# An impossible bet

Is there anyone who can help me understand this? It is a video about a so called impossible bet, but then he does the solution. I can’t understand the solution. In the solution, is he still saying I only have a 30 percent chance of winning??

It’s annoying to have to watch videos to help a person understand some shit, I know. It’s cool. But if anyone is in the mood! That’d be great!

Yes, if you (and every other person) use his method, you have a 31% chance of winning. But since the payout is 100:1, it is definitely worth it to take a 69% chance of losing.

Yes. You (and the other 99 people) have only a little over a 30% chance of winning. But the payoff is high enough that a 30% chance of winning gives you a very high expected value, and it’s hence a very good bet.

100 prisoners problem

So I just watched that, and it blew my mind–very cool!

Here’s my understanding: yes, you have a 30% chance of winning \$100, and a 70% chance of losing \$1. Decent bet. I think the final line is important: what you’re doing is clustering wins and losses through this strategy. Under most circumstances, you’ll have about half the people find their bill and half the people not find their bill. Under this, 70% of the time, fewer than half the people will find their bill, depending on the size of the largest loop. For example, if you have a loop with 70 boxes in it, 70 people won’t find their bill (I think).

Thanks for these responses !!!

That’s what he’s saying, yes. Except that “only” isn’t quite appropriate here. Betting one dollar to win a hundred at a 30% chance is a pretty sweet deal. If you could repeat the bet indefinitely, you’d make a fortune - after 1000 runs, you’d have won 300 x 100 = 30,000 bucks, and lost 700. As opposed to (practically) never winning if you (and all the others) don’t follow that strategy.

Edit: never mind!

The way I read it was if there is a loop, or cycle, that’s greater than 50 then there’s at least a 70% chance that 1 person won’t find his bill. (I think :))

From the Wiki page:

I must be missing something here. If the boxes are number why don’t the people with odd numbered bills look only in the odd numbered boxes and similarly for the evens? That way everyone wins every time.

?? An odd-numbered bill can be hidden in an even-numbered box.

See, I knew I was missing something.

Well I really rushed through that explanation video. That’s a really cool solution. By using their number as the starting point they’ll have to follow every chain in the end.

If there’s a loop longer than 50, then that’s a guarantee that anyone on that loop won’t find their bill, because everyone always finds their bill (if at all) on the last step of the loop, and you don’t have enough tries to reach the last step of a 51-box loop.

However, the chances are decent (31%) that there is no such loop.

Yeah, I should have realized that, a loop of 50 gives a 70% chance of failure, but a loop of 51 guarantees it and not just increases the chance of failure. It was on the Wikipage:

Not exactly. A loop of 50 gives a 100% chance of success, in two ways:

1. Everyone on this loop will find their bill on their last opened box, so all of them succeed.
2. Since there are only 100 boxes, there cannot be another loop that’s longer than 50, so the whole team will succeed.

The 70% chance of failure comes in because in 70% of scenarios, there’s a single loop longer than 50.

That was really cool.

Although as a math nerd who is never satisfied, I have to ask if anyone knows whether this is the optimal strategy or is there possibly another strategy that could improve your chances of winning above this?

Maybe I should think of it as if: loop of 50 or less–100% success; loop of 51 or greater–100% failure, and there’s a 70% chance you’ll get a loop of greater than 50.

Correct?

Wait a minute! When did the boxes become distinguishable or even numbered (as the explanation showed)? Say I have number 32 on my bill. How can I walk into a room of 100 identical boxes, randomly arranged and find box number 32? The puzzle proposition shows the boxes in a square array but that an assumption not explicitly stated and regardless the gamblers would have no prior knowledge of the arrangement.

Unless I’m missing something, there’s an unstated assumption being made.

They are not numbered. They are arranged in a particular way and are “reset” every time to the same configuration. So you can count to the Nth box. You can do that for any physical arrangement of boxes as long as the counting method is the same for every player.