# An impossible bet **URL:** \ **Category:** Factual Questions\ **Created:** [May 7, 2015, 3:41pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487 "2015-05-07T15:41:38Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Nzinga\_Seated](https://avatars.discourse-cdn.com/v4/letter/n/b5ac83/32.png) [@Nzinga\_Seated](https://boards.straightdope.com/u/Nzinga_Seated)\ **Post date:** [May 7, 2015, 3:41pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/1 "2015-05-07T15:41:38Z")
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! [![](https://img.youtube.com/vi/eivGlBKlK6M/maxresdefault.jpg "An Impossible Bet | The 100 Prisoners Problem") ](https://www.youtube.com/watch?v=eivGlBKlK6M) ---
**Author:** ![Terr](https://avatars.discourse-cdn.com/v4/letter/t/839c29/32.png) [@Terr](https://boards.straightdope.com/u/Terr)\ **Post date:** [May 7, 2015, 3:50pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/2 "2015-05-07T15:50:00Z")
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. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 7, 2015, 3:51pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/3 "2015-05-07T15:51:48Z")
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. ---
**Author:** ![dasmoocher](https://avatars.discourse-cdn.com/v4/letter/d/dbc845/32.png) [@dasmoocher](https://boards.straightdope.com/u/dasmoocher)\ **Post date:** [May 7, 2015, 3:53pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/4 "2015-05-07T15:53:13Z")
There was a link to this in the comments: [100 prisoners problem](http://en.wikipedia.org/wiki/100_prisoners_problem) ---
**Author:** ![Left\_Hand\_of\_Dorkness](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/left_hand_of_dorkness/32/7156_2.png) [@Left\_Hand\_of\_Dorkness](https://boards.straightdope.com/u/Left_Hand_of_Dorkness)\ **Post date:** [May 7, 2015, 3:53pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/5 "2015-05-07T15:53:22Z")
> [@Nzinga\_Seated](#): > > 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! > > [http://youtu.be/eivGlBKlK6M](http://youtu.be/eivGlBKlK6M) 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). ---
**Author:** ![Nzinga\_Seated](https://avatars.discourse-cdn.com/v4/letter/n/b5ac83/32.png) [@Nzinga\_Seated](https://boards.straightdope.com/u/Nzinga_Seated)\ **Post date:** [May 7, 2015, 3:57pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/6 "2015-05-07T15:57:53Z")
Thanks for these responses !!! ---
**Author:** ![halfreal](https://avatars.discourse-cdn.com/v4/letter/h/ccd318/32.png) [@halfreal](https://boards.straightdope.com/u/halfreal)\ **Post date:** [May 7, 2015, 4:01pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/7 "2015-05-07T16:01:34Z")
> [@Nzinga\_Seated](#): > > In the solution, is he still saying I only have a 30 percent chance of winning?? 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. ---
**Author:** ![Left\_Hand\_of\_Dorkness](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/left_hand_of_dorkness/32/7156_2.png) [@Left\_Hand\_of\_Dorkness](https://boards.straightdope.com/u/Left_Hand_of_Dorkness)\ **Post date:** [May 7, 2015, 4:12pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/8 "2015-05-07T16:12:22Z")
> [@halfreal](#): > > 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! ---
**Author:** ![dasmoocher](https://avatars.discourse-cdn.com/v4/letter/d/dbc845/32.png) [@dasmoocher](https://boards.straightdope.com/u/dasmoocher)\ **Post date:** [May 7, 2015, 4:17pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/9 "2015-05-07T16:17:52Z")
> [@Left\_Hand\_of\_Dorkness](#): > > For example, if you have a loop with 70 boxes in it, 70 people won’t find their bill (I think). 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: > [@](#): > > In the initial problem, the 100 prisoners will be successful if the longest cycle of the permutation has a length of at most 50. Their survival probability is therefore equal to the probability that a random permutation of the numbers 1 to 100 contains no cycle of length greater than 50. ---
**Author:** ![TriPolar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tripolar/32/3008_2.png) [@TriPolar](https://boards.straightdope.com/u/TriPolar)\ **Post date:** [May 7, 2015, 4:31pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/10 "2015-05-07T16:31:53Z")
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. ---
**Author:** ![Terr](https://avatars.discourse-cdn.com/v4/letter/t/839c29/32.png) [@Terr](https://boards.straightdope.com/u/Terr)\ **Post date:** [May 7, 2015, 4:34pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/11 "2015-05-07T16:34:45Z")
> [@TriPolar](#): > > 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. ---
**Author:** ![TriPolar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tripolar/32/3008_2.png) [@TriPolar](https://boards.straightdope.com/u/TriPolar)\ **Post date:** [May 7, 2015, 4:36pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/12 "2015-05-07T16:36:02Z")
> [@Terr](#): > > ?? An odd-numbered bill can be hidden in an even-numbered box. See, I knew I was missing something. 🙂 ---
**Author:** ![TriPolar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tripolar/32/3008_2.png) [@TriPolar](https://boards.straightdope.com/u/TriPolar)\ **Post date:** [May 7, 2015, 4:52pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/13 "2015-05-07T16:52:58Z")
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. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 7, 2015, 4:57pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/14 "2015-05-07T16:57:22Z")
> [@](#): > > 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 ) 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. ---
**Author:** ![dasmoocher](https://avatars.discourse-cdn.com/v4/letter/d/dbc845/32.png) [@dasmoocher](https://boards.straightdope.com/u/dasmoocher)\ **Post date:** [May 7, 2015, 5:11pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/15 "2015-05-07T17:11:38Z")
> [@Chronos](#): > > 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: > [@](#): > > In the case of eight prisoners, this cycle-following strategy is successful if and only if the length of the longest cycle of the permutation is at most 4. If a permutation contains a cycle of length 5 or more, all prisoners whose numbers lie in such a cycle will not have reached their own number after 4 steps. ---
**Author:** ![Left\_Hand\_of\_Dorkness](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/left_hand_of_dorkness/32/7156_2.png) [@Left\_Hand\_of\_Dorkness](https://boards.straightdope.com/u/Left_Hand_of_Dorkness)\ **Post date:** [May 7, 2015, 5:15pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/16 "2015-05-07T17:15:46Z")
> [@dasmoocher](#): > > 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. ---
**Author:** ![Buck\_Godot](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/buck_godot/32/6573_2.png) [@Buck\_Godot](https://boards.straightdope.com/u/Buck_Godot)\ **Post date:** [May 7, 2015, 5:27pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/17 "2015-05-07T17:27:07Z")
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? ---
**Author:** ![dasmoocher](https://avatars.discourse-cdn.com/v4/letter/d/dbc845/32.png) [@dasmoocher](https://boards.straightdope.com/u/dasmoocher)\ **Post date:** [May 7, 2015, 5:42pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/18 "2015-05-07T17:42:02Z")
> [@Left\_Hand\_of\_Dorkness](#): > > 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. 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? ---
**Author:** ![MonkeyMensch](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@MonkeyMensch](https://boards.straightdope.com/u/MonkeyMensch)\ **Post date:** [May 7, 2015, 5:44pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/19 "2015-05-07T17:44:38Z")
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. ---
**Author:** ![Terr](https://avatars.discourse-cdn.com/v4/letter/t/839c29/32.png) [@Terr](https://boards.straightdope.com/u/Terr)\ **Post date:** [May 7, 2015, 5:45pm UTC](https://boards.straightdope.com/t/an-impossible-bet/719487/20 "2015-05-07T17:45:55Z")
> [@MonkeyMensch](#): > > 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 so 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. [Next page](https://boards.straightdope.com/t/an-impossible-bet/719487.md?page=2)