1. ## prize player

Any help offered in solving the below question will be much appreciated.
Thanks
Kingman

To win a prize in a game, a player rolls a fair die until a total of three “sixes” are obtained.

A player, John ,has obtained two “sixes” and two “ones” in his first four rolls . Find the probability that he wins a prize with exactly four more roll of the die. (.0965)

2. You have to use a geometric distribution. Here's the formula.

$\displaystyle pr(X=x)=p(1-p)^{(x-1)}$

X~geometric(p=$\displaystyle \frac{1}{6}$)

X= the number of rolls until John rolls a "six" thus giving him 3 "sixes" and him winning a prize.

$\displaystyle pr(X=4)=...$
You should be able to solve the rest now.

3. ## thanks

Thanks very much for the help.
Can you please explain why you did not consider the probability of the event:

"two “sixes” and two “ones” in his first four rolls" before following up the probability of the final event:
"exactly four more roll of the die.

Thanks
Kingman

4. Simple. The goal of the game is to roll 3 "sixes." Since the player has already rolled 2 "sixes" he only needs 1 more to win. Any numbers that the player has rolled previously simply don't matter.

The part about rolling 2 "ones" is just extra information that was added to the problem to make it seem more difficult than it really
is.

5. ## thanks

thanks but if for the same we ask:

Calculate the probability that a player ,George , wins a prize on his 8th roll of the die..
this is a Binomial ~(7,1/6)

Required Prob= P(x=2).1/6 where X is the number of "sixes"

the question is why we have to consider the probability of 8th roll ( ie 1/6) but we don't have consider the probability of the first 4 rolls in the previous question which I believe it is equal to 1.
thanks

6. For the example you just gave the distribution would negative binomial.

X~negative binomial(k=3, p=$\displaystyle \frac{1}{6}$)

$\displaystyle pr(X=8){x-1\choose k-1}(p)^{k}(p-1)^{(x-k)}$

For this one, you have to consider all eight rolls because no "sixes" have happened yet. If we take this same problem that you just posted and say that George has already rolled 1 "six" and you want to find the probability that he will win on his 8th roll then we'd be looking for:

X~negative binomial(k=2, p=$\displaystyle \frac{1}{6}$)

$\displaystyle pr(X=7){x-1\choose k-1}(p)^{k}(p-1)^{(x-k)}$

Because 1 six has already occurred.

If we take this same problem that you just posted and say that George has already rolled 1 time and you want to find the probability that he will win on his 8th roll then we'd be looking for:

X~negative binomial(k=3, p=$\displaystyle \frac{1}{6}$)

$\displaystyle pr(X=7){x-1\choose k-1}(p)^{k}(p-1)^{(x-k)}$

Because 1 roll has already occurred.

I guess I'm just having a hard time explaining this for you. Sorry

7. ## Thanks

Thanks very very much for your patience .Give me some time to digest your work.
Thanks
Kingman