PROBABILITY HELP

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Merida

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Prinstan Trollner and Dukejukem are competing at the game show WASS. Both players spin a wheel which chooses an integer from 1 to 50 uniformly at random, and this number becomes their score. Dukejukem then flips a weighted coin that lands heads with probability 3/5. If he flips heads, he adds 1 to his score. A player wins the game if their score is higher than the other player’s score. The probability Dukejukem defeats the Trollner to win WASS equals m/n where m, n are coprime positive integers. Compute m + n.
 
After 19 posts you know by now how the game here works. You show no work, you get no help. Please post back showing your attempt at this problem so we know what type off help you need.
 
You should know how things work here by now!

Please show your work, so we can see where you need help, and don't think you're trying to get other people to do your work for you.

But I have to say I don't understand what the problem means. Do both put their hands on the wheel and get a score that they both get, and then D has a chance to add to his score, while PT does nothing more? Then obviously D wins 3/5 of the time. Or do they take turns, each spinning the wheel and flipping the coin? Is all this done only once, or several times, perhaps stopping when one has a higher total score than the other?
 
Solution: If the coin has probability q of landing heads, the probability of Dukejukem winning is (1 − P(tie))/2 + qP(tie), where P(tie) = 1/50 is the probability that both players spin the same number on the wheel. This is 251/500, so the answer is 751.
This is the official solution, but I understood nothing . I thought someone might help about how to start the question atleast
 
Dr.Peterson said:
You should know how things work here by now!

Please show your work, so we can see where you need help, and don't think you're trying to get other people to do your work for you.

But I have to say I don't understand what the problem means. Do both put their hands on the wheel and get a score that they both get, and then D has a chance to add to his score, while PT does nothing more? Then obviously D wins 3/5 of the time. Or do they take turns, each spinning the wheel and flipping the coin? Is all this done only once, or several times, perhaps stopping when one has a higher total score than the other?
Click to expand...
Solution: If the coin has probability q of landing heads, the probability of Dukejukem winning is (1 − P(tie))/2 + qP(tie), where P(tie) = 1/50 is the probability that both players spin the same number on the wheel. This is 251/500, so the answer is 751.
This is the official solution, but I understood nothing . I thought someone might help about how to start the question atleast .
 
First, I don’t think you gave us the problem correctly. Does D roll the die every time or just when D and T have tie?
 
Merida said:
Solution: If the coin has probability q of landing heads, the probability of Dukejukem winning is (1 − P(tie))/2 + qP(tie), where P(tie) = 1/50 is the probability that both players spin the same number on the wheel. This is 251/500, so the answer is 751.
This is the official solution, but I understood nothing . I thought someone might help about how to start the question atleast .
Click to expand...
Then that's what you should have asked in the first place! We ask you to tell us what you have tried, and that can include trying to understand their answer. That gives us at least some sense of where you are.

Since I don't even understand the problem, we can try to use this explanation (which, like most "official solutions", is very sketchy) to try to figure it out piece by piece.

Here is the problem, with extraneous nonsense removed:

PT and D are competing.​
Both players spin a wheel which chooses an integer from 1 to 50 uniformly at random, and this number becomes their score.​
D then flips a weighted coin that lands heads with probability 3/5. If he flips heads, he adds 1 to his score.​
A player wins the game if their score is higher than the other player’s score.​
The probability D wins is ...​

Now go through the solution:

If the coin has probability q of landing heads, the probability of D winning is (1 − P(tie))/2 + qP(tie), where P(tie) = 1/50 is the probability that both players spin the same number on the wheel. This is 251/500, so the answer is 751.​

We are told that the probability of heads is 3/5; they call that q, presumably to help focus on ideas rather than numbers.

Then there is something about a tie: P(tie) = 1/50 is the probability that both players spin the same number on the wheel. This tells us that each player spins the wheel, and gets that amount as an initial score. That clarifies one of my questions, which was due to bad grammar (combined with my pickiness). So here is a better version of the problem:

PT and D each spin a wheel which chooses an integer from 1 to 50 uniformly at random, and this number becomes their score.​
D then flips a weighted coin that lands heads with probability 3/5. If he flips heads, he adds 1 to his score.​
A player wins the game if their score is higher than the other player’s score.​
The probability D wins is ...​

Continuing with the solution, why is P(tie) = 1/50? That would be because whatever P spins, D has a 1/50 chance of spinning that same number. So the solution is tracking so far.

Now we're told that the probability of D winning is (1 − P(tie))/2 + qP(tie). The addition comes from there being two ways D can win: either by spinning the higher number in the first place, or by flipping the coin. Now, if D's number had been 1 less than PT's, then flipping heads would create a tie, so it looks like D only flips the coin in the case of a tie, as JeffM has suggested. Here's a new version of the problem:

PT and D each spin a wheel which chooses an integer from 1 to 50 uniformly at random, and this number becomes their score.​
In the case of a tie, D then flips a weighted coin that lands heads with probability 3/5. If he flips heads, he adds 1 to his score.​
A player wins the game if their score is higher than the other player’s score.​
The probability D wins is ...​

(That phrase alone would have made a lot of things clear!)

Now I'll let you take over. Do you see why (1 − P(tie))/2 is the probability of winning the initial toss? (I would have done it very differently, so if you prefer a different way, you're in good company. Just check that it gives the same answer.)

Then, do you see why qP(tie) gives the probability of winning in the case of a tie?
 
Dr.Peterson said:
Then that's what you should have asked in the first place! We ask you to tell us what you have tried, and that can include trying to understand their answer. That gives us at least some sense of where you are.

Since I don't even understand the problem, we can try to use this explanation (which, like most "official solutions", is very sketchy) to try to figure it out piece by piece.

Here is the problem, with extraneous nonsense removed:

PT and D are competing.​
Both players spin a wheel which chooses an integer from 1 to 50 uniformly at random, and this number becomes their score.​
D then flips a weighted coin that lands heads with probability 3/5. If he flips heads, he adds 1 to his score.​
A player wins the game if their score is higher than the other player’s score.​
The probability D wins is ...​

Now go through the solution:

If the coin has probability q of landing heads, the probability of D winning is (1 − P(tie))/2 + qP(tie), where P(tie) = 1/50 is the probability that both players spin the same number on the wheel. This is 251/500, so the answer is 751.​

We are told that the probability of heads is 3/5; they call that q, presumably to help focus on ideas rather than numbers.

Then there is something about a tie: P(tie) = 1/50 is the probability that both players spin the same number on the wheel. This tells us that each player spins the wheel, and gets that amount as an initial score. That clarifies one of my questions, which was due to bad grammar (combined with my pickiness). So here is a better version of the problem:

PT and D each spin a wheel which chooses an integer from 1 to 50 uniformly at random, and this number becomes their score.​
D then flips a weighted coin that lands heads with probability 3/5. If he flips heads, he adds 1 to his score.​
A player wins the game if their score is higher than the other player’s score.​
The probability D wins is ...​

Continuing with the solution, why is P(tie) = 1/50? That would be because whatever P spins, D has a 1/50 chance of spinning that same number. So the solution is tracking so far.

Now we're told that the probability of D winning is (1 − P(tie))/2 + qP(tie). The addition comes from there being two ways D can win: either by spinning the higher number in the first place, or by flipping the coin. Now, if D's number had been 1 less than PT's, then flipping heads would create a tie, so it looks like D only flips the coin in the case of a tie, as JeffM has suggested. Here's a new version of the problem:

PT and D each spin a wheel which chooses an integer from 1 to 50 uniformly at random, and this number becomes their score.​
In the case of a tie, D then flips a weighted coin that lands heads with probability 3/5. If he flips heads, he adds 1 to his score.​
A player wins the game if their score is higher than the other player’s score.​
The probability D wins is ...​

(That phrase alone would have made a lot of things clear!)

Now I'll let you take over. Do you see why (1 − P(tie))/2 is the probability of winning the initial toss? (I would have done it very differently, so if you prefer a different way, you're in good company. Just check that it gives the same answer.)

Then, do you see why qP(tie) gives the probability of winning in the case of a tie?
Click to expand...
Thanks for your time , got it
 
Dr.Peterson said:
Then that's what you should have asked in the first place! We ask you to tell us what you have tried, and that can include trying to understand their answer. That gives us at least some sense of where you are.

Since I don't even understand the problem, we can try to use this explanation (which, like most "official solutions", is very sketchy) to try to figure it out piece by piece.

Here is the problem, with extraneous nonsense removed:

PT and D are competing.​
Both players spin a wheel which chooses an integer from 1 to 50 uniformly at random, and this number becomes their score.​
D then flips a weighted coin that lands heads with probability 3/5. If he flips heads, he adds 1 to his score.​
A player wins the game if their score is higher than the other player’s score.​
The probability D wins is ...​

Now go through the solution:

If the coin has probability q of landing heads, the probability of D winning is (1 − P(tie))/2 + qP(tie), where P(tie) = 1/50 is the probability that both players spin the same number on the wheel. This is 251/500, so the answer is 751.​

We are told that the probability of heads is 3/5; they call that q, presumably to help focus on ideas rather than numbers.

Then there is something about a tie: P(tie) = 1/50 is the probability that both players spin the same number on the wheel. This tells us that each player spins the wheel, and gets that amount as an initial score. That clarifies one of my questions, which was due to bad grammar (combined with my pickiness). So here is a better version of the problem:

PT and D each spin a wheel which chooses an integer from 1 to 50 uniformly at random, and this number becomes their score.​
D then flips a weighted coin that lands heads with probability 3/5. If he flips heads, he adds 1 to his score.​
A player wins the game if their score is higher than the other player’s score.​
The probability D wins is ...​

Continuing with the solution, why is P(tie) = 1/50? That would be because whatever P spins, D has a 1/50 chance of spinning that same number. So the solution is tracking so far.

Now we're told that the probability of D winning is (1 − P(tie))/2 + qP(tie). The addition comes from there being two ways D can win: either by spinning the higher number in the first place, or by flipping the coin. Now, if D's number had been 1 less than PT's, then flipping heads would create a tie, so it looks like D only flips the coin in the case of a tie, as JeffM has suggested. Here's a new version of the problem:

PT and D each spin a wheel which chooses an integer from 1 to 50 uniformly at random, and this number becomes their score.​
In the case of a tie, D then flips a weighted coin that lands heads with probability 3/5. If he flips heads, he adds 1 to his score.​
A player wins the game if their score is higher than the other player’s score.​
The probability D wins is ...​

(That phrase alone would have made a lot of things clear!)

Now I'll let you take over. Do you see why (1 − P(tie))/2 is the probability of winning the initial toss? (I would have done it very differently, so if you prefer a different way, you're in good company. Just check that it gives the same answer.)

Then, do you see why qP(tie) gives the probability of winning in the case of a tie?
Click to expand...
By the way , I m having a doubt as to why we are dividing 1- P (tie) by 2 . I understand that 1 is the total probability and that p tie is 1/50 . That , 1 - p ( tie ) is the probability of getting any number other than the same number trollner got . But , why is this being divided by two or in other words , being halved ? . For instance if trollner rolled .... 7 , then out of 50 numbers , to avoid a tie and score greater ( from 8 to 50 inclusive ) , D can roll any 43 numbers out of the total 49 left , right ? So then how is the probability of rolling a grater number 1/2 of the remaining numbers ? I don't know if this doubt is senseless but I hope you understand my doubt
 
Merida said:
By the way , I m having a doubt as to why we are dividing 1- P (tie) by 2 . I understand that 1 is the total probability and that p tie is 1/50 . That , 1 - p ( tie ) is the probability of getting any number other than the same number trollner got . But , why is this being divided by two or in other words , being halved ? . For instance if trollner rolled .... 7 , then out of 50 numbers , to avoid a tie and score greater ( from 8 to 50 inclusive ) , D can roll any 43 numbers out of the total 49 left , right ? So then how is the probability of rolling a greater number 1/2 of the remaining numbers ? I don't know if this doubt is senseless but I hope you understand my doubt
Click to expand...
It does take some thought!

1- P(tie) is the probability that they do not tie. If they don't tie, then the symmetry of the problem (the fact that they are independently doing the same thing) implies that either one is equally likely to win. (For each way D wins, swapping the two spins gives a way PT wins.) So the probability that PT wins is (1 - P(tie))/2 = (1 - 1/50)/2 = 49/100.

As I indicated, you could find the same value in other ways; theirs is not the first way I thought of. My initial thought would be to imagine making a table of the 50^2 possible outcome of two spins, in a 50x50 array. The ties lie along the diagonal; on one side of that, PT wins, while on the other side D wins. Thus there are 50 ways to tie, 2450 ways not to tie, 1225 ways for D to win, out of 2500. And 1225/2500 = 49/100. Their way is quicker but equivalent.

The long way is to look at it the way you describe: P(D wins) = P(PT rolls 1, D rolls >1) + P(PT rolls 2, D rolls >2) + ... + P(PT rolls 49, D rolls 50) = 49/2500 + 48/2500 + ... + 1/2500 = (1+2+...+49)/2500 = (49*50/2)/2500 = 49/100. This is a longer path to the same answer.
 
Like Dr. Peterson, I'd do this problem in a different way.

How many ways can 2 numbers each drawn randomly from the set of positive integers 1 through 50.

[MATH]50^2 = 2500.[/MATH]
How many of those represent ties? Obviously 50.

How many of those represent immediate wins for D? Obviously

[MATH](2500 - 50) \div 2 = 1225.[/MATH]
So the probability that D wins is

[MATH]\dfrac{1225}{2500} + \dfrac{50}{2500} * \dfrac{3}{5} = \dfrac{1225 + 30}{2500} = \dfrac{1255}{2500} = \dfrac{251}{500}[/MATH]
The official solution is quite unintuitive.
 
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