SAMAY RAINA. ERIC ROSEN. DICE CHESS.

➡️ LEARN CHESS WITH ME: chessly.com/
➡️ Watch LIVE on Twitch: / gothamchess
➡️ Support via Donation: www.paypal.me/...
Finally, the collaboration we all dreamed of. Samay Raina, Eric Rosen, and GothamChess. We played Dice Chess after Anna Rudolf and Tania Sachev did the same.
↓ ↓ Chess and social media links below ↓ ↓
♛ Chess Links:
➡️ Improve your chess: bit.ly/2W85z3x
➡️ My chess.com video lesson series: bit.ly/30r8RkN
⭐️ Social Media Links:
➡️ TWITTER: / gothamchess
➡️ Discord Chess Community: / discord
➡️ THUMBNAILS BY: / karri_bird; / karri.designs

[@GothamChess]

Eric uploaded the second part... kzbin.info/www/bejne/jJvUo6igoJ2Zr9U

[@oliverqi2902]

cool! thank you gotham!

[@arborvitae4110]

Hi can you put a video on jobava london?

[@_h_i_2098]

Nice

[@joshjowo60]

"getting a queen is not easy" - But it's exact same probability as getting any other piece.

[@amineoussama8273]

yes, but actually no, you should compare the probability of getting a queen with the probability of not getting a queen witch is bigger than it, that's why it isn't easy.

[@TheSpyroplayer]

@@amineoussama8273 But that probability is just same as all other pieces, no? There’s one of each piece on the dice, nothing makes the queen special on that dice, the probability of a knight is 91/216, a pawn is 91/216, a bishop is 91/216, and so on. They even mention the fact that the probability is the same at 7:04

[@enricofaa9302]

@@TheSpyroplayer the difference is you’d rather get a queen than any other piece

[@TheSpyroplayer]

@@enricofaa9302 I disagree though. There are many cases in this video where their best move was, say, and knight move, or a rook move, because only the knight/rook could see a hanging piece. Additionally, if the probability is the same for all pieces, then that also means it's "not easy" to get a knight", and it's "not easy to get a rook" as well. So rather, wouldn't it be better to say it's not easy getting any particular piece? And also, 42% is absolutely not "not easy to get", that's almost a 50/50 chance, what the hell do you mean "not easy"??

[@matko8038]

@@TheSpyroplayer theres a greater chance of not getting the piece you want.

[@LordAJ12345]

Levy getting the probability right was so satisfying. The discussion of this has always been so frustrating.

[@rohithjacob4699]

Exactly

[@BinuratheDarkLord]

@@smalexk The (1/6)^3 would be if he names the pieces in order noh? Since it can be any order the probability should be higher if I'm not mistaken

[@rumbleroar2]

@@smalexk Calculating "at least" events is not as simple as that. The easiest way to calculate it is to substract the complement from one. Which makes 1-(5/6)^3 = 91/216

[@smalexk]

@@rumbleroar2 yeah the more i think about it, the more i realize that levy was probably right haha... i hate stats haha

[@Leo_Carrilo]

Actually: 1/6 + 1/6 + 1/6 for any piece 1/6 * 1/6 * 1/6 for 3 equal pieces

[@kormagogthedestroyer]

This is just like the game called "No Stress Chess" which is made to teach children how to play chess and just happens to be the most stressful chess variant.

[@awesomeleozejia8098]

Yea I have that game too. I totally agree that it is really stressful.

[@jacobshirley3457]

Probably stressful once you know how to play the game, lol. Ignorance is bliss, as they say.

[@neko_aple]

"We're from US. We don't do math." ah yes. American patriotism.

[@quixotix9540]

o7

[@bandanasaikia6048]

😂

[@biharek7595]

'MURICA 🇱🇷

[@johnjackson9767]

@@quixotix9540 Nailed it right on the nose

[@rishabharya3329]

@@johnjackson9767 absolutely 🔫🔫

[@JordanMetroidManiac]

For anyone wondering, here are the odds of certain types of rolls occurring. Please like the comment! At least one of any specific piece (e.g. one, two, or three knights in a single roll): 1 - (5/6)^3 = 91/216 ~ 42% All three rolled pieces are different (i.e. no doubles or triples): (6/6) * (5/6) * (4/6) = 5/9 ~ 56% All three rolled pieces are the same (e.g. three queens): 6 * (1/6)^3 = 1/36 ~ 2.8% At least one of two specific pieces (e.g. at least one knight, or at least one bishop): 1 - (4/6)^3 = 19/27 ~ 70% Now, some probabilities of successive rolls… One roll doesn’t contain a specific piece, and the next roll does contain a specific piece (e.g. how Levy attacked Eric’s queen with a bishop on h3, this is the probability that the bishop could actually capture the queen on Levy’s next turn, assuming no intermediate check can be given): (1 - 91/216) * (91/216) = 11375/46656 ~ 24% Note that the probability above does not account for more than two turns. For infinitely many turns (i.e. the *true* theoretical probability of a successful attack, again assuming no alternative event occurs that stops the attack), we sum all the powers of that probability: (11375/46656)^1 + (11375/46656)^2 + (11375/46656)^3 + ... = 11375/35281 ~ 32% One roll doesn’t contain one of two specific pieces, and the next roll does contain a specific piece (same situation as the bishop attacking queen example above, but suppose the bishop can also be taken by a knight, not only the threatened queen): (1 - 19/27) * (91/216) = 91/729 ~ 12% Similarly, the actual theoretical probability of a successful attack in that scenario is the following: (91/729)^1 + (91/729)^2 + (91/729)^3 + ... = 91/638 ~ 14% One roll contains a specific piece, the next roll doesn’t contain a specific piece, and then the roll after that contains a specific piece (e.g. you put your queen somewhere where a bishop can capture it in two moves, and you’re unable to move your queen out of the way; this is the probability that your opponent will be able to do that if they were to try to do that): (91/216) * (1 - 91/216) * (91/216) = 1035125/10077696 ~ 10% Accounting for infinitely many turns for the event above, we have the following probability: (91/216) * (11375/35281) = 1035125/7620696 ~ 14% One roll contains one of two specific pieces, the next roll doesn’t contain a specific piece, and then the roll after that contains a specific piece (e.g. there are two pieces that can potentially move twice to capture a piece, but only one of the two pieces is moving twice to capture the piece): (19/27) * (1 - 91/216) * (91/216) = 216125/1259712 ~ 17% One roll contains a specific piece, the next roll doesn’t contain one of two pieces, and the roll after that contains a specific piece (e.g. a knight jumps in to snag a piece, where it can be captured by either of two of opponent’s pieces, but they don’t get to recapture, and then the attacker is able to get the piece out to save it on the next turn): (91/216) * (1 - 19/27) * (91/216) = 8281/157464 ~ 5.3% If there’s a probability you want figured out, let me know and I will add it.

[@fermion9044]

A lot of them are wrong you can't just multiply numbers. Becouse order doesnot mather, you have to use Binomial coefficient

[@Luckyyshot]

@@fermion9044 ​ Which ones are wrong? I checked all of them until successive rolls and so far he's right.

[@pratyushsharma7649]

Dude got to give u props for such a long comment

[@JordanMetroidManiac]

@@pratyushsharma7649 It’s like answering questions on an exam

[@mightymouse652]

What about the probability that the pieces are no longer on the board and the odds have to be calculated again

[@XoIoRouge]

When Eric goes into Samay's Chat to tell Levy to DO IT. LUL

[@JuliusKingsley0]

“Doesn’t matter which religion you’re subscribed to” lol 30:08

[@srikarrao9957]

😂😂

[@cadekachelmeier7251]

Don't forget to smash that subscribe button. - The Pope

[@kkTeaz]

@@cadekachelmeier7251 lol

[@ReverbRippleSounds]

@@cadekachelmeier7251 😂🤣🤣

[@em4six348]

24:43 Levy's face 😭

[@tranced42]

i would love to see engine analysis of this just “x percent chance +4 and y% -10 and z% m2”

[@TheZekriamLP]

Saw half of this yesterday, already know It's gonna be good

[@ayush2sept]

love your channel levy been watching ur stuff before the chess boom happened in india,you are a hardworking individual and i have learnt a lot of stuff from u and eric rosen lots of love from india

[@Tkdriverx]

I've played this variant before, but with two dice, instead of 3, and if there were no legal moves from the roll, then your turn is skipped. If you roll a double, you can play whatever legal move you want. To win, must capture opponent king

[@donraider999]

13:48 "And it was as of move number 3 we have a completely new game" XD good one

[@shitpostnation3069]

lmao my name is areg, so every time he was calling eric i was looking with excitement

[@Shreyank01]

Love from India Happy that Samay introduced you to me!!

[@stuartpenwarden253]

I'm 5 minutes into this and already I love it. My friday nights are so much different now

[@omkarghagofficial]

Thanks for coming we all are very happy to see u on samay bhai's stream . You are also like him thinking about the viewers and content . Your chemistry with Samay bhai was OP and all of us would love to see u come again on stream not just for among us but for chess. ❤️

[@ajiteshjoshi2492]

27:51 3/216 equals 1/72 not 1/36,,,, Levy, that's what printing engineers are😀

[@obelustilde9427]

4:21 The probability of both players getting the same pieces on their first move should be 83/3888 or about 2.135%. It's not 1/36 because the same piece can occur multiple times. If for example the first player gets a queen on all three dice, then the probability of the second player getting that same result is (1/6)³ = 1/216. So in order to account for this, we have to calculate: • the probability of the first player rolling three different pieces, which is 6/6*5/6*4/6 = 5/9 • the probability of the first player rolling the same piece on all three dice, which is 6/6*1/6*1/6 = 1/36 • and the probability of the first player rolling the same piece on two dice and a different piece on the other dice, which is 1 -5/9 -1/36 = 5/12. The probability of the second player getting the same pieces as the first player is: • 6/216, given that the three pieces of the first player are all different • 1/216, given that the three pieces of the first player are all the same • 3/216, given that the first player had two pieces that were the same and one piece that was different. Combining these probabilities yields, that the probability of the second player getting the same pieces as the first player is: 5/9*6/216 + 1/36*1/216 + 5/12*3/216 = 83/3888. Congrats, if you read all this mess :P

[@alexismiller2349]

yess someone got the right answer!

[@ihavenotcameupwithgoodhandle]

this is tooo muh 4 me

[@racheline_nya]

26:35 if it has to be in order, it's 1/216 if the order doesn't matter, there are 6 ways to order it when all 3 are different, so if you guessed 3 different pieces, you have a 6/216 probability of being right, which is 1/36

[@alexismiller2349]

If the pieces may or may not be the same then the probability is = P(same 3 dice as last guy) = P(same 3 dice as last guy given that the last guy had three different pieces)* P(last guy had 3 different pieces) + P(same 3 dice as last guy given that the last guy had 2 same pieces)*P(last guy had 2 same pieces) + P(same 3 dices as last guy given that the last guy had all the same 3 pieces) * P(last guy had same 3 pieces) We can easily (ish) determine that the corresponding probabilities are 1/36, 5/9, 1/72, 15/36, 1/216, 1/36 Hence we get the result P(same 3 dices as last guy) = 1/36 * 5/9 + 1/72 * 15/36 + 1/216 * 1/36 = 83/3889 ~= 2.1235% which is less than 1/36 ~= 2.7778% I was a bit unsure so I simulated it and got after 20 million trials a number different by just 0.000008 so that's seems good Sorry if you don't care since you posted this 4 months ago but I just felt that this would be cool to share :)

[@singh.nitin_15]

Watched it live , loved how they imitated Agadmator! Sorry about that....😆😅

[@tommybahama4418]

And it was in this position that we have a completely new game.

[@emilynam7842]

btw the total number of combinations is actually 56; 6 combinations with the same piece three times, 30 combinations with one piece repeated (6 possibilities for the repeated one * 5 possibilities for the second), and 20 combinations for the all unique ones (6*5*4 for all unique combinations divided by 6 because there are six ways to arrange 3 numbers) which totals to 6+30+20 combinations. the reason people ended up getting 36 a lot is because they made the assumption you can just divide the whole thing (216) by 6 since there are six ways of arranging three numbers, but it doesn’t work because when you have numbers with repeats there are less than six ways of arranging them

[@chrisgascoyne2958]

Eric is so innocent man, I love the combination of your personalities

[@johnmeo1532]

4:25 is actually a more involved question than it first appears. If each die has to show the same piece, then the answer is (1/6)^3 as each die has a (1/6) chance of rolling the same piece on the second roll as it did on the first roll. However, as shown in the video, the pieces were rearranged from the first roll to the second roll, but gave the same result. We can break down all 216 rolls into 3 categories. All unique pieces, all same pieces, and 2 of a kind with 1 odd-ball piece. There are 6*5*4 of the first (6 possible pieces for first die, second die can't have the same so only 5 possibilities, third die can't have either first or second so only 4 possibilities), 6*1*1 of the second (6 possibilities for first die, second and third dice must be whatever first die is), and 6*5*1*3 of the third (6 possibilities for first die, second die must be different so 5 possibilities, third die must be same as first so 1 possibility, and then we can rearrange these dice in 3 ways -- position of odd-ball). = 120 (first category), 6 (second category), 90 (third category). If your first roll is in the second category, you have a 1/216 chance of rolling the same, because there is only one way to roll the same thing again and you must hit that. However, if your first roll is in the third category, you have a 3/216 chance of rolling the same on the second roll, because you can rearrange the dice in 3 ways (3 different positions of odd-ball) and get the same result from your roll. Finally, if your first roll is in the first category you have a 6/216 chance or rolling the same on the second roll, because there are 3! = 6 ways to rearrange 3 unique dice. Putting this all together, we get the probability of rolling the same pieces in 2 consecutive rolls equals: (120/216)*(6/216) + (90/216)*(3/216) + (6/216)*(1/216) ~ 0.0213 = 2.13% which is about 1 in 47.

[@shubhamchauhan696]

🤯

[@shadowChrist0]

Despite an Indian I found Samay's channel through Levy's and Eric's channels 🤙❤️

[@DanielKondratiuk]

HOW DID IM ROSEN GET AN IM TITLE I LITERALLY DIED

[@EIIlast2791]

?????

[@linaetlinda5832]

@@EIIlast2791 i think because he was fm ?

[@amanrai7965]

Yes, my brain stopped working thinking about that

[@EIIlast2791]

@@linaetlinda5832 no Eric is actually an IM lol

[@ezycuberz4563]

Eric fan huh? 😂 Bro they both r so good friends, let's not be villains to each other, let's too be friends like Levy n Rosen. You know, just roast eachother n stuff😂

[@rahulmistry5019]

1:32 lmao savage XD as a math phd student, your answer levy, was on point and satisfying to hear.

[@pythontron8710]

His answer was wrong tho lmao. You are definitely not a math phd

[@davidvidales2209]

14:00 Eric be like: Mr. Levy I don't feel so good

[@akyt67]

Levy OD 😂🔥

[@caffreyyy]

Samay and you make an awesome duo man! 🔥 You both always look for content. Looking forward to many more collabs.

[@Pantomime0709]

I almost screamed when I saw this in my recommendations. Was waiting for so long for this.

[@christophecornet2919]

I actually thought of doing dice chess recently, but differently. Depending your roll on a 6 faced die, you can only move a piece that is on a: 1 - even row 2 - odd row 3- even file 4- odd file 5- white square 6- black square

[@thetechnovoid]

make it two dice and add

[@PhilipVels]

Idk why he thought the WWE reference would make sense haha

[@anujyouthoob]

26:35 Any order is allowed. That reduces total possible outcomes to _only_ *56* Case 1: All pieces different 6C3=20 Case 2: Two pieces same and one different 6C1×5=30 Case 3: All pieces same 6C1=6 Probability of predicting correctly =1/56

[@Maks-ji7de]

at 33:19 samay says right now there are 3 kings so you can be sure that 3 kings won't come next turn. this is the gambler's fallacy, just because you got heads on 10 coin flips in a row the next coin flip is still 50% heads 50% tails. but the probability of 3 equal pieces is much less than 3 different pieces because the order does not matter so there are more ways of getting the pieces e.g. {rook, king, queen} = {king, queen, rook} whereas there is only one way of getting (king, king, king)

[@R2Cv1]

Levy has a stats degree.

[@doctorbees4886]

"do you believe in the heart of the cards"

[@kalebwhitmore8600]

4:35 1/36; there is a 3/6 chance the first die matches one of the others, and 2/6 for the second and a 1/6 for the third

[@Andrewkosche]

The question was (I think) the odds of “the first two” being the same. Thus it would just be 1/6

[@llawliet9327]

This stream was OP

[@clashkingdom3647]

L Lawliet please rest in peace . And let me Light yagami rule the world.

[@AkshayWalia]

Yagami Lightaaa

[@clashkingdom3647]

@@AkshayWalia yes what happened

[@Alireza-yw8yq]

oh i love this "if you believe in the heart of the card anything can happen " you are a genius man

[@foxtro7]

4:23 - The probability of getting the same roll twice in a row (where order doesn’t matter) is 1/56. (Note: For this calculation, rolling [N, P, R] and then [P, N, R] would count as repetition, but rolling [K, K, Q] and then [K, Q, Q] would not, even though the latter two rolls would be functionally identical in the game.) edit: or maybe it’s not idk lol

[@TrickShotKoopa]

How did you get 1/56? I thought it is 1/36.

[@foxtro7]

@@TrickShotKoopa 1/36 would be the odds with no repetition (where [Q Q Q] would be impossible) The way I got 1/56 was as follows: A list of selections can be written as [X X Y] or, alternatively [X X | Y], with the vertical line separating the two types. A list of 6 selections, one of each type, could be written [A | B | C | X | Y | Z]. (Note that the number of verticals is the number of types minus one.) We can cut out all the unused letters and just leave the verticals. So [X X | Y] would become [ | | | X X | Y | ]. Now the specific letters are unnecessary, as the position of a letter relative to the verticals is enough to determine to which type it belongs. So our list can be rewritten again as [ | | | - - | - | ]. Now each unique list represents one unique combination of three selections from the six types. If we read the verticals as symbols, we can see that the lists can be understood as 8 symbols long (number of types - 1 + number of selections = 6 - 1 + 3 = 8). We can create every possible combination of three pieces by choosing where to place the 3 dashes in the 8-long list. This part is a typical combination problem: 8 choose 3 or 8C3 = 56. (nCr = n!/(r!(n-r)!)) I assumed that the probability of getting the same combination of some average roll would just be 1 divided by the number of combinations, 1/56, but now I’m not so sure that’s at all correct. I am pretty confident there are 56 combinations tho.

[@fermion9044]

@@foxtro7 you got it right it is really easy and straight foward . In my coutry we learn it at 3 year at high school (usslaly 17 y.o. students)

[@alexismiller2349]

@@foxtro7 This is pretty convincing however there is a very subtle mistake, the number 56 is correct I think however not every combination out of those 56 is equally likely to occur, for instance [Q Q R] is twice as likely to appear as [P K N], I suppose one could do a weighted sum over 56 to get the right result however I did it differently, by considering individually the probability of getting 1, 2 or 3 of the same piece I got at the end 83/3888. I also did a simulation over 20 million trials and it seems to agree with me up to 0.8*10^(-5). Idk if you care but I can show off the calculations I did

[@YashSingh-td1bz]

Eagerly waiting for among us

[@RedMushroom4216]

30:07 “It doesn’t matter what religion you’re 💀subscribed💀 to”

[@andrewcavallo1877]

3:20 Remember kids, consent is everything.

[@GeorgeDCowley]

Me the whole video: "Goooo!"

[@yasersyria1608]

"We're from US , we don't do math here" Levy.

[@jangamecuber]

Calculation for the chance: There are 6 options for 3 slots so there are 6^3 or 216. He said at least 1 so it can be 1, 2, or 3 queens. 1 outcome has 3 queens, 15 have 2 queens (3 possibilities for which dice is not the queen and 5 options for what else it could be), so 16/216 is the chance of 2 or 3 queens. For exactly one queen the are 75/216 (3 pairs for what five aren’t the queen and 25 total options for that pair). The total is 75+16/216 or 91/216. This option assumes that the same piece can roll multiple times in the three dice rolls per turn. If pieces can not roll multiple times in the three rolls per turn then the probability is 60/216 (3 options for the pair of dice that aren’t queens and 20 different ways they could roll). For the chance of 2 or the same rolls there is 6 (the number of permutations of three objects) / 216. Simplified, that would be 1/36. There, I did the math so you don’t have to.

[@rdtworld9720]

They should take out that Money in the Bank thing, but this game is awesome.

[@janmejaybarve7018]

I got a technicality here. 91/216 is the probably of getting atleast one queen on the three dices. The question however was 'ONE queen in either of the three dice'. Thats a lesser probably. Anyways, this is just a reminder that statistics differ based on how the question is framed. Logically we are interested in at least one queen, but the question specifically implied EXACTLY one queen.

[@tukriko]

I remember checking out Levy's games from the archive other day... I was so confused. This explains alot :D

[@priyankgodhat4198]

Your shirt was the highlight of the stream

[@orenawaerenyeager]

8:14 Gotham - you win the jackpot Why everyone ignored*😒

[@anirudhgovada2753]

The probability is 1/36 as Levy doesn't have to guess it in order. It a combination rather than a permutation. As there are 216 possible combinations and Levy can guess three pieces in any order, the probability is actually 1/36. If he had to guess it in order then it would be 1/216.

[@aryanmaggon1406]

“So look at this” then ad starts playing 😂

[@josephalonsabe7633]

This was very fun to watch; nice work guys

[@arabbit5625]

Love the agadmator references

[@zacharyadler4071]

4:35 the prob of getting the same 3 things as you did in the previous roll, ignoring order, is 3/6 * 2/6 * 1/6, assuming the roll is 3 different pieces. An alternative way to calculate it is simply the number of ways to achieve 3 pieces divided by the number of ways total. 6^3 = 216 possible outcomes and 6 of those are the 3 pieces you desire because it is 3 factorial, 3 of the outcomes work for the 1st spot then you have 2 left then 1 left

[@alexismiller2349]

without the assumption that the three pieces are different you get the number 83/3889 instead :)

[@pairot01]

The chance of guessing the next roll is would be 1/216 if order mattered (1/6 for each die), but because it doesn't you have to multiply that by 3! (factorial, 3 positions for your first guess, 2 positions for your second guess and 1 for the last guess, 3*2*1=6). So it's 1/36

[@eklektikTubb]

Brainking has "10x10" variant with one dice and three kings.

[@rmschad5234]

Chat immediately got the probability of at least one queen, but forgot that the dice are not ordered when it came to the same first rolls.

[@rmschad5234]

"I'll probably get a pawn." Eric, we've been over this 3 times.

[@ankitghosh1524]

Love your content bro. - a samay fan

[@revoe5522]

24:31 rolls 3 queens *surprised pikachu face

[@DexSybaris]

ITS THE PARTY SHIRT IT GLOWS IN THE DARK

[@darebrained]

Always move pawn when you get one, at least until you can move all other pieces, otherwise you come into situations when you are forced into some moves like Rb1

[@jonathanryals9934]

OP = Optimus Prime

[@xnick_uy]

The chance of guessing the outcome of the 3 die *in order* is 1/6^3 = 1/216 ~ 0.46%. But since *the order of the pieces doesn't matter* (any of the 6 possible arrangements of the 3 pieces is valid) the chance is _6 times higher_ : 6/6^3 = 1/6^2 = 1/36 ~ 2.78%.

[@alexismiller2349]

6 doesn't work always because sometimes you have two of the same pieces or even three of the same pieces, then you get 3 or 1 possible arrangements correspondingly. The actual answer is surprisingly 83/3888 ~= 2.135%, :)

[@joelkinet1391]

levy I'm from flushing and I've never heard anyone say "I'm od tired" XD

[@paristhewabbit]

14:45 He should have played Nxc7 since Eric might roll queen and not king

[@sonan333]

When I heard dice chess, my immediate thought was: You have 2 d8's. One is for 1-8, the other for A-H. You roll both of them and if you have a piece on the spot that comes up, you must play it if able.

[@meatvirginia8454]

Eric just came off a 36 hour bob ross binge and you can tell it really affected him

[@tusharsurankar5958]

Good to see agad getting cameo on your channel 😆

[@TaoDragon_]

If you are in a losing position you should guess the roll that gives your opponent a really good advantage to eliminate that possibility.

[@dreamsolutions3037]

Yoooo I guessed the same pieces as Levy for the money guess. That was crazy

[@sohail12feb]

That's an amazing shirt, Levy. I loved it!!! ❤️

[@joemama0]

I missed stream :(

[@kishanjethwani5565]

Watch on samay's channel

[@SK-pj8mg]

u can watch whole thing on samay channel on yt

[@abhiveersingh3817]

Loved you in Samay's stream

[@milehighslacker4196]

OP = Ocean Pacific

[@fatitankeris6327]

4:19 Disregarding position in the dice row, the probability that Levi and Eric will have the same piece choices in a row is 1/46 656 ≈ 0.00214%.

[@markjreed]

The probability of guessing right depends on what you guess. If you guess 3 of the same piece, that's very unlikely to be right - only a 1/216 chance. If you guess a pair of the same piece with a different piece on the third die, that's a 3/216 chance (3 ways it can happen, with the different piece on the 1st, 2nd, or 3rd die), which is 1/72. If you guess three different pieces there are 6 ways that can happen (the six ways to rearrange three distinct items into different orders) so that's 6/216 or 1/36.

[@okayuvirus133]

"doesn't matter which religion you're subscribed to" - Levy Rozman, GothamChess

[@pjrousche1600]

This same thing but instead of pieces it's files and you must move a piece to one the the files you rolled

[@lazydk2654]

So i would argue knight takes on c7 in the start was winning, because it’s queen or king move. Either way it could be good for white. If king, you got decent odds of knight xa8, if queen, u are forced to take and bishop has the same odds, so xc7

[@seto749]

Why move a knight on the first roll when one gets a pawn as well? The next roll could be King/Rook/Bishop, for instance, and then one would have to move the rook instead of being able to bring the bishop out.

[@Justdifferent666]

this was a fun concept

[@akashkhente3335]

Levy is so good at immitations 😂

[@darshtailor7030]

When is the next collab??

[@CosmoSafdie]

33:32 c5=best move

[@claurox5380]

R.I.P the gotham lamp in the back

[@1807xt]

Great content guys!

[@dhushy63]

2:36 Wow Levy is smart

[@suryanshsingh4261]

LevyOD ~ Samay Raina (Hindi speaking people may get it)

[@alexfisher1467]

Levy’s shirt makes him look interesting

[@MarcusFenixKO]

14:45 Levy please...Knight C7. (Moves to F8) noooo...

[@synister_2168]

We want more of these collaborations

[@aryanbansal624]

it would be more comfortable if you could adjust your camera position on screen so that it matches your position on board like if gotham is white, his camera would be below erics camera so it makes it easier 👍👍

[@whatsinaname7289]

Stop bullying Rosen! 😂😂😂

[@mbosskdawg-vlogsandgames2979]

24:44 YAHTZEE!!!!!!

[@user-en7dx1qp3k]

"It was as of move three that we have a completely new game"