Determing the probabilities of all of the outcomes of an event

12BS

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On a game show there are five identical closed boxes. ONE box contains $10,000, TWO boxes contain $1,000 each, ONE box contains $1 and the remaining box contains a red card. A player is allowed to select a box, open it and keep the contents. The player is allowed to repeat this process until the box with the red card is opened; at this point the game ends. Find the probability of each possible outcome (so 0, 1, 1000, 1001, 2000, 2001, 10000, 10001,etc). As a check, remember that the sum of these probabilities should equal 1.

So for each X value and it corresponding P(X=x) I was trying to find them by saying all selection orders must end with a red card.
So I started by
X=0 --> P(x=X)= 1/5

X=1--> P(x=X)= 1C1*1C1/ 5C2 = 1/10
x=1000--> P(X=x)= 2C1*1C1/ 5C2= 2/10
x=1001--> P(X=x)= 2C1*1C1*1C1/ 5C3= 2/10


etc. but I do not get the probabilities to sum to one. (also trying this at X= 12,001 gives 1 which is obviously incorrect.) Is there a better way to go about this?
Any help would be greatly appreciated.
 
On a game show there are five identical closed boxes. ONE box contains $10,000, TWO boxes contain $1,000 each, ONE box contains $1 and the remaining box contains a red card. A player is allowed to select a box, open it and keep the contents. The player is allowed to repeat this process until the box with the red card is opened; at this point the game ends. Find the probability of each possible outcome (so 0, 1, 1000, 1001, 2000, 2001, 10000, 10001,etc). As a check, remember that the sum of these probabilities should equal 1.

So for each X value and it corresponding P(X=x) I was trying to find them by saying all selection orders must end with a red card.
So I started by
X=0 --> P(x=X)= 1/5

X=1--> P(x=X)= 1C1*1C1/ 5C2 = 1/10
x=1000--> P(X=x)= 2C1*1C1/ 5C2= 2/10
x=1001--> P(X=x)= 2C1*1C1*1C1/ 5C3= 2/10

etc. but I do not get the probabilities to sum to one.
What was contained in what you've denoted as "etc"?

Maybe, instead of the "combintations" formulas (which I think allows, for instance, for the red card to be selected before the one-dollar box), maybe just list your options:

$0: red box immediately (so one way of doing this)

$1: $1 box, then red box (so one way of doing this)

$1000: $1000 box, then red box (so one way of doing this)

$1001: $1000 box, then $1 box, then red box; or $1 box, then $1000 box, then red box (so two ways of doing this)

$2000: $1000 box, then other, then red box (so one way of doing this)

$2001: both $1000 boxes, then $1 box, then red box; or $1000 box, then $1 box, then other $1000 box, then red box; or $1 box, then two $1000 boxes, then red box (so three ways of doing this)

$10,000: $10,000 box, then red box (so one way of doing this)

...and so forth. Then count up the numbers of ways to get the various results. Then divide each result's number of ways by this total to get the individual probabilities. They will, by nature of the computations, add up to 1. ;)
 
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