Dice Probability

[soggles]

Hi there.
Me and my friend found this problem in a newspaper and were trying to give it a go but we haven't gotten anywhere so thought I would post it on here for help!
This is the question:
A game involves rolling a 6-sided die up to 6 times. You win the game if you roll a 1 on the 1st roll, or a 2 on the 2nd roll or ...or a 6 on the 6th roll.
What is the probability of winning this game?
Would love some help!
Thank you!

 

[Steven G]

This is the place to come for help, but not for answers. We prefer to help you answer your own problem. What have you tried? Where are you stuck?

Hint. What is the probability that the 1st roll is a 1? The 2nd roll is a 2? .... The 6th roll is a 6? What do you do with those 6 probabilities?

 

[pka]

This is the question: A game involves rolling a 6-sided die up to 6 times. You win the game if you roll a 1 on the 1st roll, or a 2 on the 2nd roll or ...or a 6 on the 6th roll.

This is very well-known problem. It is known as a derangement: a list of \(n\) ordered objects no one of which is in the correct place,
\(245361\) is a derangement of the first six digits. If \(D(n)\) is the number of derangement on \(n\) distinct items then \(\dfrac{n!}{e}\) is a good approximation for \(D(n)\). So \(D(6)\approx \dfrac{6!}{e}=264.8\) If a die is tossed six times there are \(6^6=46656\) possible outcomes yet there are only approximately \(625\) of those cases in which no die value matches the position.

 

[Cubist]

This is very well-known problem. It is known as a derangement: a list of \(n\) ordered objects no one of which is in the correct place,
\(245361\) is a derangement of the first six digits.

I think a derangement would be appropriate for drawing 6 numbers from a bag without replacement. However with 6 dice rolls the same number could occur multiple times, for example you could roll 211111. To solve this using combinatorics, isn't the number of loosing combinations just 5^6 ?

If you prefer to think in terms of probabilities then...

Hint. What is the probability that the 1st roll is a 1? The 2nd roll is a 2? .... The 6th roll is a 6?

After doing this then flip your thinking and calculate the probability of rolling a loosing number (2,3,4,5 or 6) on the 1st roll. Then, what's the probability loosing on all 6 rolls? Then, use the previous answer to work out the probability of a win.

 

[soggles]

This is the place to come for help, but not for answers. We prefer to help you answer your own problem. What have you tried? Where are you stuck?

Hint. What is the probability that the 1st roll is a 1? The 2nd roll is a 2? .... The 6th roll is a 6? What do you do with those 6 probabilities?

So I realise that the probability of rolling a 1 on the first roll is 1/6 (16.67%). And it is the same for rolling a 2 on the second etc. But what do i do with that information now?

 

[HallsofIvy]

The probability of rolling a 1 on the first roll is 1/6 as you say. So the probability of winning on the first roll is 1/6. But the probability of not rolling a 1 is 5/6 and then you roll again.

Now the probability of rolling a 2 on the second roll is again 1/6 so the probability of winning on the second roll is (5/6)(1/6)= 5/36 because in order to have a second roll you must not have won on the first roll.

Similarly the probability of winning on the third roll is (5/6)(5/6)(1/6)= (5/6)^2(1/6), the probability of winning on the fourth roll is (5/6)^3(1/6), the probability of winning on the fifth roll is (5/6)^4(1/6), and the probability of winning on the sixth roll is (5/6)^5(16).

The probability of winning at all is 1/6+ (5/6)(1/6)+ (5/6)^2(1/6)+(5/6)^3(1/6)+ (5/6)^4(1/6)+ (5/6)^5(1/6)= (1/6)(1+ 5/6+ (5/6)^2+ (5/6)^3+ (5/6)^4+ (5/6)^5).

That last sum is a "geometric sum". Let S= 1+ 5/6+ (5/6)^2+ (5/6)^3+ (5/6)^4+ (5/6)^5. It can be written as S= 1+ (5/6)(1+ (5/6)+ (5/6)^2+ (5/6)^3+ (5/6)^4)= 1+ (5/6)(1+ (5/6)+ (5/6)^2+ (5/6)^3+ (5/6)^4+ (5/6)^5- (5/6)^5)= 1+ (5/6)S- (5/6)^6.
From S= 1+ (5/6)S- (5/6)^6. (1- 5/6)S= (1/6)S= 1- (5/6)^6 so S= 6(1- (5/6)^6)= 6- (5/6)^5= (6^6- 5^6)/6^6 so the probability of winning is (6^6- 5^6)/6^7= 0.11085, approximately.

 

[JayJay]

HallsofIvy - You say the probability of winning in six throws is 0.11085 which is less than the probability you have for winning on the first throw, 0.16667!
A more straight forward solution:
The probability that any one throw loses is 5/6; that all six are losers is (5/6)^6. The probability of at least one winning throw (enough to win the game) is 1 - (5/6)^6 = 0.66510.

 

[Dr.Peterson]

That last sum is a "geometric sum". Let S= 1+ 5/6+ (5/6)^2+ (5/6)^3+ (5/6)^4+ (5/6)^5. It can be written as S= 1+ (5/6)(1+ (5/6)+ (5/6)^2+ (5/6)^3+ (5/6)^4)= 1+ (5/6)(1+ (5/6)+ (5/6)^2+ (5/6)^3+ (5/6)^4+ (5/6)^5- (5/6)^5)= 1+ (5/6)S- (5/6)^6.
From S= 1+ (5/6)S- (5/6)^6. (1- 5/6)S= (1/6)S= 1- (5/6)^6 so S= 6(1- (5/6)^6)= 6- (5/6)^5= (6^6- 5^6)/6^6 so the probability of winning is (6^6- 5^6)/6^7= 0.11085, approximately.

There's a little mistake there. Since we need to divide S by 6, the answer is just 1- (5/6)^6 = 0.6651. But that's just what JayJay got.

A more straight forward solution:
The probability that any one throw loses is 5/6; that all six are losers is (5/6)^6. The probability of at least one winning throw (enough to win the game) is 1 - (5/6)^6 = 0.66510.

What this needs is an explanation of why this is equivalent to the problem, to silence skeptics (which I was on my first read-through). One way to see it is that if we keep tossing even after winning, it won't make us win any more or less than we did, so winning under the indicated rules is equivalent to tossing at least one winner if you kept going.