The Billionaire Club
Bill Gates, and Warren Buffett were sitting together in the Billionaire Club with 56 and 52 billion dollars respectively. Suddenly Gates hit upon a new idea.
"We'll throw six dice together," he said.
"If we get two pairs I win, otherwise I lose."
"But what about four or more dice show the same digit?" enquired Warren.
"That makes two pairs; so that too gives me a win."
"And if only three dice have the same digit?"
"You win in that case because it does not make two independent pairs."
"Each games of chance will cost 2 Billion dollars to the looser" added Gates. "You want to play?"
Warren, a little confused calls YOU and asks "Who is more likely to win in this scenario?" What will you say ? Assuming they do play, how many games they can play at the least ?
"We'll throw six dice together," he said.
"If we get two pairs I win, otherwise I lose."
"But what about four or more dice show the same digit?" enquired Warren.
"That makes two pairs; so that too gives me a win."
"And if only three dice have the same digit?"
"You win in that case because it does not make two independent pairs."
"Each games of chance will cost 2 Billion dollars to the looser" added Gates. "You want to play?"
Warren, a little confused calls YOU and asks "Who is more likely to win in this scenario?" What will you say ? Assuming they do play, how many games they can play at the least ?
Labels: friday special, logic, puzzle
posted by Rajesh Lal at
8:50 AM
23 Comments:
if each game cost 2 Billion dollars to the losser, (and the winner gets nothing) then at least 26 games. (52/2)
The most likely winner of the game would be Warren Buffet because his chances 5/72 and bill gates is 1/1296 and anonymous is right least amount is 26 games
Eric could you please expalain as to how you figured your odds/chances. Other than that I agree with everyone that the least amount of games would be 26.
you huys r all really stupid. there is a 50/50 chance cuz there are 3 odd numbers and 3 even numbers on a dice. as long as the dice arn't fixed, then the chances will be 50/50 no matter how many dice there is. and there is the same amount of chance to get a pair of numbers, than there is not to.
you huys r all really stupid. there is a 50/50 chance cuz there are 3 odd numbers and 3 even numbers on a dice. as long as the dice arn't fixed, then the chances will be 50/50 no matter how many dice there is. and there is the same amount of chance to get a pair of numbers, than there is not to.
Gates is not too smart.. out of the 46656 possible combinations of the 6 dice, 921 of them are wins and 45735 are loses.
The least number of games, if they played, would be 1. It didn't say they couldn't quit before they were out of money.
no, there is a 50/50 chance of rolling one dice, but the more the dice, the more chance a getting a pair. even if you get a pair of odd numbers it doesn't matter for the game they are playing. bill gates has more of a chance than warren buffet.
no, there is a 50/50 chance of rolling one dice, but the more the dice, the more chance a getting a pair. even if you get a pair of odd numbers it doesn't matter for the game they are playing. bill gates has more of a chance than warren buffet.
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Oh sorry I probably should have explained my math So bill gates chances are based on 1/6 on one die, so to get at least 4 his chances 1/6 * 1/6 * 1/6 * 1/6 so the product would lead to 1/1296 and for the warren guy would be 5/6 because he would have to get any number but the number bill gets, so he would have to get at least 3 so the math is 5/6 * 5/6 * 5/6 which is 15/216 which reduces to 5/72 and anonymous guy you're calling everyone stupid but you made a lot of spelling errors
This problem is not as easy as it seems (are they not all ?) Bill Gates is a smart @$$ and will win
Euclid's Brother, I'm not sure the 921 is correct. If the denominator is 6^6, then you're dealing with the count of ordered tosses: (1,1,1,1,6) is distinct from (1,1,6,1,1). If that's the case, then you have to multiply the number of winning combinations by the number of permutations: 6!=720.
So, going back a step, a winning combination might be die#1=die#2 and die#3=die#4 (the other 2 dice are irrelevant), and there are 6*6=36 ways for this to happen. Then we need to multiply this by the 720 permutations to account for other winning events like die#2=die#6 and die#4=die#1.
Bottom line: P(getting 2 pairs of matching dice after throwing 6 dice) = 6!*6^2 / 6^6 = 25920/46656 = 5/9. So Gates isn't so dumb after all (if i'm correct. pass on the second part of the question)
Nice problem. I worked it out using a tree diagram treating the six dice as sequential rolls. To avoid the tree getting out of hand, each split was "A match, creating a pair" or "No new pair was created" instead of branching by specific numbers on the dice. Note: only needed 5 sets of branchings since the first dice could be anything. I ended up with 14 pathways (I was economical by not branching after two pairs).
By multiplying the probabilites along each pathway and adding all the winning pathways, I got that Bill Gates has an approximately .5988 chance of winning each toss. Therefore, Warren would be wise not to play.
And yeah, 26 games would be the minimum assuming Warren lost every toss.
BTW, anyone who starts a post with "You guys r all really stupid" is certainly an idiot and likely to be wrong. :p
Who cares, if i had that much money, i would keep it.
To answer the question, one game. Like Euclid's Brother said: "It didn't say they couldn't quit before they were out of money."
Furthermore, as far as odds go, how many dice are they playing with? Bill said, "We'll throw six dice together.", which could mean twelve total, which significantly changes the permutations compared to six dice.
Finally, the other Anonymous can't even spell, and shouldn't be calling people stupid based on allegedly faulty math...By the way, where did these puzzle writers learn English anyhow?
I stand corrected. 6^6 is not the number of combinations. Out six dice thrown together, the maximum possible combinations is 462. Out of those 371 of them contain at least 2 pair. 91 of them are loses. I guess Bill wins after all with an 80.30303030% chance of winning.
Now to me, it seems the questions may be a bit ambigous. If all 6 dice thrown in together, and you don't know which dice came from which persion (or it doesn't matter) then the above is true.
If each throws 3 dice at the same time, then, do the two pairs have to be in the same sets of 3, or paired accross the sets of 3.
example 1 (is this a win?):
Bill 1,1,2
Warren 3,3,4
example 2 (is this a win?):
Bill 1,2,3
Warrent 2,3,4
If both of the examples above are wins, then my 371 out of 462 is correct.
I'm thinking Bill is gonna win 'bouts 64-65% of the time. Warren will win 5 out of 54 times because they will roll 1,2,3,4,5,6. Of the other 49 of 54 times there will be at least one pair by definition so lets disregard that pair, 5 out of 18 time 4 dice will not form a pair. So whatever percentage 5/54 + (5*49)/(18*54) is that Warren's winning % (35.5% or so I truncated some).
Euclid's Brother: not exactly sure how you got the 462 possible combinations, but I'm guessing you counted (1,1,2,3,4,5) the same as (2,3,4,5,1,1). In any case, when you got 80.3030%, you treated each combination as equally likely to occur. Surely some combinations like (all 1s) is less likely than (four 1s and a 2) since the latter has 6 ways it could occur.
Vlad: I like your reasoning, and I can't find a fault with it other than it underestimates my answer. However, I've triple checked my methodology and math, and I am 99% sure of .5987654321. This is the fraction 97/162 or 27936/(6^6) in case that holds some clue to someone. When I have time, I may just have to program a simulation to confirm this. Peace.
Iff,
Full House (1 pair + 1 Three of a kinds) = 2 Pair,
then,
I agree with JediTony = 97/162.
Otherwise,
Warren Buffett will WIN.
anonymous 2, U r really stupid
ok, first of all, they are asking about 2 pairs of numbers, not 2 pairs of even and odd numbers.
And don't worry, you aren't the only one who can't spell, i have never seen loser's spelled like looser's.
Here's what I got.
The total number of possible combination is: 6x6x6x6x6x6=46656
The number of possible combination without any pairs is: 6x5x4x3x2x1=720
The number of possible combination with one and only 1 pair is:
First, choose 2 from 6 to be the pair: 6x5/2=15
Then, there are 6 possible numbers for the two same dice, and 5 choices for the 3rd dice, 4 choices for the 4th dice, etc. hence the number of possible combination is: 15x6x5x4x3x2=10800
So, total number possible combination with which Warren will win is: 720+10800= 11520
Hence, his chance of wining is:
11520/46656=.24
Therefore, I think he shouldn't play with Bill.
I've been waiting months to get a confirmation of the answer. I have not read or posted on the new puzzles since I feel like why try if there is not going to be an answer at some point.
Please post the answer already.
ANSWER : Who is more likely to win in this scenario?" CUNNING BILL GATES
----------------------
Let me first evaluate all the possible cases.
6 = 1 + 1 + 1 + 1 + 1 + 1
= 2 + 1 + 1 + 1 + 1
= 3 + 1 + 1 + 1
Gates wins in all other distributions.
(1) All 6 dices showing different values:
Probability = (6/6)*(5/6)*(4/6)*(3/6)*(2/6)*(1/6) = 720/46656
==> Buffett wins
(2) 1 pair and the other 4 all different:
Probability = [{C(6,2)}*(6/6)*(1/6)]*(5/6)*(4/6)*(3/6)*(2/6) = 10800/46656
==> Buffet wins
(3) 1 number occuring thrice and all the other are different.
Probability = [{C(6,3)}*(6/6)*(1/6)*(1/6)]*(5/6)*(4/6)*(3/6) = 7200/46656 ==>
Buffet wins
Probability that Buffet wins = (720 + 10800 + 7200)/(46656) = 18720/46656 = 65/162
= 0.40123456790123456790123456790123...
Probability that Gates wins = 1 - (65/162) = 97/162
= 0.59876543209876543209876543209877
Thus, we can infer that Gates is more likely to win.
Assuming they do play, how many games they can play at the least?
Least number of games = {minimum(52, 56)}/2 = 52/2 = 26 and the probability of this happening is = (97/162)^26 = 0.0000016168685471489664446756270647124...
-----------------------
GOT RIGHT First: JediToni,
JediToni, I apologize for the delay.
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