I How to calculate the probabilities here? (3-person card game)

[murshid_islam]

TL;DR Summary

Probabilities of winning in a 3-person card game

Suppose I have a shuffled pack of cards, out of which 4 red cards have been lost or removed. So now I have 22 red cards and 26 black cards in random order. I'm dealing cards in pairs. If a pair has 2 red cards, they are dealt to Mr. Red, if the pair has 2 black cards, they're dealt to Mr. Black, and if the pair contains 1 red and 1 black card, they are dealt to Mr. Mismatch.

After all the cards have been dealt in pairs, whoever has the most cards wins. Now, Mr. Mismatch will almost always win because there are more ways to get a mismatched pair (card-1 is red and card-2 is black or vice versa) than a pair of reds or blacks. Also, since there are 4 fewer red cards, Mr. Red will never win.

I want to know the probabilities of Mr. Black and Mr. Mismatch winning. I already ran a simulation (in Python) 1,000,000 (one million) times to see what percentage of games are won by the three players. Here's my code in case anyone wants to check it out

This is the result I got:

Mr. Red won = 0 times = 0.00%
Mr. Black won = 93799 times = 9.38%
Mr. Mismatch won = 906201 times = 90.62%

No matter how many times I run my code, I get similar results: Mr. Black winning little more than 9% and Mr. Mismatch winning a little more than 90%. However, Someone claimed that Mr. Mismatch should win about 95.6% of the time. I was wondering if it's my code that is wrong. So, I wanted to calculate the probabilities using math to check if my results are correct or not. Any ideas on how I can calculate the probabilities of their winning?

 

[nuuskur]

Equivalent formulation. An urn contains 22 red balls and 26 black balls. The balls are collected at random two at a time. The score starts at zero.
In the event of

• two black balls, add -1 to score.
• two red balls, add +0 to score.
• mixed colours, add +1 to score.

Determine the probability that the score is positive after the urn is emptied. Note that if there occurs +1 odd number $n$ of times, then +1 will have occurred at least $n+1$ times.

 

[PeroK]

Summary:: Probabilities of winning in a 3-person card game

Suppose I have a shuffled pack of cards, out of which 4 red cards have been lost or removed. So now I have 22 red cards and 26 black cards in random order. I'm dealing cards in pairs. If a pair has 2 red cards, they are dealt to Mr. Red, if the pair has 2 black cards, they're dealt to Mr. Black, and if the pair contains 1 red and 1 black card, they are dealt to Mr. Mismatch.

After all the cards have been dealt in pairs, whoever has the most cards wins. Now, Mr. Mismatch will almost always win because there are more ways to get a mismatched pair (card-1 is red and card-2 is black or vice versa) than a pair of reds or blacks. Also, since there are 4 fewer red cards, Mr. Red will never win.

I want to know the probabilities of Mr. Black and Mr. Mismatch winning. I already ran a simulation (in Python) 1,000,000 (one million) times to see what percentage of games are won by the three players. Here's my code in case anyone wants to check it out

This is the result I got:

Mr. Red won = 0 times = 0.00%
Mr. Black won = 93799 times = 9.38%
Mr. Mismatch won = 906201 times = 90.62%

No matter how many times I run my code, I get similar results: Mr. Black winning little more than 9% and Mr. Mismatch winning a little more than 90%. However, Someone claimed that Mr. Mismatch should win about 95.6% of the time. I was wondering if it's my code that is wrong. So, I wanted to calculate the probabilities using math to check if my results are correct or not. Any ideas on how I can calculate the probabilities of their winning?

It might be diffcult to calculate the probabilities in this case. I'll have a think about it, but a Python simulation would be the way to go.

Your code looks okay at first glance.

 

[PeroK]

Here's an idea. You only need to count the red pairs. If there are more than 6 red pairs, then black wins; otherwise, mismatch wins.

You would need to calculate the probability that there are 7-11 red pairs in your 48 random cards.

That might be not too hard, although I haven't thought about it yet.

 

[anuttarasammyak]

I made a table for the cases and confirmed that RR has no chance to win.

BB=RR+2
BR=24-BB-RR=22-2*RR

 

[PeroK]

Here's an idea. You only need to count the red pairs. If there are more than 6 red pairs, then black wins; otherwise, mismatch wins.

You would need to calculate the probability that there are 7-11 red pairs in your 48 random cards.

That might be not too hard, although I haven't thought about it yet.

This is not too hard. Suppose there are precisely $k$ red pairs. That means there are precisely $22-2k$ mismatch pairs and $2 + k$ black pairs. The number of ways of achieving this is:
$$n(r = k) = \binom{24}{k}\binom{24-k}{2+k}2^{22-2k}$$Note that there are two ways of getting every mismatch pair.

The total number of random arrangements is $N = \binom{48}{26}$. And the probability that mismatch wins is:
$$p(m) = \frac{1}{N}\sum_{k = 0}^{6}n(r = k)$$Checking this on a spreadsheet confirms that$$p(m) \approx 0.9065$$

 

[PeroK]

N	2.73857E+13​
k	(24, k)	(24-k, 2+k)	2^(22-2k)	n(k)	p(k)
0​	1​	276​	4194304​	1157627904​	4.22713E-05​
1​	24​	1771​	1048576​	4.4569E+10​	0.001627446​
2​	276​	7315​	262144​	5.2925E+11​	0.019325919​
3​	2024​	20349​	65536​	2.6992E+12​	0.098562189​
4​	10626​	38760​	16384​	6.748E+12​	0.246405473​
5​	42504​	50388​	4096​	8.7724E+12​	0.320327115​
6​	134596​	43758​	1024​	6.031E+12​	0.220224892​
					0.906515306​
7​	346104​	24310​	256​	2.1539E+12​	0.078651747​
8​	735471​	8008​	64​	3.7694E+11​	0.013764056​
9​	1307504​	1365​	16​	2.8556E+10​	0.001042731​
10​	1961256​	91​	4​	713897184​	2.60683E-05​
11​	2496144​	1​	1​	2496144​	9.11479E-08​
					0.093484694​

 

[PeroK]

I made a table for the cases and confirmed that RR has no chance to win.

View attachment 301050

BB=RR+2
BR=24-BB-RR=22-2*RR

If there are more black cards, then you can never have more red pairs than black pairs. Or, even as many red pairs as black pairs.

 

[PeroK]

Note that as long at the number of black cards is not a multiple of $3$ there can never be a tie between black and mismatch in a game.

And, as long as the number of black cards is less than three times the number of red cards, then mismatch has a chance.

 

[murshid_islam]

Thanks, @PeroK. Based on your ideas, I made a spreadsheet myself and got the same results.

Edit: Here's my spreadsheet in case anyone wants to have a look