2025-12-16 17:13 UTC
Replies (17)
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@christianp@mathstodon.xyz 2025-12-16 17:20
@alisonkiddle@mathstodon.xyz my first guess was to notice that 1/36 is a little bit less than 3/100, so (1/6)⁴ is a bit less than 9/10000. (1/2)^10 = 10/10240, and I can't decide if that's more or less. Then I realised that 10 flips vs 4 rolls has the same relationship as 5 flips vs 2 rolls, and 1/32 > 1/36.
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@Taneb@hacksrus.xyz 2025-12-16 17:33
@alisonkiddle@mathstodon.xyz my solution: I note that those are one in 2^10 and one in 6^4 respectively, so it's whether 2^10 is greater or less than 6^4. While I do know 2^10 from memory, I don't know 6^4 and I've got a little bit of a cold and would rather not work it out. But I can divide both by 2^4, to get 2^6 and 3^4, and I know those! Is 64 greater than or less than 81? It's less than! So 2^10 is less than 6^4 and 1/2^10 is greater than 1/6^4. So flipping 10 heads in a row is more likely. Which wasn't what I expected!
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@johncarlosbaez@mathstodon.xyz 2025-12-16 19:52
@alisonkiddle@mathstodon.xyz - The chances are 1/2¹⁰ and 1/6⁴, respectively. I know 2¹⁰ = 1024, but what's 6⁴? I have a vague memory that it's 1296, but let me think. It's 36 squared. I guess just multiply it out in my mind. 6 times 6 is 36, 6 times 30 is 180, 30 times 30 is 900, 900 + 180 + 180 + 36 = 1080 + 180 + 36 = 1260 + 36 = 1296. Okay, good. So it's more likely to roll 10 heads than to roll four 6's. I'm worrying about 6³ and worked out that it's 216. Sounds very familiar. I used to know these numbers better! I like powers of numbers and feel happy knowing a bunch of them. They have a nice feeling to them.
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@joshuagrochow@mathstodon.xyz 2025-12-16 20:05
@alisonkiddle@mathstodon.xyz First is 1/2¹⁰. I know the value of 2^{10} like the back of my hand, but since I don't know the value of 6^4 and don't want to calculate it that's not very helpful to me atm. But I *do* know the value of 6^2=36. And 36 is just a little more than 32=2^5 (again, another fact I know without having to think about it). So (6^2)^2 > (2^5)^2 = 2^{10}, so when we take reciprocals we get that 1/6^4 < 1/2^{10}.
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@pozorvlak@mathstodon.xyz 2025-12-16 20:23
@alisonkiddle@mathstodon.xyz It's 1/2^10 versus 1/6^4; I know 2^10 = 1024 because it's the number of bytes in a kilobyte. 6^4 I don't know, but it's 36^2 - nope, that sounds hard. I've got 216 in my head - no, that's 180 + 36, so it's 6^3. But then 6^4 is 6*216, which is at least 1200, so 1/6^4 < 1/2^10 and we're done without having to do the full calculation.
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@rojun@mastodon.social 2025-12-16 20:25
@alisonkiddle@mathstodon.xyz Probabilities are: (½)¹⁰ vs (⅙)⁴ For concise notation, I'll invert them: 2¹⁰ vs 6⁴ We can decompose the right side: (2 * 3)⁴ = 2⁴ * 3⁴ We have 2s exp. so again can reduce both sides: 2⁶ vs 3⁴ Again decomposing: 2³ * 2³ = 8 * 8 vs 9 * 9 We see that the right side is bigger, so when we invert them, the left side is bigger i.e. 10 coin tosses for same side has higher probability: (½)¹⁰ > (⅙)⁴
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@zenben@c.im 2025-12-16 20:33
@alisonkiddle@mathstodon.xyz this was fun! I did it Baez' way. Fascinating to see the range of workings for such a small problem!
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@mra@mathstodon.xyz 2025-12-16 20:55
@alisonkiddle@mathstodon.xyz the probability of the coins is, as others have noted, just 1/2¹⁰, and the probability of the dice is 1/6⁴. now, i want to know which of 2¹⁰ and 6⁴ is larger. first, note that 6⁴ = 2⁴3⁴, so i just need to know which of 2⁶ and 3⁴ is larger. then, i observed that since √2 ≈ 1.4, this means that 2√2 < 3, but (2√2)⁴ = 2⁶, ergo 2⁶ < 3⁴, and the coin flips are likelier! this seems to be a more circuitous route than some other people have taken, but it was fun to do it in a way that doesn't require any mental multiplication or knowing a bunch of powers off the top of your head, especially since i have a terrible memory for that sort of thing
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@gbargoud@masto.nyc 2025-12-16 20:57
@alisonkiddle@mathstodon.xyz I would break it in half since I know 6^2 is 36 and 2^5 is 32 and that means that the dice rolls are less likely I could then get the actual probabilities by squaring and inverting the numbers but this is good enough for comparison.
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@rachelplusplus@tech.lgbt 2025-12-16 20:59
@alisonkiddle@mathstodon.xyz Took a different angle to many of the other answers I see - definitely influenced by having worked in data compression! Started thinking about how much entropy the two options are "worth" - more formally, what is -log2(probability) ? A heads on a regular coin flip is "worth" 1 bit of entropy, so ten heads is worth 10 bits. Meanwhile, four sixes on four dice is "worth" 4 * log2(6) bits. But how do we compare those? Well, I know that sqrt(2) is 1.4[something], and in particular is less than 1.5. So 22.5 = 4 * sqrt(2) 10. Converting back to normal people terms, four sixes being "worth" more entropy means that the outcome is less likely. So 10 heads is slightly more likely than 4 sixes.
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@skua@mastodon.social 2025-12-16 21:05
@alisonkiddle@mathstodon.xyz Two to the 10th power as the denominator. vs Six to the 4th power as the denominator. With same numerator, larger denominator is less likely. Blind to what I know already.
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@hjhornbeck@freethought.online 2025-12-16 21:13
@alisonkiddle@mathstodon.xyz I'm in CompSci, so I already knew 2^10 = 1,024. I don't know my powers of six, but I do know 2*3 = 6, so I could break that down into 2^4 * 3^4. 2^4 = 16, and some squaring got me 3^4 = 9^2 = 81. So now it was down to calculating 16*81. I struggled with breaking up 16 or 81, and eventually settled on calculating 10*81 + 6*81. The first half is 810, the second half... 480 + something? 486. At that point, I realized 480 + 810 > 1,024, so there were more potential outcomes for rolling a dice four times. Thus the odds of getting one specific outcome was less, so flipping ten heads is more likely.
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@lahosken@hachyderm.io 2025-12-16 21:58
@alisonkiddle@mathstodon.xyz 10 heads, 1 in 1024: instantaneous. I, a stereotypical computer nerd have that one ingrained in the brain four sixes, 36 * 36. I'm pretty sure that's over a thousand, we're done. Oh wait, I, a stereotypical computer nerd, have gotten sloppy in distinguishing between 1000 and 1024. Now I am confident about nothing uh, okay, let's buckle down. 3*36 is uh 108 so yeah 36*36 > 1024. Whew! OK, we're done for reals this time a few seconds pass o hey, of _course_ 32^2 < 36^2. Why did I do that arithmetic in my head, ugh
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@isaackuo@spacey.space 2025-12-16 22:03
@alisonkiddle@mathstodon.xyz My head knows 2^10 is 1024, but my head does NOT feel like doing 36*36. So, I took the square root of both sides, so I only needed to calculate 32 vs 6*6. (My brain knows powers of 2 for ... reasons.)
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@Martin_Whitworth@mathstodon.xyz 2025-12-16 22:48
@alisonkiddle@mathstodon.xyz I compared 1/(2^10) with 1/(6^4). I know 2^10 by heart. 6^4 = 36^2, which I also know by heart. I subsequently realised that 2^10 = (2^5)^2 = 32^2, so I could have compared 32^2 with 36^2 without having to evaluate the latter.
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@notsoloud@expressional.social 2025-12-17 06:26
@alisonkiddle@mathstodon.xyz By head, in about a minute: 2^10=1024 (knew already) 6^4 = 36*36 which is...err.. 30*30+6*6+2*30*6 so 936+2*180 and it's already clear that it's more than 1024 so coin flip is more likely.
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@anshthewad@mathstodon.xyz 2025-12-17 13:32
@alisonkiddle@mathstodon.xyz The first has 1 in 2^10 chance of happening, and the second has 1 in 6^4 chance of happening. I want to compare them, so I can square root both sides, now I'm comparing 2^5 and 6^2. But I already know that 2^5 = 32 < 6^2 = 36, which means that the first is more likely.