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ThatDonGuy
ThatDonGuy
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December 8th, 2021 at 8:02:35 AM permalink
Quote: Gialmere

...what color would your Lamborghini be?

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I grew up next door to someone who owned one. It kept breaking down.

Actually, I would take the other option - because this is a genie we are talking about, and if I did take the car, my choice would be the lead story on pretty much every 24/7 news source immediately, and the front page of every newspaper in the country the next day.
Then again, for all I know, the genie would "end world hunger" by making half of the population disappear - the female half. "Hey, at least you won't have to worry about the next generation starving!"
Gialmere
Gialmere
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December 8th, 2021 at 8:34:44 AM permalink
Here's a Thanksgiving leftover from Riddler....



On the Food Network’s latest game show, Cranberries or Bust, you have a choice between two doors: A and B. One door has a lifetime supply of cranberry sauce behind it, while the other door has absolutely nothing behind it. And boy, do you love cranberry sauce.

Of course, there’s a twist. The host presents you with a coin with two sides, marked A and B, which correspond to each door. The host tells you that the coin is weighted in favor of the cranberry door — without telling you which door that is — and that door’s letter will turn up 60 percent of the time. For example, if the sauce is behind door A, then the coin will turn up A 60 percent of the time and B the remaining 40 percent of the time.

You can flip the coin twice, after which you must make your selection. Assuming you optimize your strategy, what are your chances of choosing the door with the cranberry sauce?

Extra credit: Instead of two flips, what if you are allowed three or four flips? Now what are your chances of choosing the door with the cranberry sauce?
Have you tried 22 tonight? I said 22.
gordonm888
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gordonm888
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December 8th, 2021 at 11:18:00 AM permalink
Cranberries or bust



I would flip the coin once and note the result, and then flip it again. If I get the same result on both flips, I assume that that side is the cranberry door. If I get differing results for the two flips, I will pick the door I got on the first flip,

Let's say that A is 0.60 and B is 0.4.

The frequency of outcomes is:

AA: 0.36
AB: 0.24
BA: 0.24
BB: 0.16

So, when the result is the same for both flips, I will be correct with a frequency of 36/52 or about 69.2307692% When i get different outcomes for the two flips, I claim I should be correct 60% of the time using my strategy.

My total chance of being correct is 0.52*(0.36/0.52)+0.48*0.6 = 0.648 or 64.8%
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
ThatDonGuy
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December 8th, 2021 at 11:18:58 AM permalink

For the 2-flip problem, I'll go with "the obvious answer": use just the first flip. You have a 60% chance of success.

I am half-expecting the "QI Klaxon" at this point, for those of you who know what that means.

unJon
unJon
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December 8th, 2021 at 11:20:56 AM permalink
Quote: gordonm888

Cranberries or bust



I would flip the coin once and note the result, and then flip it again. If I get the same result on both flips, I assume that that side is the cranberry door. If I get differing results for the two flips, I will pick the door I got on the first flip,

Let's say that A is 0.60 and B is 0.4.

The frequency of outcomes is:

AA: 0.36
AB: 0.24
BA: 0.24
BB: 0.16

So, when the result is the same for both flips, I will be correct with a frequency of 36/52 or about 69.2307692% When i get different outcomes for the two flips, I claim I should be correct 60% of the time using my strategy.

My total chance of being correct is 0.52*.692307692+0.48*0.6 = 0.648 or 64.8%

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I don’t follow:
When it comes AB or BA, you should be right 50% not 60% of the time.

ETA: isn’t your strategy functionally equivalent to flipping the coin once and going with that door? It sounds like you will always pick the first coin flip door no matter what the second coin flip is. So you should be right 60% of the time with your strategy.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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Wizard
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December 8th, 2021 at 11:32:03 AM permalink

I get 60%. I show this to be true for any weighting of the correct side of the coin.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
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Wizard
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Thanks for this post from:
Gialmere
December 8th, 2021 at 11:38:31 AM permalink

1 flip = 60.000%
2 flip = 60.000%
3 flip = 64.800%
4 flip = 64.800%
5 flip = 68.256%
It's not whether you win or lose; it's whether or not you had a good bet.
charliepatrick
charliepatrick
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December 8th, 2021 at 11:41:40 AM permalink
With two possible flips you have no reliable information after the first flip; if you stick there's a 60% of being correct and if you twist (flip!) then there's still a 60% of being correct. Thus if you've only got two flips to go you might as well flip once and take that decision.

With three flips you have more information if the first two are both the same. If so, there's a 16% chance of it being wrong and 36% chance of being correct, so you stick.
The remaining 48%, one of each, at this stage there's only a 50% chance of the second flip being correct (as it's equally likely you got XY or YX), so you flip again. The third flip has a 60% of being correct and so adds 60%*48%=28.8% chance of winning. Total chance = 36% + 28.8% = 66.8%.

With four flips use the above logic but if you need a third flip then this is equivalent to starting with two flips - so it doesn't add any help having a fourth flip available. Also say you're first two flips were the same; if you continued the only time it would give a better result would be it was X X Y Y where X was wrong (5.76%), this is offset by creating the loss for YYXX. Other cases XXX XXYX YYY YYXY leave you on the same decision. Thus it's just as easy to use the 3-flip logic.

With 5-flips it gets to 68.256% if you always flip 3-times (slightly better 67.104% than stopping if the first two are the same) Basically if you see the third occurrence of the same side you stand. I wonder if this is the best strategy for large numbers, wait for any side to exceed half the total number of flips available.
charliepatrick
charliepatrick
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December 8th, 2021 at 11:54:17 AM permalink
It's obvious that with an odd number of flips you pick the majority, so you can always stop flipping when one side gets past the halfway post; hence my previous answer would apply.
With an even number you would try and pick the majority so will always stick on 3 out of 4. What is interesting if if the first two are the same then you are better off to stick since the only occasion that can change your mind is if the rest are opposite (in which case it's 50/50, so you might as well pick the first two that came out). So in practice you might as well stick as soon as you see 2 the same.
Thus it doesn't matter if you finish all the flips or if you stop flipping once you see the current leader cannot be beaten. This means you can always stop flipping and make that flip the side you are choosing.
Gialmere
Gialmere
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December 8th, 2021 at 5:31:01 PM permalink
Quote: Wizard


1 flip = 60.000%
2 flip = 60.000%
3 flip = 64.800%
4 flip = 64.800%
5 flip = 68.256%

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Correct!!

Examining the responses, everyone obviously gets the logic but, unless I'm mistaken, only the Wizard has all the correct values in all the correct locations.

Before working through this scenario with two coins, let’s take a step back and look at one coin. Since the coin was slightly weighted in favor of the cranberry door, your best strategy was to choose whichever door was indicated by the coin. Then you’d be right — and win your delectable cranberry sauce — 60 percent of the time.

Surely, you had better odds of winning with two flips instead of one. Right?

Wrong. To see why that was, solver Rebecca Harbison looked closer at the possible results of the two flips. Instead of A and B, let’s relabel the coin C (for “Cranberry sauce is behind this door”) and D (for “Dangit, no cranberry sauce behind this door”). As stated by the problem, the coin had a 60 percent chance of landing on C and a 40 percent chance of landing on D.

With two flips, there were four possible outcomes:

The first flip was C and the second flip was also C, which occurred with probability (0.6)(0.6), or 0.36.
The first flip was C and the second flip was D, which occurred with probability (0.6)(0.4), or 0.24.
The first flip was D and the second flip was C, which occurred with probability (0.4)(0.6), or 0.24.
The first flip was D and the second flip was D, which occurred with probability (0.4)(0.4), or 0.16.

Putting these together, there was a 48 percent chance that the two flips were different. When this happened, you had absolutely no information how the coin was weighted, since the coin came up either side an equal number of times. In other words, you had to guess, which meant you’d be right half the time.

The other 52 percent of the time the two flips were the same. The majority of the time (i.e., with probability 36/52, or 9/13) that meant both flips were C. So when the two flips were the same, your best move was to guess the door both flips corresponded to.

So then what were your chances of winning the cranberry sauce? Well, 48 percent of the time you had a one-half chance of winning, while the other 52 percent of the time you had a 9/13 chance of winning. Numerically, these combined to (12/25)(1/2) + (13/25)(9/13), which simplified to 6/25 + 9/25, or 15/25 — in other words, 60 percent.

Shockingly (to me, at least), that second flip didn’t improve your chances one bit. You might as well have simply flipped the coin once and chosen the resulting door.

For extra credit, you looked at what happened when you were allowed three or four flips, instead of just two. For three flips, the optimal strategy was to choose whichever door was indicated by a majority of the flips. All three flips came up C with probability (0.6)3, or 0.216. Meanwhile, the probability of two Cs and one D was 3(0.6)2(0.4), or 0.432. Adding these together meant you’d win the cranberry sauce 64.8 percent of the time — an improvement over your chances with just one or two flips.

But just as two flips were no better than one, four flips were no better than three. That fourth flip either confirmed your decision based on the first three flips, or changed things so that now there were two flips for one door and two flips for another (opening up twice as many possibilities, and forcing you to take a random guess).

Solvers Jake Gacuan and Emily Boyajian both found a general formula for your chances of winning the cranberry sauce as a function of the number of flips N. Sure enough, having an even number of flips was no better than having the preceding odd number of flips. And as N increased, your chances of choosing the correct door approached 100 percent.

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Have you tried 22 tonight? I said 22.

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