tag:blogger.com,1999:blog-36880087.comments2021-06-05T09:33:58.327-04:00Poker GrumpRakewellhttp://www.blogger.com/profile/15873391354585352712noreply@blogger.comBlogger10018125tag:blogger.com,1999:blog-36880087.post-24257385658876120092021-05-26T08:53:20.079-04:002021-05-26T08:53:20.079-04:00The possibility of collusion has not been mentione...The possibility of collusion has not been mentioned. I'm in middle position with AA. In response to a raise I broadcast to the table that I have AA. This alerts my friend in the cut-off to fold JJ. Is this not something that should be considered when players are okay to broadcast their actual hands mid play?Anonymoushttps://www.blogger.com/profile/13358514792371774916noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-91355850723687335582021-05-25T00:25:25.476-04:002021-05-25T00:25:25.476-04:00Thanks for posting the very interesting Scientific...Thanks for posting the very interesting Scientific American article and the (17/20) game. Boghosian is a math professor at Tufts University, so I'm certain that he and the other international heavyweights cited in the article are familiar with the combinatorics behind the game. (17/20) however is presented as a suitable model for wealth trickling up, and the creation of oligarchs. This is borne out by your simulations - lots of folks going broke and a few getting obscenely wealthy.Mr Subliminalhttps://www.blogger.com/profile/03695144984530658902noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-83147218812178640282021-05-24T12:16:48.259-04:002021-05-24T12:16:48.259-04:00See the addendum I wrote to the original post. See the addendum I wrote to the original post. Rakewellhttps://www.blogger.com/profile/15873391354585352712noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-59060239438062420792021-05-24T09:42:59.943-04:002021-05-24T09:42:59.943-04:00Have you tried running the simulation a bunch more...Have you tried running the simulation a bunch more times?Pokerdogghttps://www.blogger.com/profile/12006774484023250909noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-79436276094271124732021-05-24T01:56:18.543-04:002021-05-24T01:56:18.543-04:00This true statement, "any number of coin toss...This true statement, "any number of coin tosses greater than 1 will result in a net loss if the heads and tails are equal in number, regardless of what order the heads and tails come in", has led you astray.<br /><br />An equal number of heads and tails does not happen all of the time or even a majority of the time. In a 4-round game, only 6 of the 16 outcomes have 2 heads and 2 tails. Four outcomes have 1 head and 3 tails, and one outcome has 4 tails. These 11 outcomes all lose money, but the 5 other outcomes (3+ heads) win more total money than these 11 negatives lose.<br /><br /><br />[BTW, to make sure that Kevin Nealon and I weren't going crazy, I did create a spreadsheet before my last comment. When the averages (with no resets) came out to +1.50 for 1 round, +3.02 for 2, +4.57 for 3, and +6.14 for 4, I stopped as those are exactly +1.5^1%, +1.5^2%, +1.5^3%, and +1.5^4% as expected. I also double-checked to make sure it wasn't April 1 but was unable to confirm if aliens had taken over your mind (I do think you're not playing enough poker now though).]THETA Pokerhttps://www.blogger.com/profile/12480036640392059594noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-29038382613926986392021-05-23T20:40:08.894-04:002021-05-23T20:40:08.894-04:00Looking back, there was a typo in my post of 4:40 ...Looking back, there was a typo in my post of 4:40 PM :<br /><br />"When you combine HH, HT, TH, and TT, you are in essence dealing with 4 coin tosses."<br /><br />All references to 4 coin tosses should be to <b>8</b> coin tosses.Mr Subliminalhttps://www.blogger.com/profile/03695144984530658902noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-37042578096995083482021-05-23T20:24:17.031-04:002021-05-23T20:24:17.031-04:00"When all the money is left on the table with...<b>"When all the money is left on the table without resetting, any number of coin tosses greater than 1 will result in a net loss if the heads and tails are equal in number, regardless of what order the heads and tails come in.<br /><br />I assert that that is a true and provable statement. Do you agree?"</b><br /><br />Yes. That's why I suggested looking at 3 tosses.Mr Subliminalhttps://www.blogger.com/profile/03695144984530658902noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-74029656713331764042021-05-23T20:12:46.244-04:002021-05-23T20:12:46.244-04:00"No, it's not 8 tosses. It's the 4 po..."No, it's not 8 tosses. It's the 4 possible, equiprobable results of 2 tosses, and #3 is just the expected value of those 2 tosses." <br /><br />I understand your point. Your point is that playing a two-toss game has a positive EV, because the sum of the wins and losses resulting from the four possible H/T two-toss combinations, all of which are equally probable, is a net win. <br /><br />I'll say this: There's something to what you're saying. I've played around with some numbers for the last hour, and got some results I wasn't expecting. I'll write more in the morning about what I've found. Rakewellhttps://www.blogger.com/profile/15873391354585352712noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-48823237242843223192021-05-23T19:29:46.890-04:002021-05-23T19:29:46.890-04:00"And mixing up the order of a symmetrical H a..."And mixing up the order of a symmetrical H and T count (HT or HHTT or HHTHHTTT, where number of H = number of T) will always result in a loss. No argument there." <br /><br />I'm not quite sure what you're saying here. Let me rephrase it and see if you agree: <br /><br />When all the money is left on the table without resetting, any number of coin tosses greater than 1 will result in a net loss if the heads and tails are equal in number, regardless of what order the heads and tails come in. <br /><br />I assert that that is a true and provable statement. Do you agree? Rakewellhttps://www.blogger.com/profile/15873391354585352712noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-45087554866116005352021-05-23T18:55:27.001-04:002021-05-23T18:55:27.001-04:00"Let's take the order of 8 tosses as HHHT...<b>"Let's take the order of 8 tosses as HHHTTHTT, to match what you wrote here."</b><br /><br />No, it's not 8 tosses. It's the 4 possible, equiprobable results of 2 tosses, and #3 is just the expected value of those 2 tosses.Mr Subliminalhttps://www.blogger.com/profile/03695144984530658902noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-11217698331718208812021-05-23T18:18:37.840-04:002021-05-23T18:18:37.840-04:00Firstly my 3 sentences above were about 2 coin tos...Firstly my 3 sentences above were about 2 coin tosses, so let's stick to that. I don't know why you bring up 8 coin tosses.<br /><br />And mixing up the order of a symmetrical H and T count (HT or HHTT or HHTHHTTT, where number of H = number of T) will always result in a loss. No argument there.<br /><br />Here's an idea - how about looking at 3 coin tosses, an uneven number.Mr Subliminalhttps://www.blogger.com/profile/03695144984530658902noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-6284286411451335722021-05-23T18:07:15.532-04:002021-05-23T18:07:15.532-04:00The problem is with #3, because it is not arithmet...The problem is with #3, because it is not arithmetically equivalent to running 8 coin tosses in a row while leaving the money on the table the whole time. <br /><br />Let's take the order of 8 tosses as HHHTTHTT, to match what you wrote here. Starting with $100, what you'll have left after that sequence of tosses is $98.41. That's trivially easy to check. Do you get a different result? <br /><br />Assuming you get the same result that I do, then mix up the order of the Hs and Ts. I believe you'll find that it makes no difference. Agreed? <br />Rakewellhttps://www.blogger.com/profile/15873391354585352712noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-62306060981330586472021-05-23T17:14:26.453-04:002021-05-23T17:14:26.453-04:00"As long as they are evenly divided into head...<b>"As long as they are evenly divided into heads and tails, they can be in any order you like."</b><br /><br />Unfortunately this isn't the case in real life.<br /><br />Before I reluctantly spend time on a simulation on this beautiful Sunday afternoon, let me know the first sentence you disagree with below, and why :<br /><br />1) For 2 coin tosses with initial bet of $100, HH = +$44, HT = -$0.40, TH = -$0.40, TT = -$31.11<br /><br />2) For 2 coin tosses, the probability of HH = HT = TH = TT = 0.25, and there are no other possible permutations.<br /><br />3) The expectation of 1) above therefore is $11 - $0.10 - $0.10 - $7.7775 = $3.0225.Mr Subliminalhttps://www.blogger.com/profile/03695144984530658902noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-69035329251640745512021-05-23T17:02:05.510-04:002021-05-23T17:02:05.510-04:00BTW, I just did exactly that trial, with randomiza...BTW, I just did exactly that trial, with randomization. Incredibly, about halfway through the 1000 simulated coin tosses, I was up to $53,000, because of several heads-heavy runs. But the law of large numbers caught up to me. It finally dipped back below $100 on toss #883, and after 1000, I was down to $3.09. <br />Rakewellhttps://www.blogger.com/profile/15873391354585352712noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-69404506516953553282021-05-23T16:44:59.343-04:002021-05-23T16:44:59.343-04:00OK, so run a spreadsheet simulation of the game th...OK, so run a spreadsheet simulation of the game that isn't strictly 50/50, but randomly chooses H or T for each toss. Let it run 1000 trials or more so that the outcome isn't overly influenced by short-term lopsidedness in luck. What is the result? <br /><br />Rakewellhttps://www.blogger.com/profile/15873391354585352712noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-80602538650764556052021-05-23T16:40:56.974-04:002021-05-23T16:40:56.974-04:00I agree that "if you run through HH, HT, TH, ...I agree that "if you run through HH, HT, TH, and TT once each--consecutively (in any order you like), leaving the money on the table every time ... you'll find that it's $98.41--a loss, not a win". This only confirms the commutative property of multiplication. However, I believe you are making the same error here as you did in the 2 coin toss scenario - ignoring all the other permutations.<br /><br />When you combine HH, HT, TH, and TT, you are in essence dealing with 4 coin tosses. So running through only HH, HT, TH, and TT in any order will lead to a loss, just as running through only HT and TH in the 2 coin toss scenario leads to a loss. However in 4 coin tosses, we don't always have the perfect symmetry of 4 heads and 4 tails. Sometimes we'll have 5 heads and 3 tails for a profit of $42.28, which eclipses the $31.93 loss of the opposite 3 heads and 5 tails. Not to mention the other permutations.<br /><br /> Mr Subliminalhttps://www.blogger.com/profile/03695144984530658902noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-37015626973348392262021-05-23T16:35:56.502-04:002021-05-23T16:35:56.502-04:00For now, let's set aside what you say your the...For now, let's set aside what you say your theoretical expectations should be. Make the rubber hit the road. Use a spreadsheet, or grind out the math with a calculator one toss at a time, and find out the net win or loss from whatever number of tosses you choose (10, 50, 100, 1000). As long as they are evenly divided into heads and tails, they can be in any order you like. Assuming that you're leaving all the money on the table for the entire run, I am very confident that you'll find a net loss at the end. <br /><br />Go actually do this. I'm obviously not going to convince you on the basis of theory, so go run the numbers and see what happens. If you find a net win, tell me how you got there. If you find a net loss, I will accept your groveling apology with humility and grace. :-) Rakewellhttps://www.blogger.com/profile/15873391354585352712noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-91080464795279911462021-05-23T15:41:47.886-04:002021-05-23T15:41:47.886-04:00I'm sure the light bulb will turn on for you a...I'm sure the light bulb will turn on for you any moment, but until then, I will happily play this game with you for any amount of money for as long as you like.<br /><br />The averages are absolutely what you want to know long-term. For example, if I play one round, I expect to make $1.50. If I play two rounds (without resetting), I expect to make $3.02. Three rounds get me $4.57 and four $6.14.<br /><br />It's probably not intuitive that HHTT and TTHH are both losers, or that 11 of the 16 four-round outcomes are negative, but the remaining 5 outcomes will fill my pocket (HHHH = +107.36, HHHT/HHTH/HTHH/THHH = +43.42).THETA Pokerhttps://www.blogger.com/profile/12480036640392059594noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-17079577423305355982021-05-23T12:27:39.262-04:002021-05-23T12:27:39.262-04:00Blogger FTW!!Blogger FTW!!BloodyPhttp://bloodyp.blogsot.comnoreply@blogger.comtag:blogger.com,1999:blog-36880087.post-32681790838798592822021-05-23T11:54:38.846-04:002021-05-23T11:54:38.846-04:00I used Excel to lengthen out the calculation. Star...I used Excel to lengthen out the calculation. Starting with $100, I made it win 50 times in a row (each new cell being 1.2 x the previous cell), then lose 50 times in a row (each new cell being 0.83 x the previous cell. Net result is that $100 turns into $81.84. I tried the reverse just to be sure (50 losses followed by 50 wins), and it's the same result. <br /><br />If you take the four possibilities for the first two tosses and arrange them consecutively (HH, HT, TH, TT) in any order, the net result after 8 tosses (four heads and four tails) is $98.41, a loss of $1.59. If you repeat this pattern more times, you just get more net loss. <br /><br />The problem with your math is in taking an average of the four. That's not how the game is played. The effect of how you're calculating it is almost (but not quite) the same as how I pointed out that you could manipulate the betting so that you're starting with $100 for each coin toss. If I employ the strategy I described over the course of eight tosses, four of which are heads and four of which are tails (in any order), I will have four wins of $20 each, for +$80, and four losses of $17 each, for -$68. That's a net gain of $12, which translates to an average of $1.50 win per toss. <br /><br />The difference in your math is that it's like you're playing the game twice, then taking the money off the table and starting over, repeated four time. After your HH, you have $144. You take the profit off the table and start again. You get HT, for a loss of $0.40. You add back money from your pocket to reset the bet to $100, play again, and get TH. Another loss of $.40. You reset to $100 again, and play two more tosses with TT, losing $31.11. Your total wins are $44. Your total losses are $31.91. Your net gain is $12.09 over 8 tosses. With the "reset after each bet" strategy, you will have a net gain of $1.50 per toss over the long run. They're very close to the same. <br /><br />Go ahead--do the arithmetic on what happens to your initial $100 if you run through HH, HT, TH, and TT once each--consecutively (in any order you like), leaving the money on the table every time. What do you end up with? I think you'll find that it's $98.41--a loss, not a win. <br /><br />Agree? Rakewellhttps://www.blogger.com/profile/15873391354585352712noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-77273928236493448132021-05-22T20:18:14.654-04:002021-05-22T20:18:14.654-04:00I don't consider it to be a paradox at all. It...I don't consider it to be a paradox at all. It's just counterintuitive that HT and TH both lose. But the game is obviously and definitely +EV long-term because HH makes up for HT/TH/TT.<br /><br />HH = +44.00<br />HT = -0.40<br />TH = -0.40<br />TT = -31.11<br /><br />Average = +3.0225THETA Pokerhttps://www.blogger.com/profile/12480036640392059594noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-76260559844768256212021-05-22T19:52:05.884-04:002021-05-22T19:52:05.884-04:00Unless I'm missing something, I don't see ...Unless I'm missing something, I don't see any paradox here. The game has a positive EV of 1.5% per coin toss, for each and every toss. I would be happy to sit down and play it as long as I'm allowed to.<br /><br />You mention the Win Loss (WL) and Loss Win (LW) scenarios, but forget the equally probable WW and LL outcomes when tossing a coin twice. <br /><br />For a $100 initial bet :<br /> WW ends with $44 profit<br /> WL ends with $0.40 loss<br /> LW ends with $0.40 loss<br /> LL ends with $31.11 loss<br /><br />for a total of $12.09, so expected gain after 2 tosses is $12.09/4, or $3.0225.<br /><br />This corresponds to $100 * 1.015 * 1.015 = $103.0225.Mr Subliminalhttps://www.blogger.com/profile/03695144984530658902noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-12186105100869602662021-05-01T14:31:03.241-04:002021-05-01T14:31:03.241-04:00There is another possibility. The odds is 100% i...There is another possibility. The odds is 100% if Troy had slipped in a cold deck somehow. Pokerdogghttps://www.blogger.com/profile/12006774484023250909noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-19095378707379154112021-05-01T14:29:18.805-04:002021-05-01T14:29:18.805-04:00You don't factor that in, because the odds do ...You don't factor that in, because the odds do not depend on what people wish for.Pokerdogghttps://www.blogger.com/profile/12006774484023250909noreply@blogger.comtag:blogger.com,1999:blog-36880087.post-86923055843135228782021-04-27T16:18:14.375-04:002021-04-27T16:18:14.375-04:00Interesting analysis.
Now, how do you factor in ...Interesting analysis. <br />Now, how do you factor in the fact that 90% of all poker players would have called the same flop cards, based on the two hands.<br />LOL.riddlernoreply@blogger.com