brianu@ada-uts.UUCP (12/19/85)
>***** ada-uts:net.puzzle / glacier!bhayes / 10:33 pm Dec 16, 1985 >It may not make sense to talk about "choosing randomly" one of >an infinite number of possibilities. Just watch... > >Let's play a game. I have, in this large and ornate paper bag, >an infinite number of poker chips. Each poker chip has one integer >on each side, and, in fact, the two integers on any poker chip >differ by exactly one. The game is played as follows: > >1) I reach into the bag, and choose a random poker chip, without > peeking at the chip. >2) I hold it between us so that I can see one side, and you can see > the other, but neither of us sees both sides. >3) Either of us can reject the chip, just from seeing the side we see. >4) We bet some fixed amount. >5) The person with the low side wins. > >The distribution of the chips is as follows: >Chip numbers how many >1/2 1 >2/3 2 >3/4 4 >4/5 8 >and so on > >I calculate the odds as follows: >If I see "1" on my chip [a very unlikely event] I must win. >For any other number [say "4"] there are twice as many chips which >will win for me [8 "4/5" chips] as will lose [4 "3/4" chips]. Therefore, >I should always bet, and I will win 2/3 of the time. > >You of course can make the same calculations and will also win 2/3 >of the time. > > -Barry "at no time do my fingers leave my hands" Hayes > bhayes@glacier I don't see how you have proved your hypothesis, that it doesn't make sense to discuss random numbers in an infinite domain. First, the number of points in the range 0-1 is the same as the number of points in the range -oo to +oo. Secondly, lets play another little game: Since I don't have enough extra cash on hand to buy an infinite number of poker chips, let's say we stop after the chips with 9 on one side and 10 on the other. Now let's play the game. Like you said, if I see a 1 then I know I've won. If on the other hand, I see a 10 I know I've lost. Now, for all the other possibilities the same reasoning as above applies. Suppose I see a 5, the other side of the chip could have a 4 or a 6. Since there are twice as many cases where I win (16) as where I lose (8) you conclude that I have a 2/3 chance of winning. And of course my opponent reasons the same way and decides HE has a 2/3 chance of winning. Since we both have a 2/3 chance of winning, by your reasoning we must conclude that it makes no sense to discuss random numbers on a finite domain. Right. What this really means is that the reasoning is faulty. The game can be broken into two random acts: first the selection of the disk, second the orientation of the selected disk. Since each disk has a winning side and a losing side, they are all equivalent. Thus only the orientation matters. Seeing the number imparts no information about this, so the odds are just fifty/fifty, which is what you would expect. Brian Utterback Intermetrics Inc. 733 Concord Ave. Cambridge MA. 02138. (617) 661-1840 UUCP: {cca!ima,ihnp4}!inmet!ada-uts!brianu LIFE: UCLA!PCS!TELOS!CRAY!I**2
tan@ihlpg.UUCP (Bill Tanenbaum) (12/24/85)
> > >***** ada-uts:net.puzzle / glacier!bhayes / 10:33 pm Dec 16, 1985 > >It may not make sense to talk about "choosing randomly" one of > >an infinite number of possibilities. Just watch... > > > >Let's play a game. I have, in this large and ornate paper bag, > >an infinite number of poker chips. Each poker chip has one integer > >on each side, and, in fact, the two integers on any poker chip > >differ by exactly one. The game is played as follows: > > > >1) I reach into the bag, and choose a random poker chip, without > > peeking at the chip. > >2) I hold it between us so that I can see one side, and you can see > > the other, but neither of us sees both sides. > >3) Either of us can reject the chip, just from seeing the side we see. > >4) We bet some fixed amount. > >5) The person with the low side wins. > > > >The distribution of the chips is as follows: > >Chip numbers how many > >1/2 1 > >2/3 2 > >3/4 4 > >4/5 8 > >and so on > > > >I calculate the odds as follows: > >If I see "1" on my chip [a very unlikely event] I must win. > >For any other number [say "4"] there are twice as many chips which > >will win for me [8 "4/5" chips] as will lose [4 "3/4" chips]. Therefore, > >I should always bet, and I will win 2/3 of the time. > > > >You of course can make the same calculations and will also win 2/3 > >of the time. > > > > -Barry "at no time do my fingers leave my hands" Hayes > > bhayes@glacier -------- > I don't see how you have proved your hypothesis, that it doesn't make > sense to discuss random numbers in an infinite domain. First, the number > of points in the range 0-1 is the same as the number of points in the > range -oo to +oo. Secondly, lets play another little game: > Since I don't have enough extra cash on hand to buy an infinite > number of poker chips, let's say we stop after the chips with 9 on one > side and 10 on the other. Now let's play the game. Like you said, if > I see a 1 then I know I've won. If on the other hand, I see a 10 I know > I've lost. Now, for all the other possibilities the same reasoning as > above applies. Suppose I see a 5, the other side of the chip could have > a 4 or a 6. Since there are twice as many cases where I win (16) as where > I lose (8) you conclude that I have a 2/3 chance of winning. And of course > my opponent reasons the same way and decides HE has a 2/3 chance of > winning. Since we both have a 2/3 chance of winning, by your reasoning > we must conclude that it makes no sense to discuss random numbers on a > finite domain. Right. > What this really means is that the reasoning is faulty. The game can > be broken into two random acts: first the selection of the disk, second > the orientation of the selected disk. Since each disk has a winning side > and a losing side, they are all equivalent. Thus only the orientation > matters. Seeing the number imparts no information about this, so the odds > are just fifty/fifty, which is what you would expect. > Brian Utterback -------- Barry Hayes is correct (at least in his analysis of the game) and Brian Utterback is incorrect. In the infinite version, each player calculates correctly, independent of what number he sees, a probability of two thirds of winning, an obvious contradiction. In the finite game, if one player sees a five, he and his opponent are both CORRECT in calculating their winning odds at 2/3. Fifty-Fifty is wrong! There is no contradiction, as each has different information! To see this, in draw poker, if I have four aces and you have four kings, both of us would be quite correct in calculating a greater than 50% chance of winning. This is no contradiction in a game of imperfect information! Here are my odds of winning Brian's game. I see a 1. (Probability 1/1022) Win probability 1 I see a 2-9. (Probability 765/1022) Win probability 2/3 I see a 10. (Probability 256/1022) Win probability 0 ----------------------------------------------------- Total probability of winning Brian's game is 1 x (1/1022) + 2/3 x (765/1022) = 511/1022 = 1/2 as expected. No contradiction. In the infinite game, the TOTAL probability is 2/3 on both sides. A contradiction. -- Bill Tanenbaum - AT&T Bell Labs - Naperville IL ihnp4!ihlpg!tan