The answer is completely bogus. You can’t make everyone in the room choose a different set of answers, even if you pair them up. You can have 100 people choose TTTTTHHHHH, for example, or THTHTHTHTH, just because it’s easy to choose. Please make a retraction / correction for your puzzler, because I think you can do better

# Bill Gates Coin Toss Puzzler

As stated on the radio, it’s not bogus. Each pair will always have a “winner,” either the one who correctly called it, or the other one who wins when the caller is wrong. After each toss, the half that are winners goes to the next toss. At the end of 10 tosses, there will be one winner with ten wins in a row. Unfortunately, the written version on cartalk.com didn’t transcribe it correctly, and is indeed bogus.

Insightful, thank you for helping me understand the part I was missing. The ten tosses aren’t called out ahead of time, but rather one at a time before the next toss. Isn’t it funny how the procedure makes all the difference?

The way I imagined everyone calling all ten at once, you could indeed have a room full of people who all call TTTTTHHHHH, and the other half of the room can be forced to choose HHHHHTTTTT, and then they could all lose

But you’re saying in this case, that only the winners then pair up with new people to call it again, forcing there to be a certain number of winners each round. That is a bit of tricky there!

OK. There are 2 ** 10 possible coin flip combinations, or 1024. How do we know that? Well, suppose heads = 1 and tails = 0 . We know from the puzzler that each guest must choose a string of 10 coin flips, where all tails would be 0000000000 and all heads would be 1111111111 . So you have a calculator on your computer; switch the mode of that calculation to scientific and select the “BIN” button, then enter 10 1s . Now select the “DEC” button to convert. Ha! 1023 ( add in that all zeros choice and you have a total of 1024 ) . There are 1024 possible different combinations of 10 coin flips. If Bill Gates were to ask each guest as he of she came in the door to choose a string of 10 coin flips, and each guest were to choose such a string at random, many possible coin flip combinations would remain unselected when all guests had arrived, due to duplication. Sorry, I must be missing something. Yeah, there are ways to restate the puzzler to justify an answer.

There was indeed some ambiguity between how this was stated on the radio program and how it was stated on the website. Of course if Bill Gates and the guests were all in on it, it would still work as stated on the web site. Each guest would be pre-assigned their own sequence. With 1024+ guests every one of the 1024 possible sequences would be covered, so someone would have to “guess” the right answer even if it were done with just Bill Gates tossing a single coin ten times. The method used – as explained on the radio show – basically accomplishes the same thing. As long as there are enough guests to cover all 1024 possible sequences, one of the guests will get it right.

Here’s another question for this puzzler. Say it was in fact just Bill Gates tossing a coin 10 times, and the guests were not in on the trick; i.e. each of them made an independent guess of the sequence, and some of them might make the same choice, so not every sequence would be covered. What’s the probability that at least one of the guests would get it correct?

So, what is the probability (p’y) that at least one of 1100 guests would independently correctly guess 10 coin flips by Bill Gates (no tricks)?

P’y of 1 guest guessing correctly = 1/1024 = 0.00097656

P’y of 1 guest NOT guessing correctly = 1 - 0.00097656 = 0.99902344

P’y of 1100 guests NOT guessing correctly = 0.99902344**1100 = 0.3414

So, the p’y of at least 1 guest guessing correctly = 1 - 0.3414 = 0.6586

Q.E.D.

Good job @insightful , that looks correct. I would have thought it would be closer to p= 0.5 as an intuitive guess, that at least one guest would get it right. With the chance of a win at 1/1000 and 1000 guests guessing I mean. Seems like it would be even-steven. Maybe the probability of *exactly* one guest guessing correctly, that might be closer to 0.5. Your calculation is higher than 0.5, maybe b/c that figure includes winning combinations where more than one guest gets the answer correct.

The explanation on the website is bogus because if you pair up two people for each coin toss, only ONE guesses, the other gets the flip side. Of course someone WINS every time, but not necessarily the guesser. The puzzler bet says guess correctly 10 times, not win 10 times.

To win this bet without cheating Bill would not have said, “If we flipped a coin 10 times in a row, do you think there’s anyone in this room who could call it correctly each time?" Bill would need to have said, “How likely is it, do you think, that somebody in this room could win 10 coin tosses in a row?”

Winning ten in a row is not the same as calling ten in a row, so…Bo-oh-gus!

I think pretty much everyone agrees that the language on the web site for this particular puzzler isn’t an accurate representation of the actual puzzler as stated by T&R on the show. I have to admit I didn’t really understand what they said on the show either. Both the web-site and the show needed some add’l clarification for me to understand what’s allowed and what isn’t I guess.

OK, I could have quoted Bill Gates better. Going back to the actual audio version of the puzzler, here is what Bill is said to have said about calling 10 flips correctly (minus some extraneous Tom/Ray crosstalk) …

You left of an important part, the add’l constraint where they say “Here’s what calling correctly consists of”. I listened to the puzzler question again from the puzzler section of this website, but still can’t really understand what they are trying to say in that constraint section, how the interaction of the two people work and what exactly it means to “win”.

A betting person with a month’s pay in the bank would have to think that betting $1000 against 10 million at 3 to 10 or so odds might not be such a bad bet. But watch out! The reason for the high wager may be due to the devil in the incomprehensible details.

the puzzler is my least favorite part of the show.

it s a car show, make the puzzlers car related if you ever get new hosts

I take to be simply like a coin toss before a football game. The guests pair up and one calls the toss. If he’s right, he wins. If he’s wrong, his partner wins. Each toss produces one winner in each pair. The winners go on to the next toss and pair up with new partners. After 10 tosses, there is *guaranteed* to be a winner who won 10 coin tosses in a row.

They key from the show is that it was Bill Gates’ millions vs ten thousand from me. Sure, the odds are in Bill Gates favor, by almost 2:1, but he’s putting up way more money than me. I would definitely take the bet, and then just keep doubling down until I win, which is likely to happen every the third time or so.

If done as stated in my previous post, you are a guaranteed loser. You’re right, though, using the version transcribed on this web site.

I think you are right @insightful . To make this puzzler a bit more clear, I wouldn’t have the guests to toss the coins. Bill Gates would do the tossing, toss one coin ten times in a row is all he’d have to do. As long as everyone could see his toss outcome, heads or tails, I mean. One person of the pair would be assigned to guess the toss result. If they are correct, they win, if not the other person wins. After each toss the winner of that round would advance to the next round, and be paired with another winner.

Each round half of the contestants are eliminated. If you start with 1024 contestants, after the first round only 512 remain, second, 256 remain, etc … after the 9th round only two contestants remain. One of them would be guaranteed to win on the 10th toss.

In the explanation of “what calling correctly consists of” Ray says, “I toss the coin and you call it. You say, ‘heads.’ If it comes up heads, I lose. If it’s tails, I win, even though I didn’t call it.” The single coin theory would seem the intuitively obvious approach, where everyone calling heads is asked to go to one side of the room and everyone calling tails asked to go to the other side of the room. Groupings would not, in that case, be in even 512, 256, 128, etc. divisions for each toss, but would be close to those figures. But I can see that was not the approach that Bill had in mind. Maybe after securing the bet Bill said, “Hang on a jiff. I’ll be right back. I happen to have 13 rolls of quarters in my desk which I just happen to have on hand for just such an occasion.” You would need at least 512 coins for the first toss. No?