How High Would You Go

[CrazyJoe 0]

i know it would suck super hard nuts to roll a 6 on the first roll.

[RodReynolds 87]

even at x=478 (477 dbls)EV = 6,635,193,109,306,990,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 lol.that's nonsenseI"m missing sometthing

I don't think you are. Perform the experiment n = (5/6)^478 times. You expect one success. Divide your earnings of 2^478 by n. You get your number. (I think, I didn't actually do the math... I'm just reassuring your troubled mind.)Discussions with you really pad my post count. Thanks for that.Edit: n=1/(5/6)^478 = (6/5)^478

[Actuary 3]

chime in dorks:13 Members: kkcountry, joshrogo, Vyszion, RodReynolds, FatBurger, mase_gotsem, davezz5, madams01, TheCinciKid, Viking145, bursell, pragtyro

[davezz5 0]

I don't think you are. Perform the experiment n = (5/6)^478 times. You expect one success. Divide your earnings of 2^478 by n. You get your number. (I think, I didn't actually do the math... I'm just reassuring your troubled mind.)Discussions with you really pad my post count. Thanks for that.Edit: n=1/(5/6)^478 = (6/5)^478

I would roll untili i hit a reasonable amount i.e 5k +. Then on the second roll i would take a chance to see if i could reach a life changing amount of money, knowing i can still take 5k as a consolation if I hit that six.

[76169 0]

I can't prove we will go broke eventually, but surely we do.any math wizards?

Chance of going broke each time we play is:(1/6) + (5/6)(1/6) + (5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(1/6) + ...= (1/6)(1/(1-(5/6))) = (1/6)(1/(1/6)) = 1 = 100%And EV of the game is positive infinity

[Actuary 3]

Chance of going broke each time we play is:(1/6) + (5/6)(1/6) + (5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(1/6) + ...= (1/6)(1/(1-(5/6))) = (1/6)(1/(1/6)) = 1 = 100%And EV of the game is positive infinity

I think chance probability of going broke is Limit 1-(5/6)^n as n goes to infinity. Which equal 1-0 = 1.And the EV = Integral 2^n * (5/6)^n = (5/3)^n as n goes from 0 to Infinity. Which, although I don't remember my integrals for powers, has to be Infinity.

[Dratj 0]

???

[GWCGWC 83]

I'd roll until I went broke3 times

[SabaAba 0]

LOL, you guys would NEVER reach 16,000 let alone 1,000,000.Think about it, you roll a 6 once every 6 times, so if you are lucky, you will roll say 12 times before rolling a 6 which is only 2,048.The question is all about greed. When you say you want to win 64,000, you are really saying I'm willing to lose 32,000 16.67% of the time for a chance to win another 32,000. Try this at home and see if you can even roll 11 times to get to 1,024 without rolling a 6. Not very likely.

[troyomac 0]

LOL, you guys would NEVER reach 16,000 let alone 1,000,000.Think about it, you roll a 6 once every 6 times, so if you are lucky, you will roll say 12 times before rolling a 6 which is only 2,048.The question is all about greed. When you say you want to win 64,000, you are really saying I'm willing to lose 32,000 16.67% of the time for a chance to win another 32,000. Try this at home and see if you can even roll 11 times to get to 1,024 without rolling a 6. Not very likely.

Yes but if you're lucky enough to go 12 times without rolling a 6. Then your next roll is still a 5/6 chance to double up. I'd risk it!

[Actuary 3]

LOL, you guys would NEVER reach 16,000 let alone 1,000,000.

true.But would you stop at $8it's not what you expect to win, it's where you would stop.can you figure any reason why the EV is not Infinity? It just seems wrong, but mathematically looks correct

[onlyme386 0]

true.But would you stop at $8it's not what you expect to win, it's where you would stop.can you figure any reason why the EV is not Infinity? It just seems wrong, but mathematically looks correct

Damn you all! You've set my brain into motion and it's firing 100% now. I shall return in 15-20 minutes with the ultimate answer. Maybe, if you're real lucky, I'll show my work on paper, scan it, and post it here. That way you can all see how it works.

[onlyme386 0]

OK, so here's what I've come up with.Roll # = n Start Amount = SEnd Amount = EEV = 5E/6 - S (for any single roll)n...S...E...EV1...1....2... .66672... 2 ... 4 ... 1.3333 ... 4 ... 8 ... 2.6674 ... 8 ... 16 ... 5.3335 ... 16 ... 32 ... 10.6676 ... 32 ... 64 ... 21.3337 ... 64 ... 128 ... 42.6678 ... 128 ... 256 ... 85.3339 ... 256 ... 512 ... 170.66710 ... 512 ... 1024 ... 341.33311 ... 1024 ... 2048 ... 682.66712 ... 2048 ... 4096 ... 1365.33313 ... 4096 ... 8192 ... 2730.66714 ... 8192 ... 16384 ... 5461.33315 ... 16384 ... 32768 ... 10922.66716 ... 32768 ... 65536 ... 21845.33317 ... 65536 ... 131072 ... 24690.66718 ... 131072 ... 262144 ... 87381.33319 ... 262144 ... 524288 ... 174762.66720 ... 524288 ... 1048576 ... 349525.33321 ... 1048576 ... 2097152 ... 699050.66722 ... 2097152 ... 4194304 ... 1398101.33323 ... 4194304 ... 8388608 ... 2796202.66724 ... 8388608 ... 16777216 ... 5592405.33325 ... 16777216 ... 33554432 ... 11184810.667So, the question becomes, how much are you willing to risk on one dice throw? For me, I'd have to stop at $1,024 the first game. In fact, it's most likely that you'd never even reach that point. Only 13% of the time would you reach it. That next roll has an expected value gain of $1365 but as a college student, I'd say that $1k on a 1 minute gamble would be enough to make my day. I might even stop below that, maybe at 512 or even 256. The second game and third, if I was able to get there I would stop at $4,096. That next roll has way too much of a risk for me.Anyways, where do you stop given this?

[Actuary 3]

Anyways, where do you stop given this?

um.what new info did you provide tha would afect answers?I don't follow your EV = 5E/6 - S (for any single roll) equationoh, wait, E is amount you'd have after the roll, if not a 6.If I"m interpreting that correctly, the EV = 5/6 * ENo need to subtract S

[onlyme386 0]

um.what new info did you provide tha would afect answers?I don't follow your EV = 5E/6 - S (for any single roll) equationoh, wait, E is amount you'd have after the roll, if not a 6.If I"m interpreting that correctly, the EV = 5/6 * ENo need to subtract S

Expected Value is the probability of winning, multiplied by the difference if you win MINUS the probability of losing, multiplied by the amount lost.EV = (5/6)(E-S) - (1/6)SEV = (5/6)E - (5/6)S - (1/6)SEV = (5/6)E - SThat clarifies things, I hope.

[Actuary 3]

Expected Value is the probability of winning, multiplied by the difference if you win MINUS the probability of losing, multiplied by the amount lost.EV = (5/6)(E-S) - (1/6)SEV = (5/6)E - (5/6)S - (1/6)SEV = (5/6)E - SThat clarifies things, I hope.

you are calculating the EV of the Gain, ok, as opposed to the EV of the final value after the nth roll.so, either way, you agree EV approaches infinity, right?

[onlyme386 0]

you are calculating the EV of the Gain, ok, as opposed to the EV of the final value after the nth roll.so, either way, you agree EV approaches infinity, right?

Yes, EV ---> Infinity as n ---> infinity.Also, as n ---> infinity, probability of loss = 1/6. Statiscally, you will lose 1/6 of the time. I guess the argument could be made that you should roll forever, because you always win more. But, obviously you never win if you never stop.

[SilentSnow 1]

true.But would you stop at $8it's not what you expect to win, it's where you would stop.can you figure any reason why the EV is not Infinity? It just seems wrong, but mathematically looks correct

Well the EV is infinity but that might not be as exciting as it sounds.Here is another infinite EV game-Someone hands you 10 million dollars. But wait, it gets even better! They give you the chance to play roulette as the house with it. Every time you do not land on black you double your money. Of course if black hits you lose everything.As infinite EV as this is, you would have to be insane to spin the wheel even once.

[onlyme386 0]

Well the EV is infinity but that might not be as exciting as it sounds.Here is another infinite EV game-Someone hands you 10 million dollars. But wait, it gets even better! They give you the chance to play roulette as the house with it. Every time you do not land on black you double your money. Of course if black hits you lose everything.As infinite EV as this is, you would have to be insane to spin the wheel even once.

Very, very different. Half the time you double up, half the time you lose everything. EV = (1/2)(E-S) - (1/2)(S)EV = (1/2)E - (1/2)S - (1/2)SEV = .5E - SSo, first time... E = 2 mill, S = 1 millEV = 1 - 1 = 0 Therefore, the net change for 1 time is 0. Doesn't really matter. The difference is in the amount of money you start with. It's obviously much easier to risk $1 than it is to risk $1 million. Especially when you were GIVEN the $1 million, not put it on the table yourself. The difference is huge...

[SilentSnow 1]

Very, very different. Half the time you double up, half the time you lose everything. EV = (1/2)(E-S) - (1/2)(S)EV = (1/2)E - (1/2)S - (1/2)SEV = .5E - SSo, first time... E = 2 mill, S = 1 millEV = 1 - 1 = 0 Therefore, the net change for 1 time is 0. Doesn't really matter. The difference is in the amount of money you start with. It's obviously much easier to risk $1 than it is to risk $1 million. Especially when you were GIVEN the $1 million, not put it on the table yourself. The difference is huge...

Youre forgetting that you are the house, so your chances are better than 50 percent. But my main point is that large amounts of money make EV calculations almost irrelevant. I suppose a better example would be the dice roll, but this time start with a billion dollars.Oh, and whether you were given the money or put it up yourself is absolutely irrelevant assuming your net worth is the same in either case.

[Merby 3]

Just refer to the chart and decide what odds you are willing to play for:n = Number of dice rolls$ = Value of prize payout on a given roll% = Percent chance of successfully reaching this roll without going bustn ......... $ ........ %1 ......... $2 ....... 83%2 ......... $4 ....... 69%3 ......... $8 ....... 58%4 ........ $16 ...... 48%5 ........ $32 ...... 40%6 ........ $64 ...... 33%7 ....... $128 ..... 27%8 ....... $256 ..... 23%9 ....... $512 ..... 19%10 ..... $1,024 ... 16%11 ..... $2,048 ... 13%12 ..... $4,096 ... 11%13 ..... $8,192 ... 9%14 ... $16,384 ... 8%15 ... $32,768 ... 6%16 ... $65,536 ... 5%Part of me would want to go until I am (roughly) a 50/50 chance of successfully walking away with money: $16......The other part of me says, "To hell with it! I have a 1 in 20 chance of walking away with over $65,000... For only a dollar risked, I'm GOING FOR IT!" ...$65,000 would probably satisfy me (for now) so I would stop there.

[iggymcfly 0]

You didn't miss anything Actuary; the expected value does approach infinity if you keep rolling. The thing is that even though your likelihood of going broke approaches zero, the money you gain when you don't go broke is increasing faster. If you plug in an actual number, say 100,000 rolls, even though you'd be broke the 99.9999999% of the time or so, that 0.00000001 time when you're not broke, you'd have some ridiculous amount like 6*10^78 or something. The thing that makes the game worth stopping at some point is that the utility of money earned does not increase linearly with money. The difference between having $16 million and $8 million is not the same as the difference between having $8 million and $0. Once the utility of the doubled amount is no longer worth at least 1.2 times the utility of the original amount you should quit.

[BuffDan 0]

You didn't miss anything Actuary; the expected value does approach infinity if you keep rolling. The thing is that even though your likelihood of going broke approaches zero, the money you gain when you don't go broke is increasing faster. If you plug in an actual number, say 100,000 rolls, even though you'd be broke the 99.9999999% of the time or so, that 0.00000001 time when you're not broke, you'd have some ridiculous amount like 6*10^78 or something. The thing that makes the game worth stopping at some point is that the utility of money earned does not increase linearly with money. The difference between having $16 million and $8 million is not the same as the difference between having $8 million and $0. Once the utility of the doubled amount is no longer worth at least 1.2 times the utility of the original amount you should quit.

I was going to add to the conversation, but this said everything I was going to say. Infinity does some really weird/cool things that can be counterintuitive at times.

[Actuary 3]

Just refer to the chart and decide what odds you are willing to play for:n = Number of dice rolls$ = Value of prize payout on a given roll% = Percent chance of successfully reaching this roll without going bustn ......... $ ........ %1 ......... $2 ....... 83%2 ......... $4 ....... 69%3 ......... $8 ....... 58%4 ........ $16 ...... 48%5 ........ $32 ...... 40%6 ........ $64 ...... 33%7 ....... $128 ..... 27%8 ....... $256 ..... 23%9 ....... $512 ..... 19%10 ..... $1,024 ... 16%11 ..... $2,048 ... 13%12 ..... $4,096 ... 11%13 ..... $8,192 ... 9%14 ... $16,384 ... 8%15 ... $32,768 ... 6%16 ... $65,536 ... 5%Part of me would want to go until I am (roughly) a 50/50 chance of successfully walking away with money: $16...

Merby,I'm not saying you'd have to announce where you are a stopping ahead of time. So, each roll you would have a 5/6 chance of doubling. Your post implies that you have to announce a predetermined stopping point.**************Iggy,yep. Utility THeory for sure comes inIn Limits this is a case of Infinity * 0 with the Infinity out racing the 0.

[sprewell 0]

$16 and I'm done. lolWell...maybe I'd take a shot at $32...then I'm really done.

LOL indeed...