WWTBAM Bored

Two players go on a hot new game show called “Higher Number Wins.” The two go into separate booths, and each presses a button, and a random number between zero and one appears on a screen. (At this point, neither knows the other’s number, but they do know the numbers are chosen from a standard uniform distribution.) They can choose to keep that first number, or to press the button again to discard the first number and get a second random number, which they must keep. Then, they come out of their booths and see the final number for each player on the wall. The lavish grand prize — a case full of gold bullion — is awarded to the player who kept the higher number. Which number is the optimal cutoff for players to discard their first number and choose another? Put another way, within which range should they choose to keep the first number, and within which range should they reject it and try their luck with a second number?

silverscreenselect wrote:They've started a weekly contest on Friday at http://www.fivethirtyeight.com in which they have a puzzle that requires math or logic to solve. They will post the answer the next Friday. You don't get any prize for winning other than a shout out on their website if you are the "big" winner, but this week's puzzle at least is challenging.

Two players go on a hot new game show called “Higher Number Wins.” The two go into separate booths, and each presses a button, and a random number between zero and one appears on a screen. (At this point, neither knows the other’s number, but they do know the numbers are chosen from a standard uniform distribution.) They can choose to keep that first number, or to press the button again to discard the first number and get a second random number, which they must keep. Then, they come out of their booths and see the final number for each player on the wall. The lavish grand prize — a case full of gold bullion — is awarded to the player who kept the higher number. Which number is the optimal cutoff for players to discard their first number and choose another? Put another way, within which range should they choose to keep the first number, and within which range should they reject it and try their luck with a second number?

Intuitively, you would think the answer should be .5 because at that point your odds are 50/50 of improving or not. But if your opponent adopts that strategy, he's going to wind up with a number higher than .5 some 75% of the time, so if you stay on .5, you rate to lose. So your answer should be some number higher than .5, but I'm not sure what it is right now.

Kind of sounds like how "Card Sharks" worked, except for the knowing what the opposition has.


Here's the link to the puzzle: http://fivethirtyeight.com/features/can ... game-show/

silverscreenselect wrote: Intuitively, you would think the answer should be .5 because at that point your odds are 50/50 of improving or not. But if your opponent adopts that strategy, he's going to wind up with a number higher than .5 some 75% of the time, so if you stay on .5, you rate to lose.

silverscreenselect wrote:So your answer should be some number higher than .5, but I'm not sure what it is right now.

jarnon wrote:I submitted this answer:

Spoiler

(√5 - 1)/2 = 0.618, the Golden Ratio

.

plasticene wrote:

jarnon wrote:I submitted this answer:

Spoiler

(√5 - 1)/2 = 0.618, the Golden Ratio

.

I believe the answer is

Spoiler

√5 - 1

. If you use that as your threshold, there is no way for your opponent to gain an advantage by choosing a different threshold. If I use my answer and you use your answer, I'll win about 52.5% of the time.

I used a little bit of brute force to come up with that. I won't submit it unless I can figure out how to derive it from first principles.

plasticene wrote:

jarnon wrote:I submitted this answer:

Spoiler

(√5 - 1)/2 = 0.618, the Golden Ratio

.

I believe the answer is

Spoiler

√5 - 1

. If you use that as your threshold, there is no way for your opponent to gain an advantage by choosing a different threshold. If I use my answer and you use your answer, I'll win about 52.5% of the time.

I used a little bit of brute force to come up with that. I won't submit it unless I can figure out how to derive it from first principles.

smilergrogan wrote:

plasticene wrote:

jarnon wrote:I submitted this answer:

Spoiler

(√5 - 1)/2 = 0.618, the Golden Ratio

.

I believe the answer is

Spoiler

√5 - 1

. If you use that as your threshold, there is no way for your opponent to gain an advantage by choosing a different threshold. If I use my answer and you use your answer, I'll win about 52.5% of the time.

I used a little bit of brute force to come up with that. I won't submit it unless I can figure out how to derive it from first principles.

Huh? You must be understanding the question in a different way than I am.

If I choose 0.5 as the cutoff, then half the time the first number will be higher and on average that number will be 0.75 (evenly distributed between 0.5 and 1). The other half of the time it will be less than 0.5 so I will get a 2nd number which will be on average 0.5 (evenly distributed between 0 and 1). So my expectation value is 50% of 0.75 + 50% of 0.5 = 0.625

If I choose 0.618 as the cutoff, then 38.2% of the time the first number will be higher and on average that number will be 0.809 (evenly distributed between 0.618 and 1). The other 61.8% of the time it will be less than 0.618 so I will get a 2nd number which will be on average 0.5 (evenly distributed between 0 and 1). So my expectation value in that case is 38.2% of 0.809 + 61.8% of 0.5 = 0.618. How does that beat the first option?

silverscreenselect wrote:They've started a weekly contest on Friday at http://www.fivethirtyeight.com in which they have a puzzle that requires math or logic to solve. They will post the answer the next Friday. You don't get any prize for winning other than a shout out on their website if you are the "big" winner, but this week's puzzle at least is challenging.

Two players go on a hot new game show called “Higher Number Wins.” The two go into separate booths, and each presses a button, and a random number between zero and one appears on a screen. (At this point, neither knows the other’s number, but they do know the numbers are chosen from a standard uniform distribution.) They can choose to keep that first number, or to press the button again to discard the first number and get a second random number, which they must keep. Then, they come out of their booths and see the final number for each player on the wall. The lavish grand prize — a case full of gold bullion — is awarded to the player who kept the higher number. Which number is the optimal cutoff for players to discard their first number and choose another? Put another way, within which range should they choose to keep the first number, and within which range should they reject it and try their luck with a second number?

Intuitively, you would think the answer should be .5 because at that point your odds are 50/50 of improving or not. But if your opponent adopts that strategy, he's going to wind up with a number higher than .5 some 75% of the time, so if you stay on .5, you rate to lose. So your answer should be some number higher than .5, but I'm not sure what it is right now.

Here's the link to the puzzle: http://fivethirtyeight.com/features/can ... game-show/

Bob78164 wrote:I don't have time to work through the analysis right now, but I think the right approach is to assume that you choose x, fixed, as your cutoff and then let your opponent choose y, variable, as hers. You want to choose x that maximizes your minimum (which will occur when your opponent also chooses x). I believe Nash's Theorem still applies in this setting, in which case that choice of x is the best you can do.

You then have 4 possibilities to consider -- you and your opponent each do or do not end up with a first number below the chosen cutoff. In each case, with a little care you can figure out the likelihood that you win. The values of x and y tell you how likely you are to fall into each of the four cases, and your likelihood of winning then falls out as a function of x and y.

You probably have to do the analysis twice, once assuming x < y, and once assuming x > y. --Bob

Bob78164 wrote:

Bob78164 wrote:I don't have time to work through the analysis right now, but I think the right approach is to assume that you choose x, fixed, as your cutoff and then let your opponent choose y, variable, as hers. You want to choose x that maximizes your minimum (which will occur when your opponent also chooses x). I believe Nash's Theorem still applies in this setting, in which case that choice of x is the best you can do.

You then have 4 possibilities to consider -- you and your opponent each do or do not end up with a first number below the chosen cutoff. In each case, with a little care you can figure out the likelihood that you win. The values of x and y tell you how likely you are to fall into each of the four cases, and your likelihood of winning then falls out as a function of x and y.

You probably have to do the analysis twice, once assuming x < y, and once assuming x > y. --Bob

To take this a step farther, if you and your opponent are both above your cutoffs on the first draw and x < y, then you will certainly lose with probability y - x (because your number is below your opponent's cutoff) and you have 50-50 odds of winning the remainder of the time you're in this subcase, so your odds of winning this subcase are (1 + x - y)/2. You'll be in this subcase with probability (1 - x)(1 - y).

If you and your opponent both end up below your respective cutoffs (so that you both redraw), your odds of winning are 50-50. You'll be in this subcase with probability xy. The remaining two subcases I leave as an exercise for the reader. --Bob

silverscreenselect wrote:The answer actually is .5 (I think), and I did the calculations.

For any value Y, your expected result is:

(.5)(Y) + ((1+Y)/2) x (1-Y)

Do all the calculations and you get (2 + Y - Y*2)/2. To get the maximum value, you take the derivative, which is -Y + .5 and set it to zero, which makes Y = .5

Bob78164 wrote: But as I noted above, maximizing the expectation value does not necessarily solve the problem. You really need to walk through the process outlined in my posts above. --Bob

silverscreenselect wrote:

Bob78164 wrote: But as I noted above, maximizing the expectation value does not necessarily solve the problem. You really need to walk through the process outlined in my posts above. --Bob

Yes it does, because you don't know what your opponent has or what his strategy is. He doesn't start with the same number you do. His number might be higher or lower. He might stay or spin again. But you have no way of knowing. If you adopt a strategy with an expected value of .625, then the worst you can do is have a 50/50 chance of winning the game. If your opponent adopts any other strategy than that, he's going to have a lower expected value and you will have a greater than 50% chance of winning.

Bob78164 wrote:If my calculations are correct, for any fixed value of x, the best possible choice of y is 1/(1 + x^2). --Bob

Bob78164 wrote:

plasticene wrote:

jarnon wrote:I submitted this answer:

Spoiler

(√5 - 1)/2 = 0.618, the Golden Ratio

.

I believe the answer is

Spoiler

√5 - 1

. If you use that as your threshold, there is no way for your opponent to gain an advantage by choosing a different threshold. If I use my answer and you use your answer, I'll win about 52.5% of the time.

I used a little bit of brute force to come up with that. I won't submit it unless I can figure out how to derive it from first principles.

Isn't your answer larger than 1? --Bob

silverscreenselect wrote:

Bob78164 wrote: But as I noted above, maximizing the expectation value does not necessarily solve the problem. You really need to walk through the process outlined in my posts above. --Bob

Yes it does, because you don't know what your opponent has or what his strategy is. He doesn't start with the same number you do. His number might be higher or lower. He might stay or spin again. But you have no way of knowing. If you adopt a strategy with an expected value of .625, then the worst you can do is have a 50/50 chance of winning the game. If your opponent adopts any other strategy than that, he's going to have a lower expected value and you will have a greater than 50% chance of winning.

Bob78164 wrote:

Bob78164 wrote:I don't have time to work through the analysis right now, but I think the right approach is to assume that you choose x, fixed, as your cutoff and then let your opponent choose y, variable, as hers. You want to choose x that maximizes your minimum (which will occur when your opponent also chooses x). I believe Nash's Theorem still applies in this setting, in which case that choice of x is the best you can do.

You then have 4 possibilities to consider -- you and your opponent each do or do not end up with a first number below the chosen cutoff. In each case, with a little care you can figure out the likelihood that you win. The values of x and y tell you how likely you are to fall into each of the four cases, and your likelihood of winning then falls out as a function of x and y.

You probably have to do the analysis twice, once assuming x < y, and once assuming x > y. --Bob

To take this a step farther, if you and your opponent are both above your cutoffs on the first draw and x < y, then you will certainly lose with probability y - x (because your number is below your opponent's cutoff) and you have 50-50 odds of winning the remainder of the time you're in this subcase, so your odds of winning this subcase are (1 + x - y)/2. You'll be in this subcase with probability (1 - x)(1 - y).

If you and your opponent both end up below your respective cutoffs (so that you both redraw), your odds of winning are 50-50. You'll be in this subcase with probability xy. The remaining two subcases I leave as an exercise for the reader. --Bob

plasticene wrote:I'm starting to think I messed up my probability formula too. Maybe jarnon's answer was right after all! I've got more work to do.