.Also banana.

Scott says:

Jul 24, 2011

1) 2 cards – For the rule to hold true, the Ken Ham card must have a dinosaur (modus tollens) and the narwhal card must have a non-religioso (modus tollens).2) 50 lbs. – If the 100 lb watermelon is 99% water, then 1 lb is not water. Now with some of the water gone, that 1 lb of non-water is now 2% of the weight. Therefore, 50 lbs.3) You have a 0.5 probability of winning any given spin (actually, slightly less if there are 1 or more “0” slots on the wheel). So, to lose 67 times in a row would be a 1 in 2^67 longshot at best. Statistically speaking, any occurrance with a 0.05 probabilty of happening by chance is considered to be “statistically significant”. This occurance is far below that threshold. A broken table shouldn’t matter (unless, as a previous poster mentioned, the coin is also broken, or unless the table is broken to always land on one of the “0” slots). So no matter how srtrict the laws regulating gambling are enforced (they can never be 100% due to human error), there is still a far greater chance that the casino is cheating, as opposed to the next most reasonable option (the wheel AND the coin are “broken” or the wheel is broken to always land on a “0” slot), and the distant third place option (shitty luck). The most rational choice is F.

Scott says:

Jul 24, 2011

If the water didn’t leave the watermelon, then how is it now only 98% of the weight? ;)

Scott says:

Jul 24, 2011

Bonus: You got #3 wrong.

Trevor Burn says:

Jul 24, 2011

Since I’ve seen the right answer for the first two a half dozen times (Ken Ham & narwhal; 50 lbs), I’m curious as to the logic behind the 3rd question. Here’s mine. Banana banana banana terracotta pie!Let’s look at this scientifically. Let’s take alpha = 5% level of significance. The null hypothesis is that the roulette wheel is truly random. A further assumption is that your coin is fair: it gives heads / tails at a 1:1 ratio. Finally, we assume that the coin and roulette wheel are independent.As discussed above, the probability that the coin correctly predicts the outcome of the roulette wheel is 50% (let’s disregard the 0 and 00 slots). Then again, as discussed above, the probability of exactly 0 correct predictions is choose(67, 0) * pow(1/2, 0) * pow(1/2, 67) = 1/2^67.Thus the probability that this occurs due to random chance (i.e. the probability that the null hypothesis is correct) is 1/2^67, or MUCH lower than 5%. Thus we would reject the null hypothesis.So the alternate hypothesis is that the roulette wheel isn’t random. Or more particularly, the roulette wheel is *not* independent of the coin flip. Thus, without any further information, I would assume that the table operators are somehow cheating. That being said, I would run some clandestine experiments to see if I can somehow prove this.

David Byars says:

Jul 24, 2011

Question 1: 2. Ken Ham, and a narwhal. You are trying to disprove if a card has a religious person on it’s famous person side, it has a dinosaur on its animal side. But a nonreligious person could still have a dinosaur, so the dino and Dawkins are irrelevant (as well as irreverent, in the latter case) Whereas Ham’s reverse being a non dino or the narwhal having a religious person (like Banana Man) as its backside would disprove the notion.Question 2: To the nearest pound, what does the watermelon now weigh? 50 pounds.At first, 99% water weight is 99 lbs. 1 lb non water weight for 1%. 100 lbs total.1/1%=100After, 98% water weight, 1 lb non water weight for 2%.1/2%=50 lbs totalQuestion 3: F. You can’t reasonably decide which of the listed options are more likelyA. You can’t expect to win 50%B. It’s a random generator, so flipping a coin works as well (badly) as any other strategyC. Failing previous rolls doesn’t make the same strategy more likely in the futureD. No reason to think it is brokenE. No reason to suspect fraudso, F. You can’t reasonably decide which of the listed options are more likely

Ira Hanson says:

Jul 24, 2011

1. Ken Ham, Narwhal2. 50 lbs3. E

Mark Entel says:

Jul 24, 2011

#1 – 2 cards, Ken Hamm & the narwhal#2 – watermelon weighs 50 pounds#3 – First off, though, ‘most rational’ is too open-ended a criteria, unless you means it satisfies Occam’s Razor (i.e., requires fewest assumptions). I want to say F, but possibly my answer is E, depending on a definition of the word “luck” in option AAt any rate my answert depends on what the word “luck” means. Does it refer to an intrinsic quality of myself that made it more likely that I was the one to experience this event, as opposed to any other rando? If so, then I answer E, the people are cheating, no evidence that an individual can alter the probability of events happening (apart from interfering, in this case with the ball/spin).if “luck” just means that I witnessed a crazy rare event because, well, someone had to, and I am applying the term “luck” to emphasize how unlikely it felt, then I would answer F, can’t decide.Flipping a coin is an irrelevant strategy, there is a 1/4 chance on any spin (minus odds it hits 0/00) that spin & coin-flip match up. This is same odds for the strategy of always choosing red or black, coin flip should be no different than just choosing one or the other, in any case there is a 1/2 odds you choose black, 1/2 ball lands black, 1/4 you choose correctly (minus odds of 0/00 hitting).based on my iPhone calculator (which is really based on what little I remember of long-aso math skills), the odds that you lose every hand are on the order of 4 x 10^-9 (.76^67), so like 4 in a billion? Those are actually pretty good odds, all things considered and given several decades of roullette games. So I wouldn’t know if I was being cheated or if this was just some fluke event,Is there a better than 4 in a billion chance casinos are cheating me? Whew, I tend to think Nevada is pretty serious about their regulation of casinos, but this impression is based mostly on the film Ocean’s 11, so that hardly counts. And regulating bodies are made of people, so that leaves room for error or corruption, both extremely common-place characteristics of people & their institutions. So I think it is too hard for me to tell how likely the odds of a cheatign casino are, maybe it is just a lack of info, but that is what I am working with

Scott says:

Jul 24, 2011

Ugh. I meant (as my explanation shows) that the most rational choice is E.

Leo Buzalsky says:

Jul 24, 2011

1. Two cards. The Ken Ham card is fairly obvious. We need to make sure a dinosaur is on the other side. The other card is the narwhal. We need to make sure the other side is NOT a religious person.We do not need to flip the Richard Dawkins card as he is not religious. We also do not need to flip the T-Rex card, as the rule only works one way. In other words, the rule does not state that a dinosaur card will have a religious person on the other side.2. The watermelon now weighs 50 pounds. Initially, the watermelon was 99 lbs water, 1 lbs other. The watermelon still has that 1 lbs of other weight. Let x = water weight and t = total weightx/t * 100 = 98%Substitute in (x+1) for t.x/(x+1) = 0.98 –> x = 0.98 (x + 1) –> x – 0.98x = 0.98 –? 0.02x = 0.98 –> x = 0.98 / 0.02 –> x = 49Now, remember, x is just the water weight. We still need to add in that 1 lbs of other weight.49 + 1 = 50 lbs.3. I have to say the answer is B. It is a bad strategy to flip a coin. There is a 50/50 chance of getting a certain color, but there is also a 50/50 chance of getting a certain coin flip. The probability that your coin flip matches the roulette color is 1/2 * 1/2 = 1/4. By adding in the coin flip, you have effectively reduced the probability of getting the desired result by 1/2. Now, going back to that 1/4 probability, if you play 67 times, you would expect to get the desired result 67 * 1/4 = 16.75 (16 after rounding down) times. It would seem that there is some “shitty luck” here (making A look like reasonable answer), until you consider, using your skepticism, that probability is not a guarantee of the outcome. (To clarify, just because probability predicts that the desired result should happen 16 times, we know that it does not have to, and this is where we get this absurd notion of “luck”.)I now need a banana to rejuvenate from this mental exercise.

MidwayPete says:

Jul 24, 2011

So, are we going to get an official ruling about the answers?

ElGatoCello says:

Jul 24, 2011

Question 1: In front of you are four cards. You know that each card has a photo of a famous person on one side, and a photo of an animal on the other. The four sides that are visible to you are as follows: Ken Ham, Richard Dawkins, a narwhal, and a T-Rex. I let you know that all of these cards follow the same rule – that if a card has a religious person on it’s famous person side, it has a dinosaur on its animal side. What’s the lowestnumber of cards you’d need to flip to determine if this rule is true or false for these cards, and which cards would you flip? The rule is a logical statement of the form A —> B (“A implies B”) where A stands for “the card has a religious person on its famous side” and B stands for “The card has a dinosaur on its animal side. In order for this statement to be true, if A is true, B must be true. (If A is false, it doesn’t matter, because then the situation doesn’t need to follow the rule, and if Bis false and A is true, then it breaks the rule and we’ve learned that you are a liar liar pants of fire…these are the situations we are hunting down: statements where we know A is true and B is potentially false.)This means that if we have a non-religious person on the card, we don’t need to be concerned with the other side.From this, we know that we definitely need to check the Ken Ham card but not the Richard Dawkins card. What is important to note (of course, this is not *strictly* necessary, but I find that, for me, it is easier to think about it this way.) is that A —> B is equivalent to ~B —> ~A (That is, “NOT B implies NOT A” or “If there is not a dinosaur on the animal side, then there is not a religious figure on the famous person side.” This is called the contrapositive of the statement.) Because we need to make sure that both of these equivalent statements hold in the same fashion as before, we now know we must also check the narwhal card because that has a chance of breaking the rule. The T-Rex card doesn’t need to be checked, because we have no statements that assert that having adinosaur on the card implies anything…we can still have a non-religious figure on the card and have it follow the rule. However, the Narwhal card must be checked because since this is not a dinosaur, it satisfies ~B and we need to make sure that it doesn’tbreak the rule by having a religious figure on the other side. Tl;dr, you must turn over the Narwhal and Ken Ham.Question 2: Because I’m super nice, I give you a giant one hundred pound watermelon as a gift. You determine that this giant watermelon is ninety-nine percent water byweight. Unfortunately you let the watermelon sit out in the sun, and some water evaporates. Now the watermelon is only ninety-eight percent water by weight. To the nearest pound, what does the watermelon now weigh?What we know is that only water has evaporated. We can also say that “water + non-water = 100 lbs”Since we know that precisely 1 percent of the weight is non-water watermelon stuff, that is 1 pound of watermelon stuff. After we accidentally let the watermelon sit out in the sun, the watermelon stuff has not changed, but it now makes up a greater proportion of the watermelon!We know enough about the watermelon to set up ratios into a proportion:Let x be the total weight of the new watermelon, then:2%/100% = 1 lb/x lbsWhich, if we solve for x, gives us x = 50, which means the watermelon now weighs 50 lbs!Question 3: While you were at TAM9, you decided to suspend skepticism and gamble – specifically, by playing roulette. But since you want to have some sort of strategy,you decide to flip a coin before each bet to decide whether to place a bet on red or on black (which should have a 50/50 chance of happening). Sadly, you lose sixty seven times in a row – that is, the ball always lands on the opposite color that you pick. If you turned your skepticism back on, it would be most rational to think:A. You just have shitty luckB. It’s terrible strategy to flip a coin to pick what color to bet on in rouletteC. You should keep up this strategy because you’ve really likely to win the next betD. The roulette table is obviously broken, but you can’t assume that’s intentionalE. The casino or the staff are dirty crooks who have rigged the game against you somehowF. You can’t reasonably decide which of the listed options are more likelyThe thing about gambling is that so many people fall victim to what is called “The Gambler’s Fallacy” or that “If you keep playing, eventually your number will come up!No really, I’ll make it ALLLLL back eventually.” This drove gamblers banana for many years. However, the problem is that every single one of these roulette spins is independent of each other! You don’t get to stack up your losses and hope that you’ll get the next one because dice and roulette wheels don’t have a memory! Each rollor spin is a fresh chance to lose all of your hard earned money. (C is out.)When it comes to strategy, this one is particularly interesting because it doesn’t actually change the outcome! As with dice and roulette wheels, coins have no memory either. (Coins are, really, just two-sided dice, after all.)Buuuut, because we here are all about skepticism and showing our work here, let’s draw out all the different possibilities:Let’s break this down into 2 different categories, since heads means we choose red and tails means we choose black, (you could switch that around, but I’m the one playingroulette, damnit, so we’re using Jersey Rules) we don’t have to worry about making a category for the coin flip too.So let’s do the math:Let’s write out our Sample Space (all the possible outcomes) and see what’s the probability that we will win (I’ll write them out in the form “(color I picked, color thatcame up)”):{(Red, Red), (Red, Black), (Black, Red), (Black, Black), (Red, Green), (Black, Green)}Note there are 18 red spaces, 18 black spaces, and 1 green space (or 2 if you’re playing American Roulette) so there are thirty six winning situations i.e. (Red, Red) and(Black, Black) of which there are 18 each and (38 or 40) losing combinations.Notice that 36 times out of 74 (48.6%) (or out of 76, 47.36%) the color I picked came up on the roulette. What is different about my strategy than just picking a color on my own?Let’s look again at the sample set:{(Red, Red), (Red, Black), (Black, Red), (Black, Black), (Red, Green), (Black, Green)}What’s missing? That’s right! With the old strategy, you could NEVER win if a 0 or a 00 came up because you were restricted to Red and Black! So we need to compare this to an unencumbered random choice:{(Red, Red), (Red, Black), (Red, Green), (Black, Red), (Black, Black), (Black, Green), (Green, Red), Green, Black), (Green, Green)}Which is exactly a 1/3 chance (38 wins/113 possible). Of course, this could be achieved equally well by just not playing the greens, but it does suggest that, in terms of “how many times you should win” this is more than 10% better than playing ALL the colors.Of course, Red and Black only pay back 1 to 1, so getting that green is much more lucrative if you land it.The funny thing is, though, because of the payout, the expected value of each roll (a loss of 5 cents) is equal for red, black, and green. This suggests that, depending on what we’re going for, our strategy actually allows us to win more often than simply picking a color!So…B is out.As for the casino being dirty crooks…that is probably more likely than not in general…but again, assumption is not the way to go here. Roulette payouts areclearly listed, so they’re not trying to obfuscate the fact that they’re trying to cheat you by awarding you less money than the expected value of the roll.Let’s analyze the presence of foul play:If we are defining “luck” as some sort of subjective measure of “how much we have won” then we have INCREDIBLY bad luck. The probability of this happening is somethinglike 6.8 x 10^-21 which is absurd. It is pretty safe to assume that, if everything was in working order and/or no foul play was involved, that this particular outcome would be very improbable.So, let’s strike A for the time being.So, between D and E, we have a conundrum: Is it possible to abstract any information?It is somewhat unlikely for the roulette wheel to be broken in a way that would cause a system like this to lose continuously without a staff member being forced to close the roulette wheel down or there to be a massive run on the wheel to collect on the green tiles. While this observation does not diametrically oppose the likelihood of D, it does present a very strong possibility. With E, if they have, indeed rigged the game, they aren’t going to just have me lose 67 times in a row…most people will have certainly given up long before that and I KNOW my attention span isn’t that long. Studies have shown that you can keep people playing for much longer by slowly increasing the payment intervals. This is the reason why WoW and other RPG games tend to be so addictive.D and E are equally unlikely; but, they ARE the best answers out of the five, so we can probably safely assume the answer to be F.As for me, I say, “The only winning move is not to play.”

Spgcell says:

Jul 24, 2011

Ken Ham, narwhal, 50, banana.Why is the coin toss included in problem 3? It is necessary to establish two things: that you the bettor are switching randomly between red and black; and that the roulette wheel is also switching between red and black, and always the opposite color from the one you bet on (never green). That means the wheel is not broken, it is rigged.Imagine the following sequence: you bet 7 times, and lose all 7. Suspecting something, you decide to keep track for 30 bets, starting with the 8th bet. As the ball drops for the 30th time in the opposite color from your bet, you realize that the odds of that happening are about 1 in a billion.Astonished, you hail James Randi, and explain what has happened. He says that is extremely unlikely, and then asks the skeptic’s question: is that reproducible?So with Randi sitting at your side, you then do another 30 bets, again switching randomly between red and black based on your coin toss. Once again, the ball drops in the opposite color of your bet each and every time. You look over at Randi to see what he thinks, and he has just one word to say:”Rigged.”

Cripdyke says:

Jul 24, 2011

Ken Ham – a total of ONE card flipped50 poundsE – the game is riggedBonus: You got the watermelon’s weight wrong

Cripdyke says:

Jul 24, 2011

it’s too late, but I realize now that you would have to flip TWO cards, Ken Ham & the Narwhal. Frack.

Spgcell says:

Jul 25, 2011

As a comparison to the long odds of losing on red or black 67 times in a row, you are at least 25 times more likely to win the Powerball Jackpot twice in a row.

Frank says:

Jul 25, 2011

Flip 2 cards: Ken Ham and narwhal. Theonly way the rule can be violated is if a card has a religious personand a non-dinosaur. Richard Dawkins is not a religious person, and aT-Rex is a dinosaur, so neither of those cards could possibly violatethe rule. If the Ken Ham card has a non-dinosaur, or the narwhal cardhas a religious person, then the rule is violated, so both of thosecards need to be checked.50 pounds. Before it had 99 pounds ofwater and 1 pound of non-water. Now it still has 1 pound ofnon-water, which accounts for 2% of its weight. 2% * x pounds = 1pound, divide both sides by 2% and we have x pounds = 1 pound / 0.02= 50 pounds.E, by process of elimination. It’s notA, because the probability of loosing 67 coin flips is 2^-67, whichis less than 10^-20. In the entire history of humans gambling, almostcertainly nobody has had luck that is that bad. It’s not B, becauseflipping a coin is no better or worse than any other method ofdeciding what to bet on. It’s not C, because the chances of winningthe next round have nothing to do with how many rounds one has justlost. There is no magic ledger of luck that makes sure that people ona loosing streak start winning. It’s not D, because there is no way atable can be broken in just such a way as to always go against a cointhat it isn’t interacting with. All that leaves is E.

Gary Usleaman says:

Jul 25, 2011

But, you forget to factor in that I was drinking!

Dave Wright says:

Jul 26, 2011

However, if there is a religious person on the Narwhal card it would disprove the rule.

Adam says:

Jul 26, 2011

Where are you, answeeeeeers? I would have thought you have to turn over 3 cards (Ham, Dawkins, Narwhal). Because the Dawkins side we’re seeing could actually be the Animal side of a card, not the Famous Person side. For all we know, Fred Phelps could be lurking under Dawkins.The watermelon now weighs 50 pounds. 1 pound of non-water mass with the new 98% water ratio gives 50 total.And for question 3, I would never gamble. I’m far too cheap. But presuming someone spiked my banana-tini, I would have to go with Option E. I can’t say they’re for sure cheating me, but it’s probably a reasonable enough assumption that it’s time for me to get back to my banana-tinis.

UrsaMinor says:

Jul 27, 2011

And on the off chance that I won a prize, it will have to wait until I get back from vacation in a couple of weeks.