A funny probability problem for the younger ones

  • #1
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Homework Statement



This is a problem that I really liked and that I want to share with you. Firstly because of the story around it, secondly because of the unexpected solution, and finally because it can be investigated with a computer for those who are the least comfortable with maths.
Everybody is invited to participate, but don't forget this is for the 16 to 18 years old.


problem:
The chief of the forty thieves is finally captured and brought before the sultan, chained, and surrounded by 12 janissaries (https://en.wikipedia.org/wiki/Janissaries). The sultan tells him 'you can still save your life. Here are five black balls, and five white balls. Distribute them as you wish between these two boxes. I will then make them indistinguishable, and will shake them. You will then be blindfolded by one of my janissaries, and you will have to choose a box, and pick a ball. If the ball is white, I will spare you. If it is black ... ' and he shows him the shiny blade of his guard's scimitar (https://en.wikipedia.org/wiki/Scimitar).

How should the chief of the forty thieves make his choice in order to maximize his chances of survival ?
 
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Answers and Replies

  • #2
Jonathan Scott
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A couple of requests for clarification:

1. Does "distribute them as you wish" include the possibility of zero in one of the boxes? If so, this conflicts with the subsequent instruction to "pick a ball" as this will not be possible, so it seems likely that this was not intended.

2. And does "choose a box" permit handling it in advance in such a way that one might be able to tell whether it contains zero, one or more balls?
 
  • #3
BiGyElLoWhAt
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Wow, you can get surprisingly high probability for choosing white.
 
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  • #4
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A couple of requests for clarification:

1. Does "distribute them as you wish" include the possibility of zero in one of the boxes? If so, this conflicts with the subsequent instruction to "pick a ball" as this will not be possible, so it seems likely that this was not intended.

2. And does "choose a box" permit handling it in advance in such a way that one might be able to tell whether it contains zero, one or more balls?
1- Any choice is allowed, knowing that the only thing that will save him is to pick a white ball. An empty box leads to his execution
2- No handling possible after the distribution
 
  • #5
RUber
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The best I can find is about 72%. Is there a better way?
 
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  • #6
BiGyElLoWhAt
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72.2222222222, that's a little better, no?
 
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  • #7
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I feel that @BiGyElLoWhAt won't resist long before giving away his solution ;-)
 
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  • #8
BiGyElLoWhAt
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I'm trying... so ever hard... haha =D
 
  • #9
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Distribute the balls as one wishes. I would be tempted to disregard black balls to the floor / trashcan , and distribute 2 white balls into box A and 3 white balls into box 2...
100%

If that is illegal, then I suppose that 72.22% recurring is correct. ( as a fraction it would be 13/18)
that woudl be the overall chances of living
 
  • #10
35,258
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Distribute the balls as one wishes.
between the boxes.

13/18 is indeed optimal.

If we don't have to first pick a box and then a ball, 100% are possible.
 
  • #11
mathman
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Hint: 13=4+9.
 
  • #12
Ray Vickson
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The best I can find is about 72%. Is there a better way?
The answer 13/18 ≈ 0.7222 is provably optimal.
 
  • #13
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Ok, thank you everybody for your participation. I think that this game can end now as it does not interest the public it is for. I hope that you liked the problem as much as I did.
@BiGyElLoWhAt , if you still want to give away the solution, the last word belongs to you.
 
  • #14
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I used tree diagram and pen-and-paper. That seemed like a good way to begin the problem because of tthe either-or-choice in the beginning with the choice of box.
 
  • #15
301
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for either-or-probability... the sultan seems to give this kind of choice

the initial choice would be random at 50% 50%

Only by the act of picking a white ball could we be saved. Evidently all of the 10 balls must be utilized in the two boxes.

It leads to believe that the best way to distribute 5 white and 5 blacks would be in such a manner that both of two boxes A and B would have individually good chances of surviving (because only one box can be chosen, and it will be random choice at 50%)
If the choice was not completely random, and you could feel how the box sounds, when you rattle it. Then you could always win this kind of challenge. Because 1 ball inside a box, sounds totally different compared to multiple balls inside a box.

In plain language, you have plan A of surviving. And then you have plan B of surviving.
plan A involves being lucky with a coin toss and receiving the 1 white ball inside the box...The box only contains one ball, which will be automatically chosen. This is also incidentally the best possible plan individually for a box to contain the surviving odds. 1 ball chosen from 1 ball= 100%
This is "the cheapest" way to have the winning ticket in one box. The other balls would go to the other box.
0.5*[(1 favorable)/(1 total)]= 0.5

plan B involves being unlucky with a coin toss. But... you still have some chances. And those chances are 4 white balls amongst (9 balls). You have to win an "unfavorable coin toss" at this stage to survive.

overall chances of living =
P(A) + P(B) = 0.5*(1/1) + 0.5*(4/9) =13/18
 
  • #16
535
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The chief of the forty thieves has to devise a strategy that will give him the highest possible probability of survival.

Let us call strategy ##(\alpha,\beta)## the strategy that consists in placing ##\alpha## white and ##\beta## black balls in box 1, and let us define the events
##A##: 'blindfolded, the chief picks box 1'
##B##: 'blindfolded, the chief picks a white ball '

Let us call ##p(\alpha,\beta)## the probability of survival associated to the strategy ##(\alpha,\beta)##, for ##0\le \alpha,\beta \le 5##.
Since ##A## and ##\bar A## form a complete system of events, the law of total probability tells us that ##p(\alpha,\beta) = P(B) = P(A) P(B/A) + P(\bar A) P(B/\bar A) ##

We find ##p(0,0) = p(5,5) = 1/4 ##, and ##p(\alpha,\beta) = \frac{1}{2} ( \frac{\alpha}{\alpha + \beta} + \frac{5-\alpha}{10 - \alpha - \beta }) ## otherwise. Using a spreadsheet to calculate and record these probabilities, we see that the strategy ##(1,0)## is, by far, the best strategy. Note that the strategies consisting in placing only white balls in a box, this is to say the ##(\alpha,0)##-strategies, offer better chances of survival.

For those who liked it, you can show that the strategy ##(1,0)## remains the best choice in case the sultan does not give five balls of each color, but any number of each.
 
  • #17
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The (0,0) and (5,5) cases leave one box empty, which leads to an undefined situation if that box is chosen.
 
  • #18
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The (0,0) and (5,5) cases leave one box empty, which leads to an undefined situation if that box is chosen.
It doesn't according to post 4.
 
  • #19
301
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I suppose so

when you plug into the function

alfa= 0
beta= 0

0.5* (0/0) + 0.5* (5/10)

0/0 is undefined in a standard sense. Though when one thinks about it in common sense. It is clear that this is simply an empty box entirely. Empty box was allowed as per Sultan's instructions, but it would not really help us a lot in terms of surviving.
 
  • #20
35,258
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It doesn't according to post 4.
Oh, I missed that.

It is quite easy to show that:
- if one box does not contain a ball at all, moving a white ball into it increases the chance to survive
- moving all black balls to the box with more white balls increases the chance to survive
- moving a white ball from a box with at least two white balls and no black balls increases the chance to survive
Combining those three rules always leads to the same optimal final state, one box has a single white ball and the other box has everything else.
 
  • #21
301
15
I reckon that the method of proofing (with spreadsheet) would be a proof by exhaustion essentialy

When there is only a fixed limit on the balls. then I reckon that the possible combinations of white and black balls is also limited.
Check the probabilities for each combo and pick the highest probability.
 
  • #22
Buzz Bloom
Gold Member
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What was left out of the question is whether one could do something with boxes, other than opening them, before choosing a box. For example, pickup one box with the left hand and the other with the right hand before choosing. Would that be legal?
 

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