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Consider an array of $L$ binary values all initially set as False. I want to flip them all to True by doing a $T$-block flip, where $T$ consecutive values are flipped to be True (already True values stay True).

What is the expected number of $T$-block flips until the entire array is True (assuming the position of each $T$-block flip is chosen randomly with equal probability for each site)?

Example: $L = 10$, $T = 4$: $FFFFFFFFFF$. Flip $1$ at position $3$: $FFTTTTFFFF$. Flip $2$ at position $5$: $FFTTTTTTFF$, and so on.

TShiong
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cjdjpj
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  • Do flips wrap? For instance, in your example is there a flip at position 9, and if so does it affect positions 9, 10, 1, 2 or just positions 9, 10? – paw88789 Jan 24 '25 at 23:27
  • No they don't. In my example, if a flip occurs at position 10, the other three flips are wasted. Although if it does wrap and be therefore circular, might be related to this?: https://math.stackexchange.com/questions/325141/probability-that-n-points-on-a-circle-are-in-one-semicircle. I'm also mainly interested for when L >> T, but a general solution would be nice too. – cjdjpj Jan 24 '25 at 23:30
  • @QiaochuYuan The expected time is much less than $2^L$ here because you never flip to False. When $T=1$ the expectation is a coupon collector's $L,H_{L}^{,}\approx L\log_e(L)+L\gamma+\frac12$, and falls as $T$ increases; the expectation is $L$ when $T=L$. – Henry Jan 25 '25 at 00:21
  • @QiaochuYuan When $T=L$ all you need to do and must do is flip the first value, which seems to have probability $\frac1L$ and so you have a geometric distribution. – Henry Jan 25 '25 at 00:26
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    For what it is worth, using simulations for $L=10$ and $T=1,2,\ldots, 10$, I get expectations of about (last digit not guaranteed) $29.3$, $16.0$, $12.6$, $11.3$, $10.7$, $10.4$, $10.3$, $10.2$, $10.1$, $10.0$. The first should be $29.28968\ldots$ and the last should be $10$ exactly, which is encouraging for the others. – Henry Jan 25 '25 at 01:13
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    Using simulations for larger $L=100$ and again $T=1,2,\ldots, 10$, I get expectations of about (last digit again not guaranteed) $519$, $264$, $185$, $152$, $134$, $124$, $117$, $113$, $111$, $108$, which is falling fast towards the expectation of $100$ only achieved when $T=100$. – Henry Jan 25 '25 at 01:33

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