Smuggling

[Guest]

A man steals and Idol. He hands it off to a group of men designed to lose the item. The first man has a 90% chance of handing it off to the second man and a 10% chance of keeping it. The second man has an 80% chance of handing it off and a 20% chance of keeping it. The third man has a 70% chance handing it off and a 30% chance of keeping it. This pattern continues until the tenth man has a 100% chance of keeping it. What are the odds that any one man has it or what are the greatest odds that any one man can have of possessing the item?

[Guest]

Now, this really depends on what you mean by "possessing" the idol. If you mean possessing as in holding the idol, then the formula for the probability of being handed the idol is [tex]\frac{9!}{(10-x)!}* 10^{-(x-1)}[/tex]. If by "possessing" the idol you mean keeping the dol, then you would multiply the given formula by [tex]\frac{x}{10}[/tex]. Let me explain.
The formula for the probability of the first person being handed the idol is [tex]1[/tex], the second person [tex]0.9[/tex], the third person [tex]0.9 * 0.8[/tex], and so on until person 10 is [tex]0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 * 0.1[/tex]. Person 10's formula is where the [tex]9![/tex] comes from, and the [tex](10-x)![/tex] comes from the taking away of numbers until you get to the required formula.
Of course, this formula would get you a whole-number probability greater than 1, which is where the [tex]10^{-(x-1)}[/tex] comes in, to multiply each of the parts of the factorial by [tex]0.1[/tex].
Then, you would multiply by [tex]\frac{x}{10}[/tex], which is the person's probability of keeping the idol.
Using this formula, the person with the greatest chance of being handed the item is person 1, of course, and the person with the greatest chance of keeping the item is person 3.
Hope I could help!

[HallsofIvy]

It is specifically said that the first man has a 10% chance of keeping it and a 90% chance of passing it on. The second man has a 90% of getting it and then a 20% chance of keeping it so, overall, a 0.9(0.2)= 0.18= 18% chance of keeping it and a 1- 0.18= 0.72= 72% chance of passing it on. The third man has a 72% chance of getting it and, if he gets it, a 30% chance of passing it on so, overall, a 0.72(0.3)= 0.216= 21.6% chance of keeping it and a 1- 0.216= 0.784= 78.4% chance of passing it on.

For the other 7 people do the same thing.

[Guest]

Now, this really depends on what you mean by "possessing" the idol. If you mean possessing as in holding the idol, then the formula for the probability of being handed the idol is [tex]\frac{9!}{(10-x)!}* 10^{-(x-1)}[/tex]. If by "possessing" the idol you mean keeping the dol, then you would multiply the given formula by [tex]\frac{x}{10}[/tex]. Let me explain.
The formula for the probability of the first person being handed the idol is [tex]1[/tex], the second person [tex]0.9[/tex], the third person [tex]0.9 * 0.8[/tex], and so on until person 10 is [tex]0.9 * 0.8 * 0.7 * 0.6 * 0.5 * 0.4 * 0.3 * 0.2 * 0.1[/tex]. Person 10's formula is where the [tex]9![/tex] comes from, and the [tex](10-x)![/tex] comes from the taking away of numbers until you get to the required formula.
Of course, this formula would get you a whole-number probability greater than 1, which is where the [tex]10^{-(x-1)}[/tex] comes in, to multiply each of the parts of the factorial by [tex]0.1[/tex].
Then, you would multiply by [tex]\frac{x}{10}[/tex], which is the person's probability of keeping the idol.
Using this formula, the person with the greatest chance of being handed the item is person 1, of course, and the person with the greatest chance of keeping the item is person 3.
Hope I could help!

sorry yes, i should have worded it better. I was looking for the highest odds any person ENDED UP with the item. but you answered that so thank you. I'm not even sure why I came up with this question. it just occurred to me, but thanks for explaining it in a way I could understand