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There are two tables in a room. On the one to the right there are a 100 coins, 20 with H facing up and the rest (80) with the T side facing up. There are no coins on the other table.
The goal is to somehow move the coins so that there will be an equal number of coins with H facing up, on both tables.
You cannot see the coins (dark room) or touch them to tell if they're facing "up" or "down". - Dan from Tel Aviv
Go to my other site, mathproblems.info for the solution (spoiler warning!)
June 9, 2006
In your last column you said that "The probability of 5500 coins not being a 55 can be very closely approximated as 0.9864012865500 = 1 in 507,033,772,284,213,000,000,000,000,000,000." I assume "approximated" because of the effect of removal as you go through the 5500 coins. Talk about a minuscule effect of removal! This is a good example of the target coins becoming LESS likely as you remove non-targets, because the effect of removal is so small compared to the much larger probability of a crooked game, i.e. the target coins have been removed. – Pete from NY
Yes, I said “closely approximated” because there are only so many pennies in the world. Remove one non-55 from the bag and the effect of removal increases the probability that every other penny in the bag is a 55. If I hadn’t said “closely approximated” at least three people would have written in to correct me. It is of course an extremely minute effect, but many of my readers are perfectionists, and will jump all over me for the slightest of errors.
June 9, 2006
Hello, my name is Patty. You have a very nice site and you seem like a very knowledgeable man. Definelty the kind of man I want by my side in a Casino!!!!!!! I was wondering if you could help me. I told my boy friend that I would look for an answer to a problem on the internet. If you were to help me, it would really make me look good. My boy friend is a coin collector. He bought a bag of wheat pennies. I don't know much about coins myself. (He is teaching me as time goes by) But he said that he was amazed that one certain year wasn't represented in the bag because they are so common. He said the odds of that happening had to be a billion to one. I told him I would try to ask people in my office (the self proclaimed geniuses!!) and if they didn't know, I would try to do a little research online. I came across you. Anyway if you can help. I would greatly appreciate it. The bag had approximately 5,500 pennies. The total amount of wheat pennies minted by all the US mints was 24,267,000,000 The number of 1955's that were minted (The one he was looking for) was 330,000,000. Some of the guys in my office say that there are other factors like; demographics, the fact that the MInts may not have distributed all the pennies etc. ...................... I would assume that they are correct but, I (and I'm sure my boy friend also) would settle on knowing the approximant odds!!!!!!!!! I hope you can help.
Thank you in advance, Patty
Your mintage numbers are close to those of Mountain View Coins. Assuming that every wheat penny ever minted has the same probability of being in the bag the probability of any one penny not being a 55 is (24,267,000,000-330,000,000)/24,267,000,000 = 0.986401286. The probability of 5500 coins not being a 55 can be very closely approximated as 0.9864012865500 = 1 in 507,033,772,284,213,000,000,000,000,000,000.
My father is a coin collector so I asked him for help on this one. Here is what he said, Here is my guess. In the year 1955, there were a small number of Lincoln cents struck in Philadelphia with the date struck twice. no one knows exactly how many They were mixed with other cents for circulation before the error was discovered. An uncirculated specimen today is worth about $2000.to $6000. I suspect that the bag of "wheats" had already been culled of all of its 1955's by someone looking for double-die specimens. Here is a picture of one: 1955 Doubled Die Obverse One Cent. Note that this website is selling "wheats" and you can bet that some culling of dates has already occurred after the coins were collected by the dealer. I would have thought that the 1955's that were not double-die would have been returned to the collection, but perhaps they are sold separately, or melted down. The copper in wheat pennies is worth much more than one cent today. That is why they switched to copper-plated zinc cents a few decades ago. There is a possibility that the mint itself decided not to distribute many of the 1955's, and melted them down after minting to avoid a frantic scramble for the rare double-die specimens. The mint and Post Office has always been embarrassed by printing errors, and tries to keep them out of circulation. May 31, 2006
A friend chooses a 3-flip sequence of heads or tails, and gives me the option to choose my own (different) 3-flip sequence. We flip a fair coin as many times as necessary until one of our sequences comes up. If he picks H-H-H, which sequence should I pick, and what is my advantage on this bet? How do I calculate which sequence to pick based on his chosen sequence? - Pepe from Philadelphia The following table shows the probability of player A winning according to all possible chosen patterns of player A and player B.
A mnemonic device to select the optimal pattern is his first and second picks should be your second and third respectively. Your first pick should be the opposite of your third. For example if your opponent chooses HTT your second and third picks should be HT. Your last pick is T so your first should be H, for an HHT pattern. Following this strategy your probability of winning will be 2/3 to 7/8, depending on what pattern your opponent picks. April 5, 2006
In your March 13, 2006 Ask the Wizard, you gave three formulae for the "rupee" game. While the solution is mathematically right, I can't for the life of me understand how the three equations model the problem. Can you give any insight on how you came up with the three equations? - Rick from Covington, LA I had a number of people ask me to expand on my answer. The solution requires basic matrix algebra. Start by defining x as the answer, or the average number of flips until the disparity between heads and tails is 3. Let y be the expected number of flips from a point where one side is up by one flip. Let z be the expected number of flips from a point where one side is up by two flips. After the first flip one side will be in the majority by one flip. So x=1+y. When either side is one flip ahead another flip will result in either the initial tied state, or one side being up by two flips. Both outcomes are equally likely. So y=1+0.5*x + 0.5*z When either side is two flips ahead another flip will result in either one side being up by one flip, or the end of the game. Again, both outcomes are equally likely. So z=1+0.5*y So we have three equations and three unknowns: (1) X= 1+y (2) Y = 1+ 0.5x + 0.5z (3) Z = 1+ 0.5y To solve lets first get rid of the decimals by multiplying the last two equations by 2. (1) X= 1+y (2) 2Y = 2+ x + z (3) 2Z = 2+ y Let's substitute 1+y, from (1) for x in (2). 2Y = 2 + 1 + y + z (4) y = 3 + z No substitute 3+z for y in (3) 2z = 2 + 3 + z z = 5 Now substitute 5 for z in (4) to get (5) y = 3+ 5 = 8 No substitute y = 8 in (1) to get (6) x = 9 March 27, 2006
Suppose we have a gambling game. An unbiased coin flipped repeatedly. For each flip, we have to pay 1 rupee. There are two possible outcomes H or T. If difference between head tossed and tail tossed becomes 3, we will get 8 rupee from gambler. Should we play the game & why? How much our probability of winning that 8 rupee? What should be impact on probability of winning when we are getting 7 or 9 rupees? -- utpal from lucknow
Lets call x the expected number of flips from the starting point.
E(x) = 1 + E(y) It is then easy matrix algebra to see that E(x) = 9, E(y) = 8, and E(z) = 5. So on average it will take 9 flips for the disparity between heads and tails to be 3. So at 8 rupees it is a good bet for the person collecting the one rupee per flip, because he will receive on average 9 rupees, but pay back only 8. At 9 rupees it is a fair bet, at 7 the person collecting the one rupee per bet has an even stronger advantage. March 13, 2006I believe I may have an answer to the spinning penny. Over a decade ago I did a science project about penny spinning in sixth grade. I read in Omni magazine that spinning a penny very fast will nearly always stop tails up because the sides slope towards heads. I tried it hundreds of times and gotten nearly unbiased results, except twice when it stood on its edge. After numerous hours wasted I finally found that I was spinning it too fast, and a slower spin gave me the results desired, that is tails up. Also the penny isn't completely even and starting the spin on the thinnest part seemed to add to the consistency. A few charts full of bs and a giant cardboard circle decorated like a penny got me an A for Science and failing marks for every other class as I ignored all my homework. So I've proven* conclusively with one crudely done experiment done over ten years ago that I barely remember anything about and had no real understanding what I was doing at the time, that you perhaps maybe might just be spinning your penny a tad little too fast. - Jon Alright, I tried this again spinning the penny slowly 100 times. By slowly I mean the time between flicking and when the outcome was obvious was at least two seconds but less than five. I used a nice shiny 2004-D penny. My results were 52 heads and 48 tails. So I still remain unconvinced a spinning penny at any speed is highly biased towards tails. Jan. 2, 2005 Do you have any advice for betting on the flip of a coin? Yes! My advice is bet on the face up side at the start of the flip. According to Science News Online the probability that a coin will land on the same side it started on is 51%. The article says the reason is because a flipped coin does not spin perfectly around its axis and sometimes appears to be flipping when it actually isn't. The hypothesis only applies if coin is caught in the palm of the hand, so that bouncing is not an issue. The article also says that a spinning penny will land on tails 80% of the time, due to the heavier head side gravitating towards falling down first. However I'm skeptical of this. I tried this 20 times and got 11 heads and 9 tails. The probability of getting 9 or fewer tails in 20 spins with a probability of success of 80% is 1 in 1775. Dec. 13, 2004 I have a bag of 100 coins, one of those coins is a two-headed coin. I randomly pick a coin and then I observe the coin flipping 10 heads in a row. What is the probability that I picked the two-headed coin? This is a textbook Bayesian conditional probability question. In general the probability of A given B is the probability of A and B divided by the probability of B. In this case A is flipping 10 heads in a row and B is picking the two-headed coin. The probability of A and B is 1/100. This is because there is a 1 in 100 chance of picking the two-headed coin, and if you do the probability is 100% of flipping 10 heads in a row. The probability of flipping 10 heads in a row, assuming a randomly picked coin, is (1/100)*1 + (99/100)*(1/2)10. That is because there is a 1% chance of picking the two-headed coin, which has a 100% of getting 10 heads, and a 99% of picking a fair coin, which has a (1/2)10 chance of flipping 10 heads in a row. So, the probability that you picked the 2-headed coin, given that you flipped 10 heads in a row, is 0.01/(0.01*1 + 0.99* 0.000977) = 0.911843. Aug. 23, 2004 Mr. Wizard, if 50 different people toss a coin in the air 8 different times. What percentage of the 50 people will toss 8 consecutive heads or tails? Thanking you in advance. The probability of any given person tossing 8 heads or tails is 2*(1/2)8 = 1 in 128. If 50 people did this on average 0.39 of them will get all heads or tails. The probability of at least one person getting all heads or tails is 32.44%. Aug. 2003 If a coin was flipped 1000 times what is the probability the total number of heads would fall in the range of 452 to 548? For this problem we can use the normal approximation to the binomial distribution. The variance of the number of heads is 1000*(1/2)*(1/2)=250. So the standard deviation is 2501/2=15.8114. The probability of less than 548 heads is normdist((548+0.5-500)/15.8114) = 0.998920, where normsdist is the Excel function for the probability a random variable with a normal distribution of mean 0 and standard deviation 1 will fall under the given Z score. Next we subtract the probability of less than 452 heads. This is normdist((452-0.5-500)/15.8114) = 0.001080. So the answer is 0.99892-0.00108 = 0.997840. Again, this is an approximation. The actual answer is 0.997856, but is more tedious to derive. June 27, 2003
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