added not on slots vs blocks

random/randao_analysis.ipynb

@@ -272,7 +272,9 @@
   {
    "cell_type": "code",
    "execution_count": 12,
-   "metadata": {},
+   "metadata": {
+    "scrolled": true
+   },
    "outputs": [
     {
      "data": {
@@ -289,11 +291,63 @@
     "# Or 6 epochs?\n",
     "expectedoptions(tm*tm*tm*vector({16:1,63:0}))"
    ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## What if randomness were sampled at a block number not a slot number?\n",
+    "\n",
+    "As I know how to query the block number and not the slot number from the EVM, this makes smart contracts getting randomness from RanDAO even trickier. So one might have say entrants to some lottery or the inout to some interactive proof that happens at block n and want to sample RanDAO using the prevRandDAO precompile at block n+128. The problem is that thia need not occur 128 slots later. In slots where the attacker controls the block producer, even when they cannot influence the randomness directky because of a later honest slot, they can still skip producing a block and push the sampling point to a later slot. In this way they can hope to sample in the middle of a strng of adversarial slots.\n",
+    "\n",
+    "However, there is not much point them doing this until 2 epochs before, because then they don't know which slot to aim for until they know the block producers. Just how many choices of slots do they have, starting at 64 blocks before? If all block producers produce blocks, then the can be a minium of 64 slots. The maximum number of slots before 64 blocks is 64 plus the number of adversarial slots before there are 64 honest slots. If the slots assignments were random, and of course they can be biased as above, this is a sample from a negative binomial distribution with parameters r=64. p=2/3."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def negbintail(r,p,t):\n",
+    "    tail=0.0\n",
+    "    for k in range(ceil(t),2*ceil(t)+r):\n",
+    "        tail += binomial(k+r-1,k)*p^r*(1-p)^k\n",
+    "    return tail"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "7774.15709413154"
+      ]
+     },
+     "execution_count": 3,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "# The expectation of this is 33.5, but it takes almost twice that for the tail probability to become small.\n",
+    "1/negbintail(67,2/3,64)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
   }
  ],
  "metadata": {
   "kernelspec": {
-   "display_name": "SageMath 9.2",
+   "display_name": "SageMath 10.1",
    "language": "sage",
    "name": "sagemath"
   },
@@ -307,7 +361,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.8.5"
+   "version": "3.11.1"
   }
  },
  "nbformat": 4,